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Entanglement correlations, singlet spin state

  1. Apr 4, 2010 #1
    Let's say you have a pair of electrons in the singlet spin state. I thought that Alice measuring the spin of one electron (about the "z axis") corresponded to applying the operator [tex]\hat{S}_z\otimes \hat{I}[/tex] (where [tex]\hat{I}[/tex] is the identity operator) to the singlet state [tex]\frac{1}{\sqrt{2}}(\uparrow \otimes \downarrow - \downarrow \otimes \uparrow )[/tex]. However, I don't see (mathematically) why Alice measuring spin up forces the state to be [tex]\uparrow\otimes\downarrow [/tex].

    I say this because the operator [tex]\hat{S}_z\otimes \hat{I}[/tex] has eigenvectors (like [tex]\frac{1}{\sqrt{2}}(\uparrow\otimes\uparrow + \uparrow\otimes\downarrow )[/tex]) which (1) would give Alice the spin up result and (2) have a nonzero inner product with the initial (singlet) state. So why can't the state collapse to another eigenvector like the one I just mentioned (and thus Bob not necessarily measure spin down)? I think I'm missing something simple here.
     
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  3. Apr 5, 2010 #2
    Just thought about this a little more and I have another question which could solve this problem. Let me clarify the steps I was describing in my initial question first.

    Step 1: Prepare state [tex]1/\sqrt{2}(\uparrow\otimes\downarrow -\downarrow\otimes\uparrow )[/tex]

    Step 2: Alice measures spin up when measuring [tex]\hat{A}=\hat{S}_z\otimes\hat{I}[/tex].

    Step 3: State collapses to an eigenstate of [tex]\hat{A}[/tex] with eigenvalue [tex]\hbar /2[/tex], namely a state of the form [tex]\uparrow\otimes (\mu_1\uparrow + \mu_2 \downarrow) \equiv S_{\mu_1,\mu_2}[/tex].

    Step 4: Bob measures the spin of the other electron via [tex]\hat{B}=\hat{I}\otimes\hat{S}_z[/tex].

    Step 5: State collapses to an eigenstate of [tex]\hat{B}[/tex] where the probability of collapsing to any such state is the square of the inner product of that state with [tex]S_{\mu_1,\mu_2}[/tex]. This means we get [tex]\uparrow\otimes\uparrow[/tex] with probability [tex]|\mu_1|^2[/tex], otherwise [tex]\uparrow\otimes\downarrow[/tex].

    Now, if you consider steps 2 and 4 together as one step (and 3&5 as one step), I do agree that we must end up in the state [tex]\uparrow\otimes \downarrow[/tex] (or [tex] \downarrow\otimes\uparrow[/tex]) -- the normal correlations appear. But how do you justify that? Why don't we apply that logic to other cases like measuring the position and then the momentum of particle and then saying it was a simultaneous measurement of both properties, ...
     
  4. Apr 5, 2010 #3
    Good question.

    I wonder if [tex]\hat{A}=\hat{S}_z\otimes\hat{I}[/tex] can be considered as an observable for the complete state. Maybe we shouldn't expect to find an eigenstate of that kind of operator when Alice only is performing a measurement.

    I think that this operation should be considered as a projection of the initial state onto the subspace borne by the eigenstate got by Alice. The eigenstate that you think about is indeed into this subspace, but the initial state is not projected onto it.
     
  5. Apr 5, 2010 #4

    SpectraCat

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    I think this has more to do with the allowable eigenstates of the system, rather than the possible measurement results at Alice's detector. Since we are dealing with fermions, the overall wavefunction must be anti-symmetric ... since we further know that we are dealing with a singlet state, that limits the possible configurations to (up,down) and (down,up). Basically what I am saying is that Alice's measurement cannot throw the system into a configuration that has a coefficient of zero in the initial superposition state. So, in step 3 of your example, we know that the [tex]\mu_1[/tex] coefficient is zero going in (due to the specification of the singlet state), so Alice's measurement cannot change that.
     
