Verifying Equality in Scarani's "Six Quantum Pieces

[StevieTNZ]

Hi there

I'm reading through Valerio Scarani's "Six Quantum Pieces" and have hit an exercise which requires verification of an equality.

Suppose that photons D-A, B-C are entangled as follows:
photons D and A: Psi+ = 1/2 |H>|V> + |V>|H>
photons B and C: Phi+ = 1/2 |H>|H> + |V>|V>

Verify the following equality
Psi+(DA)Phi+(BC) = Phi+(AB)Psi+(CD) + Phi-(AB)Psi-(CD) + Psi+(AB)Phi+(CD) + Psi-(AB)Phi-(CD)

where the letters in the brackets indicate the photons.

How does one verify the above equality. The answer given in the book is

Expand the left side:
|H>|V>|H>|V> + |V>|H>|H>|H> + |H>|V>|V>|V> + |V>|H>|V>|V>
Next, using the definitions of the Bell bases in an equation, similarly expand and simplify the expression above to verify the equality.

That is where I am stuck. I know the different bell-states, but entangle swapping photons that are entangled differently I can't figure out what the outcomes will be.

Any help much appreciated
Stevie

 

[atyy]

I think Psi+(AB) means that you have to put labels on the kets: Psi+(AB) = 1/2 |HA>|VB> + |VA>|HB>.

So the LHS should be |HD>|VA>|HB>|HC> + |VD>|HA>|HB>|HC> + |HD>|VA>|VB>|VC> + |VD>|HA>|VB>|VC>. Then do the same on the RHS, and use brute force.

 

[StevieTNZ]

I guess my confusion lies not with expanding the right hand side, but how the author obtains those bell-states in the first place.

Any clarity on that would be great!

 

[StevieTNZ]

Apparently the right hand side is an expansion of |Psi+>{AB} |Phi+>{CD}, not of |Psi+>{DA} |Phi+>{BC}. Now I'm extra confused at how the author obtains
= Phi+(AB)Psi+(CD) + Phi-(AB)Psi-(CD) + Psi+(AB)Phi+(CD) + Psi-(AB)Phi-(CD)
from
|Psi+>{AB} |Phi+>{CD}, not of |Psi+>{DA} |Phi+>{BC}

 

[StevieTNZ]

Okay, so I expand the right hand side:
=(|H>|H>+|V>|V>(AB) |H>|V>+|V>|H>(CD)) + (|H>|H>-|V>|V>(AB) |H>|V>-|V>|H>(CD)) + (|H>|V>+|V>|H>(AB) |H>|H>+|V>|V>(CD)) + (|H>|V>-|V>|H>(AB) |H>|H>-|V>|V>(CD))

That done, I don't see how that is equivalent to |H>|V>+|V>|H>(DA) and |H>|H>+|V>|V>(BC).

If instead photons D&A are entangled as |H>|V> - |V>|H>, what impact does that have on the equality presented?

If the photons D&A and B&C are both in the bell-state |H>|H> + |V>|V> or |H>|H> - |V>|V>, would detecting photons A & B in the bell-state Phi+ mean photons C & D are in that bell-state also? Or if D&A are entangled as |H>|H> + |V>|V> and B&C are entangled as |H>|H> - |V>|V>, would we find photons C&D in the opposite bell-state that photons A & B are found in (i.e C&D in Psi-, A&B in Psi+)?

 

[atyy]

Okay, so I expand the right hand side:
=(|H>|H>+|V>|V>(AB) |H>|V>+|V>|H>(CD)) + (|H>|H>-|V>|V>(AB) |H>|V>-|V>|H>(CD)) + (|H>|V>+|V>|H>(AB) |H>|H>+|V>|V>(CD)) + (|H>|V>-|V>|H>(AB) |H>|H>-|V>|V>(CD))

Maybe try expanding it further? The same way the left hand side is expanded in post #2?

