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Mathematics of entanglement in quantum erasers

  1. Jun 26, 2015 #1

    andrewkirk

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    I am trying to understand the mathematics of quantum eraser experiments, in order to deepen my understanding of what is really happening. The paper I am currently working on is:

    "A double-slit quantum eraser" by S. P. Walborn, M. O. Terra Cunha, S. Padua, and C. H. Monken (2001)

    in which a quarter-wave plate is placed behind each slit, with the fast axis of one plate being perpendicular to that of the other. The photons passing through a slit have polarisations corresponding to either ##\vert x \rangle## or ##\vert y \rangle##, where the directions of those two polarisations are orthogonal. Each photon has been, prior to approaching the slit, entangled with a twin that has the other polarisation out of ##\vert x \rangle## and ##\vert y \rangle##.

    The paper says, just after equation 9 on page 3, that the interference pattern, which is there if the quarter-wave plates are not in place, disappears when the quarter-wave plates are in place. Unfortunately it does not show the derivation by which that is deduced, contenting itself with the vague throwaway line that 'since ##\psi_1## and ##\psi_2## [the states representing transit through the first or second slit] have orthogonal polarizations, there is no possibility of interference'.

    I have tried to confirm this result by mathematical derivation from the equations 1-9 in the paper. But in order to reach a result of non-interference, I have to make a couple of assumptions with which I am not comfortable. I am not very confident working with these states because I have not read about how to represent entangled states in bra-ket equations. My two texts - Shankar and Cohan-Tannoudji - do not mention entanglement at all, or at least the word does not appear anywhere in the index of either one.

    I would be grateful if anybody could:
    * point me to a derivation of the result that the interference disappears; and/or
    * give me some hints that might help with my derivation; and/or
    * point me towards a detailed explanation of how entangled states are represented by bra-ket equations.

    Thank you
     
  2. jcsd
  3. Jun 26, 2015 #2

    bhobba

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    I suggest you read all of the following:
    http://quantum.phys.cmu.edu/CQT/index.html

    The specific chapter relevant to the quantum eraser is:
    http://quantum.phys.cmu.edu/CQT/chaps/cqt20.pdf

    Here is the math of entanglement and the related idea of a partial trace.

    Suppose we have two systems A and B that can only be in state |a> and state |b>. If system A is in state |a> and system B in state |b> that is written as |a>|b>. Similarly if system A is in state |b> and system B in state |a> that is written as |b>|a>. But from the principle of superposition any superposition of |a>|b> and |b>|a> is also a state eg 1√2|a>|b> + 1√2|b>|a>. Such systems are called entangled. Neither system is in a definite pure state. Now lets say you observe system A, then since its the only two states it can be in you will get |a> or |b>. But because of the superposition if system A is in state |a> the total system A+B is in state |a>|b> ie you have immediately determined and know the state of system B. This is the spooky action at a distance that is sometimes talked about. Really its just a funny type of correlation.

    Suppose we have the following superposition |p> = 1/√2|b1>|a1> + 1/√2|b2>|a2>. This is obviously an entangled system where system A is entangled with system B. It's a pure state. It remains in a pure state until observed ie until its interacted with. Neither system is in a pure state but the interesting thing is each system, if you just observe that system, behaves as if it was a mixed state.

    We will demonstrate this by doing an observation on just system A with the observable A.

    E(A) = <p|A|p> = 1/2 <a1|<b1|A|b1>|a1> + 1/2 <a1|<b1|A|b2>|a2> + 1/2 <a2|<b2|A|b1>|a1> + 1/2 <a2|<b2|A|b2>|a2>

    Now here is the kicker - since you are only observing system A the observable A has no effect on the B system or its states. So we have:

    <p|A|p> = 1/2 <Aa1|<b1|b1>|a1> + 1/2 <Aa1|<b1|b2>|a2> + 1/2 <Aa2|<b2|b1>|a1> + 1/2 <Aa2|<b2|b2>|a2> = 1/2 <a1|A|a1> + 1/2 <a2|A|a2>
    = Trace((1/2|a1><a1| + 1/2|a2><a2|) A) = Trace (p' A)

    Here p' is the mixed state 1/2|a1><a1| + 1/2|a2><a2|. Thus observing system A is equivalent to observing a system in the mixed state p' - which by definition is the state from |p> by doing what's called a partial trace over B. The observation will of course give |a1> or |a2> and the entanglement will be broken so that if you get |a1> system B will be in |b1> and conversely. We still have collapse if you like that language - but now it has a different interpretation - you are not observing a pure state - but a mixed one. Its not a proper mixed state because its not prepared the way a proper mixed state is prepared - but the state is exactly the same. Any observable A will not be able to tell the difference. This means we, in a sense, can kid ourselves and say, somehow, its a proper mixed state. If it was a proper mixed state then prior to observation it is in state |a1> or state |a2> with probability of half. Prior to observation its in superposition - after it isn't. Until observed it continues in superposition - its simply because of the entanglement it can now be interpreted differently. By observing 'inside' the system - ie only observing system A - it is in a mixed state - not a proper one - but still a mixed state. Because of that it allows a different and clearer interpretation that avoids a lot of problems. This is the origin of decoherence as an explanation for what's called effective collapse. It doesn't explain actual collapse - but in a sense is a partial solution to the so called measurement problem. In fact it has morphed the issue somewhat - but that would be another thread.