  6. Apr 5, 2010 #5
    I suppose one way of looking at it is, when Alice measures one particle she projects BOTH particles into one of the states in the initial superposition.

    Why should it end up in any eigenstate of [itex]S_z \otimes 1[/itex]? If a system is in the state (1 0) an Sz measurement won't find it in any eigenstate of Sz, else you could get (0 1) which doesn't make sense.
     
  7. Apr 5, 2010 #6
    Pio2001: Yes, I was assuming that if Alice measures Sz (and no measurement is performed on Bob's end), then the appropriate operator should be [tex]\hat{S}_z\otimes \hat{I}[/tex]. If that's not correct, does anyone know what the appropriate operator would be? Also, the state I mentioned does have a nonzero projection onto the initial state -- their inner product is [tex]\mu_2[/tex].

    SpectraCat: My question is what limits the states to (up,down) and (down,up) -- the the normal collapse rules don't seem to enforce that in this situation (although it looks like they DO if the operator [tex]\hat{S}_z\otimes \hat{S}_z[/tex] was applied instead). What do you mean the configuration has a coefficient of zero in the initial state -- the inner product of the two states is non-zero ([tex]\mu_2[/tex]) right?

    Tomsk: What I'm wondering is why the other particle must collapse...maybe [tex]\hat{S}_z\otimes \hat{I}[/tex] is not Hermetian and any Hermetian operator would enforce this? I'll try to check that. On your second point, it should be an eigenstate of that operator if that's the correct operator, the reason your example is different is because in your case the inner product of (0 1) with the initial state (1 0) is 0, so there is 0 probability of ending up in (0 1) after the measurement. In the example I gave the inner product is not 0.
     
  8. Apr 5, 2010 #7

    SpectraCat

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    Well, that's just the point ... the only states that can be projected out of a superposition are the ones that were in it to begin with. You cannot create probability density for the (up,up), or (down,down) just by making a measurement.

    Think of it this way .. imagine you have a superpostion of just the v=0 and v=1 states of a harmonic oscillator. Now you make a measurement of the energy. What possible values can you observe? You will measure energy eigenvalues to be sure, but only those that were in the superposition to begin with (i.e. either 0.5 or 1.5 times hv).

    The situation you are talking about here are is no different ... the only states in the initial superposition (i.e. the singlet), are (up,down) and (down,up) ... therefore those are the only possible states that can be projected out by Alice's measurement.
     
  9. Apr 5, 2010 #8
    SpectraCat: If I look at a simpler example in which a system is in a state [tex]1/\sqrt{2}(\psi_1 - \psi_2)[/tex], where [tex]$\psi_i$[/tex] are eigenstates of a nondegenerate operator O, then I understand that measurement of O will make the state become either [tex]\psi_1[/tex] or [tex]\psi_2[/tex], with an equal probability of either result. I.e. you have to end up with one of the 2 states in the initial superposition as you said. This seems to be what you are relating to or thinking of, but I think there is a crucial difference here -- the fact that O is nondegenerate implies that no other eigenstates have a nonzero inner product with [tex]\psi_1[/tex] and [tex]\psi_2[/tex], and therefore no other eigenstates can result from measurement of O. However, in the problem I was talking about the degeneracy of the operator is such that other states (like the one I mentioned) can both (1) be eigenstates of the operator and (2) have a nonzero inner product (and therefore a nonzero probability of resulting from the measurement, presumably).
     
  10. Apr 6, 2010 #9

    SpectraCat

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    Ok .. I guess I see what you are saying. I am not sure that you are correct, but let's assume for the moment you are. The state you constructed (up,up) + (down,up), is not a valid eigenstate for any system of two identical particles, because it is neither antisymmetric nor symmetric with respect to exchange of the particles. Thus I don't think it is possible for Alice's measurement to throw the system into such a state, even though it does seem to be a possible eigenstate of her measurement operator.

