Photonic Quantum Information ProcessingOPTI 647: Lecture 3

Saikat GuhaSeptember 03, 2019

Associate ProfessorCollege of Optical SciencesMeinel 523

Outline for today

• Coherent states and linear optics• Quantizing the field• Distinguishing pure states

General pure state of a single mode

If we do ideal direct detection of mode , the total number of photons K is a Poisson random variable of mean N

�(t)

Mode , a quantum system, is excited in a coherent state�(t) |↵i ,↵ 2 C

Mode , a quantum system, is excited in a number state�(t) |ni, n 2 {0, 1, . . . ,1}

If we do ideal direct detection of mode , the total number of photons K = n (exactly so; K is not a random variable).

�(t)

A mode of ideal laser light is in a coherent state. Number (Fock) state of a given mode is very hard to produce experimentally

There are infinitely many other types of “states” of the mode . Coherent state and Fock state are just two example class of states

�(t)

|ni, n 2 {0, 1, . . . ,1} Fock states of a mode are special: they form an orthonormal basis that spans any general quantum state of that mode

hm|ni = �mn and

| i

,1X

n=0

|cn|2 = 1| i =1X

n=0

cn|ni

This orthogonality (in the Hilbert state) is different from that of modes (in L2norm space)

Coherent state as a quantum state

|↵i =1X

n=0

e�

|↵|22 ↵n

pn!

!|ni

cn

Ideal photon detection is a von Neumann quantum measurement described

by projectors,{|nihn|} , n = 0, 1, . . . ,1

|0i =

2

6664

100XXX

3

7775|1i =

2

6664

010XXX

3

7775|2i =

2

6664

001XXX

3

7775

Fock states can

be thought of as

infinite-length

unit-length

column vectors

corresponding to

the orthogonal

axes of an

infinite-

dimensional

vector space

. . .

Ideal direct detection on a coherent state produces outcome “n”

(i.e., n “clicks”) with probability, pn = |hn|↵i|2 = |cn|2 =e�NNn

n!

|↵i

Poisson detection statistics in a laser pulse is a result of the projection of the

quantum state of the laser pulse—a coherent state–on to one of the Fock states

Coherent states and “linear-optical transformations” (beam-splitters)

|↵1i

|↵2i

⌘ 2 (0, 1)

|�1i

|�2iTransmissivity,

Mach Zehnder Interferometer (MZI):

An arbitrary 2-mode linear optical unitary can be realized with a MZI – two 50-50 beam-splitters and two phases

Complex Unitary matrix,

Arbitrary N-mode linear optical unitary• Any N-by-N unitary U can be realized with M = N(N-1)/2 Mach

Zehnder Interferometers, . So, we need N(N-1) 50-50 beam-splitters and N(N-1) tunable phases to realize any N-mode linear optical unitary U

1

N

2

... Reck et al., PRL 73, 1 (1994)

Clements et al., Optica 3 (12), 1460-1465 (2016)

U

...

A coherent state is always single mode

• By an appropriate choice of modal basis, any “multi-mode” coherent state can always be expressed as a single mode coherent state

– In other words… if we have a deterministic field in any spatio-temporal shape (of any given polarization), we can always represent that as a single-mode coherent state of an appropriate normalized mode

– We will see later, this is not true for other quantum states in general. For example, a multimode thermal state or a multimode squeezed state, etc.

|↵1i|↵2i . . . |↵Ki ⌘ |�i|0i . . . |0i

Slicing a coherent state pulse (in time)

t0 T|

E

|↵i, ↵ =pN, N = E2T

t0 T|

p1/T

Single-mode coherent state of this mode: �(t)(1)

�(t)

|�i|�i . . . |�i ,� =

rN

M

(2) M-mode coherent state of the modes: k(t), k = 1, . . . ,M

tT|

tT|

tT|

tT|

⌧

p1/⌧

p1/⌧

p1/⌧

p1/⌧

M =T

⌧Orthogonal temporal modes

1(t)

2(t)

M (t)

Slicing a coherent state pulse (in space)

t0 T

|

E

|0i|0i

|0i

|↵i |�i

|�i|�i

|�i

......