    Now getting back to the quantum eraser. What's actually going on is decoherene can be undone in simple cases. I don't know the detail in your particular set-up - I will have to leave that to experimental types. However basically that's all there is to it.

    Thanks
    Bill
     
  4. Jun 26, 2015 #3

    bhobba

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    By definition the interference of two states is |a> and |b> is <a|b>. If the states are orthogonal, then again by definition <a|b> =0 so there is no interference. Its really saying the same thing. Physically its from the fact orthogonal states correspond to different eigenvalues hence you get no collapse. That is suppose you observe the system to see if its in state |a> so you use the observable |a><a|. Then applying the Born rule you get state |a> with 100% certainty and never state |b>.

    Thanks
    Bill
     
  5. Jun 26, 2015 #4

    andrewkirk

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    Thank you Bill. What you've written looks like exactly the start I need. It'll take me a while to work through it. But I'm encouraged by the fact that the representation you've written for an example entangled state is the same as the one in Shankar for the symmetric state of a system of two identical particles, which is something I have covered (he doesn't use the word 'entanglement' though).

    I expect I'll have a few questions as I work through it. I hope you don't mind.

    My first question may just be a matter of terminology but I thought I'd better make sure, so I don't head off in the wrong direction.

    You say p' is the mixed state 1/2|a1><a1| + 1/2|a2><a2|.

    I'm puzzled by the use of the word 'state' here. So far, all states I have encountered have been linear sums of kets. The above is not though. In fact it seems to describe an operator. In what sense is p' a state?

    Thank you

    Andrew
     
  6. Jun 26, 2015 #5

    bhobba

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  7. Jun 27, 2015 #6

    andrewkirk

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    Ah, I see! Shankar introduces density matrices but doesn't explain the point of them, and doesn't do anything important with them, so I didn't pay much attention at the time.

    I like the fact that your link explains at point 20 the point of them (or at least a point) which is that a density matrix |x><x| contains all the physical info about the state represented by the ket |x>, but in a way that gives a bijection between physical states and DMs, compared to a one-to-many mapping between states and kets.

    So if I've understood correctly, the density matrix you showed: p'= 1/2|a1><a1| + 1/2|a2><a2| corresponds to the state that has ket
    1/√2 (|a1> + |a2>).

    Thank you

    Andrew
     
  8. Jun 27, 2015 #7

    bhobba

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    No. 1/√2 (|a1> + |a2> is shorthand for 1/2(|a1 + a2><a1 + a2|) which is what you get when you write it out as an operator. Its not the same as 1/2|a1><a1| + 1/2|a2><a2|.

    Here is another way of looking at it. The correct statement of the Born Rule is given an observable O, a positive operator of unit trace, P, exists, such that the expected value of the observation associated with O, E(O), is E(O) = Trace( PO). By definition P is called the state of the system.

    Note - the state is not an element of a vector space - it is a positive operator. Let that sink in. Forget the vector space stuff.

    Now a state of the form |u><u| is called pure. Since |u> is an element of a vector space that's where the vector space thing comes from - but it only applies to pure states. Also note that it is invariant to a phase transformation because |u><u| remains the same. States that are convex sums of pure states are called mixed ie are of the form ∑pi |ui><ui|. It can be shown (and it's not too hard - I will leave it as an exercise) that all states are mixed or pure.

    Now, hopefully, you can see the fundamental difference between 1/√2 (|a1> + |a2>) and 1/2|a1><a1| + 1/2|a2><a2|

    Just as an aside to deepen your understanding the Born Rule can in fact be derived from the axiom about observables - see post 137 of the following:
    https://www.physicsforums.com/threads/the-born-rule-in-many-worlds.763139/page-7

    Thanks
    Bill
     
  9. Jun 27, 2015 #8

    andrewkirk

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    Thank you Bill. I see the distinction now, and that the state represented by DM 1/2(|a1><a1| + |a2><a2|) cannot be represented by any linear sum of the kets |a1> and |a2> because it is a mixed state. There are two questions that arise for me out of that.

