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PROFESSOR: OK.

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00:00:22,550 --> 00:00:27,980
Now, before we start with
the prepared questions,

10
00:00:27,980 --> 00:00:29,610
I have another question for you.

11
00:00:29,610 --> 00:00:33,940
It's sort of taking a straw
poll before and after discussing

12
00:00:33,940 --> 00:00:34,730
something.

13
00:00:34,730 --> 00:00:36,920
And the question for
you, I want to ask

14
00:00:36,920 --> 00:00:41,060
you is, what is the
nature of shot noise?

15
00:00:41,060 --> 00:00:44,100
The whole set of
questions will build up

16
00:00:44,100 --> 00:00:47,180
to teach you something
about what is shot noise.

17
00:00:47,180 --> 00:00:51,740
In other words, when
you measure in photons,

18
00:00:51,740 --> 00:00:54,010
you start with coherent
beams, you always

19
00:00:54,010 --> 00:00:55,970
have sort of square root n.

20
00:00:55,970 --> 00:00:58,800
And the question
I want you to ask

21
00:00:58,800 --> 00:01:02,060
is, when we measure square
root n fluctuations,

22
00:01:02,060 --> 00:01:05,920
when we detect n
photons, is this

23
00:01:05,920 --> 00:01:11,270
caused by the nature of
the measurement process--

24
00:01:11,270 --> 00:01:16,320
sort of how we detect
photons-- or is it

25
00:01:16,320 --> 00:01:18,865
an intrinsic quantity
of the quantum fields?

26
00:01:21,400 --> 00:01:24,630
So I just want to get your
intuition, your gut feeling,

27
00:01:24,630 --> 00:01:28,880
with the shot noise-- it's
maybe not black and white,

28
00:01:28,880 --> 00:01:32,950
but if you think it's more
related to the measurement--

29
00:01:32,950 --> 00:01:34,472
Well, let me write it down.

30
00:01:34,472 --> 00:01:35,470
Shot noise.

31
00:01:40,890 --> 00:01:52,290
OK, so the question
is, shot noise,

32
00:01:52,290 --> 00:01:55,440
is it caused by the quantum
measurement process, projection

33
00:01:55,440 --> 00:01:58,880
of whatever you know about it?

34
00:01:58,880 --> 00:02:07,180
Second part, is it a property
of the quantum fields?

35
00:02:07,180 --> 00:02:09,710
For instance, if you
don't make a measurement,

36
00:02:09,710 --> 00:02:12,920
if you run the quantum fields
through a beam splitter,

37
00:02:12,920 --> 00:02:13,980
we can split it.

38
00:02:13,980 --> 00:02:16,750
We can split the shot noise
even before we measure it.

39
00:02:16,750 --> 00:02:18,700
That would mean quantum fields.

40
00:02:18,700 --> 00:02:22,640
And maybe then, third is both.

41
00:02:22,640 --> 00:02:26,240
And the fourth is
none of the above.

42
00:02:26,240 --> 00:02:29,130
So let's just take an opinion.

43
00:02:29,130 --> 00:02:32,870
And I'm not discussing
the answer at this point.

44
00:02:32,870 --> 00:02:36,370
I will come back to that at the
end of those 10 or 15 minutes

45
00:02:36,370 --> 00:02:38,270
and ask you the question again.

46
00:02:38,270 --> 00:02:47,250
So I just want to see what your
intuitive understanding is when

47
00:02:47,250 --> 00:02:50,760
you measure the number
of photons in a laser

48
00:02:50,760 --> 00:02:55,490
beam in an optical field
and you see fluctuations.

49
00:02:55,490 --> 00:02:59,650
Quantum measurement versus
properties of a quantum field.

50
00:02:59,650 --> 00:03:01,680
Quantum measurement
means mainly it

51
00:03:01,680 --> 00:03:03,890
caused the moment you detected.

52
00:03:03,890 --> 00:03:06,950
And quantum fields means
it's there already before,

53
00:03:06,950 --> 00:03:09,820
and it can be,
whatever, modified.

54
00:03:15,880 --> 00:03:17,840
Quantum fields or both.

55
00:03:17,840 --> 00:03:18,460
All right.

56
00:03:18,460 --> 00:03:21,250
Good.

57
00:03:21,250 --> 00:03:21,750
Fine.

58
00:03:21,750 --> 00:03:23,166
Let's now go through
the question.

59
00:03:23,166 --> 00:03:26,920
We first have to make sure
that we use the same definition

60
00:03:26,920 --> 00:03:29,170
and measure things
in the same unit.

61
00:03:29,170 --> 00:03:30,860
We want to talk about
coherent states.

62
00:03:30,860 --> 00:03:33,950
So the first is just
little, quick checks.

63
00:03:33,950 --> 00:03:37,030
Coherent states are
eigenstates of the Hamiltonians

64
00:03:37,030 --> 00:03:38,310
of a or a dagger?

65
00:03:42,334 --> 00:03:44,625
For the easy question, I will
just give you 10 seconds.

66
00:03:49,180 --> 00:03:49,980
All right.

67
00:03:49,980 --> 00:03:50,710
Stop.

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00:03:50,710 --> 00:03:51,720
Display.

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00:03:51,720 --> 00:03:53,170
Yes.

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00:03:53,170 --> 00:03:53,810
OK.

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00:03:53,810 --> 00:03:54,330
That's good.

72
00:03:58,590 --> 00:04:00,650
Next question.

73
00:04:00,650 --> 00:04:04,980
Which states realize the
minimum quantum uncertainty

74
00:04:04,980 --> 00:04:08,350
with delta p delta q
equals h-bar over 2?

75
00:04:08,350 --> 00:04:10,970
I give you five
different states.

76
00:04:10,970 --> 00:04:13,150
And in order to use
the clicker, you

77
00:04:13,150 --> 00:04:16,576
should just answer how many of
them are minimum uncertainty

78
00:04:16,576 --> 00:04:17,075
states.

79
00:04:32,270 --> 00:04:33,285
Ready, set, go.

80
00:04:39,840 --> 00:04:44,980
So three or two.

81
00:04:44,980 --> 00:04:46,670
Let's just go through them.

82
00:04:46,670 --> 00:04:50,590
The vacuum, definitely.

83
00:04:50,590 --> 00:04:54,590
The coherent state, definitely.

84
00:04:54,590 --> 00:04:58,650
The thermal state,
definitely not.

85
00:04:58,650 --> 00:05:03,710
Now, number states,
they are eigenstates.

86
00:05:03,710 --> 00:05:05,575
But let's just take
a big number state.

87
00:05:05,575 --> 00:05:07,960
It's a big lean in
the quasi-probability.

88
00:05:07,960 --> 00:05:10,480
It has a huge delta
x and a huge delta p.

89
00:05:10,480 --> 00:05:14,100
It's definitely-- it's a
very quantum like state,

90
00:05:14,100 --> 00:05:15,350
very nonclassical.

91
00:05:15,350 --> 00:05:18,250
But it doesn't have
minimum uncertainty.

92
00:05:18,250 --> 00:05:19,870
Final question.

93
00:05:19,870 --> 00:05:22,450
x and p, are those minimum
uncertainty states?

94
00:05:25,170 --> 00:05:26,790
x and p are a
little bit cryptic,

95
00:05:26,790 --> 00:05:29,770
because one is infinite,
the other one is 0.

96
00:05:29,770 --> 00:05:33,200
And infinite times 0
is h-bar over 2, yes.

97
00:05:33,200 --> 00:05:38,320
So it's a minimum uncertainty
state, because we discussed it.

98
00:05:38,320 --> 00:05:41,780
It is the limit of squeezing,
when we perform the limit

99
00:05:41,780 --> 00:05:43,700
that the squeezing
parameter goes to infinity.

100
00:05:46,004 --> 00:05:47,420
You can send me
your lawyer, and I

101
00:05:47,420 --> 00:05:48,670
may have to step back from it.

102
00:05:48,670 --> 00:05:51,270
But the way I present it,
this is one correct answer.

103
00:05:51,270 --> 00:05:53,540
It's a limit of an
infinitely squeezed state.

104
00:05:53,540 --> 00:05:58,120
So if the squeezed vacuum--
which I forgot actually

105
00:05:58,120 --> 00:06:00,890
to ask-- if the squeezed vacuum
is a minimum uncertainty state,

106
00:06:00,890 --> 00:06:04,370
then the x and p state
are the limiting case.

107
00:06:11,386 --> 00:06:13,260
By the way, your clicks
are getting recorded,

108
00:06:13,260 --> 00:06:16,390
and the clicks are responsible
for half of your semester

109
00:06:16,390 --> 00:06:19,070
grade.

110
00:06:19,070 --> 00:06:22,050
This was a joke.

111
00:06:22,050 --> 00:06:23,970
You can actually use
clickers and record them,

112
00:06:23,970 --> 00:06:25,719
but then each student
would have a clicker

113
00:06:25,719 --> 00:06:26,850
with an already set number.

114
00:06:26,850 --> 00:06:29,530
But as you see, you just take
a random clicker out of the box

115
00:06:29,530 --> 00:06:30,590
and put it back in.

116
00:06:30,590 --> 00:06:33,050
So I have no idea what
your clicker number is.

117
00:06:33,050 --> 00:06:37,075
And to use the clicker is solely
for our joint entertainment

118
00:06:37,075 --> 00:06:38,480
here.

119
00:06:38,480 --> 00:06:40,720
OK, next question.

120
00:06:40,720 --> 00:06:43,850
What features are unique
to nonclassical light?

121
00:06:43,850 --> 00:06:48,479
And nonclassical
light means-- well,

122
00:06:48,479 --> 00:06:50,020
we have discussed
nonclassical lights

123
00:06:50,020 --> 00:06:51,510
but their different aspects.

124
00:06:51,510 --> 00:06:54,080
One is that the roots
of the two quadrature

125
00:06:54,080 --> 00:06:56,890
components-- the uncertainty of
the two quadrature components--

126
00:06:56,890 --> 00:06:59,810
the cosine and sine quadrature
components are unequal;

127
00:06:59,810 --> 00:07:04,100
negative cosine probabilities;
a correlation function, which

128
00:07:04,100 --> 00:07:08,370
is smaller than 1; and
sub-Poissonian statistics.

129
00:07:11,040 --> 00:07:12,970
So I've given you four criteria.

130
00:07:12,970 --> 00:07:20,100
How many of those four are
features which we will not

131
00:07:20,100 --> 00:07:23,040
associate with classical light,
only with nonclassical light?

132
00:07:30,940 --> 00:07:31,510
OK.

133
00:07:31,510 --> 00:07:32,815
Stop.

134
00:07:32,815 --> 00:07:33,315
Display.

135
00:07:38,550 --> 00:07:41,940
So there's a wide distribution.

136
00:07:41,940 --> 00:07:43,810
Let's quickly
discuss the answer.

137
00:07:47,060 --> 00:07:50,470
Negative cosine
probabilities, well,

138
00:07:50,470 --> 00:07:54,300
if you use w distribution
and p distribution,

139
00:07:54,300 --> 00:07:55,920
negative probabilities
are always

140
00:07:55,920 --> 00:07:57,820
associated with
nonclassical light.

141
00:07:57,820 --> 00:07:59,320
You can also say
it's the definition

142
00:07:59,320 --> 00:08:00,760
of nonclassical light.

143
00:08:00,760 --> 00:08:04,100
g 2 function of-- maybe, I
should not say smaller or equal

144
00:08:04,100 --> 00:08:07,160
-- smaller than 1 is only
possible for nonclassical

145
00:08:07,160 --> 00:08:07,870
light.

146
00:08:07,870 --> 00:08:10,400
Sub-Poissonian
statistics is also

147
00:08:10,400 --> 00:08:12,830
only possible for
nonclassical light.

148
00:08:12,830 --> 00:08:17,560
Now, the fact that the two
quadrature components are

149
00:08:17,560 --> 00:08:20,072
different, you
can just use some,

150
00:08:20,072 --> 00:08:21,530
and you can choose
classical states

151
00:08:21,530 --> 00:08:24,662
and prepare them, that
you have more certainty

152
00:08:24,662 --> 00:08:26,870
in the sine quadrature than
in the cosine quadrature,

153
00:08:26,870 --> 00:08:30,050
so definitely not.

154
00:08:30,050 --> 00:08:35,250
OK, now, we really
need prominently

155
00:08:35,250 --> 00:08:39,710
to discuss quantum noise and
shot noise, the coherent state.

156
00:08:39,710 --> 00:08:42,080
So just to make sure
that we relate--

157
00:08:42,080 --> 00:08:44,370
I will relate later
everything to photon number.

158
00:08:44,370 --> 00:08:45,890
Because photon
number is intuitive.

159
00:08:45,890 --> 00:08:48,365
You can count, you can feel
the photons fluctuation

160
00:08:48,365 --> 00:08:51,760
of square root n and
immediately tell you Poissonian,

161
00:08:51,760 --> 00:08:53,890
if it's less,
sub-Poisson, all that.

162
00:08:53,890 --> 00:08:56,920
So I want to make sure we are
correctly associating photon

163
00:08:56,920 --> 00:08:58,716
number with coherent states.

164
00:08:58,716 --> 00:09:00,090
So, therefore,
the first question

165
00:09:00,090 --> 00:09:02,570
is if you have a
coherent state, alpha,

166
00:09:02,570 --> 00:09:06,170
and we detect all the
photons with a photodiode

167
00:09:06,170 --> 00:09:08,400
of 100% efficiency,
is the number

168
00:09:08,400 --> 00:09:11,300
of detected photons
alpha or alpha squared?

169
00:09:26,600 --> 00:09:27,720
Yes.

170
00:09:27,720 --> 00:09:30,030
Alpha is, so to
speak, the field.

171
00:09:30,030 --> 00:09:33,910
And alpha squared
is the intensity.

172
00:09:33,910 --> 00:09:38,630
So, therefore, form
now one, I will always

173
00:09:38,630 --> 00:09:41,840
use units for the
photocurrent in such a way

174
00:09:41,840 --> 00:09:44,960
that the photocurrent, or
the integrated photocurrent,

175
00:09:44,960 --> 00:09:47,100
is the number of photons.

176
00:09:47,100 --> 00:09:52,900
So when I say, what is
the current associated

177
00:09:52,900 --> 00:09:57,970
with a coherent state,
it's just alpha squared.

