A5 #1

BELL’S THEOREM, ENTANGLEMENT,TELEPORTATION, QUANTUM

CRYPTOGRAPHY,QUANTUM COMPUTING AND ALL THAT

A. J. LeggettUniversity of Illinois at Urbana-Champaign

Support: NSF E1A-01-21568

1. Elementary considerations on classical EMradiation and photons.

2. Bell’s theorem3. The general notion of entanglement4. Some applications5. Superconducting qubits – recent advances

A5 #2

BIREFRINGENT CRYSTAL(“polarizer”):

Transmitted amplitude = E E cos θ

Transmitted energy µ E E cos θ(initial energy µ 𝐸 )

⇒ fraction of energy transmitted = cos θ(“Malus’s law”)

y Re const. exp i kz 𝜔𝑡

I. Light waves and photons

Classical light wave ( into screen)E Re const. x exp i kz 𝜔𝑡

reflection axis

θ

transmission axis

E ≡ E sin θ

E ≡E cos θ

Principle of (classical) superposition:

E Re E E exp i kz ωt also solutionnote: E , E may be complex, e.g. const. Re i exp i kz ωt ≡ const. cos kz ωt sin kz ωtcarries finite angular momentum.

A5 #3

|𝜓 ⟩ ≡ | 𝑦|𝜓 ⟩ ≡ |𝑥

QM: Photons

QM amplitude |𝜓⟩ ⇐⇒ classical field amplitude :in particular, if |𝜓 ⟩, |𝜓 ⟩ allowed, so is superposition

single photon polarized along the y-axis

θ

𝛼 ψ ⟩ 𝛽 ψ ⟩

e.g. if

then|𝜓⟩ cos θ |𝑥 sin θ| 𝑦

describes single photon with (linear) polarization at angle θ in the xy-plane

and

|𝜓⟩1

2| ⟩ 𝑖| ⟩

describes photon with R(L) circular polarization(angular momentum ℏ)

Single photon incident on birefringent crystal (“polarizer”):

if |𝜓⟩ |𝑥⟩, transmitted if |𝜓⟩ |𝑦⟩, reflectedbut what if |𝜓⟩= cos θ|x⟩ sin θ |𝑦⟩?photon cannot be “split”! (well, not at this level of discussion…)

θ

single photon polarizedalong the x-axis

A5 #4

With suitable choice (f() = cos 2 ( )) of random distribution, can reproduce PT() = cos2

If closer to , transmitted: if closer to , reflected. If transmitted, distribution of “hidden” variable is adjusted:

Single photon incident on birefringent crystal (“polarizer”):either transmitted or reflected.

probability of transmission PT ()= cos2

(quantum version of Malus’ law)

Digression: Can a classical probabilistic theory explain this?

YES!

nominal polarization

random distribution of“hidden” (“true”) polarization variable

“true” polarization

A5 #5

|Ψ⟩ |x iy⟩ |x iy⟩ (phase factor): |x iy⟩ |x iy⟩

2. 2-PHOTON STATES FROM CASCADE DECAY OF ATOM

z

z

“1” “2”0+

0+

ATOM

What is polarization state of photons?

(note: |𝑥⟩ i|𝑦⟩ corr. photon angular momentum ℏ)

General principle of QM: if process can happen either of two ways

and we don’t (can’t) know which, must add amplitudes!

Here, we know that total angular momentum of 2 photons is

zero, but we don’t know whether photon 1 carried off ℏ and

2 ℏ (intermediate atomic state has m = 1) or vice versa (int.

state m = +1). Hence must write: crucial!

Actually (parity ⇒) phase factor = +1, so

|Ψ⟩ x⟩ x⟩ |y⟩ |y⟩

(for normalization)

A5 #6

POLARIZATION STATE OF 2 PHOTONS EMITTED BACK TO BACK IN ATOMIC 0+ → 1- → 0+ TRANSITION (recap):

Ψ1

2|x⟩ |x⟩ |y⟩ |y⟩

So, if photon 1 is measured to have polarization x(y) so inevitably will photon 2!

But, state is rotationally invariant:

Ψ1

2|x′⟩ |x′⟩ |y′⟩ |y′⟩

so if 1 is measured to have polarization x′ y so does 2!(NOT true for “mixture” of |x⟩ |x⟩ and |y⟩ |y⟩ )

______________________Now:

What if photon 1 is incident on polarizer with “transmission” axis a. and photon 2 on one with a differentlyoriented transformation axis b ?

Since choice of axes for Ψ arbitrary, choose x a.Then:

Prob. of transmission of 1 .