    Notice that all of the other (i.e. non-singlet) properly symmetrized states for the two-particle system will have zero inner-products with the singlet state, so the issue that you have raised cannot ever occur. Thus the only possible state that Alice can measure is the singlet state, which is physically sensible, and matches with observation.
     
  11. Apr 6, 2010 #10
    Doesn't the question actually boil down to "why [tex]I|\uparrow\rangle[/tex] doesn't result in a final state other than [tex]|\uparrow\rangle[/tex] even though all states in the space [tex]C^2[/tex] except [tex]|\downarrow\rangle[/tex] has nonzero overlap with the original state?" The identity operator has only one eigenvalue, not counting multiplicity, so it can't be broken down into "smaller" projection operators. Hence the identity operator can't force the state to change. That's how I understand it.
     
    Last edited: Apr 6, 2010
  12. Apr 6, 2010 #11
    Spectracat: I see what you are saying about the symmetry/antisymmetry, but I thought that the "symmetrization requirement" was supposed to automatically be enforced by the laws of QM, so long as you start with a symmetric (or antisymmetric) state. I may be wrong about this, but this seems to be what Griffths says (p205 of Introduction to Quantum Mechanics, 2nd Ed.). I may not be interpreting this right?

    Truecrimson: I don't see how it would boil down to that because the state we're in is not just [tex]\uparrow[/tex], but maybe the answer to this would give the required insights. What do you mean the identity operator can't be broken down into smaller projection operators? I mean, isn't it true that [tex]\hat{I} = |\uparrow\rangle\langle\uparrow | + |\downarrow\rangle\langle\downarrow |[/tex]? Do you mean something different?

    Thanks guys
     
  13. Apr 6, 2010 #12
    The problem is that all the vectors of the space are eigenstates of [tex]\hat{I}[/tex]. Since their are not all orthogonal, [tex]\hat{I}[/tex] can't be an observable.

    Therefore it makes no sense to say that you are "measuring [tex]\hat{A}=\hat{S}_z\otimes\hat{I}[/tex]".
     
  14. Apr 6, 2010 #13
    Pio2001: OK, thanks this was one of the questions I asked earlier. If that is not the correct operator, then what is the correct operator? I.e. what operator represents "Alice measuring the spin of one particle about the z axis?"

    Thanks
     
  15. Apr 6, 2010 #14
    Measuring[tex]\hat{I}[/tex] is "doing nothing." [tex]I[/tex] is self-adjoint, so it's an observable in a mathematical sense. And the tensor product operator you use is indeed the correct one to represent local operation by Alice. You can find them in any quantum information book.
     
  16. Apr 6, 2010 #15
    According to the french wikipedia, in order to be qualified as an observable, an operator must meet the following requirements :
    1-It must be a linear operator.
    2-Its eigenvalues must be real.
    3-Its eigenvectors must be othogonal ( = the probability of finding another engenstate after performing the same measurement again must be zero).
    4-The eigenstates of the operator must be a basis of the Hilbert space.
    5-The eigenstates of the operators must be normalizable.

    It seems to me that [tex]\hat{I}[/tex] does not meet the third and fifth conditions.

    I didn't keep my book. Could you be more specific ? Alice is certainly not performing a tensorial product. She is performing a measurement. And she is not applying either operator to the state vector. Her result is random, while the action of an operator on the state vector is not.

    According to my memories, her measurement consists in projecting the initial state onto the subspace borne by her measurement result.
     
  17. Apr 6, 2010 #16
    In chapter 2 of Preskill's note http://www.theory.caltech.edu/people/preskill/ph229/notes/chap2.pdf

    page 15 " I would like to consider more general observables acting on qubit
    A, and I would like to characterize the measurement outcomes for A alone
    (irrespective of the outcomes of any measurements of the inaccessible qubit
    B). An observable acting on qubit A only can be expressed as
    [tex]\hat{M}_A[/tex] ⊗ [tex]\hat{I}_B[/tex] ,
    (2.49)
    where [tex]\hat{M}_A[/tex] is a self-adjoint operator acting on A, and [tex]\hat{I}_B[/tex] is the identity
    operator acting on B. "
     
  18. Apr 6, 2010 #17
    The tricky part here is "acting on qubit A only". Msumm21 is talking about a non-separable state. Therefore, in this case, "acting on Alice's particle state only" can make no sense.
     