178
00:09:57,970 --> 00:09:59,090
Just one side remark.

179
00:09:59,090 --> 00:10:01,600
If you use a laser
beam, you sort of

180
00:10:01,600 --> 00:10:03,630
have a stream of
photons, but you

181
00:10:03,630 --> 00:10:05,720
have a time constant
in your detector,

182
00:10:05,720 --> 00:10:08,140
and every millisecond
you do a measurement.

183
00:10:08,140 --> 00:10:11,220
And then, what I mean by
alpha or alpha squared

184
00:10:11,220 --> 00:10:13,740
is the number of photons
which arrive during the time

185
00:10:13,740 --> 00:10:15,160
constant.

186
00:10:15,160 --> 00:10:16,710
So that's sort of the mapping.

187
00:10:16,710 --> 00:10:18,770
We all usually talk
about coherent state

188
00:10:18,770 --> 00:10:20,080
in an isolated system.

189
00:10:20,080 --> 00:10:23,080
But if you have a beam, you
just chop the beam into pieces

190
00:10:23,080 --> 00:10:26,770
and associate that with
your coherent state.

191
00:10:26,770 --> 00:10:27,270
OK.

192
00:10:27,270 --> 00:10:30,830
So now, we are talking about the
measurement of coherent beams.

193
00:10:30,830 --> 00:10:32,780
We will have now lots
of coherent beams,

194
00:10:32,780 --> 00:10:35,760
running through beam splitters,
being mixed at beam splitters

195
00:10:35,760 --> 00:10:36,660
and such.

196
00:10:36,660 --> 00:10:39,440
And we will always
characterize the strength

197
00:10:39,440 --> 00:10:41,710
of the coherent beam
by alpha squared.

198
00:10:41,710 --> 00:10:43,470
And that means in photons.

199
00:10:46,420 --> 00:10:49,680
So now, the next
question is-- again,

200
00:10:49,680 --> 00:10:53,540
just to make sure we
are all on the same--

201
00:10:53,540 --> 00:10:56,180
we have the same basis,
what is now the variance?

202
00:10:56,180 --> 00:10:59,950
If I measure the fluctuating
current, I measure the current,

203
00:10:59,950 --> 00:11:04,550
I have a coherent state, and I
repeatedly measure the current,

204
00:11:04,550 --> 00:11:06,360
it will show fluctuations.

205
00:11:06,360 --> 00:11:09,530
And the question is, what is
not the standard deviation

206
00:11:09,530 --> 00:11:12,804
of the current, what
is the variance, which

207
00:11:12,804 --> 00:11:14,220
is the standard
deviation squared?

208
00:11:23,240 --> 00:11:26,480
Yes, the standard
deviation is square root n.

209
00:11:26,480 --> 00:11:28,600
The variance is n.

210
00:11:28,600 --> 00:11:30,370
OK.

211
00:11:30,370 --> 00:11:32,475
Stop.

212
00:11:32,475 --> 00:11:32,975
Hide.

213
00:11:36,390 --> 00:11:41,720
OK, now finally, now we can
have fun with beam splitters.

214
00:11:41,720 --> 00:11:43,856
So, so far, we have
really just repeated

215
00:11:43,856 --> 00:11:46,230
a little bit of definitions
of classical and nonclassical

216
00:11:46,230 --> 00:11:47,980
light at coherent states.

217
00:11:47,980 --> 00:11:50,210
We measured current.

218
00:11:50,210 --> 00:11:53,670
We know now in what units we
need to measure the current.

219
00:11:53,670 --> 00:11:57,890
And we characterized the
fluctuations in variance

220
00:11:57,890 --> 00:12:02,880
not in standard deviation
for the rest of this unit.

221
00:12:02,880 --> 00:12:10,140
But now, we run the coherent
state through a beam splitter.

222
00:12:10,140 --> 00:12:14,500
And we have two photodiodes,
one i1 and one i2.

223
00:12:14,500 --> 00:12:18,130
And we measure the current, i1.

224
00:12:18,130 --> 00:12:21,020
What is the variance now
in the current you measure?

225
00:12:39,070 --> 00:12:39,570
OK.

226
00:12:43,220 --> 00:12:46,050
It is A. It's n/2.

227
00:12:46,050 --> 00:12:47,770
The way how you
think about what it

228
00:12:47,770 --> 00:12:50,710
is, when you split a laser
beam with a beam splitter,

229
00:12:50,710 --> 00:12:52,390
you get two laser beams.

230
00:12:52,390 --> 00:12:54,500
Each is half the intensity.

231
00:12:54,500 --> 00:12:59,200
So therefore, each creates a
photocurrent, which is n/2.

232
00:12:59,200 --> 00:13:01,780
And since one laser beam,
which is a coherent state,

233
00:13:01,780 --> 00:13:04,470
gives two laser beams, which
are two coherent states,

234
00:13:04,470 --> 00:13:07,160
each of them has
a variance which

235
00:13:07,160 --> 00:13:09,550
is equal to the photon number.

236
00:13:09,550 --> 00:13:12,520
And this means, we have
Poissonian statistics.

237
00:13:12,520 --> 00:13:16,830
So we have n/2
photons in each beam.

238
00:13:16,830 --> 00:13:19,480
The standard deviation is
the square root of n/2.

239
00:13:19,480 --> 00:13:24,555
And the variance is n/2.

240
00:13:32,062 --> 00:13:33,520
Ask me, if there
are any questions.

241
00:13:33,520 --> 00:13:35,370
I mean, this is still
the preliminaries.

242
00:13:35,370 --> 00:13:37,480
We are not yet bringing
in squeezed light.

243
00:13:37,480 --> 00:13:39,950
Then, it gets
really interesting.

244
00:13:39,950 --> 00:13:43,490
OK, we will always
refer to that situation.

245
00:13:43,490 --> 00:13:44,430
We have a laser beam.

246
00:13:44,430 --> 00:13:45,540
We split it.

247
00:13:45,540 --> 00:13:48,315
We create-- and that's actually
your homework assignment.

248
00:13:48,315 --> 00:13:51,360
We'll actually show
wonderful operator algebra.

249
00:13:51,360 --> 00:13:54,430
If you split a coherent beam,
you get two coherent beams.

250
00:13:54,430 --> 00:13:56,080
And they are simply
characterized

251
00:13:56,080 --> 00:13:59,210
that the alpha value is
now down by square root 2.

252
00:13:59,210 --> 00:14:01,010
The number is down by 2.

253
00:14:01,010 --> 00:14:03,840
And if the number is down by 2
and you have a coherent beam,

254
00:14:03,840 --> 00:14:05,684
the variance is down by 2.

255
00:14:05,684 --> 00:14:06,600
Poissonian statistics.

256
00:14:12,770 --> 00:14:18,780
OK, next question is we
have a beam splitter.

257
00:14:18,780 --> 00:14:22,880
We have two photodiodes,
i1 and i2, and now

258
00:14:22,880 --> 00:14:27,020
we measure-- it's important
for balanced homodyne

259
00:14:27,020 --> 00:14:30,130
detection-- we can now
take the two currents,

260
00:14:30,130 --> 00:14:34,340
and we can add them or
we can subtract them.

261
00:14:34,340 --> 00:14:40,280
So the question is, if we add
the two currents, what is now

262
00:14:40,280 --> 00:14:43,030
the variance in
the two currents?

263
00:14:43,030 --> 00:14:47,650
Remember, each current
had a variance of n/2.

264
00:14:47,650 --> 00:14:50,670
So now, you are asked, when
you sum the currents of the two

265
00:14:50,670 --> 00:14:53,170
photodetectors, what happens
to the two variances?

266
00:14:56,410 --> 00:14:58,109
Do they sum up?

267
00:14:58,109 --> 00:14:58,608
Do they--?

268
00:15:13,709 --> 00:15:14,250
I don't know.

269
00:15:14,250 --> 00:15:17,350
Do people want
to-- does somebody

270
00:15:17,350 --> 00:15:20,744
want to defend
his or her choice?

271
00:15:20,744 --> 00:15:22,160
What do you think
is the variance?

272
00:15:26,328 --> 00:15:28,758
AUDIENCE: It should be
the answer, you just

273
00:15:28,758 --> 00:15:31,284
measure before
the beam splitter.

274
00:15:31,284 --> 00:15:33,950
PROFESSOR: You said it should be
as if I measure before the beam

275
00:15:33,950 --> 00:15:38,500
splitter, and then it would be
n, because we have n photons.

276
00:15:38,500 --> 00:15:42,400
Well, you can also see if
you put in a mirror here,

277
00:15:42,400 --> 00:15:46,690
and you shine the two beams
on the same photodiode.

278
00:15:46,690 --> 00:15:50,200
And then you would say
you have n photons.

279
00:15:50,200 --> 00:15:55,250
And if there are Poissonian
fluctuations, you would get n.

280
00:15:55,250 --> 00:15:58,540
But there is one thing
you should keep in mind.

281
00:15:58,540 --> 00:16:02,880
Since we have the
beam splitter here--

282
00:16:02,880 --> 00:16:04,260
this is the correct answer.

283
00:16:04,260 --> 00:16:06,790
But I want to point out
one subtlety for you, which

284
00:16:06,790 --> 00:16:08,900
will become important later.

285
00:16:08,900 --> 00:16:12,530
The beam splitter has an
open part with the vacuum.

286
00:16:12,530 --> 00:16:16,800
And so when you said we can
measure before or by sort

287
00:16:16,800 --> 00:16:19,950
of just combining the two
beams with a big photodiode,

288
00:16:19,950 --> 00:16:25,730
we can measure after, that's
now a question for you-- which,

289
00:16:25,730 --> 00:16:27,740
maybe hold the thought
and ask me later,

290
00:16:27,740 --> 00:16:30,360
if it doesn't become clear
with the following question.

291
00:16:30,360 --> 00:16:33,650
Hasn't the vacuum entered here,
and hasn't the vacuum, which

292
00:16:33,650 --> 00:16:36,206
has entered here
added fluctuations?

293
00:16:36,206 --> 00:16:37,580
So if you put in
a beam splitter,

294
00:16:37,580 --> 00:16:41,610
and we mix in the vacuum
with our coherent states,

295
00:16:41,610 --> 00:16:43,697
shouldn't that possibly
mean that if you

296
00:16:43,697 --> 00:16:47,060
use a big photodiode and
measure everything together,

297
00:16:47,060 --> 00:16:49,750
that we measure the
variance we had before.

298
00:16:49,750 --> 00:16:52,150
But isn't that
possibly a contribution

299
00:16:52,150 --> 00:16:55,470
of the vacuum, because we were
opening the door to the vacuum?

300
00:16:55,470 --> 00:16:56,530
Please hold the thought.

301
00:17:00,780 --> 00:17:02,720
OK.

302
00:17:02,720 --> 00:17:07,770
The next thing is now, you can't
have the solution that we take

303
00:17:07,770 --> 00:17:10,390
sort of a big photodiode
and combine the two beams,

304
00:17:10,390 --> 00:17:12,500
because I'm asking
you now is, what

305
00:17:12,500 --> 00:17:17,069
is the variance in i1 minus i2?

306
00:17:17,069 --> 00:17:19,960
So you subtract
the two currents.

307
00:17:19,960 --> 00:17:22,950
So the question is, if
you have technical noise,

308
00:17:22,950 --> 00:17:24,920
the two beams would
be fluctuating,

309
00:17:24,920 --> 00:17:27,170
and you would form i1 minus i2.

310
00:17:27,170 --> 00:17:30,890
All the technical noise
would cancel away.

311
00:17:30,890 --> 00:17:33,520
But what is shot noise?

312
00:17:33,520 --> 00:17:35,770
Is shot noise technical noise?

313
00:17:35,770 --> 00:17:39,710
Is shot noise that every
time you prepare an ensemble,

314
00:17:39,710 --> 00:17:44,190
there is a hiccup, and you
can split it and subtract it?

315
00:17:44,190 --> 00:17:46,230
So now, it's getting
really interesting.

316
00:17:46,230 --> 00:17:50,420
To what extent does a
beam splitter-- does

317
00:17:50,420 --> 00:17:52,990
a split coherent state,
which has shot noise,

318
00:17:52,990 --> 00:17:55,520
and you now perform the
difference of the measurement,

319
00:17:55,520 --> 00:17:57,730
can you get rid of some
of the shot noise or not?

320
00:18:02,037 --> 00:18:02,745
You vote, please.

321
00:18:20,680 --> 00:18:21,180
Yes.

322
00:18:23,950 --> 00:18:28,310
So it doesn't matter
whether we measure

323
00:18:28,310 --> 00:18:30,000
the sum or the difference.

324
00:18:30,000 --> 00:18:32,820
So it seems, we have two beams.

325
00:18:32,820 --> 00:18:38,860
And it seems that the
noise in i1 and i2

326
00:18:38,860 --> 00:18:40,859
is completely uncorrelated.

327
00:18:40,859 --> 00:18:42,900
And whether we perform
the sum or the difference,

328
00:18:42,900 --> 00:18:43,650
it doesn't matter.

329
00:18:46,060 --> 00:18:48,690
As long as you think
about classical states

330
00:18:48,690 --> 00:18:51,850
and coherent states, you can
get away with this notion.

331
00:18:51,850 --> 00:18:54,530
But now I'm going to
add some squeezed light.

332
00:18:54,530 --> 00:18:57,131
And now, it's
getting more subtle.

333
00:18:59,780 --> 00:19:03,540
So we have exactly
the same situation.

334
00:19:03,540 --> 00:19:05,150
We have our beam splitter.

335
00:19:05,150 --> 00:19:11,640
We split our coherent beam into
two beams, exactly as before.

336
00:19:11,640 --> 00:19:15,110
But now, I'm playing
with a vacuum.

337
00:19:15,110 --> 00:19:18,700
I'm playing with the open
part of the beam splitter,

338
00:19:18,700 --> 00:19:24,650
and I put in squeezed vacuum.

339
00:19:24,650 --> 00:19:28,010
And just to be specific,
I've squeezed the vacuum

340
00:19:28,010 --> 00:19:31,340
in such a way that
what is narrow now

341
00:19:31,340 --> 00:19:35,930
is the quadrature component
of the coherent state.

342
00:19:35,930 --> 00:19:39,680
In other words, the coherent
state is cosine omega t.

343
00:19:39,680 --> 00:19:42,050
In sine omega t, we haven't
been put in any photons.

