an operator approach to bell inequalities · an operator approach to bell inequalities daniel...

An operator approach to Bell inequalities

Daniel Alsina, Jagiellonian University, 16th November 2015

Based on work with José Ignacio Latorre and Alba Cervera in University of Barcelona

Summary

● Basics of entanglement● Bell inequalities● Bell operator● Experiments● Conclusions

Basics of entanglement

● Entanglement: quantum correlations between particles

Intuitive conditions for entanglement:

1) Result of a measure on A is somewhat (possibly totally) uncertain

2) A measure on B will give more (possibly total) information about A.

Paradigmatic example: Bell/EPR state

We don't know the result of a measure on A (could be 0 or 1) but a measure on B will give us the key to A (it will be 0 if B has been 0, and 1 if B has been 1)

|ψ⟩=1

√2 (|0A0B ⟩+|1A1B ⟩)


1) Result of a measure on A is somewhat (possibly totally) uncertain

2) A measure on B will give more (possibly total) information about A.

Examples of non-entangled states:

|ψ⟩=1

√2 (|0A0B ⟩+| 0A1B ⟩)

∣ψ⟩=12(∣0A 0B ⟩+∣0A 1B ⟩+∣1A 0B ⟩+∣1A1B ⟩ )

|ψ⟩=|0A0B ⟩

|ψ⟩=1

√2 (|0A1B ⟩+|1A1B ⟩)

Condition 1not fulfilled

Condition 2not fulfilled


● Non-entangled states can be decomposed as products of its constituents:

|ψ⟩=|0A0B ⟩=|0A ⟩⊗ |0B ⟩

|ψ⟩=1

√2 (|0A0B ⟩+|0A1B ⟩)=|0A ⟩⊗1

√ 2 (|0B ⟩+|1B ⟩)

|ψ⟩=1

√2 (|0A1B ⟩+|1A1B ⟩)=1

√2 (|0A ⟩+|1A ⟩)⊗ |1B ⟩

|ψ⟩=12(|0A0B ⟩+|0A1B ⟩+|1A0B ⟩+|1A1B ⟩)

=1

√2 (|0A ⟩+|1A ⟩)⊗(1

√ 2 (|0B ⟩+|1B ⟩))


● But entangled states cannot be decomposed like this!

If it could be written as a product state, it would be false that a measure on B makes an impact on A!

|ψ⟩=1

√2 (|0A0B ⟩+|1A1B ⟩)≠|ψA ⟩ ⊗|ψB ⟩


● Most general definition of entanglement:

A general (pure or mixed) state is separable (non-entangled) if it can be written as:

If it is impossible to write it like this, it is an entangled state.

However, in this talk we are only interested in pure states.

ρ=∑i

wiρAi ⊗ρB

i


|ψ⟩=1

√2 (|0A0B ⟩+|1A1B ⟩) implies that if A

measures 0, B will get 0, and if A gets 1, sodoes B. This also happens the other wayround.

This suggests there is some informationtravelling between A and B to “tell” the otherparticle which state should it collapse to.


|ψ⟩=1

√2 (|0A0B ⟩+|1A1B ⟩)

A B

|ψA ⟩=|0A ⟩

A B

A B A B

A B

|ψA ⟩=|0A ⟩

A BρB

A B

|ψA ⟩=|0A ⟩

A B A B

|ψA ⟩=|0A ⟩

A BρB

I am 0!

I am 0!

A B

|ψA ⟩=|0A ⟩

A B A B

|ψA ⟩=|0A ⟩

A B

|ψB ⟩=| 0B ⟩

A B A B

A B

ρBρA

Creation of 2 entangled particles

Separation

Measurement and collapse of A

Information travelling

Collapse of B


What if the distance between A and B is sufficiently big and the time between measurements of A and B sufficiently small such that information has no time to travel from A to B? (spacelike separation according to special relativity)

Possibility 1: B will collapse independently of A (entanglement is lost at some point)

Possibilty 2: B will collapse according to A anyway (entanglement is a non-local property)

Possibility 2 is what really happens!


Spooky action at a distance...

2 new possibilities:

1- Hidden variables: QM is incomplete and there are new variables that determine the outcome of any experiment with certainty. EPR (1935),

De Broglie-Bohm (1927-1952)

2- QM is intrinsically non-local and we have to live with it.

Is there an experiment to differentiate between those two?

Letter Einstein to Born (1947)

Bell inequalities

● The answer is YES: Bell inequalities

Correlations between measurements in a hidden variable theory have constraints violated by “conventional” quantum mechanics.