But if 1 transmitted, then polarization of 2 is a, so probability of transmission of polarizer set in direction b cos θ(Malus’s law)

P both transmitted cos θ

𝑥

𝑦𝑥′

𝑦′

⇒

A5 #7

Prob. (both deleted) = cos θ 1 cos 2θ

Prob. (neither deleted) = cos θ

Prob. (1 deleted, 2 not)

Prob. (2 deleted, 1 not)

transmission axis

transmission axis

atomic source

1 2detector detector

= sin θ

by contrast,“Isotropic mixture”

⟹14

112

cos 20

Df: If for a given pair, with polarizer 1 set at a, photon l is detected,

df. A ≡ +1: if rejected, A ≡ 1. Similarly with polarizer 2 set at b,

if photon 2 detected, df. B ≡ +1: if rejected, then B ≡ 1. Then

above is equivalent to the statement that for the average over the

ensemble of pairs,

AB cos θ sin θ cos 2θ [“Mixture”: ½ cos 2θab]

THESE ARE THE PREDICTIONS OF STANDARD

QUANTUM MECHANICS. CAN THEY BE EXPLAINED BY

A CLASSICAL PROBABILISTIC THEORY?

as special cases, for θab = 0 AB = +1, and for θab = 𝜋/2. AB = 1.(EPR). These two special cases can be accounted for by a classicalprobabilistic model. But ...

A5 #8

A C PHOTON 1 PHOTON 2

~ ~ B ATOMIC D SWITCH SOURCE SWITCH

( A , etc.) transm. axis = a~

EXPERIMENTS ON CORRELATED PHOTONS

DEFINITION: If photon 1 is switched into counter “A”, then:

If counter “A” clicks, A = + 1 (DF.)

If counter “A” does not click, A = – 1 (DF.)

NOTE:

If photon 1 switched into counter “B”, then A is NOT DEFINED.

Experiment can measure

<AC>exp on one set of pairs (1 “A”, 2 “C”)

<AD>exp on another set of pairs (1 “A”, 2 “D”)

etc.

Of special interest is

Kexp <AC>exp + <AD>exp + <BC>exp – <BD>exp

for which Q.M. makes clear predictions.

A5 #9

POSTULATES OF “OBJECTIVE LOCAL” THEORY:

(1) Local causality

(2) Normal “arrow of time”

(3) Microscopic realism OR macroscopic

“counter-factual definiteness”

BELL’S THEOREM

1. (3) For each photon 1, EITHER A = + 1 OR A = – 1, independently of whether or not A is actually measured.

2. (1) Value of A for any particular photon 1 unaffected by whether C or D measured on corresponding photon 2. : etc.

3. ∴ For each pair, quantities AC, AD, BC, BD exist, with A, B, C, D, = 1 and A the same in (AC, AD) (etc.)

4. Simple algebra then for each pair, AC + AD+ BC –BD 2

5. Hence for a single ensemble,

<AC>ens + <AD>ens + <BC>ens – <BD>ens 2

6. (2) <AC>exp = <AC>ens, hence the measurable quantity

Kexp <AC>exp + <BC>exp + <BC>exp – <BD>exp

satisfies

A5 #10

Thus, QM predicts (ideal case)

exptl. predictions of QM incompatible with those of

any theory embodying

Local causality

Arrow of time

Macroscopic counter-factual definiteness

1. “It is a fact that either A would have clicked or A would not have clicked”

2. “Either it is a fact that A would have clicked, or it is a fact that A would not have clicked”

Kexp = 2 2

OBJECTIVE LOCAL THEORY: Kexp ≤ 2.

QM: If polarizer settings are

then e.g. for a 0+ transition predict

AC = cos (2θa⋅c), etc.

⇒ for

~ ~ ~ ~a, b,c,d,

~a

22.5

A5 #11

IDEA OF “ENTANGLEMENT”:

System composed of 2 (separated) subsystems 1 and 2:

= (1,2) [general]

(a) (1,2) = (1)(2) product, nonentangled

“Properties” of 1 described by (1)

“Properties” of 2 described by (2)

Complete information on system obtainable by making measurements on subsystems separately.

(b) (1,2) (1) (2) entangled

e.g. (2 photons)

(1,2) = (|x>1 |x>2 + |y>1 |y>2 )

(“EPR pair”) (1 2 + 1 2)

Subsystems 1 and 2 do not “possess” individual properties (Bell’s theorem).

Complete information on system obtainable only by correlated measurements on 1 and 2

INFORMATION “STORAGE” IS NONLOCAL!

A5 #12

1. Quantum “Teleportation”

(e.g. of state of photon)

Rules of the game: Alice is to transmit to Bob an arbitrary

state | > of a photon , without direct physical contact (but A

(or D) may communicate with B e.g. by a classical phone

line). (Alice may not even know what state she has sent).