344
00:19:42,050 --> 00:19:44,685
Our coherent state has really
a classical oscillation

345
00:19:44,685 --> 00:19:46,880
in cosine omega t.

346
00:19:46,880 --> 00:19:50,480
And what I have squeezed
here with the vacuum

347
00:19:50,480 --> 00:19:54,880
is the noise in the cosine
omega t quadrature component.

348
00:19:54,880 --> 00:20:00,000
So in other words, the coherent
state and the short axis

349
00:20:00,000 --> 00:20:04,210
of the ellipse are the
same quadrature component,

350
00:20:04,210 --> 00:20:08,020
which means, they
can now interfere.

351
00:20:08,020 --> 00:20:10,380
So the question is
now the following.

352
00:20:10,380 --> 00:20:11,710
We split the beam.

353
00:20:11,710 --> 00:20:14,060
We have n/2 photons here.

354
00:20:14,060 --> 00:20:17,797
We have n/2 photons there.

355
00:20:17,797 --> 00:20:19,630
And what is now the
result for the variance?

356
00:20:23,056 --> 00:20:25,300
I formulated a
condition up there,

357
00:20:25,300 --> 00:20:30,240
which is the strong local
oscillator condition.

358
00:20:30,240 --> 00:20:32,280
You should assume,
if you get confused,

359
00:20:32,280 --> 00:20:35,410
that the local oscillator
is very, very strong.

360
00:20:35,410 --> 00:20:38,370
And if you have any doubts
about some smaller terms,

361
00:20:38,370 --> 00:20:40,880
just crank up the
power and make sure

362
00:20:40,880 --> 00:20:43,980
that it's really the local
oscillator which dominates,

363
00:20:43,980 --> 00:20:47,881
which is the biggest
number in all equations.

364
00:20:47,881 --> 00:20:48,380
OK.

365
00:20:48,380 --> 00:20:52,840
So the reference is
without squeezing,

366
00:20:52,840 --> 00:20:57,530
the noise, or the
variance, in i1 was n/2.

367
00:20:57,530 --> 00:21:01,510
With n/2 photons, Poissonian
statistics was n/2.

368
00:21:01,510 --> 00:21:08,230
The question is now, when
we use squeezed light,

369
00:21:08,230 --> 00:21:10,770
does the variance stay the same?

370
00:21:10,770 --> 00:21:12,010
Does it go up?

371
00:21:12,010 --> 00:21:13,850
Does it go down?

372
00:21:13,850 --> 00:21:15,210
Or does it go to 0?

373
00:21:28,176 --> 00:21:30,780
By the way, at the end
of the clicker question,

374
00:21:30,780 --> 00:21:34,260
if you come to me after
class and honestly declare

375
00:21:34,260 --> 00:21:37,990
you got 100% on all
questions, I buy you a lunch.

376
00:21:40,650 --> 00:21:41,590
OK, you're done?

377
00:21:50,500 --> 00:21:52,730
What have you learned?

378
00:21:52,730 --> 00:21:53,780
It's almost random.

379
00:21:59,190 --> 00:22:03,250
I will give you a little bit
formalism later, but I mean,

380
00:22:03,250 --> 00:22:04,660
let me say one thing.

381
00:22:04,660 --> 00:22:06,990
I had a few discussion
with students after class,

382
00:22:06,990 --> 00:22:09,350
and I came to the
conclusion after last class

383
00:22:09,350 --> 00:22:15,050
that I could teach today about
the Deutsch-Jozsa algorithms'

384
00:22:15,050 --> 00:22:17,060
wonderful applications
of all that,

385
00:22:17,060 --> 00:22:19,870
but I came to the conclusion,
you can go to the Wikipedia

386
00:22:19,870 --> 00:22:21,370
and read about yourself.

387
00:22:21,370 --> 00:22:24,200
Because this is simply
using beam splitters,

388
00:22:24,200 --> 00:22:25,160
homodyne detection.

389
00:22:25,160 --> 00:22:27,600
It's just using kind
of all the tools.

390
00:22:27,600 --> 00:22:30,460
But what I said instead is,
I'd rather spend half the class

391
00:22:30,460 --> 00:22:35,120
today in trying to give you,
with very simple examples,

392
00:22:35,120 --> 00:22:36,355
an intuitive understanding.

393
00:22:36,355 --> 00:22:37,730
I mean, everything
I'm saying you

394
00:22:37,730 --> 00:22:39,100
can do with operator algebra.

395
00:22:39,100 --> 00:22:41,480
But to some extent, I feel
when we learn physics,

396
00:22:41,480 --> 00:22:44,230
we use operators, we use
equations, but in the end,

397
00:22:44,230 --> 00:22:46,220
we want to shape our intuition.

398
00:22:46,220 --> 00:22:48,650
And what I try to sort of
challenge you today is,

399
00:22:48,650 --> 00:22:51,500
what happens to noise?

400
00:22:51,500 --> 00:22:56,060
So what I have to tell you
now is-- OK, got to hide this.

401
00:23:00,770 --> 00:23:03,580
The way you should look
at it very soon, I hope,

402
00:23:03,580 --> 00:23:07,680
is the following, that when
you split the coherent beam,

403
00:23:07,680 --> 00:23:09,630
you also split the noise.

404
00:23:09,630 --> 00:23:12,440
In other words, your
little uncertainty circle

405
00:23:12,440 --> 00:23:14,960
goes down in the amplitude
by square root of 2.

406
00:23:14,960 --> 00:23:18,080
It is split-- the reflection
coefficient of the beam spit

407
00:23:18,080 --> 00:23:21,940
and transmission coefficient
is 1 over square root 2.

408
00:23:21,940 --> 00:23:26,270
But what happens is, you
let in the vacuum here.

409
00:23:26,270 --> 00:23:29,710
And so, the vacuum comes into
the transmission coefficient

410
00:23:29,710 --> 00:23:31,750
of 1 over square root 2.

411
00:23:31,750 --> 00:23:34,990
So what I want to teach you
is that when we think back

412
00:23:34,990 --> 00:23:42,860
to this example and we had
this situation with a vacuum,

413
00:23:42,860 --> 00:23:47,290
yes, it looked to
us that we had n/2--

414
00:23:47,290 --> 00:23:54,050
so this was i1 plus
i2-- we had n/2 photons

415
00:23:54,050 --> 00:23:57,900
and the variance was n/2, it
looked completely Poissonian

416
00:23:57,900 --> 00:23:59,990
as if nothing has happened.

417
00:23:59,990 --> 00:24:03,860
But what you should understand
is, that the beam splitter

418
00:24:03,860 --> 00:24:08,400
has taken out-- in amplitude,
1 over square root 2,

419
00:24:08,400 --> 00:24:11,300
when I square it--
half of the noise out

420
00:24:11,300 --> 00:24:12,820
of the coherent state.

421
00:24:12,820 --> 00:24:15,360
But half of the vacuum
state was added.

422
00:24:15,360 --> 00:24:19,330
And half of the vacuum state
and half of the intrinsic noise

423
00:24:19,330 --> 00:24:23,130
of the coherent state give
you back this circle of unity.

424
00:24:23,130 --> 00:24:25,760
And you can pretty much
neglect all subtleties

425
00:24:25,760 --> 00:24:28,050
of the quantum field.

426
00:24:28,050 --> 00:24:32,150
But now, by messing
up with a vacuum,

427
00:24:32,150 --> 00:24:35,980
by squeezing the vacuum,
half of the shot noise

428
00:24:35,980 --> 00:24:37,300
comes from the beam splitter.

429
00:24:37,300 --> 00:24:40,510
It's a 50-50 beam splitter.

430
00:24:40,510 --> 00:24:43,740
But the other half comes
from the open part.

431
00:24:43,740 --> 00:24:47,890
But if you squeeze it strongly,
it's on the order epsilon.

432
00:24:47,890 --> 00:24:51,240
That means, it
doesn't contribute.

433
00:24:51,240 --> 00:25:02,350
And the correct answer is
now, we have-- just a second.

434
00:25:02,350 --> 00:25:05,110
It is linear or quadratic?

435
00:25:05,110 --> 00:25:08,170
Variance is quadratic.

436
00:25:08,170 --> 00:25:12,000
The correct answer is here,
because this is epsilon;

437
00:25:12,000 --> 00:25:13,170
this is small.

438
00:25:13,170 --> 00:25:15,800
We only get half of the
noise we had before.

439
00:25:15,800 --> 00:25:17,420
We had n/2, and
now, we have n/4.

440
00:25:21,030 --> 00:25:21,530
OK.

441
00:25:30,170 --> 00:25:31,955
Now, we take exactly
the same situation.

442
00:25:34,500 --> 00:25:39,890
So the thought is that you
measure the fluctuation here

443
00:25:39,890 --> 00:25:44,410
and you have only half of them,
because this squeezed vacuum,

444
00:25:44,410 --> 00:25:47,860
you've squeezed out the
noise from the vacuum part.

445
00:25:47,860 --> 00:25:50,010
And since you have a
strong coherent state,

446
00:25:50,010 --> 00:25:53,000
everything which is sine
omega t doesn't matter.

447
00:25:53,000 --> 00:25:55,725
Cosine omega t is homodyned,
everything fluctuation

448
00:25:55,725 --> 00:25:57,590
and cosine omega t.

449
00:25:57,590 --> 00:26:01,600
When we measure, the intensity
is the square of phi,

450
00:26:01,600 --> 00:26:03,140
or it's a dagger a.

451
00:26:03,140 --> 00:26:05,490
And so if you measure
a quadratic quantity

452
00:26:05,490 --> 00:26:08,100
and we have a very
strong local oscillator,

453
00:26:08,100 --> 00:26:10,470
everything is projected
on cosine omega t.

454
00:26:10,470 --> 00:26:13,400
This is why only the
cosine omega t fluctuations

455
00:26:13,400 --> 00:26:15,830
of discrete statement are here.

456
00:26:15,830 --> 00:26:17,553
It's a strong local
oscillator limit.

457
00:26:21,120 --> 00:26:21,620
OK.

458
00:26:21,620 --> 00:26:24,390
Now, I hope you've
learned something.

459
00:26:24,390 --> 00:26:30,250
Next question is now, we
do exactly the same thing,

460
00:26:30,250 --> 00:26:33,350
but now we have
the two currents,

461
00:26:33,350 --> 00:26:38,390
and the question is, what
is the variance in i-plus?

462
00:26:38,390 --> 00:26:41,050
I add the two currents.

463
00:26:41,050 --> 00:26:42,000
What do I get?

464
00:26:44,830 --> 00:26:47,780
And I give you the
same choices as before.

465
00:26:47,780 --> 00:26:52,220
Just as a reminder,
when we did not squeeze,

466
00:26:52,220 --> 00:26:54,150
the variance was n.

467
00:26:54,150 --> 00:27:00,420
So the question is now, has
this squeezing done nothing,

468
00:27:00,420 --> 00:27:02,301
or has it reduced the variance?

469
00:27:02,301 --> 00:27:03,050
What is happening?

470
00:27:27,040 --> 00:27:29,390
The correct answer is C. It's n.

471
00:27:35,700 --> 00:27:38,690
And I will show you-- I owe
you a little bit of math now.

472
00:27:38,690 --> 00:27:42,050
But the way how you should
feel is the following.

473
00:27:42,050 --> 00:27:47,080
The noise of the coherent
oscillator is equally split.

474
00:27:47,080 --> 00:27:51,690
But the amplitude-- I mean,
you can say 50% go through, 50%

475
00:27:51,690 --> 00:27:52,610
go through here.

476
00:27:52,610 --> 00:27:54,410
And if you add the
two currents, you

477
00:27:54,410 --> 00:27:58,230
get back the full noise of
the original coherent state.

478
00:27:58,230 --> 00:28:00,410
You would say what
about the squeezing?

479
00:28:00,410 --> 00:28:02,550
Well, the issue
is the following.

480
00:28:02,550 --> 00:28:05,340
The squeezed light,
the squeezed vacuum,

481
00:28:05,340 --> 00:28:09,124
is reflected with a reflection
coefficient, which is minus 1

482
00:28:09,124 --> 00:28:10,040
over square root of 2.

483
00:28:10,040 --> 00:28:13,660
You know when you have two
beams reflected, by unitarity,

484
00:28:13,660 --> 00:28:16,030
it's necessary that
one has a negative,

485
00:28:16,030 --> 00:28:18,660
the other one has a positive
reflection coefficient.

486
00:28:18,660 --> 00:28:22,640
So therefore, you have
this squeezed vacuum

487
00:28:22,640 --> 00:28:25,030
appears in this arm
with a minus sign

488
00:28:25,030 --> 00:28:27,310
and this arm with a plus sign.

489
00:28:27,310 --> 00:28:29,640
And if you sum up
the two currents,

490
00:28:29,640 --> 00:28:33,870
you lose everything which is
related to the vacuum part.

491
00:28:33,870 --> 00:28:36,950
So when you measure
i1 plus i2, what

492
00:28:36,950 --> 00:28:40,190
came from the vacuum part
entered here with a minus sign,

493
00:28:40,190 --> 00:28:41,280
here with a plus sign.

494
00:28:41,280 --> 00:28:43,030
And therefore, you
[INAUDIBLE] what you've

495
00:28:43,030 --> 00:28:45,310
done at the vacuum part.

496
00:28:45,310 --> 00:28:48,820
This is when I told you
earlier, you measure i1 plus i2,

497
00:28:48,820 --> 00:28:51,840
and we all agreed,
the variance was n.

498
00:28:51,840 --> 00:28:53,760
And I said, well, this
is exactly the same

499
00:28:53,760 --> 00:28:57,120
if you put a photodiode-- big
photodiode-- behind the beam

500
00:28:57,120 --> 00:28:58,880
splitter or before
the beam splitter.

501
00:28:58,880 --> 00:29:01,020
The vacuum has not contributed.

502
00:29:01,020 --> 00:29:02,030
That's what it seems.

503
00:29:02,030 --> 00:29:04,920
But even if you squeeze
the vacuum, it cancels out.

504
00:29:07,900 --> 00:29:08,400
OK.

505
00:29:08,400 --> 00:29:10,420
Now, the obvious
next question-- but I

506
00:29:10,420 --> 00:29:12,270
think I've given
you the answer--

507
00:29:12,270 --> 00:29:17,270
is what happens when we
now measure the difference,

508
00:29:17,270 --> 00:29:20,700
i1 minus i2?