    1. Is it also the case that there is no ket that can represent that mixed state, or just that it cannot be expressed as a linear sum of |a1> and |a2>? I looked up various formulations of the first postulate of QM and found that in some presentations it says that the state of a particle can be represented by a ket, whereas in others it says that the state of a system can be represented by a ket. Shankar takes the former approach while Cohen-Tannoudji takes the latter. If there were no possible ket representation of a mixed state, it would mean either that the latter formulations are incorrect or that they mean something different by 'system' that somehow excludes mixed states.

    2. Is an entangled pair of particles necessarily a mixed state?

    thank you

    Andrew
     
  10. Jun 27, 2015 #9

    bhobba

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    No ket can represent a mixed state - which should be obvious from its definition. A ket represents a pure state - a mixed state is the sum of pure states

    They are wrong - but its a very common error - so common most would say it's not really an error because more advanced treatments know that's what's usually done and corrects it. I think you need to see an axiomatic treatment which the link I gave to my post about where the Born Rule comes from is the start of. For that see the first three chapters of Ballentine:
    https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

    That's why me and Strangerep have what we have at the bottom of our posts:-p:-p:-p:-p:-p:-p:-p

    Yes - when entangled with other systems, systems can no longer be in pure states, and it turns out they are in mixed states. But there is subtlety here between what are called proper mixed states and improper ones:
    http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

    See section 1.2.3

    Thanks
    Bill
     
    Last edited: Jun 28, 2015
  11. Jun 28, 2015 #10

    andrewkirk

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    Bill now that I"ve read enough of http://pages.uoregon.edu/svanenk/solutions/Mixed_states.pdf to feel that I have at least a basic understanding of pure vs mixed states and the role of density matrices, traces and partial traces, I'm trying to work through your post #2 above.

    My next question is about the following part of that post:
    Here it looks like you have dropped the 2nd and 3rd terms. I'm guessing that is because you are assuming that <b1|b2>=0, ie that |b1> and |b2> are orthogonal, but I want to check because no such assumption was stated.

    If so, are we also assuming that <a1|a2>=0 (I think we might need that for the next line)?

    Thank you.

    Andrew
     
  12. Jun 28, 2015 #11

    bhobba

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    I am assuming they are orthonormal.

    Thanks
    Bill
     
  13. Jul 11, 2015 #12

    andrewkirk

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    I am working through this, doing all the proofs marked by '(P)'.

    There is one statement that is made, not marked by (P) but not proved either. In para 26 it says that any mixed-state operator ##\hat{\rho}## can be diagonalised as ##\hat{\rho}=\sum_{k=1}^M \lambda_k |\psi_k\rangle \langle \psi |## and that all the ##\lambda_k## are in [0,1].

    The diagonalisability follows directly from ##\hat{\rho}## being a sum of Hermitian operators, and hence also Hermitian. But what is the justification for the claim that all its eigenvalues are in [0,1]? The Hermitian property guarantees that they are real, but not that they are positive.

    thank you

    Andrew
     
  14. Jul 16, 2015 #13

    andrewkirk

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    Does anybody have any idea about whether the statement in the above post is right, and why?
    ie why should the density matrix of a mixed state have all non-negative eigenvalues?
     
  15. Jul 16, 2015 #14

    bhobba

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    By definition its a positive operator of unit trace. It follows from that definition easily using the spectral theorem ie from the spectral theorem P = ∑ yi |bi><bi| where the yi are the eigenvalues.

    Because its positive <bj|P|bj> is positive. Hence yj is positive.

    Thanks
    Bill
     
    Last edited: Jul 16, 2015
  16. Jul 16, 2015 #15

    andrewkirk

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    Ah, OK.
    Not actually by definition, but I can see that it follows fairly readily from the definition, which is :
    $$\hat{\rho}\equiv \sum_{k=1}^N p_k|\psi_k\rangle\langle \psi_k|$$
    where ##\{|\psi_1\rangle,...,|\psi_N\rangle\}## is a set of pure states, not necessarily orthogonal, and
    $$\sum_{k=1}^N p_k=1\wedge \forall k, 0< p_k\leq 1,$$.

    So for ##\hat{\rho}## to be a positive operator we require that for every pure state ##|x\rangle## we have ##\langle x|\hat{p}|x\rangle\geq 0##

    We have
    ##\langle x|\hat{p}|x\rangle\equiv
    \langle x|\bigg(\sum_{k=1}^N p_k|\psi_k\rangle\langle \psi_k|\bigg)|x\rangle
    =\sum_{k=1}^N p_k\langle x| \psi_k\rangle\langle \psi_k|x\rangle
    =\sum_{k=1}^N p_k \langle x| \psi_k\rangle \langle x| \psi_k\rangle^*
    =\sum_{k=1}^N p_k |\langle x| \psi_k\rangle|^2
    ##
    which is greater than or equal to zero because the ##p_k## are all positive by definition and ##|\langle x| \psi_k\rangle|## is real.

    The unit trace is part of the definition.

    Thank You
     
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