t0 T

|

EpM

t0 T

|

EpM

t0 T

|

EpM

t0 T

|

EpM

Examples of optical qubits

• Single-rail qubit

• Dual-rail qubit

• Cat-basis qubit

|0i = |0i, |1i = |1i

|0i = |0, 1i, |1i = |1, 0i

|0i = N+(|↵i+ |� ↵i),|1i = N�(|↵i � |� ↵i)

Prove that the cat-basis qubit states are mutually orthogonal, and find the normalization constants N+ and N- in terms of ↵

Problem 4

Binary pure-state discrimination

– Assume equal priors:

=1p2

p1 + �p1� �

�

=1p2

p1 + �

�p1� �

�

|w2i =1p2

1�1

�

|w1i =1p2

11

�Consider a von Neumann projective measurement:⇧1 = |w1ihw1|⇧2 = |w2ihw2|

| 1i (?vTQi?2bBbH1) pbX | 2i (?vTQi?2bBbH2)

| 1i

| 2i

h 1| 2i = �

Inner product between the two states

|0i =✓

10

◆

|1i =✓

01

◆

p1 = p2 =1

2

Pe = P (H1)P (H2|H1) + P (H2)P (H1|H2)

=1

2|hw2| 1i|2 +

1

2|hw1| 2i|2

Pe =1

2

h1�

p1� |�|2

iShow that: and find the expression for minimum average error probability for Problem 5P (H1) = p, P (H2) = 1� p

Coherent states are not orthogonal

• Distinguishability of two coherent states

– Recall:

– Inner product between two coherent states:

|↵i =1X

n=0

e�

|↵|22 ↵n

pn!

!|ni

Binary phase shift keying (BPSK) coherent-state modulation

• Optimal measurement operators are cat states• Minimum probability of error

| 1i

| 2i

= |↵i

= |� ↵i|0i = N+(|↵i+ |� ↵i)

|1i = N�(|↵i � |� ↵i)

|w1i

|w2i

Pe =1

2

h1�

p1� e�4|↵|2

i

� = h↵|� ↵i = e�2↵

T0

E

T0

�E

Kennedy receiver: a suboptimal receiver

• Displace the BPSK states, then use direct detection

Im

Re|↵i|�i

⌘ ⇡ 1�����p1� ⌘

�|� + �i|�i

X = 1

X = 2 |↵i Y = 2

Y = 1|� ↵i e��2

1� e��2

e�(2↵+�)2

1� e�(2↵+�)2

Pe(N) = min�

1

2e�(2↵+�)2 +

1

2

⇣1� e��2

⌘�

=1

2e�4N , � = 0 (exact nulling case)

BPSK error probability

N

Prob

abilit

y of

erro

r

Optimize (minimize) the probability of error of the optimal-nulling Kennedy receiver (find optimal ) and plot the probabilities of error as function of N, as above� Problem 6

1

2e�4N

1

2

h1�

p1� e�4N

i

How to design a structured receiver (Dolinar receiver) that achieves the minimum error probability. See OPTI 495B/595B [Dolinar, MIT Ph.D. Thesis, 1976]

Density operator – pure and mixed states

16

• State of a quantum system– Complete knowledge is a pure state– Incomplete knowledge is a statistical (classically-

random) mixture of pure states• density operator: positive and unit trace,

| i

⇢X =X

x

pX(x)| xih x| =X

i

�i|�iih�i|X

x

pX(x) = 1X

i

�i = 1

h�i|�ji = �ij

Spectral decompositionMixture of pure states,

the states don’t have to be orthogonal

⇢ = | ih |

Take the statistical mixture intuition with a pinch of salt -- same density operator can be expressed as different mixtures

Projective measurement on mixed state

• Measure state, with projective measurement

Probability of outcome j

Conditional post-measurement state

Projective measurement on mixed state• Consider ensemble of pure states,

– Density operator,• Measurement projectors,• Assume the state in the ensemble was

– Post-measurement states:

–• If we get outcome j, we have conditional ensemble

– with

E = {pX(x), | xi}⇢ =

X

x

pX(x)| xih x|

| xi

Projective measurement on mixed state

• Density operator of this post-measurement ensemble of states Ej

Projective measurement on mixed state

• Probability of outcome j

Statistical mixture of coherent states

• Classical state: P function representation

• Statistical field in classical EM theory

• Multimode classical state

– cannot in general be written as a single mode state unlike a multimode coherent state