509
00:29:20,700 --> 00:29:26,800
Again, without squeezing,
the variance was n.

510
00:29:26,800 --> 00:29:30,350
So now, we measure i1
minus i2, and what happens?

511
00:29:45,780 --> 00:29:47,760
OK.

512
00:29:47,760 --> 00:29:50,190
The correct answer is this.

513
00:29:58,025 --> 00:30:01,200
Well, when I say 0, it's
always epsilon squared terms

514
00:30:01,200 --> 00:30:01,910
which I neglect.

515
00:30:01,910 --> 00:30:03,780
I just assume strong squeezing.

516
00:30:03,780 --> 00:30:06,390
I mean, I just want to give you
the most conceptional problem.

517
00:30:06,390 --> 00:30:08,840
What happens is the following.

518
00:30:08,840 --> 00:30:10,264
And I will show
you the equation.

519
00:30:17,360 --> 00:30:21,520
The noise of the coherent
state is split equally.

520
00:30:21,520 --> 00:30:25,510
So we have a delta a--
a is the field operator,

521
00:30:25,510 --> 00:30:28,970
or the quadrature operator
here-- and a delta a here.

522
00:30:28,970 --> 00:30:31,470
And therefore, the
original shot noise

523
00:30:31,470 --> 00:30:34,850
is common mode, because the
reflection and transmission

524
00:30:34,850 --> 00:30:36,310
coefficients are the same.

525
00:30:36,310 --> 00:30:39,000
And if you subtract the
current, you completely

526
00:30:39,000 --> 00:30:41,730
get rid of all noise,
which was originally

527
00:30:41,730 --> 00:30:43,230
associated with
your coherent state.

528
00:30:46,460 --> 00:30:51,446
Now, what happens is
the small noise we get,

529
00:30:51,446 --> 00:30:53,320
but now, we have to
consider the vacuum part.

530
00:30:53,320 --> 00:30:55,820
But what comes in on the
same quadrature component

531
00:30:55,820 --> 00:31:00,510
point on the vacuum
tube is strongly split.

532
00:31:00,510 --> 00:31:06,800
And because the beam
splitter has different signs

533
00:31:06,800 --> 00:31:10,260
for reflection and transmission,
the noise is anti-correlated.

534
00:31:10,260 --> 00:31:12,230
And if you take
the difference, you

535
00:31:12,230 --> 00:31:19,160
will actually measure the noise
of the input of the open part

536
00:31:19,160 --> 00:31:27,110
of the beam splitter, which
would mean, in terms of order

537
00:31:27,110 --> 00:31:31,400
epsilon-- I didn't work it out.

538
00:31:31,400 --> 00:31:33,290
I think it's epsilon squared.

539
00:31:33,290 --> 00:31:34,760
But it's a power of epsilon.

540
00:31:40,540 --> 00:31:44,750
So anyway, it seems your
intuition fails here.

541
00:31:44,750 --> 00:31:48,570
So let me try to give you a few
equations which explain that.

542
00:31:48,570 --> 00:31:51,870
In your homework and
in most of the course,

543
00:31:51,870 --> 00:31:53,740
we really want to
work with operators.

544
00:31:53,740 --> 00:31:56,370
We do transformation
of operators and such.

545
00:31:56,370 --> 00:31:59,570
But I've found-- for
shaping my intuition--

546
00:31:59,570 --> 00:32:04,130
I found a simplified
approach very helpful.

547
00:32:04,130 --> 00:32:07,050
And I've posted two
references for you.

548
00:32:07,050 --> 00:32:09,180
One is the article
by Schumacher,

549
00:32:09,180 --> 00:32:11,560
and one are some
older lecture notes,

550
00:32:11,560 --> 00:32:14,360
where I summarize some
results of this article.

551
00:32:14,360 --> 00:32:16,520
And this goes as follows.

552
00:32:16,520 --> 00:32:19,110
When you have a
coherent input, you

553
00:32:19,110 --> 00:32:23,480
can say you have a
deterministic classical field.

554
00:32:23,480 --> 00:32:28,180
And then, you have some noise
in one quadrature component,

555
00:32:28,180 --> 00:32:30,730
which is cosine omega t.

556
00:32:30,730 --> 00:32:32,710
And the i is a reminder
that there's also

557
00:32:32,710 --> 00:32:34,450
noise in sine omega t.

558
00:32:34,450 --> 00:32:37,750
So you can really say
the electric field has

559
00:32:37,750 --> 00:32:39,520
a sharp value,
but then, there is

560
00:32:39,520 --> 00:32:42,529
some fluctuation in the
coefficient of cosine omega t

561
00:32:42,529 --> 00:32:45,070
and a little bit fluctuation in
the coefficient of sine omega

562
00:32:45,070 --> 00:32:46,720
t.

563
00:32:46,720 --> 00:32:51,140
And what is important now is, if
you shine that on a photodiode,

564
00:32:51,140 --> 00:32:52,330
this is sort of the field.

565
00:32:55,260 --> 00:32:58,962
The current is proportional
to the square of it.

566
00:32:58,962 --> 00:33:00,670
You have to be a little
bit careful where

567
00:33:00,670 --> 00:33:03,920
you put complex conjugation,
but it is the square of it.

568
00:33:03,920 --> 00:33:11,190
And now, if we use the strong
local oscillator assumption,

569
00:33:11,190 --> 00:33:13,260
we only take alpha
squared, which

570
00:33:13,260 --> 00:33:14,940
gives us a number of photons.

571
00:33:14,940 --> 00:33:18,160
But then, those
fluctuations are homodyned.

572
00:33:18,160 --> 00:33:19,845
We only take-- we
don't consider that

573
00:33:19,845 --> 00:33:21,330
the square of the fluctuation.

574
00:33:21,330 --> 00:33:22,590
We only homodyne it.

575
00:33:22,590 --> 00:33:28,350
We use, at the cross term,
where we multiply with alpha.

576
00:33:28,350 --> 00:33:36,380
So if I now put in
photon numbers for alpha,

577
00:33:36,380 --> 00:33:37,540
we get n photons.

578
00:33:37,540 --> 00:33:38,690
Yes, OK.

579
00:33:38,690 --> 00:33:41,500
And now, we get some noise.

580
00:33:41,500 --> 00:33:47,190
Well, the noise is
the current squared.

581
00:33:47,190 --> 00:33:53,710
The noise is-- we take the
average of the current squared

582
00:33:53,710 --> 00:33:58,600
and subtract the average
of the current squared.

583
00:33:58,600 --> 00:34:02,920
So if you do that--
this is the current

584
00:34:02,920 --> 00:34:05,625
squared-- the current
squared average,

585
00:34:05,625 --> 00:34:08,690
and the average current
squared takes that away.

586
00:34:08,690 --> 00:34:11,400
So we get this here
as the variance.

587
00:34:11,400 --> 00:34:14,810
And now, I have to tell you that
in the units I'm using here,

588
00:34:14,810 --> 00:34:18,540
this delta a1 squared
for coherent state--

589
00:34:18,540 --> 00:34:20,580
so if I show the
standard circle--

590
00:34:20,580 --> 00:34:22,570
the standard
Heisenberg uncertainty

591
00:34:22,570 --> 00:34:26,045
circle-- this quantity is 1/4.

592
00:34:26,045 --> 00:34:28,250
I've just kept track
of my constant.

593
00:34:28,250 --> 00:34:35,290
So therefore, we obtain in this
picture that the variance is n.

594
00:34:35,290 --> 00:34:38,199
And this is Poissonian
light with a coherent state.

595
00:34:38,199 --> 00:34:41,679
And the fluctuation
squared of the current, n,

596
00:34:41,679 --> 00:34:43,900
this is Poissonian
statistics, square root

597
00:34:43,900 --> 00:34:45,530
in fluctuations of the current.

598
00:34:45,530 --> 00:34:47,340
The variance is n.

599
00:34:47,340 --> 00:34:49,989
But what you should
realize is one thing.

600
00:34:49,989 --> 00:34:53,409
The moment I have a
strong coherent state,

601
00:34:53,409 --> 00:34:57,190
the sign quadrature component
doesn't matter at all.

602
00:34:57,190 --> 00:34:59,740
Because if you multiply
cosine with sine and average,

603
00:34:59,740 --> 00:35:00,470
you get 0.

604
00:35:00,470 --> 00:35:03,740
So we only consider the
quadrature component,

605
00:35:03,740 --> 00:35:06,130
which is given by
the coherent state.

606
00:35:06,130 --> 00:35:07,600
This is a principle of homodyne.

607
00:35:07,600 --> 00:35:09,556
You set your
quadrature component

608
00:35:09,556 --> 00:35:10,680
with your local oscillator.

609
00:35:17,210 --> 00:35:18,010
OK.

610
00:35:18,010 --> 00:35:24,320
But now, I have a beam splitter.

611
00:35:24,320 --> 00:35:26,810
And the way you
can think about it

612
00:35:26,810 --> 00:35:30,270
is, you can get everything
in operator algebra,

613
00:35:30,270 --> 00:35:35,210
but just take my word or
follow the procedure now

614
00:35:35,210 --> 00:35:37,200
that we are saying,
we can always

615
00:35:37,200 --> 00:35:38,780
think about a
quantum field that it

616
00:35:38,780 --> 00:35:41,770
has an average value
and some fluctuations.

617
00:35:41,770 --> 00:35:44,300
And if you pass it through
an optical element or a beam

618
00:35:44,300 --> 00:35:46,860
splitter, we have a
transmission coefficient,

619
00:35:46,860 --> 00:35:49,930
but it multiplies the
mean value of the field.

620
00:35:49,930 --> 00:35:53,580
But it also multiplies the
fluctuations of the field.

621
00:35:53,580 --> 00:35:57,860
So that's why I'm saying the
beam splitter is actually

622
00:35:57,860 --> 00:36:01,030
splitting the
fluctuations of the field.

623
00:36:01,030 --> 00:36:06,205
And therefore, you get reduced
fluctuations in either beam.

624
00:36:06,205 --> 00:36:09,680
But we have to consider
that there may be vacuum,

625
00:36:09,680 --> 00:36:12,260
and the vacuum has
also some fluctuations

626
00:36:12,260 --> 00:36:16,025
and will reflect in or transmit
through parts of the vacuum.

627
00:36:22,720 --> 00:36:31,150
So now, we want to
understand photocurrents.

628
00:36:31,150 --> 00:36:45,070
For photocurrents, we take those
quantities and square them.

629
00:36:48,430 --> 00:36:52,280
So which one should I take?

630
00:36:55,636 --> 00:36:57,610
Well, let me take
the first line.

631
00:36:57,610 --> 00:36:59,950
So this is i1.

632
00:36:59,950 --> 00:37:07,530
So if I calculate now i1, I
obtain t squared alpha squared,

633
00:37:07,530 --> 00:37:12,450
plus the cross product
between the first two terms

634
00:37:12,450 --> 00:37:19,510
is 2 times t alpha
times t delta a1.

635
00:37:19,510 --> 00:37:22,340
And then, I get a second
term from the vacuum,

636
00:37:22,340 --> 00:37:30,580
which is 2tr alpha delta b2.

637
00:37:30,580 --> 00:37:38,270
And let me assume that we have
a balanced beam splitter, 50-50.

638
00:37:38,270 --> 00:37:41,280
So t squared is simply 1/2.

639
00:37:41,280 --> 00:37:44,970
Alpha squared is the
number of photons.

640
00:37:44,970 --> 00:37:46,190
So this term becomes n/2.

641
00:37:57,120 --> 00:37:58,500
t squared is 1/2.

642
00:37:58,500 --> 00:37:59,610
It cancels with the 2.

643
00:37:59,610 --> 00:38:01,110
We are left with alpha.

644
00:38:01,110 --> 00:38:04,660
And this is shot noise.

645
00:38:04,660 --> 00:38:07,510
Here, appears our square root n.

646
00:38:07,510 --> 00:38:10,510
So you can say the
shot noise comes

647
00:38:10,510 --> 00:38:14,150
because we have homodyned
the quantum noise.

648
00:38:14,150 --> 00:38:17,950
The quantum noise, which is the
same for all coherent states,

649
00:38:17,950 --> 00:38:20,082
has now been multiplied
with square root n

650
00:38:20,082 --> 00:38:21,290
through the homodyne process.

651
00:38:29,935 --> 00:38:30,435
OK.

652
00:38:34,120 --> 00:38:35,860
So I lost my line.

653
00:38:35,860 --> 00:38:38,620
That's delta a2.

654
00:38:38,620 --> 00:38:41,630
And then-- and
this is important--

655
00:38:41,630 --> 00:38:44,820
because of the mixing
at the beam splitter,

656
00:38:44,820 --> 00:38:49,160
the quantum noise off the
vacuum gets also homodyned

657
00:38:49,160 --> 00:38:50,560
by a factor of square root n.

658
00:38:55,270 --> 00:38:59,980
So now, I think you see
what I told you before.

659
00:38:59,980 --> 00:39:05,310
The transmitted beam has
half of its noise coming

660
00:39:05,310 --> 00:39:09,060
from the coherent state,
and half of the noise

661
00:39:09,060 --> 00:39:13,570
comes from the vacuum.

662
00:39:13,570 --> 00:39:15,750
So if I would now
take the current,

663
00:39:15,750 --> 00:39:18,140
square it, calculate
the variance,

664
00:39:18,140 --> 00:39:22,650
I would find that I have n/2
photons with a variance of n/2.

665
00:39:22,650 --> 00:39:26,360
But an identical
contribution of the noise

666
00:39:26,360 --> 00:39:28,950
will come from the
original coherent state

667
00:39:28,950 --> 00:39:30,840
and from the vacuum.

668
00:39:30,840 --> 00:39:32,900
So therefore, when
I replace the vacuum

669
00:39:32,900 --> 00:39:38,760
by a strongly squeezed vacuum,
the beam, the current i1,

670
00:39:38,760 --> 00:39:40,800
lost half of its shot noise.

671
00:39:40,800 --> 00:39:43,930
We had n/2 photons
with a noise--

672
00:39:43,930 --> 00:39:45,530
with a variance, which was n/4.

673
00:39:48,087 --> 00:39:48,670
Any questions?

674
00:39:48,670 --> 00:39:49,170
Yes.

