bell’s theorem, entanglement, teleportation, quantum
TRANSCRIPT
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!