 

[naima]

Do not use labels. Always write things in the same order A B C D i.e HVHH for A = H, B = V C = H and D = H.
It is easy for le right side (they are in the good order. For the left side you ave DA and BC terms, take Psi+(DA)Phi+(BC)
you have a HH term for BC, you write it and a HV for DA. You write V at the left and H at the right of the HH and you get VHHH for ABCD and so on for the four terms of the left side.

Here we begin with a pair of entangled photons BC and DA they are in Phi+ and Psi+ Next we make a bell state measurement on AB. there are four possibles results. According to the result you get, the equality tells you in which Bell State is the CD pair.

Does anybody know how one can say BC are Phi+ and AD in Psi+? I suppose that it can not be known before AB Bell measurement. But how are they post selected?

 

[atyy]

I agree with naima. The idea of using the labels is to first write the expansion in the "wrong*" order, since that's how it's presented in the book. Then rearrange every term so that the labels are in the order ABCD.

*It's not wrong, since labels instead of ordering are used to indicate ABCD. One can notate either way, which is why I put "wrong" in quotes.

 

[StevieTNZ]

Maybe try expanding it further? The same way the left hand side is expanded in post #2?

I'm not entirely sure how I would start going about that. Any guidance must appreciated.

 

[naima]

First write the 16 terms of the right side (it is easy) and then the four of the left side the way i gave you.

We have here Psi+Phi+ swapping. In another thread we had Psi-Psi- swapping.
How many other swapping possibilities?

 

[atyy]

Okay, so I expand the right hand side:
=(|H>|H>+|V>|V>(AB) |H>|V>+|V>|H>(CD)) + (|H>|H>-|V>|V>(AB) |H>|V>-|V>|H>(CD)) + (|H>|V>+|V>|H>(AB) |H>|H>+|V>|V>(CD)) + (|H>|V>-|V>|H>(AB) |H>|H>-|V>|V>(CD))

As naima says, there are 16 terms on the right. For example the first term is expanded into 4 terms:

(|H>|H>+|V>|V>(AB))( |H>|V>+|V>|H>(CD))
= |HA>|HB>|HC>|VD>+|HA>|HB>|VC>|HD>+|VA>|VB>|HC>|VD>+|VA>|VB>|VC>|HD>

 

[StevieTNZ]

I expanded further and after cancelling I get the following:
|H>(A)|H>(B)|H>(C)|V>(D) + |V>(A)|V>(B)|V>(C)|H>(D) + |H>(A)|V>(B)|H>(C)|H>(D) +
|V>(A)|H>(B)|V>(C)|V>(D)

 

[atyy]

I expanded further and after cancelling I get the following:
|H>(A)|H>(B)|H>(C)|V>(D) + |V>(A)|V>(B)|V>(C)|H>(D) + |H>(A)|V>(B)|H>(C)|H>(D) +
|V>(A)|H>(B)|V>(C)|V>(D)

I think Psi+(AB) means that you have to put labels on the kets: Psi+(AB) = 1/2 |HA>|VB> + |VA>|HB>.

So the LHS should be |HD>|VA>|HB>|HC> + |VD>|HA>|HB>|HC> + |HD>|VA>|VB>|VC> + |VD>|HA>|VB>|VC>. Then do the same on the RHS, and use brute force.

Taking the LHS:
|HD>|VA>|HB>|HC> + |VD>|HA>|HB>|HC> + |HD>|VA>|VB>|VC> + |VD>|HA>|VB>|VC>

Arranging each term to be in ABCD order:
|VA>|HB>|HC>|HD> + |HA>|HB>|HC>|VD> + |VA>|VB>|VC>|HD> + |HA>|VB>|VC>|VD>

which doesn't seem to be the same :(

Can you double check your RHS, or is my LHS wrong?

 

[StevieTNZ]

Find attached my right hand side calculation. As far as I can tell, it is correct.

 

[atyy]

Is there a typo in the book, should it be this?

Psi+(DA)Phi+(BC) = Phi+(AB)Psi+(CD) + Phi-(AB)Psi-(CD) + Psi+(AB)Phi+(CD) - Psi-(AB)Phi-(CD)

 

[StevieTNZ]

Is there a typo in the book, should it be this?