675
00:39:49,170 --> 00:39:53,376
AUDIENCE: Why are the subscripts
on the delta operator of

676
00:39:53,376 --> 00:39:56,340
[INAUDIBLE]?

677
00:39:56,340 --> 00:39:59,100
PROFESSOR: Because I should
have dropped them all together,

678
00:39:59,100 --> 00:40:02,590
saying we are only using
the quadrature component

679
00:40:02,590 --> 00:40:04,510
of the local oscillator now.

680
00:40:04,510 --> 00:40:08,360
Everything else is orthogonal
to it and cancels out.

681
00:40:08,360 --> 00:40:12,800
But if you want me to put them
back, they should all be equal.

682
00:40:12,800 --> 00:40:17,317
I compiled this writeup from
different lecture notes, which

683
00:40:17,317 --> 00:40:19,275
one had the cosine, one
had the sine component.

684
00:40:19,275 --> 00:40:21,930
I just made a
[? stitching error. ?]

685
00:40:21,930 --> 00:40:24,000
OK.

686
00:40:24,000 --> 00:40:26,240
So this was i1.

687
00:40:29,260 --> 00:40:36,450
If I would now look at--
let me change the color--

688
00:40:36,450 --> 00:40:42,340
at the current, i2, the only
anything which changes is

689
00:40:42,340 --> 00:40:47,260
that because the quantum
noise is transmitted

690
00:40:47,260 --> 00:40:50,070
with a positive
coefficient, that I

691
00:40:50,070 --> 00:40:53,460
have to put a plus sign here.

692
00:40:53,460 --> 00:40:56,390
And now, you realize the magic.

693
00:40:56,390 --> 00:41:01,510
When you form i1
plus i2, whatever

694
00:41:01,510 --> 00:41:04,430
came through the vacuum part,
or through the open part,

695
00:41:04,430 --> 00:41:06,430
cancels out.

696
00:41:06,430 --> 00:41:09,990
Therefore, when we calculated--
or when I asked you

697
00:41:09,990 --> 00:41:13,060
for the variance
in i1 plus i2, it

698
00:41:13,060 --> 00:41:15,990
was independent, whether
there was squeezing,

699
00:41:15,990 --> 00:41:18,530
whether whatever was
at the vacuum part.

700
00:41:18,530 --> 00:41:23,680
And therefore, it doesn't matter
whether I put my power meter,

701
00:41:23,680 --> 00:41:26,220
measuring both currents--
both beams-- before the beam

702
00:41:26,220 --> 00:41:27,720
splitter or after
the beam splitter.

703
00:41:31,050 --> 00:41:32,630
But-- yes, a question?

704
00:41:32,630 --> 00:41:35,020
AUDIENCE: This
last really depends

705
00:41:35,020 --> 00:41:38,323
on the convention
of where we put

706
00:41:38,323 --> 00:41:39,656
the negative sign in reflection?

707
00:41:39,656 --> 00:41:41,641
Because if I had put
the negative sign here

708
00:41:41,641 --> 00:41:43,890
and a positive there, then
I would have been canceling

709
00:41:43,890 --> 00:41:46,194
my delta a's instead of b's?

710
00:41:49,470 --> 00:41:51,320
[INAUDIBLE] the process?

711
00:41:51,320 --> 00:41:52,110
PROFESSOR: Yes.

712
00:41:58,085 --> 00:42:00,760
Well, so you should
[INAUDIBLE] conventions here.

713
00:42:00,760 --> 00:42:02,950
It's clear that you
have to be careful.

714
00:42:02,950 --> 00:42:05,490
We're talking about
[? face-sensitive ?] detection.

715
00:42:05,490 --> 00:42:08,300
So what happens
is-- number one is--

716
00:42:08,300 --> 00:42:09,990
there is really
physics associated.

717
00:42:09,990 --> 00:42:13,940
When you have a
beam splitter, one,

718
00:42:13,940 --> 00:42:16,320
let's assume you have
a piece of glass,

719
00:42:16,320 --> 00:42:19,040
which has some coating on it.

720
00:42:19,040 --> 00:42:24,850
One beam is reflected when
it goes from the vacuum

721
00:42:24,850 --> 00:42:26,410
to material interface.

722
00:42:26,410 --> 00:42:28,445
The other one is reflected
from the material

723
00:42:28,445 --> 00:42:29,610
to vacuum interface.

724
00:42:29,610 --> 00:42:31,953
AUDIENCE: So then, it depends
on which face of the beam

725
00:42:31,953 --> 00:42:33,030
that will be put which--?

726
00:42:33,030 --> 00:42:35,390
PROFESSOR: But one thing
is absolutely clear.

727
00:42:35,390 --> 00:42:39,190
The two reflected beams, one
of them has a minus sign.

728
00:42:39,190 --> 00:42:47,228
So therefore, when I do i1 minus
i2, the contribution of the--

729
00:42:47,228 --> 00:42:49,036
AUDIENCE: --If a's will cancel.

730
00:42:49,036 --> 00:42:50,510
PROFESSOR: Oh.

731
00:42:50,510 --> 00:42:52,680
It is clear that
in i1-- OK, yes.

732
00:42:52,680 --> 00:42:56,052
But the question
is whether we have

733
00:42:56,052 --> 00:43:00,120
to perform i1 minus
i2 or i1 plus i2.

734
00:43:00,120 --> 00:43:01,780
Well, I don't know
if it helps you,

735
00:43:01,780 --> 00:43:06,170
but what happens is we have
a propagating laser beam.

736
00:43:06,170 --> 00:43:10,270
If you would simply move your
mirror by half a wavelength,

737
00:43:10,270 --> 00:43:11,840
you get a minus sign.

738
00:43:11,840 --> 00:43:15,960
So eventually, all of the
phases-- the beam splitter

739
00:43:15,960 --> 00:43:18,880
phase, the propagating
phase-- all the phases

740
00:43:18,880 --> 00:43:20,140
have to be controlled.

741
00:43:20,140 --> 00:43:23,200
And in essence, what you do is
you put an experiment together,

742
00:43:23,200 --> 00:43:25,705
you measure some noise, and
then you move your mirror

743
00:43:25,705 --> 00:43:27,070
or you change your phase.

744
00:43:27,070 --> 00:43:30,524
And then, you really figure out
which noise is now subtracting

745
00:43:30,524 --> 00:43:32,440
and which noise is
constructively interfering.

746
00:43:36,019 --> 00:43:36,685
Other questions?

747
00:43:39,860 --> 00:43:40,630
OK.

748
00:43:40,630 --> 00:43:43,060
But the other thing
is now obvious.

749
00:43:43,060 --> 00:43:49,620
When we measure i1
minus i2, everything

750
00:43:49,620 --> 00:43:54,740
related to the coherent
input, to the original noise

751
00:43:54,740 --> 00:43:57,380
of the beam, cancels out.

752
00:43:57,380 --> 00:43:59,360
And what we are
measuring is only

753
00:43:59,360 --> 00:44:02,420
the noise which came in
through the other part.

754
00:44:02,420 --> 00:44:06,080
So this is the reason why, if
you have interesting light you

755
00:44:06,080 --> 00:44:08,790
want to measure it, you
use a local oscillator.

756
00:44:08,790 --> 00:44:11,140
But then, you do i1 minus i2.

757
00:44:11,140 --> 00:44:14,570
And this is what I presented to
you at the end of last class,

758
00:44:14,570 --> 00:44:17,970
is the way how
people have measured

759
00:44:17,970 --> 00:44:19,095
the signature of squeezing.

760
00:44:31,920 --> 00:44:32,600
Any questions?

761
00:44:36,780 --> 00:44:40,850
And finally, I think this is
one of the coolest things.

762
00:44:40,850 --> 00:44:46,432
Now, we want to talk about
the displaced squeezed vacuum.

763
00:44:46,432 --> 00:44:48,510
Now, we put everything together.

764
00:44:48,510 --> 00:44:50,970
Remember, we have
learned by applying

765
00:44:50,970 --> 00:44:54,520
the squeezing operator, how
we can squeeze the vacuum.

766
00:44:54,520 --> 00:44:56,740
And then, with the
displacement operator,

767
00:44:56,740 --> 00:45:01,370
I can move the displaced vacuum,
that there is an expectation

768
00:45:01,370 --> 00:45:04,450
value, alpha, for the field.

769
00:45:04,450 --> 00:45:09,070
So it now looks like a
squeezed coherent state.

770
00:45:09,070 --> 00:45:12,720
So this is now the
state we have generated.

771
00:45:12,720 --> 00:45:18,170
And I explained to
you that such a state

772
00:45:18,170 --> 00:45:23,330
can be generated by using a
beam splitter in the limit where

773
00:45:23,330 --> 00:45:25,970
the transmission
coefficient goes to 1.

774
00:45:25,970 --> 00:45:31,500
So the squeezed vacuum is pretty
much transmitted without loss.

775
00:45:31,500 --> 00:45:34,200
But now, we have a very
strong local oscillator.

776
00:45:36,790 --> 00:45:40,000
Also, the reflection
coefficient goes to 0.

777
00:45:40,000 --> 00:45:43,830
We can just crank up the
power in alpha in such a way

778
00:45:43,830 --> 00:45:47,380
that r alpha squared-- the
number of photons which get

779
00:45:47,380 --> 00:45:48,890
reflected-- is n.

780
00:45:48,890 --> 00:45:53,130
So the limit we are looking at
it is small n and large alpha.

781
00:45:53,130 --> 00:45:55,960
But what we keep constant--
throughout actually

782
00:45:55,960 --> 00:45:59,390
all those questions-- is that
we have n photons in that beam.

783
00:46:01,900 --> 00:46:06,690
And I explained
to you in class--

784
00:46:06,690 --> 00:46:08,690
and I showed it to
you with operators--

785
00:46:08,690 --> 00:46:11,850
that this setup
is really creating

786
00:46:11,850 --> 00:46:16,430
the displeased squeezed vacuum.

787
00:46:16,430 --> 00:46:17,100
OK.

788
00:46:17,100 --> 00:46:20,790
Now, we have one
laser beam, which

789
00:46:20,790 --> 00:46:25,670
has an average value
of the field, alpha.

790
00:46:25,670 --> 00:46:26,770
It has certain properties.

791
00:46:29,700 --> 00:46:33,440
And now, I want to ask you, what
is the variance of this beam?

792
00:46:33,440 --> 00:46:35,820
We measure the
photocurrent of this beam.

793
00:46:35,820 --> 00:46:39,790
We measure-- of course, the
photocurrent is n photons.

794
00:46:39,790 --> 00:46:41,045
But what is this variance?

795
00:46:45,650 --> 00:46:59,440
So the choices are
we have n photons.

796
00:46:59,440 --> 00:47:02,440
Is the variance n?

797
00:47:02,440 --> 00:47:05,510
We learned before, with
the 50-50 beam splitter,

798
00:47:05,510 --> 00:47:07,990
that we were able, by
squeezing the vacuum,

799
00:47:07,990 --> 00:47:11,540
to eliminate half
of the shot noise.

800
00:47:11,540 --> 00:47:13,375
Maybe what I've done
is wrong, and we get 2

801
00:47:13,375 --> 00:47:15,500
and we increase the noise?

802
00:47:15,500 --> 00:47:19,540
Or does the noise
really go to 0,

803
00:47:19,540 --> 00:47:23,850
which means it's on the order
of epsilon, whatever I define

804
00:47:23,850 --> 00:47:29,150
for the squeezing for the short
axis of this squeezed vacuum.

805
00:47:29,150 --> 00:47:30,980
So now, we are not
subtracting currents.

806
00:47:30,980 --> 00:47:32,200
No i1 minus i2.

807
00:47:32,200 --> 00:47:35,680
We have one beam hitting
straight the photodiode.

808
00:47:35,680 --> 00:47:39,100
There are n photons
per unit time.

809
00:47:39,100 --> 00:47:42,465
And the question is, what are
the fluctuations of this beam?

810
00:47:42,465 --> 00:47:43,340
What is the variance?

811
00:47:59,780 --> 00:48:03,520
Well, the correct
answer is this one.

812
00:48:07,200 --> 00:48:10,190
What happens is the following.

813
00:48:10,190 --> 00:48:12,180
In terms of fields--
you should go

814
00:48:12,180 --> 00:48:13,680
through that again
a little bit more

815
00:48:13,680 --> 00:48:17,050
slowly-- but what happens
is the coherent state is

816
00:48:17,050 --> 00:48:18,350
sort of this disk.

817
00:48:18,350 --> 00:48:20,930
But when we make alpha
larger and larger,

818
00:48:20,930 --> 00:48:23,930
the disk of the coherent
state is the same.

819
00:48:23,930 --> 00:48:26,360
But now, we have of a
reflection coefficient,

820
00:48:26,360 --> 00:48:30,260
which goes to 0, such that
the product of r alpha

821
00:48:30,260 --> 00:48:31,550
stays the same.

822
00:48:31,550 --> 00:48:34,280
But that means the
reflected beam,

823
00:48:34,280 --> 00:48:37,787
we have divided the disk by
the reflection coefficient--

824
00:48:37,787 --> 00:48:39,620
multiplied with the
reflection coefficient--

825
00:48:39,620 --> 00:48:43,430
so therefore, the original
noise of the coherent beam

826
00:48:43,430 --> 00:48:47,810
has been completely eliminated
because of the small reflection

827
00:48:47,810 --> 00:48:48,900
coefficient.

828
00:48:48,900 --> 00:48:54,620
So the only noise which is
there, is from the other part.

829
00:48:54,620 --> 00:48:56,635
But we have strongly
squeezed the noise

830
00:48:56,635 --> 00:48:58,210
in this quadrature component.

831
00:48:58,210 --> 00:49:00,580
And therefore, the
noise goes really down

832
00:49:00,580 --> 00:49:04,830
to 0 or to I think it's
epsilon squared when epsilon

833
00:49:04,830 --> 00:49:09,560
is the normalized widths of
the narrow part of the ellipse.

834
00:49:24,110 --> 00:49:31,360
I've written down the
math for you here,

835
00:49:31,360 --> 00:49:33,510
which is the following.

836
00:49:33,510 --> 00:49:36,270
We have a reflection coefficient
for the coherent beam

837
00:49:36,270 --> 00:49:39,860
and the reflection coefficient
for the fluctuating fields.