  19. Apr 6, 2010 #18
    I forgot that my class notes are online. Here it is. http://www.maths.bris.ac.uk/~manl/handouts10_im1.pdf [Broken]
    On the second to last page, you can see that writing the operator in this form is consistent with the partial trace operation and the calculation of expectation values.

    For a bit more detailed discussion, take a look at Nielsen and Chuang page 107 box 2.6 "Why the partial trace?"
     
    Last edited by a moderator: May 4, 2017
  20. Apr 6, 2010 #19
    When we measure the z-component of the spin there are only two possible results for the superposition state given:

    Result 1 Alice obtains spin up, which tells us that if Bob measures the same component, he gets spin down.
    Result 2 Alice obtains spin down, which tells us that if Bob measures the same component, he gets spin up.
    Alice up, Bob down always go together, The two values constitute a single result. Likewise for Alice down, Bob up. No other results are possible.
    We are dealing with entangled states here and the non-separability principle of quantum mechanics implies that we talk about the pair of particles, not individual particles.

    Assuming you are looking for projection operators, the projection operator for result 1 is [tex]\left| { \uparrow \downarrow } \right\rangle \left\langle { \uparrow \downarrow } \right|[/tex]. The projection operator for result 2 is [tex]\left| { \downarrow \uparrow } \right\rangle \left\langle { \downarrow \uparrow } \right|[/tex].

    Best wishes
     
  21. Apr 6, 2010 #20
    Pio2001: What do you mean when you say that "acting on Alice's particle only" doesn't make sense? Alice can measure the spin of her particle while Bob does not measure the spin of his particle, right? And, from Truecrimson's links, it does appear that the operator I mentioned is the correct operator for such a situation. Now, according to the stuff I read on the internet Bob's particle somehow collapses to the state with spin opposite to Alice's (and maybe that's the action on Bob's particle you are talking about), but I still just don't see mathematically what enforces that. I agree that if we make a new fundamental rule stating that "Bob's particle must collapse when Alice's does (and vice versa)" then we could prove that his particle does collapse to the opposite spin state (cause the inner product of the other state with the initial state is 0), but I don't think that's supposed to be a fundamental rule, right? I mean, I thought that was supposed to be an implication of more fundamental rules, right? So I think there's some other fundamental rule that I'm not including when doing the math, just not sure what it is and I don't see it in places I'd expect to see it, like the postulates of QM listed on Wikipedia.

    eaglelake: The question I was asking is what forces the result to be one of the 2 you mentioned. In my previous posts I explained how I came to the conclusion that other states could result, while _seemingly_ following the basic rules of QM. Evidentially I did break a fundamental rule somewhere in that logic, I just don't see where. I know people have pointed out that the example state I mentioned breaks the symmetrization rules and is contrary to the literature which says the state must collapse to one of the two you mentioned, ... but I THINK those things are supposed to follow from the fundamental rules of QM, right? I don't think those things themselves are supposed to be the fundamental QM rules. In the case of the symmetrization requirement Griffths seems to imply that once a system is in an antisymmetric state (like the initial one) the state will automatically remain antisymmetric when evolving under the QM collapse/evolution rules (i.e. symmetrization shouldn't have to be added as a new rule, it should be automatic, a consequence of the more fundamental rules).
     
    Last edited: Apr 6, 2010
  22. Apr 7, 2010 #21
    Yes, but [tex]\frac{1}{\sqrt{2}}(\uparrow \otimes \downarrow - \downarrow \otimes \uparrow )[/tex] is not the state of her particle, nor is it the state of Bob's particle.

    I'm lost here too.