Psi+(DA)Phi+(BC) = Phi+(AB)Psi+(CD) + Phi-(AB)Psi-(CD) + Psi+(AB)Phi+(CD) - Psi-(AB)Phi-(CD)

Should it be that, if the way to verify the equality is to expand as I have done and cancel like terms, and re-arrange the photons as they're originally entangled?

 

[atyy]

Should it be that, if the way to verify the equality is to expand as I have done and cancel like terms, and re-arrange the photons as they're originally entangled?

I think so, but my algebra is terrible. I was going to check that the method is correct by trying to verify Eq 2.12 of http://arxiv.org/abs/0910.4222.

 

[StevieTNZ]

I think so, but my algebra is terrible. I was going to check that the method is correct by trying to verify Eq 2.12 of http://arxiv.org/abs/0910.4222.

Did you manage to check if your method is correct by verifying that equality?

 

[naima]

There is a typo in the answer given by the author:

The answer given in the book is
Expand the left side:
|H>|V>|H>|V> + |V>|H>|H>|H> + |H>|V>|V>|V> + |V>|H>|V>|V>

Phi+ * Psi+ is (HH+VV)(HV+ VH)
The result cannot contain HVHV

 

[atyy]

Did you manage to check if your method is correct by verifying that equality?

Yes, seems to work.

 

[StevieTNZ]

There is a typo in the answer given by the author:

Phi+ * Psi+ is (HH+VV)(HV+ VH)
The result cannot contain HVHV

An error on my part.

The answer should be |H>|V>|H>|H> + |V>|H>|H>|H> + |H>|V>|V>|V> + |V>|H>|V>|V> for the left hand side.

 

[naima]

When i expand the left side, i find
HHHV +VHHH + HVVV + VVVH
the BC terms in the middle are HH and VV they come from Phi+
the other H and V come from the AD Psi+ terms.

Let us verify the equality (i do not write the 1/2 factor)
HHHV + VVVH come from (HH+VV)(HV+VH) + (HH-VV)(HV-VH)
According to the exercise VHHH + HVVV should come from (HV + VH)(HH+VV)+ (HV -VH)(HH-VV)
but they cancel. I think that there is a typo in the LHS of the equality: there must be a minus sign before Psi-Phi-
(HV + VH)(HH+VV) - (HV -VH)(HH-VV) gives the good result.

 

[StevieTNZ]

Say we have two pairs of entangled photons such as:
AB: |H>|H> - |V>|V>
CD: |H>|V> - |V>|H>

and we detect B and C in the bell-state |H>|H>+|V>|V>. How would A and D be entangled?

I can't seem to figure it out due to not knowing whether I place a minus sign before Phi+(BC)Psi+(AD) etc.

 

[naima]

In the exercise we do not start with 2 pairs AB and CD and we do not detect BC in a given Bell state.
We start with 2 pairs BC and AD (Phi+ and Psi+) and we detect AB in a given Bell state (say Psi-)
the other 3 terms of the RHS disappear and we know that CD will be -Phi-
I think that Phi- and -Phi- are the same physical state. One equality with minus sign is correct and the other is false.

Say we have two pairs of entangled photons such as:
AB: |H>|H> - |V>|V>
CD: |H>|V> - |V>|H>

and we detect B and C in the bell-state |H>|H>+|V>|V>. How would A and D be entangled?

It would be another exercise with another equality to be verified.
There must be a rule which answers the general question but i do not know it.

 

[naima]

look at theorem 1
It gives the general rule.

 

[naima]

Say we have two pairs of entangled photons such as:
AB: |H>|H> - |V>|V>
CD: |H>|V> - |V>|H>

and we detect B and C in the bell-state |H>|H>+|V>|V>. How would A and D be entangled?

If B is H in the AB pair A will be H
if C is H in the CD pair then D will be V
so if BC is HH then AD wil be HV

If B is V then A will be -V
if C is V then D will be -H
so if BC is VV then AD will be (-V)(-H)= VH

by linearity if BC is HH + VV then AD will be HV + VH (Psi+)
If no typo!