838
00:49:39,860 --> 00:49:42,920
Whereas, the squeezed
fluctuations,

839
00:49:42,920 --> 00:49:44,990
they are fully transmitted.

840
00:49:44,990 --> 00:49:46,710
And what I'm
completely neglecting

841
00:49:46,710 --> 00:49:49,300
is-- and you'll realize it
already-- the other quadrature

842
00:49:49,300 --> 00:49:50,110
component.

843
00:49:50,110 --> 00:49:53,610
Because the moment I square
it to obtain the current,

844
00:49:53,610 --> 00:49:57,420
I'm only considering
parts, which

845
00:49:57,420 --> 00:49:59,117
are in the right
quadrature component,

846
00:49:59,117 --> 00:50:00,450
that they get enhanced by alpha.

847
00:50:03,470 --> 00:50:09,110
So if I square that and
neglect all terms which do not

848
00:50:09,110 --> 00:50:14,080
get multiplied with
alpha, I get this line.

849
00:50:14,080 --> 00:50:18,580
And now, you see that
this term here-- the one

850
00:50:18,580 --> 00:50:20,980
which had the fluctuations
of the coherent beam--

851
00:50:20,980 --> 00:50:22,460
is what I just said.

852
00:50:22,460 --> 00:50:25,600
r times alpha is n.

853
00:50:25,600 --> 00:50:27,140
So this is n.

854
00:50:27,140 --> 00:50:29,060
It would be the normal variance.

855
00:50:29,060 --> 00:50:31,110
But now, we have
another power of r.

856
00:50:31,110 --> 00:50:33,907
So therefore, delta a1
really got diminished.

857
00:50:33,907 --> 00:50:35,740
I mean, you see that
actually already better

858
00:50:35,740 --> 00:50:36,950
in the previous line.

859
00:50:36,950 --> 00:50:40,270
So therefore, in the limit of
r, in the limit considered,

860
00:50:40,270 --> 00:50:42,480
there is no contribution
from the fluctuations

861
00:50:42,480 --> 00:50:43,830
of the coherent beam.

862
00:50:43,830 --> 00:50:46,930
The only fluctuations
come from the squeezing.

863
00:50:46,930 --> 00:50:51,370
And therefore, the variance
is, well, 0 or epsilon squared.

864
00:50:54,000 --> 00:50:58,070
So you can create a beam, which
has no fluctuations-- or almost

865
00:50:58,070 --> 00:51:02,760
no fluctuation-- in
the photon number

866
00:51:02,760 --> 00:51:07,810
by mixing a strong local
oscillator with a squeeze

867
00:51:07,810 --> 00:51:08,310
vacuum.

868
00:51:12,220 --> 00:51:16,110
And that would mean now,
if you take this beam

869
00:51:16,110 --> 00:51:18,360
and send it through
a cesium cell

870
00:51:18,360 --> 00:51:23,640
and do spectroscopy of cesium,
you have reduced noise,

871
00:51:23,640 --> 00:51:27,105
and you are more sensitive
to obtain, to observe,

872
00:51:27,105 --> 00:51:30,360
a small quantity of cesium.

873
00:51:30,360 --> 00:51:32,800
I've posted it actually--
this is exactly what

874
00:51:32,800 --> 00:51:36,740
I described here, how
the first spectroscopy

875
00:51:36,740 --> 00:51:38,250
with squeezed light was done.

876
00:51:38,250 --> 00:51:40,560
And I've posted the paper
of Jeff Kimble's group

877
00:51:40,560 --> 00:51:41,370
on our website.

878
00:51:50,290 --> 00:52:01,860
So I'm wondering, by going
through these examples,

879
00:52:01,860 --> 00:52:06,923
if I would ask you that the
shot noise, which is square root

880
00:52:06,923 --> 00:52:11,430
n in the fields and
n in the variance,

881
00:52:11,430 --> 00:52:14,200
after looking at the beam
splitter and the modes which

882
00:52:14,200 --> 00:52:18,700
are involved, how would you
now answer this question?

883
00:52:18,700 --> 00:52:23,380
Is the shot noise related
to quantum measurement,

884
00:52:23,380 --> 00:52:27,840
is it related to quantum
fields, both, or none?

885
00:52:45,176 --> 00:52:46,550
That's interpretation
of physics;

886
00:52:46,550 --> 00:52:48,820
I think there is
no right and wrong.

887
00:52:48,820 --> 00:52:51,800
I would actually very
strongly-- my opinion is,

888
00:52:51,800 --> 00:52:53,830
it's B. It's really
the quantum fields.

889
00:52:53,830 --> 00:52:55,850
The quantum fields
which are split.

890
00:52:55,850 --> 00:52:58,010
If the noise were done
in the measurement,

891
00:52:58,010 --> 00:53:01,615
you cannot have constructive
and destructive interference due

892
00:53:01,615 --> 00:53:04,310
to the positive and negative
reflection coefficient.

893
00:53:04,310 --> 00:53:07,320
So for me, the noise, which
is responsible for shot noise,

894
00:53:07,320 --> 00:53:09,710
is a field which
can be split where

895
00:53:09,710 --> 00:53:13,490
you have constructive and
destructive interference

896
00:53:13,490 --> 00:53:15,590
even before the measurement
process takes place.

897
00:53:18,530 --> 00:53:21,337
But this is sort of
my interpretation.

898
00:53:21,337 --> 00:53:26,960
And let me maybe point out why
I think that for those cases,

899
00:53:26,960 --> 00:53:30,920
the quantum measurement
process is not relevant.

900
00:53:30,920 --> 00:53:34,170
Here, we've always talked
about square root n

901
00:53:34,170 --> 00:53:36,820
fluctuations, where n is big.

902
00:53:36,820 --> 00:53:39,880
So the fluctuations of
the coherent field--

903
00:53:39,880 --> 00:53:43,070
all of the fluctuations
we were discussing here--

904
00:53:43,070 --> 00:53:44,980
had many, many photons.

905
00:53:44,980 --> 00:53:49,010
So to some extent that we have
maybe just a single photon,

906
00:53:49,010 --> 00:53:51,300
and we can only get one
click and not two clicks,

907
00:53:51,300 --> 00:53:52,130
doesn't measure it.

908
00:53:52,130 --> 00:53:53,180
That doesn't matter here.

909
00:53:53,180 --> 00:53:55,160
All the fluctuations
we considered

910
00:53:55,160 --> 00:53:58,020
here were fluctuations
in the field, where

911
00:53:58,020 --> 00:54:00,370
many, many photons
were involved.

912
00:54:00,370 --> 00:54:02,380
So, therefore, I think
the quantum character

913
00:54:02,380 --> 00:54:06,630
of the measurement process
does not really show up.

914
00:54:06,630 --> 00:54:10,740
If I would construct
something-- other examples--

915
00:54:10,740 --> 00:54:14,860
where we wouldn't always have
this strong local oscillator,

916
00:54:14,860 --> 00:54:16,860
this strong homodyning.

917
00:54:16,860 --> 00:54:18,520
Then I think we
would find quantum

918
00:54:18,520 --> 00:54:21,460
noise of a single photon,
and then I would say,

919
00:54:21,460 --> 00:54:24,530
we're getting closer, that's
really quantum measurement.

920
00:54:24,530 --> 00:54:27,620
But here, I would say
the short noise related

921
00:54:27,620 --> 00:54:31,210
to a strong local
oscillator-- at least,

922
00:54:31,210 --> 00:54:33,540
I find it easier for go
through those questions

923
00:54:33,540 --> 00:54:37,080
if I think about it as a quantum
field, which can be split,

924
00:54:37,080 --> 00:54:37,850
added up.

925
00:54:37,850 --> 00:54:39,600
Constructive and
destructive interference.

926
00:54:42,240 --> 00:54:42,910
Good.

927
00:54:42,910 --> 00:54:43,650
Questions?

928
00:54:43,650 --> 00:54:45,284
Yes?

929
00:54:45,284 --> 00:54:47,700
AUDIENCE: So, when we're talking
about [INAUDIBLE], can we

930
00:54:47,700 --> 00:54:49,780
actually talk about it
without referring to i,

931
00:54:49,780 --> 00:54:51,613
because that makes me
think of measurements.

932
00:54:58,144 --> 00:54:59,830
PROFESSOR: You mean
i, as the current?

933
00:54:59,830 --> 00:55:00,455
AUDIENCE: Yeah.

934
00:55:03,000 --> 00:55:06,380
PROFESSOR: But we
have to do something.

935
00:55:06,380 --> 00:55:09,860
OK, what happens is
i was nothing else

936
00:55:09,860 --> 00:55:14,260
than the expectation
value of a dagger a.

937
00:55:23,840 --> 00:55:29,660
In other words, whenever we
had a photodiode-- I mean,

938
00:55:29,660 --> 00:55:33,770
we learn how to
transform modes, and i

939
00:55:33,770 --> 00:55:36,920
was synonymous to
performing a measurement,

940
00:55:36,920 --> 00:55:41,200
and the measurement
is a dagger a.

941
00:55:41,200 --> 00:55:43,880
And when we talked
about i squared,

942
00:55:43,880 --> 00:55:47,010
in order to get the noise,
the operator involved

943
00:55:47,010 --> 00:55:49,590
was a a dagger, a a dagger.

944
00:55:57,780 --> 00:56:01,110
I think I will ask you
something about the displaced.

945
00:56:01,110 --> 00:56:02,760
Yeah, it's too nice a problem.

946
00:56:02,760 --> 00:56:05,960
I should ask about the
displaced squeezed vacuum

947
00:56:05,960 --> 00:56:07,030
in the next problem set.

948
00:56:07,030 --> 00:56:10,050
I really want you to calculate,
maybe with a higher order

949
00:56:10,050 --> 00:56:14,060
term-- it's a three-liner-- in
operators, what is the noise?

950
00:56:14,060 --> 00:56:18,440
And in your calculation,
photodiode will not appear.

951
00:56:18,440 --> 00:56:21,840
What will appear is
"Calculate the expectation

952
00:56:21,840 --> 00:56:24,440
value of a dagger a.

953
00:56:24,440 --> 00:56:26,472
If you call it a
measurement process,

954
00:56:26,472 --> 00:56:28,680
it's [? inobservable ?] and
it's an expectation value

955
00:56:28,680 --> 00:56:29,680
of an emission operator.

956
00:56:44,209 --> 00:56:44,875
Other questions?

957
00:56:47,730 --> 00:56:53,208
AUDIENCE: [INAUDIBLE]
understand the [INAUDIBLE]

958
00:56:53,208 --> 00:57:01,176
for a stream of particles,
which [INAUDIBLE]

959
00:57:01,176 --> 00:57:04,662
with a constant
probability per unit time?

960
00:57:04,662 --> 00:57:08,646
[INAUDIBLE] The detector
is also [INAUDIBLE].

961
00:57:16,614 --> 00:57:18,606
I think you also
get the square root.

962
00:57:24,090 --> 00:57:27,970
PROFESSOR: Well, a lot of
classical counting statistics

963
00:57:27,970 --> 00:57:29,380
needs to short noise.

964
00:57:29,380 --> 00:57:30,800
If you have
independent particles

965
00:57:30,800 --> 00:57:34,290
with a certain probability,
then you get short noise.

966
00:57:34,290 --> 00:57:38,620
But I would sort of
press you hard to make

967
00:57:38,620 --> 00:57:43,930
a classical explanation for
the last example I gave you,

968
00:57:43,930 --> 00:57:46,220
where we had a strong
local oscillator.

969
00:57:46,220 --> 00:57:50,990
We reflected it and superimposed
it with squeezed vacuum.

970
00:57:50,990 --> 00:57:54,710
At that point, all the
short noise has disappeared.

971
00:57:57,640 --> 00:58:02,050
So therefore, I would be
hard-pressed to connect it

972
00:58:02,050 --> 00:58:05,950
with the classical probability
of detecting a photon.

973
00:58:05,950 --> 00:58:09,093
Of course, classically
speaking, the empty part

974
00:58:09,093 --> 00:58:12,040
of a beam splitter is
empty, and squeezed

975
00:58:12,040 --> 00:58:16,120
vacuum doesn't
exist classically.

976
00:58:16,120 --> 00:58:17,850
Yes, [INAUDIBLE].

977
00:58:17,850 --> 00:58:18,760
AUDIENCE: So it might be
kind of a silly question,

978
00:58:18,760 --> 00:58:20,426
but we say we have a
minimum uncertainty

979
00:58:20,426 --> 00:58:22,962
state when delta x delta p is 0.

980
00:58:22,962 --> 00:58:24,336
But a lot of times,
we talk about

981
00:58:24,336 --> 00:58:27,051
the number-phase uncertainty.

982
00:58:27,051 --> 00:58:29,050
So, I'm just curious,
because the coherent state

983
00:58:29,050 --> 00:58:32,790
has the same circle
of-- the area

984
00:58:32,790 --> 00:58:34,692
of the circle is
the same, right?

985
00:58:34,692 --> 00:58:36,150
I know this is
wrong, but can't you

986
00:58:36,150 --> 00:58:39,880
always make the delta
m delta phi uncertainty

987
00:58:39,880 --> 00:58:42,666
very small by having
a very large n?

988
00:58:42,666 --> 00:58:46,080
Delta n stays the same, but
the angular spread always

989
00:58:46,080 --> 00:58:51,980
goes down as you make the circle
farther away from the origin?

990
00:59:01,240 --> 00:59:04,030
PROFESSOR: So you're talking
about an uncertainty relation

991
00:59:04,030 --> 00:59:07,910
we have not discussed.

992
00:59:11,180 --> 00:59:13,840
In the last homework,
you looked at it.

993
00:59:13,840 --> 00:59:17,570
So what happens is if you
have a coherent state,

994
00:59:17,570 --> 00:59:21,680
you would say the phase angle
phi has this uncertainty.

995
00:59:24,280 --> 00:59:28,465
And then there's an
uncertainty relation.

996
00:59:32,027 --> 00:59:34,402
AUDIENCE: Maybe you should
have talked about this problem

997
00:59:34,402 --> 00:59:35,848
yesterday.

998
00:59:35,848 --> 00:59:39,704
I think the uncertainty
and the size of the disc

999
00:59:39,704 --> 00:59:44,668
is constant,
[INAUDIBLE] in alpha.

1000
00:59:44,668 --> 00:59:49,608
But alpha is proportional not
to n, but the square root of n.