⇢ =

Z

CP (↵)|↵ih↵|d2↵

⇢ =

Z

Cn

P (↵)|↵ih↵|d2n↵

|↵i = |↵1i . . . |↵niwhere

What is the P function of a coherent state ?|�i

Single-mode thermal state

• Gaussian mixture of coherent states

– Probability distribution for photon counting,

– Show that

– Show that,

– Hence,

⇢ =

Z

CP (↵)|↵ih↵|d2↵ P (↵) =

e�|↵|2/N

⇡N

P (n) = Tr(|nihn|⇢) = hn|⇢|ni{⇧n = |nihn|}

P (n) =Nn

(1 +N)1+n;n = 0, 1, . . .

hn|⇢|mi = 0, B7m 6= n

⇢ =1X

n=0

P (n)|nihn|Problem 7

Phase scrambled coherent state

• Application of a phase,• Consider the state after application of a random

phase to a coherent state:

– Show that:– and that– So, is diagonal in the number basis,

• Circularly-symmetric states are diagonal in the number basis (we will revisit this later)

U✓|↵i = |↵ei✓i

⇢ =

Z 2⇡

0U✓|↵ih↵|U †

✓d✓

⇢hn|⇢|mi = 0, B7m 6= n

⇢ =1X

n=0

P (n)|nihn|

P (n) = hn|⇢|ni = e�NNn

n!;n = 0, 1, . . .

Problem 8

Annihilation operator of a mode

• Recall field quantization:

• Annihilation operator of a single mode– Eigenstate is a coherent state– “Annihilates” photon number,

– Number operator,

– Show that,

E(t) =KX

k=1

ai�i(t)E(t) =KX

k=1

ai�i(t)

Coherent state of modes with complex field amplitudes, ai

Field operator for the set of modes , with modal annihilation operators ai

�i(t)�i(t)

aa|↵i = ↵|↵i

N = a†aN |ni = n|ni

[a, a†] = aa† � a†a = I Problem 9

The phase operator

• Application of a phase,• The phase operator,

• Show that random phase scrambling of any pure state leads to a number diagonal state

– With

U✓|↵i = |↵ei✓i

U✓ = ei✓N = ei✓a†a

⇢ =

Z 2⇡

0U✓| ih |U †

✓d✓ =1X

n=0

P (n)|nihn|

P (n) = |hn| i|2 Problem 10

Coherent states resolve the identity

• Recall:• Prove that:

• Coherent states are an “over-complete” basis• They are not orthogonal (and hence cannot be

distinguished perfectly)• They form a positive operator valued measure

(POVM) – the most general description of a quantum measurement

1

⇡

Z

C|↵ih↵|d2↵ = I

I =1X

n=0

|nihn|

Problem 11

Quantization of the field: summary

• Classical (deterministic) field (coherent state)

• Quantum description of the field:– Field becomes an operator– Field described by a quantum state of constituent modes– Modal annihilation operator:– Classical field is a special case: each mode i is excited in

a coherent state– Classical statistical field is a mixture of coherent states,

density operator ,

E(t) =KX

k=1

ai�i(t)

⇢ =

ZP (↵)|↵ih↵|d↵ |↵i = |↵1i|↵2i . . . |↵Ki

,↵i = ai

E(t) =KX

k=1

ai�i(t)

ai

|↵ii

Recap of what we learnt today

• Coherent state is always single mode• “splitting” a coherent state: product of coherent states• Classical state is a mixture of coherent states• Coherent states are not orthogonal , yet

they resolve the identity,• Distinguishing equally-likely states,

the minimum average Pr(error),• The coherent state is an eigenstate of the “field”

operator, ; • Canonical commutation relation,• Applying random phase to a pure state gives us a

“circularly symmetric” state, which is number diagonal

1

⇡

Z

C|↵ih↵|d2↵ = I

a|↵i = ↵|↵i a|ni =pn|n� 1i, a†|ni =

pn+ 1|n+ 1i

|↵i =1X

n=0

e�

|↵|22 ↵n

pn!

!|ni

cn = hn|↵i

[a, a†] = aa† � a†a = I

Pe =1

2

h1�

p1� |�|2

i{| 1i, | 2i} , h 1| 2i = �

Upcoming topics

• Single mode quantum optics– Phase space, Characteristic functions, Wigner functions,

Entanglement– Squeezed states