    In the basis [tex]\{| \uparrow_A \rangle, | \downarrow_A \rangle \}[/tex], the operator [tex]\hat{S}_z[/tex] is

    Code (Text):
    1  0
    0 -1
    Thus in the basis [tex]{| \uparrow_A \otimes \uparrow_B \rangle, | \uparrow_A \otimes \downarrow_B \rangle, | \downarrow_A \otimes \uparrow_B \rangle, | \downarrow_A \otimes \downarrow_B \rangle}[/tex], the operator [tex]\hat{S}_z \otimes \hat{I}[/tex] is

    Code (Text):
    1  0  0  0
    0  1  0  0
    0  0 -1  0
    0  0  0 -1
    The state [tex]\frac{1}{\sqrt{2}}(\uparrow\otimes\uparrow + \uparrow\otimes\downarrow )[/tex] that you talk about is

    Code (Text):
    0.707
    0.707
    0
    0
    And it is indeed an eigenstate of this operator :

    Code (Text):
    1  0  0  0 | 0.707   | 0.707
    0  1  0  0 | 0.707   | 0.707
    0  0 -1  0 | 0     = | 0
    0  0  0 -1 | 0       | 0
    However, since all eigenstates are not orthogonal, you can built several different basis from the eigenstates of [tex]\hat{S}_z \otimes \hat{I}[/tex].

    The rule says that when you make a measurement, your system ends up in one of the eigenstates of your observable. With such an operator, which is not an observable according to the french wikipedia, this statement is unclear and we can't decide what kind of result is got from Alice's measurement.
     
  23. Apr 7, 2010 #22
    The only rule I have learnt is that if we do a measurement on a state [tex]|\psi\rangle[/tex] and get an eigenvalue [tex]\lambda[/tex], the state we end up with is
    [tex]P_{\lambda}|\psi\rangle[/tex] (apart from the normalization factor) where [tex]P_{\lambda}[/tex] is a projection operator corresponding to the eigenvalue.

    In msumm21's example, the projection operator of [tex]S_z[/tex][tex]\otimes[/tex][tex]I[/tex] corresponding to the eigenvalue +1 (Alice got spin up) is [tex]|\uparrow\rangle\langle\uparrow|[/tex] [tex]\otimes[/tex][tex]I[/tex] ([tex]I[/tex] itself is the projection operator we've to consider here) So the final state is

    [tex]|\uparrow\rangle\langle\uparrow|[/tex] [tex]\otimes[/tex][tex]I[/tex]([tex]|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle[/tex]=[tex]|\uparrow\rangle|\downarrow\rangle[/tex] without any ambiguity.

    Edit: I can't get the last equation in LaTeX properly for some reason, but you know what I mean.

    Edit 2: Thank you SpectraCat.
     
    Last edited: Apr 7, 2010
  24. Apr 7, 2010 #23

    SpectraCat

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    I fixed the LaTeX in the quote above, and also clarified the notation for Alice and Bob's measurements. I also agree with what you have written .. it is basically what I was trying to say qualitatively in some of my earlier posts.
     
  25. Apr 7, 2010 #24
    That seems logical to me.
     
  26. Apr 7, 2010 #25
    OK, so the eigenstate resulting from a measurement is the projection of the initial state into the eigenspace of the observed eigenvalue? I guess that is what Spectracat was trying to say earlier and I see that now in Wikipedia. I agree that this rule prevents the state I mentioned earlier. I never considered this because my QM book gives a different rule, it says measurement gives a particular eigenvalue with probability equal to the square of the inner product of a corresponding eigenstate with the initial state and then the STATE COLLAPSES TO THE CORRESPONDING EIGENSTATE. This is p106 of Intro to QM 2nd Ed by Griffths. So I think this explanation in Griffths is bad -- it allows states like the one I mentioned in earlier posts, which are NOT allowed by this rule yall just told me. I.e. the rule Griffths is explaining does not seem to be equivalent.

    I just looked at my other QM book (Intro QM 4th Ed by Liboff) and on the top of page 74 it has the same problem as Griffths. Am I missing something?

    Thanks for the info!
     
    Last edited: Apr 7, 2010
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