1001
00:59:49,608 --> 00:59:55,042
[INAUDIBLE] an arrow propagation
[INAUDIBLE] square root of n.

1002
00:59:58,006 --> 01:00:00,970
If we move the
circle further away,

1003
01:00:00,970 --> 01:00:05,416
the uncertainty alpha stays the
same, but the uncertainty of n

1004
01:00:05,416 --> 01:00:06,404
goes actually up.

1005
01:00:06,404 --> 01:00:08,380
But we don't see
this [INAUDIBLE],

1006
01:00:08,380 --> 01:00:10,356
because we only have
alpha, not [INAUDIBLE].

1007
01:00:16,290 --> 01:00:17,915
PROFESSOR: Yes, you
have to be careful.

1008
01:00:20,450 --> 01:00:23,500
That's another comment
I wanted to make,

1009
01:00:23,500 --> 01:00:25,990
but thanks for
reminding of that.

1010
01:00:25,990 --> 01:00:29,212
When we have coherent states,
it looks that the uncertainty

1011
01:00:29,212 --> 01:00:32,360
is always the same,
because it's a disk.

1012
01:00:32,360 --> 01:00:37,880
But we learned today,
and in the last class

1013
01:00:37,880 --> 01:00:41,520
that the uncertainty of the
coherent state when you measure

1014
01:00:41,520 --> 01:00:44,840
the photocurrent
is square root n.

1015
01:00:44,840 --> 01:00:50,500
What happens is the following:
the uncertainty-- and this is

1016
01:00:50,500 --> 01:00:53,510
what I was referring
to-- the uncertainty,

1017
01:00:53,510 --> 01:00:56,920
let's say delta a in one
[? quadrature ?] component, is

1018
01:00:56,920 --> 01:01:00,260
constant for the coherent
state, independent of n.

1019
01:01:00,260 --> 01:01:02,440
But that would mean
what you really

1020
01:01:02,440 --> 01:01:06,120
calculate, when you
calculate the photocurrent,

1021
01:01:06,120 --> 01:01:09,330
is alpha plus delta
a which is constant.

1022
01:01:09,330 --> 01:01:11,080
But then you square it.

1023
01:01:11,080 --> 01:01:13,180
In the limit of a
strong coherent state,

1024
01:01:13,180 --> 01:01:16,150
you get alpha squared, which
is the number of photons,

1025
01:01:16,150 --> 01:01:19,940
plus 2 alpha times delta a.

1026
01:01:19,940 --> 01:01:22,290
So the delta a,
which is constant,

1027
01:01:22,290 --> 01:01:26,800
gets multiplied with
square root n here.

1028
01:01:26,800 --> 01:01:29,925
So therefore when you go
to launch a coherent state,

1029
01:01:29,925 --> 01:01:32,480
you increase the
number uncertainty,

1030
01:01:32,480 --> 01:01:36,070
but by the same factor, you
reduce the phase uncertainty.

1031
01:01:50,670 --> 01:01:52,950
Well, I thought this
would take half an hour.

1032
01:01:52,950 --> 01:01:58,110
It took an hour, but I hope
it was time well spent.

1033
01:01:58,110 --> 01:02:02,280
So let's get started
on the next unit, which

1034
01:02:02,280 --> 01:02:07,430
is fully dedicated
to single photons.

1035
01:02:07,430 --> 01:02:11,630
So after learning
about different aspects

1036
01:02:11,630 --> 01:02:14,020
of the electromagnetic
field, we now

1037
01:02:14,020 --> 01:02:18,270
want to focus on
the purest quantum

1038
01:02:18,270 --> 01:02:20,810
states, which are
single photons.

1039
01:02:20,810 --> 01:02:24,030
And this unit it is
definitely motivated

1040
01:02:24,030 --> 01:02:26,060
by quantum information
processing.

1041
01:02:26,060 --> 01:02:29,110
We want to use control
over single photos

1042
01:02:29,110 --> 01:02:33,120
to realize interesting devices.

1043
01:02:33,120 --> 01:02:36,200
We talk about beam splitters
and phase shifters,

1044
01:02:36,200 --> 01:02:39,250
but then we put beam splitters
and phase shifters together

1045
01:02:39,250 --> 01:02:42,050
to make Mach-Zehnder
interferometers.

1046
01:02:42,050 --> 01:02:45,210
And then eventually
I can show you

1047
01:02:45,210 --> 01:02:49,200
how Mach-Zehnder interferometers
with a non-linearity

1048
01:02:49,200 --> 01:02:53,760
can be used to
realize quantum gates.

1049
01:02:53,760 --> 01:02:57,280
But let me make maybe
one general comment here.

1050
01:03:00,110 --> 01:03:05,530
This unit here on single photons
has actually two purposes.

1051
01:03:05,530 --> 01:03:08,530
It tries to give you some
fundamentals, which are really

1052
01:03:08,530 --> 01:03:11,720
necessary to follow recently
literature in quantum

1053
01:03:11,720 --> 01:03:14,660
information science, so
we introduce some concepts

1054
01:03:14,660 --> 01:03:18,820
of gates and truth
tables and such.

1055
01:03:18,820 --> 01:03:21,890
But for those of
you who are maybe

1056
01:03:21,890 --> 01:03:24,110
less interested in quantum
information science,

1057
01:03:24,110 --> 01:03:29,687
I have to say that as a more
traditional atomic physicist,

1058
01:03:29,687 --> 01:03:32,020
I've never worked in quantum
information science myself.

1059
01:03:32,020 --> 01:03:34,325
I started with atomic
molecular spectroscopy,

1060
01:03:34,325 --> 01:03:37,680
and advanced from laser
cooling to quantum gases.

1061
01:03:37,680 --> 01:03:43,430
I feel that using the
quantum logic formalism

1062
01:03:43,430 --> 01:03:47,250
is a very elegant way of
organizing your thoughts.

1063
01:03:47,250 --> 01:03:50,315
So I do feel even for the
rest of the world, who is not

1064
01:03:50,315 --> 01:03:52,190
doing kind quantum
information process, which

1065
01:03:52,190 --> 01:03:54,630
is a pretty big
fraction of AMO science

1066
01:03:54,630 --> 01:03:59,740
these days, even
for the rest of us,

1067
01:03:59,740 --> 01:04:04,250
to use concepts of quantum
logic and quantum information

1068
01:04:04,250 --> 01:04:05,900
is very, very powerful.

1069
01:04:05,900 --> 01:04:08,030
So in other words, I
would say right now

1070
01:04:08,030 --> 01:04:13,970
every atomic physicist should
have some knowledge in what

1071
01:04:13,970 --> 01:04:16,440
happens to beam splitters, what
happens to singular photons

1072
01:04:16,440 --> 01:04:20,570
when they're manipulating,
and the way, the approach,

1073
01:04:20,570 --> 01:04:23,690
how we formulate it using
concepts of quantum information

1074
01:04:23,690 --> 01:04:26,430
processing is
extremely powerful.

1075
01:04:26,430 --> 01:04:28,610
Like in quantum
mechanics, you need

1076
01:04:28,610 --> 01:04:30,530
to know the Schrodinger
representation

1077
01:04:30,530 --> 01:04:32,340
and the Heisenberg
interpretation.

1078
01:04:32,340 --> 01:04:33,850
You can't get away with one.

1079
01:04:33,850 --> 01:04:38,090
Similarly, for some aspects
of the propagation of quantum

1080
01:04:38,090 --> 01:04:41,040
states, you should just use the
language of quantum information

1081
01:04:41,040 --> 01:04:42,870
processing.

1082
01:04:42,870 --> 01:04:45,300
Just sort of to give you
already an outlook where

1083
01:04:45,300 --> 01:04:47,820
I think everything
comes together,

1084
01:04:47,820 --> 01:04:50,790
I've taught for 15
years, 20 years,

1085
01:04:50,790 --> 01:04:53,720
about decoherence
and master equation.

1086
01:04:53,720 --> 01:04:56,540
And I know how to derive it
from general quantum physics,

1087
01:04:56,540 --> 01:04:58,960
and I will show you that
later in this course.

1088
01:04:58,960 --> 01:05:00,910
But then I learned from
Professor Ike Chuang,

1089
01:05:00,910 --> 01:05:02,980
when I co-taught
the course with him,

1090
01:05:02,980 --> 01:05:06,630
that he could actually make
a model for decoherence

1091
01:05:06,630 --> 01:05:10,540
by just passing light through
many, many beams splitters.

1092
01:05:10,540 --> 01:05:12,120
So there is a beam
splitter model

1093
01:05:12,120 --> 01:05:14,500
for decoherence, which
is just beautiful.

1094
01:05:14,500 --> 01:05:18,010
You can give derive a master
equation by just sending light

1095
01:05:18,010 --> 01:05:20,120
through many, many
beams splitters.

1096
01:05:20,120 --> 01:05:22,410
So that taught me something,
that actually those

1097
01:05:22,410 --> 01:05:27,520
was elementary devices for
quantum information science can

1098
01:05:27,520 --> 01:05:31,160
really act as conceptual
models for more complicated,

1099
01:05:31,160 --> 01:05:33,940
more general things.

1100
01:05:33,940 --> 01:05:36,370
So I've decided now,
over the last few years,

1101
01:05:36,370 --> 01:05:38,530
to make quantum
information language

1102
01:05:38,530 --> 01:05:42,370
an essential ingredient
of this course.

1103
01:05:42,370 --> 01:05:44,770
But I should say over
the last couple of years,

1104
01:05:44,770 --> 01:05:49,220
I've de-emphasized some of the
more formal algorithmic things,

1105
01:05:49,220 --> 01:05:51,240
but I'm focusing
on the concepts.

1106
01:05:51,240 --> 01:05:52,740
And I think some
concepts are really

1107
01:05:52,740 --> 01:05:57,230
highlighted by using
quantum information aspects.

1108
01:05:57,230 --> 01:05:59,170
So with that
introduction, we want

1109
01:05:59,170 --> 01:06:02,360
to talk about single photons.

1110
01:06:02,360 --> 01:06:10,060
And I want to introduce for
you the single photon qubit.

1111
01:06:10,060 --> 01:06:15,610
Now, [INAUDIBLE] would think, if
you talk about single photons,

1112
01:06:15,610 --> 01:06:18,850
you automatically
have one qubit.

1113
01:06:18,850 --> 01:06:20,910
0 is no photon.

1114
01:06:23,740 --> 01:06:25,740
One is the single photon state.

1115
01:06:34,510 --> 01:06:37,120
We make in addition
later on, this

1116
01:06:37,120 --> 01:06:39,740
is not the most
robust qubit, because

1117
01:06:39,740 --> 01:06:43,110
if, due to unavoidable losses
or limited quantum efficiency,

1118
01:06:43,110 --> 01:06:45,380
you have one photon,
and you lose it,

1119
01:06:45,380 --> 01:06:47,970
you think the bit
has been flipped.

1120
01:06:47,970 --> 01:06:49,900
So you don't want
to really use that,

1121
01:06:49,900 --> 01:06:55,376
but we will use, in a
moment, a single photon

1122
01:06:55,376 --> 01:06:57,190
in two different quantum states.

1123
01:06:57,190 --> 01:07:00,010
And then, if you detect zero,
you know you've lost it.

1124
01:07:00,010 --> 01:07:03,520
We will talk about one photon,
which is either here or there.

1125
01:07:03,520 --> 01:07:06,260
And this is zero of
our qubit, and this

1126
01:07:06,260 --> 01:07:07,750
would be one of our qubit.

1127
01:07:07,750 --> 01:07:09,110
But we will come to that later.

1128
01:07:09,110 --> 01:07:11,720
I first need the Hilbert
space of zero one

1129
01:07:11,720 --> 01:07:14,120
to introduce a few
things, and then we'll

1130
01:07:14,120 --> 01:07:23,930
use what we've learned to define
the dual-rail single photon

1131
01:07:23,930 --> 01:07:26,050
state.

1132
01:07:26,050 --> 01:07:28,985
But for now we don't have
to make this distinction.

1133
01:07:31,775 --> 01:07:33,150
The question we
want to answer is

1134
01:07:33,150 --> 01:07:42,940
what happens when
those states pass

1135
01:07:42,940 --> 01:07:44,090
through optical components.

1136
01:07:53,110 --> 01:07:56,800
I want to show you how
beam splitters and phase

1137
01:07:56,800 --> 01:08:01,940
shifters can be used-- very
simple optical components--

1138
01:08:01,940 --> 01:08:05,690
they can be used to realize
an arbitrary single qubit

1139
01:08:05,690 --> 01:08:08,040
operation.

1140
01:08:08,040 --> 01:08:10,370
And then we introduce
non-linear interferometers,

1141
01:08:10,370 --> 01:08:13,890
and we get two qubit operations.

1142
01:08:13,890 --> 01:08:23,869
So let's start out very simple
and talk about phase shifters

1143
01:08:23,869 --> 01:08:24,660
and beam splitters.

1144
01:08:33,010 --> 01:08:34,779
And this is pretty
much how far we

1145
01:08:34,779 --> 01:08:38,950
will get today in this chapter.

1146
01:08:38,950 --> 01:08:52,399
Well, if you have a single
photon and time passes on,

1147
01:08:52,399 --> 01:08:55,479
the single photon
gets phase shifted.

1148
01:08:59,590 --> 01:09:19,706
If you pass it through
a dispersive medium,

1149
01:09:19,706 --> 01:09:21,080
it will have
another phase shift.

1150
01:09:24,200 --> 01:09:31,649
So this is how phase
shifts can be realized.

1151
01:09:31,649 --> 01:09:35,950
But there is one caveat, and
that immediately tells us

1152
01:09:35,950 --> 01:09:39,640
we can't get away with just
a single photon in one fiber

1153
01:09:39,640 --> 01:09:42,140
and phase shifting it,
because everything phase

1154
01:09:42,140 --> 01:09:43,040
shifts a photon.

1155
01:09:43,040 --> 01:09:47,120
Time and everything which
is in the beam path.

1156
01:09:47,120 --> 01:09:50,109
So whenever we talk about the
phase, we need a reference.

1157
01:09:59,090 --> 01:10:02,220
So therefore, we're
immediately drawn

1158
01:10:02,220 --> 01:10:06,720
that one mode and one
photon is not enough,

1159
01:10:06,720 --> 01:10:09,216
even if you're interested
in one-photon gates,

1160
01:10:09,216 --> 01:10:09,965
we need two modes.