"If [a hidden variable theory] is local it will not agree with quantum mechanics, and if it agrees with quantum mechanics it will not be local. This is what the theorem says”. (John Bell, 1987)

1+E (bc )⩾| E(ab)− E(ac)|Original Bell inequality (1964)

a,b,c = +1,-1

E(x): Expected value

Bell inequalities

● CHSH Inequality (Clauser et al., 1969)

| E(ab)+E (a ' b)+E(ab ' )− E(a ' b ')|⩽2

a ,a ' , b , b ' stand for 4 different variables with values {+1,-1}. A observes randomly a or a', B observes randomly b or b'

a

a'

b

b'

A B

In QM applied to a spin 1/2 system, is to be interpreted aswhere is the Pauli vector.a σ⃗ · a⃗

σ⃗={σ x ,σ y ,σ z }

Bell inequalities

We had to guess first which state would violate the BI:

Compute the expected value (which can be more complicated for other states):

and then optimize the directions:

Useful for experiments where we have a concrete state (up to uncertainties) but not from an analytical point of view.

|ψ⟩=1

√2(|0A1B ⟩−|1A0B ⟩)

E(ab)=−cos( a⃗ , b⃗)

a=90 º ,b=45º , a '=0 º , b'=135 º

Bell operator

C=ab+ab '+a ' b− a ' b ' Bell operator

The maximal eigenvalue of C gives the maximum violation of the BI, and the corresponding eigenvector will be the state responsible for it. We only have to maximize over the directions now.

Let's look for a more deductive way to find the classical and quantum limits of a BI and the states that saturate the quantum limit.

Bell operator

We can still do better with a little trick: computing C²

C2=4 I a I b− [a ,a ' ][b ,b ' ] ([x , y ]=xy− yx)

From this expression we can obtain a lot of information:- Classically, all commutators are 0, so- In QM, It is thus easy to see that in order to maximize the commutators, it's enough to impose that a and a' are perpendicular ( the same for b and b' ). Each commutator will then give a maximum value of 2, and:

C clas2

=4[σi ,σ j]=2 i ϵijkσ k .

Cquant2

⩽8

Bell operator

But mathematically:

so we readily deduce that

and the states responsible for the maximum quantum violation of C will be the same as those of C², so we got all information almost for free!

Avi=λ i v i→A2 v i=A (λ i v i)=λi Avi=λ i

2 v i

| C clas | =2 |Cquant |⩽2√2

Bell operator

C² corresponds actually to a different experiment:

C²=C(t1)C(t2)

C

t1t2 t1 t2

{A1,A'1}{B1,B'1} {A2,A'2}{B2,B'2}Source

But mathematically it gives us the correct results for C!

Bell operator

Can the trick be extended to other BI?

General BI: BI(m,n,d)

● m: number of settings: a,a',a''…● n: number of parties: a,b,c,…● d: local dimension: 2,3,4…

CHSH is the most simple BI: it's a BI(2,2,2)

Bell operator

For BI (2,n,2) the trick works very well!

If we don't know the form of C, we can work the other way round: impose a form of identity+commutators for C² and deduce C.

BI(2,3,2):

C=−abc+ab' c '+a' bc '+a' b ' cC2

=4 I a I b− ([a ,a ' ] [b ,b ' ]+[a ,a ' ][c , c ' ]+[b ,b' ][c ,c ' ])

|C clas|=2 |Cquant |⩽4

Bell operator

Quantum/classical rate for a BI(2,n,2)

Mermin-type inequalities (1990)

Expressions obtained by other means by Werner and Wolf (2001)

Bell operator

● Which states saturate the quantum bound of a BI(2,n,2) of Mermin type?

Always states equivalent to GHZ:

These states are maximally entangled for n=2,3 but not for n>3 !Max. entanglement does not imply max. violation, although violation needs entanglement.Entanglement related to Nonlocality, but NOT EQUIVALENT!!

Bell operator

● The C² trick does not work so well for other BI.

For example the B(3,2,2):

Collins-Gisin (2004)

In operator form reads:

But it does not have an easy C² expression

Bell operator

Working the other way round, our best effort produced the following result:

which needs to be optimized. Our try produced:

Bell operator

● BI (2,2,3) is even more difficult. Now we have qutrits instead of qubits, with values {1,0,-1}.

The operator basis needs to be expanded to {a,a2} (because now , )

The generators are no longer the 3 Pauli Matrices of SU(2) but the 8 Gell-Mann matrices of SU(3)

Bell operator

BI(2,2,3): (Collins et al., 2002)

Its Bell operator is:

Again the form of C² is too complicated. But we can just optimize over the directions to find:

Bell operator

for the directions:

And the corresponding state is:

Again not the maximally entangled state!!

Experiments

BI have been tested experimentally for quite a while now, albeit generally just in the CHSH form with the singlet state.

Aspect et al. (1982) found first confirmation of CHSH violation

Now already quite close to the quantum bound: 10-3 in Poh et al., (2015)

Experiments

Possible loopholes in experiments● Detection loophole● Locality loophole● Freedom of choice loophole

Many recent experiments are claiming having closed all loopholes:

Hensen et al. (August 2015)

Shalm et al. (November 2015)

Conclusions

● Bell inequalities are a crucial tool to falsify hidden variable theories and strengthen our belief in quantum mechanics

● The Bell operator is a useful tool to analyze mathematically all properties of Bell inequalities

● Lots of recent experiments are closing the CHSH, but still lot of scope to probe other BI

Thanks for your attention!

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