Some applications of entanglement:

A5 #13

Solution: C emits “EPR pair.” D then measures combined

state of R and G photons. If D finds EPR pair, then angular

momentum conservation state received by B = that sent

by A (and D phones B to tell him so). If D finds a different

state, can “rotate” into EPR pair. Then B must perform

inverse of this rotation (and D so instructs him).

A5 #14

S emits a string of EPR pairs. A measures in basis , or in basis

, in a random way: B measures similarly, also at random. At the

end, A and B inform one another by phone which basis they have used

for each measurement, discard those for which they used different

bases, and compare notes on a subset of the rest. If they always agree,

they can be sure of no eavesdropping, and so use the rest for the key.

If Eve tries to “listen in” on the quantum channel…

2. Quantum Cryptography:

Key distribution problem: Alice must be able (a) to communicate the “key” (a string of 1’s and 0’s) to Bob, and moreover (b) know if Eve is listening in. Classically, no 100% secure way of ensuring this is known.

Solution:

A

E

B

E ?

BA

S

E

A5 #15

3. Quantum Computation

EX: particles of spin ½.

A single particle of spin ½ in a pure state is parametrized

by 2 independent variables, e.g. corresponding to the angles

(θ, φ) made by its spin with the z- and x-axes.

A collection of N particles of spin ½ in a product state is

parametrized similarly by 2 independent variables for each,

i.e. 2N in total.

But the same collection of N particles in a generic

entangled state requires for its parameterization 2N variables!

e.g. (N=3)

Ψ 1,2,3 a ↑ ↑ ↑ ⟩ a ↑ ↑ ↓ ⟩ a |↑ ↓ ↑ ⟩

… a |↓ ↓ ↓ ⟩

⇒ MASSIVELY PARALLEL COMPUTATION!

(Shor, 1994: factorization of N-digit no. takes time which for

classical computer is exponential in N, but for quantum

computer is power-law).

A5 #16

(SIMPLEST) DESIGN FOR A QUANTUM COMPUTER:

couple together N 2-state systems(“QUBITS”) in such a way that we

can reliably perform “1-qubit” operations(non-entangling) and “2-qubit” operations (entangling)

e.g. Heisenberg interaction Jσ1∙ σ2 induces

↑1↓2 ⟶ 2-½ (↑1↓2 + i↓1↑2)

non-entangled entangled

Principal requirements for qubit:2 states onlyeasy initialization and readoutscalable

* decoherence-free

“Figure of merit” for (lack of) decoherence:

𝑄 ≡ 𝜔 𝑇

char. freq. of “dephasing” time2-state system

Current “designs” require Qφ ≳ 104

A5 #17

A5 #18

PHYSICS OF SUPERCONDUCTIVITY

Electrons in metals: spin ½ fermions

But a compound object consisting of an even no. of fermions has spin 0, 1, 2 ... boson .

(Ex: 2p 2n 2e 4He atom)

can undergo Bose condensation

Fermi energy

↓

“Fermi sea”

kT

No.

of

part

icle

s pe

r st

ate

→

2

1

← “Bose condensate”

“Spin” of elementary = n2 ℏ

particles

0, 1, 2 … bosons12 ,

32 ,

52 … fermions

At low temperatures:

A5 #19

Pairing of electrons:

“di-electronic molecules” Cooper Pairs

In simplest (“BCS”) theory, Cooper pairs, once formed, must automatically undergo Bose condensation!

must all do exactly the same thing at the same time (also in nonequilibrium situation)

A5 #20

𝐾 ≡ 𝑣 · 𝑑𝑙ℎ

2𝑚n Φ Φ⁄

SUPERCONDUCTING RING IN EXTERNAL MAGNETIC FLUX:

Note: state with 50% and 50%

Φ

Quantization condition for “particle” of charge 2e (Cooper pair):

integer

“circulation”“flux quantum”

h/2e

E K2

A. Φ = 0: groundstate unique (n = 0)

all pairs at rest.

B. Φ = ½ Φ0 : groundstate doubly degenerate (E (n – ½ )2)

(n = 0 or n = 1)

Either all pairs rotate clockwise

or all pairs rotate anticlockwise

strongly forbidden by energy considerations

(n – Φ/Φ0)2

A5 #21

Josephson “qubit”

Ψ 2 ⁄ | ↺⟩ | ↻⟩

Trapped flux

Φext

Bulk superconducting ring

Josephson junction

~ 1μA

ΔΕ ↑

Φext →

Evidence: a) spectroscopic:

(SUNY, Delft 2000)

b) real-time oscillations (like NH3)

between ↺ and ↻

(Saclay 2002, Delft 2003) (Qφ ~ 50-100)

by 2018: Q > 104 !and~50 qubits entangled!