1161
01:10:15,340 --> 01:10:17,995
Just think about two wave
guides or two optical fibers.

1162
01:10:21,810 --> 01:10:31,310
We call them mode A and B.
If you put one photon in--

1163
01:10:31,310 --> 01:10:35,530
and this is now the symbol for
a phase shifter, a little box

1164
01:10:35,530 --> 01:10:45,670
with the letter
phi-- then we get

1165
01:10:45,670 --> 01:10:48,900
a phase shift of the state one.

1166
01:10:48,900 --> 01:10:52,380
And the second mode A
acts as a reference mode

1167
01:10:52,380 --> 01:10:55,050
to detect the shift
of phase, let's say

1168
01:10:55,050 --> 01:10:58,590
by interfering the two photons.

1169
01:10:58,590 --> 01:11:05,650
So in general, we
have an input state,

1170
01:11:05,650 --> 01:11:08,580
and we obtain an output state,
which has a phase shift.

1171
01:11:11,440 --> 01:11:13,315
We'll come back to
phase shifters later.

1172
01:11:28,230 --> 01:11:30,110
So we'll have two modes.

1173
01:11:30,110 --> 01:11:36,220
The next thing we can do
is we can combine modes,

1174
01:11:36,220 --> 01:11:38,850
and this is done
with a beam splitter.

1175
01:11:38,850 --> 01:11:40,810
So we've talked
about beam splitters

1176
01:11:40,810 --> 01:11:43,280
in the context of
homodyne detection.

1177
01:11:43,280 --> 01:11:45,300
We've talked about
beams splitters most

1178
01:11:45,300 --> 01:11:47,650
of the first hour with
a clicker question,

1179
01:11:47,650 --> 01:11:51,590
and I was mainly referring
to the beam splitter

1180
01:11:51,590 --> 01:11:54,000
before in a rather
non-formal way.

1181
01:11:54,000 --> 01:11:58,720
If you've a 50:50 beam splitter,
and you have two inputs,

1182
01:11:58,720 --> 01:12:01,230
you get a symmetric and
antisymmetric combination

1183
01:12:01,230 --> 01:12:03,200
of the two, if it's a
50:50 beam splitter.

1184
01:12:03,200 --> 01:12:04,790
What else can you get?

1185
01:12:04,790 --> 01:12:10,410
Or, I used the concept of just
reflection and transmission

1186
01:12:10,410 --> 01:12:11,240
coefficients.

1187
01:12:11,240 --> 01:12:13,270
I think that's all important
to get an intuitive

1188
01:12:13,270 --> 01:12:14,820
feel for the beam splitter.

1189
01:12:14,820 --> 01:12:17,550
But now I use a more
rigorous approach.

1190
01:12:17,550 --> 01:12:21,960
I have a beam splitter,
which is characterized

1191
01:12:21,960 --> 01:12:22,955
by an angle theta.

1192
01:12:25,830 --> 01:12:28,630
We will later find
out what theta is.

1193
01:12:28,630 --> 01:12:35,060
And we have an input mode A
and a second input mode B,

1194
01:12:35,060 --> 01:12:37,540
and those modes get mixed now.

1195
01:12:37,540 --> 01:12:43,260
We have output modes B prime
and output modes A prime.

1196
01:12:43,260 --> 01:12:47,360
And we can use the
Schrodinger picture

1197
01:12:47,360 --> 01:12:50,270
and talk about the photon
states in those modes,

1198
01:12:50,270 --> 01:12:52,790
but I prefer to use
the Heisenberg picture,

1199
01:12:52,790 --> 01:12:55,770
so when I put A, A
prime, B and B prime in,

1200
01:12:55,770 --> 01:12:59,760
just think about
annihilation operators which

1201
01:12:59,760 --> 01:13:03,730
annihilate photons
in those modes.

1202
01:13:03,730 --> 01:13:08,301
So what I postulate
is that this system

1203
01:13:08,301 --> 01:13:10,050
is described by the
following Hamiltonian.

1204
01:13:21,680 --> 01:13:24,450
And I'm sort of telling
you, this Hamiltonian

1205
01:13:24,450 --> 01:13:26,500
describes this beam splitter.

1206
01:13:26,500 --> 01:13:27,910
You will say, why and how?

1207
01:13:27,910 --> 01:13:32,020
Well, I will tell you what this
Hamiltonian does to the modes,

1208
01:13:32,020 --> 01:13:34,090
and then we have the
quantum description

1209
01:13:34,090 --> 01:13:37,840
of the beams splitter, and you
will then see in your homework

1210
01:13:37,840 --> 01:13:40,870
that this quantum description
of the beam splitter

1211
01:13:40,870 --> 01:13:45,230
does exactly what it is expected
to do to coherent states.

1212
01:13:45,230 --> 01:13:47,230
It just splits a
laser beam into two,

1213
01:13:47,230 --> 01:13:49,800
with a reflection and
transmission coefficient

1214
01:13:49,800 --> 01:13:53,510
which are cosine squared
theta and sine squared theta.

1215
01:13:53,510 --> 01:13:57,120
And I will also show you that
this beam splitter is doing

1216
01:13:57,120 --> 01:13:59,640
the same, splitting
with cosine squared

1217
01:13:59,640 --> 01:14:03,210
and sine squared theta
probabilities, a single photon.

1218
01:14:03,210 --> 01:14:05,760
So in essence, I give
you the Hamiltonian,

1219
01:14:05,760 --> 01:14:08,470
I discuss that the
transfer matrix, which

1220
01:14:08,470 --> 01:14:12,670
is built upon this Hamiltonian,
does everything you ever

1221
01:14:12,670 --> 01:14:15,460
wanted a beam splitter to do,
and therefore you can say,

1222
01:14:15,460 --> 01:14:19,470
this defines the beam splitter.

1223
01:14:19,470 --> 01:14:21,610
OK, if you have a
Hamiltonian, that

1224
01:14:21,610 --> 01:14:30,990
means after propagating
through the Hamiltonian,

1225
01:14:30,990 --> 01:14:33,810
we have the following transfer.

1226
01:14:33,810 --> 01:14:38,170
You may be familiar with
putting the unitary time

1227
01:14:38,170 --> 01:14:40,790
evolution here, and the
time which has elapsed,

1228
01:14:40,790 --> 01:14:43,760
but we are always
using a fixed time

1229
01:14:43,760 --> 01:14:46,300
after the light has
transformed the beam splitter.

1230
01:14:46,300 --> 01:14:49,600
So you can see the
time t is absorbed

1231
01:14:49,600 --> 01:14:55,830
in the definition of
h, the Hamiltonian,

1232
01:14:55,830 --> 01:14:59,930
and b is really the unitary
time evolution from [? being ?]

1233
01:14:59,930 --> 01:15:01,980
before and [? being ?]
after the beam splitter.

1234
01:15:06,050 --> 01:15:09,740
So the transformation,
which transforms the modes

1235
01:15:09,740 --> 01:15:15,070
at this beam splitter, has now
the following unitary operator.

1236
01:15:20,540 --> 01:15:26,250
And we have modes a,
b, mode one and two,

1237
01:15:26,250 --> 01:15:27,700
before the beam splitter.

1238
01:15:27,700 --> 01:15:29,370
After the beam
splitter, they have

1239
01:15:29,370 --> 01:15:32,680
been transformed to
a prime and b prime.

1240
01:15:32,680 --> 01:15:37,840
And we obtain that by
taking the input modes

1241
01:15:37,840 --> 01:15:40,910
and doing unitary
time evolution.

1242
01:15:40,910 --> 01:15:44,930
You do simply the
operator transformation

1243
01:15:44,930 --> 01:15:47,560
for unitary transformation.

1244
01:15:47,560 --> 01:15:51,130
So we can now transform
the operators a and b.

1245
01:15:57,130 --> 01:15:59,056
I'm not going with you
through the exercise

1246
01:15:59,056 --> 01:16:00,305
how you transform an operator.

1247
01:16:03,050 --> 01:16:06,740
I think in undergraduate
quantum mechanics classes,

1248
01:16:06,740 --> 01:16:09,300
you do the
Baker-Campbell-Hausdorff

1249
01:16:09,300 --> 01:16:14,680
formula, how to use with
operators in the exponent.

1250
01:16:14,680 --> 01:16:19,560
And if you do with that,
you find-- not surprisingly,

1251
01:16:19,560 --> 01:16:22,420
a beam splitter is a linear
element-- that what you get

1252
01:16:22,420 --> 01:16:25,640
is a linear
combination of a and b.

1253
01:16:45,760 --> 01:16:49,770
So if you don't like me
postulating an operator

1254
01:16:49,770 --> 01:16:52,400
and saying, "Believe me,
this is a beam splitter,"

1255
01:16:52,400 --> 01:16:54,240
you can reverse engineer.

1256
01:16:54,240 --> 01:16:58,020
You can say, this really looks
like the modes, the operators

1257
01:16:58,020 --> 01:17:01,720
a b get linearly superposed
with cosine and sine.

1258
01:17:01,720 --> 01:17:05,100
This is what a beam splitter
does, is it mixes two modes.

1259
01:17:05,100 --> 01:17:07,320
And then you can reverse
engineer and figure out

1260
01:17:07,320 --> 01:17:09,440
that it's exactly the
Hamiltonian above which

1261
01:17:09,440 --> 01:17:09,940
does it.

1262
01:17:12,520 --> 01:17:17,110
So your homework will show
that this happens exactly

1263
01:17:17,110 --> 01:17:18,580
for coherent states.

1264
01:17:18,580 --> 01:17:21,350
So if you've a coherent
state with alpha,

1265
01:17:21,350 --> 01:17:24,545
you get a coherent state with
alpha cosine theta in one arm,

1266
01:17:24,545 --> 01:17:27,020
and alpha sine theta
in the other arm,

1267
01:17:27,020 --> 01:17:31,090
and you will see that
everything works beautifully.

1268
01:17:31,090 --> 01:17:40,380
I have to use one convention
now for a 50:50 beam splitter.

1269
01:17:40,380 --> 01:17:43,980
And I use a convention
that I chose theta

1270
01:17:43,980 --> 01:17:48,670
to be pi over 2
b and 90 degrees.

1271
01:17:48,670 --> 01:17:52,980
And we want to use
symbolic language,

1272
01:17:52,980 --> 01:17:56,600
because we want to draw
diagrams of multiple beam

1273
01:17:56,600 --> 01:17:58,630
splitters and such.

1274
01:17:58,630 --> 01:18:10,130
So this tilted square is the
symbol for beam splitter.

1275
01:18:10,130 --> 01:18:15,040
We have input ports a
and b on the left side.

1276
01:18:15,040 --> 01:18:22,440
We have output ports, which
are now b prime and a prime.

1277
01:18:22,440 --> 01:18:27,450
And for the 50:50 beam
splitter, the modes

1278
01:18:27,450 --> 01:18:38,310
are symmetric and
antisymmetric superpositions.

1279
01:18:38,310 --> 01:18:43,930
So this is the symbol for
the unitary transformation b.

1280
01:18:43,930 --> 01:18:47,640
And since one output port has a
minus sign, it's antisymmetric,

1281
01:18:47,640 --> 01:18:49,410
and one is positive.

1282
01:18:49,410 --> 01:18:52,824
We usually put a dot there
which distinguishes the two

1283
01:18:52,824 --> 01:18:53,490
different ports.

1284
01:19:01,960 --> 01:19:05,020
So this is the operator B.

1285
01:19:05,020 --> 01:19:10,840
We have already used, for
the unitary transformation,

1286
01:19:10,840 --> 01:19:11,860
the operator b dagger.

1287
01:19:22,350 --> 01:19:34,360
So for the operator b dagger,
using the equations above,

1288
01:19:34,360 --> 01:19:38,140
we have input modes a
b, but what happens now

1289
01:19:38,140 --> 01:19:42,290
is for the operator b
dagger, the output modes

1290
01:19:42,290 --> 01:19:44,350
are interchanged.

1291
01:19:44,350 --> 01:19:47,200
So we have a plus b over
square root a, a minus b

1292
01:19:47,200 --> 01:19:50,680
over square root 2, and
therefore the symbol

1293
01:19:50,680 --> 01:19:52,435
for b dagger has the dot here.

1294
01:20:01,840 --> 01:20:10,150
I just want to warn you
that there are other phase

1295
01:20:10,150 --> 01:20:13,670
conventions in the literature
which are equivalent.

1296
01:20:13,670 --> 01:20:16,440
But sometimes if you read a
paper and all of the minus

1297
01:20:16,440 --> 01:20:19,170
signs appear somewhere
else, well, people

1298
01:20:19,170 --> 01:20:20,950
have a different
phase convention.

1299
01:20:20,950 --> 01:20:27,190
You can postulate that
you have a beam splitter

1300
01:20:27,190 --> 01:20:32,020
without having the i here.

1301
01:20:32,020 --> 01:20:34,530
And everything works similarly.

1302
01:20:34,530 --> 01:20:38,590
It's just that, for
instance, in this equation,

1303
01:20:38,590 --> 01:20:40,640
you get an i here.

1304
01:20:40,640 --> 01:20:43,900
So imaginary units
and plus-minus signs

1305
01:20:43,900 --> 01:20:46,640
and that busyness are
phase conversations.

1306
01:20:46,640 --> 01:20:48,850
If you'd say, hey, but if
you another definition,

1307
01:20:48,850 --> 01:20:51,080
don't you have
another beam splitter?

1308
01:20:51,080 --> 01:20:53,030
Yes, it is another
beam splitter,

1309
01:20:53,030 --> 01:20:56,740
but I can always get it out
of my beam splitter-- which

1310
01:20:56,740 --> 01:20:59,370
now is our beam splitter--
by putting a phase

1311
01:20:59,370 --> 01:21:00,670
shifter afterwards.

1312
01:21:00,670 --> 01:21:04,390
So it doesn't limit the
generality of the discussion

1313
01:21:04,390 --> 01:21:08,200
by defining our beam
splitter in this way.

1314
01:21:08,200 --> 01:21:09,520
OK, time is over.

1315
01:21:09,520 --> 01:21:13,120
Reminder, we have another
class this week on Friday,

1316
01:21:13,120 --> 01:21:15,576
and finally, your
graded homework is here.

1317
01:21:15,576 --> 01:21:17,980
Have a good afternoon.