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I Physical significance of a.σ in expectation -E(a.σ b.σ)?

  1. Mar 8, 2017 #1

    N88

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    Admins: Please excuse my E and brackets in the title, and correct if possible.

    My questions are these, please:

    1. What is the physical significance of [itex] (\hat{a}\cdot\boldsymbol{\sigma}_{1})
    [/itex]
    in [itex]
    \left\langle (\hat{a}\cdot\boldsymbol{\sigma}_{1})(-\boldsymbol{\sigma}_{2}\cdot\hat{b})\right\rangle
    [/itex]?

    2. This QM formulation appears (to me) to be local and realistic: but I take it that that is not the mainstream view?

    The questions arise in calculating the expectation for the famous EPRB experiment.

    I understand the orientations of the detectors and representing the intrinsic spin of a spin-half particle by s = (hbar/2) [itex]\boldsymbol{\sigma}[/itex] where [itex]\boldsymbol{\sigma}[/itex] is the intrinsic spin vector. So the QM formulation looks (to me) to be local and realistic.

    PS: I have a derivation of the correct result which looks to be local and realistic. But I have not found the same approach in any textbook, so my derivation may be non-mainstream. That's why I would like to check the physical significance of the mainstream result first (and in detail).

    Thank you.
     
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  3. Mar 8, 2017 #2

    PeterDonis

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    Where are you getting this equation from? What is the context?

    The symbol ##\boldsymbol{\sigma}## refers to a "vector" of matrices, the Pauli matrices: i.e., a "vector" with components ##\sigma_1##, ##\sigma_2##, ##\sigma_3##, which are the Pauli matrices:

    $$
    \sigma_1 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
    $$
    $$
    \sigma_2 = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}
    $$
    $$
    \sigma_3 = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
    $$

    So the notation ##\hat{a} \cdot \boldsymbol{\sigma}## refers to the "dot product" of some vector ##\hat{a}## with the "vector" of matrices given above. If, as I suspect, ##\hat{a}## is meant to represent a unit vector in some particular direction (the direction of the axis along which the spin is to be measured), then ##\hat{a} \cdot \boldsymbol{\sigma}## would give an operator which would represent the proper "mixture" of the three Pauli spin matrices to denote measuring the spin of a spin-1/2 particle about the axis ##\hat{a}##. But an operator is not the same thing as a state, which is what "the intrinsic spin of a spin-1/2 particle" should be. So I'm confused about this notation and wondering where it comes from.
     
  4. Mar 8, 2017 #3

    N88

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    The expectation equation is from Bell (1964), eqn (3).
    http://cds.cern.ch/record/111654/files/vol1p195-200_001.pdf

    [itex]\hat{a}[/itex] is a detector unit-vector; the orientation of the principal axis.

    The intrinsic spin equation (in which I replaced his m with s),* is from p.149, eqn (48):
    Dirac, P. A. M. (1982). The Principles of Quantum Mechanics (4th ed., rev.). Clarendon, Oxford.

    * To have s related to intrinsic spin just better suits the way I think.

    I will certainly benefit from discussion of the physical significance of the 'vector'**
    [itex]\boldsymbol{\sigma}[/itex] being defined by [itex]\boldsymbol{\sigma}\equiv(\sigma_{x}, \sigma_{y}, \sigma_{z}).[/itex]

    ** I would like to call it by its correct name, more fully reflecting its physical significance (so that we don't need to put in "scare quotes" -- which makes it appear even scarier to a beginner.)

    Thank you.
     
    Last edited: Mar 8, 2017
  5. Mar 9, 2017 #4

    stevendaryl

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    I don't understand in what sense you would call that "local".

    When Alice measures the spin of her particle along axis [itex]\vec{a}[/itex], she is measuring the observable
    [itex]\vec{a} \cdot \vec{\sigma_1}[/itex]. When Bob measures the spin of his particle along axis [itex]\vec{b}[/itex], he is measuring the observable [itex]\vec{b} \cdot \vec{\sigma_2}[/itex]. Quantum-mechanically, the expectation value for the product [itex](\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})[/itex] is given by:

    [itex]\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle[/itex]

    where [itex]|\Psi\rangle[/itex] is the two-particle spin-state, which in the case of anti-correlated entangled spins is the state [itex]|\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle)[/itex]. Since the first operator only acts on the first particle, and the second operator only acts on the second particle, we can calculate:

    [itex]\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = - \vec{a} \cdot \vec{b}[/itex]

    That's not the explanation, that's the fact that needs to be explained. What Alice actually gets from her measurement of spin is not [itex]\vec{a} \cdot \vec{\sigma_1}[/itex], but one of the eigenvalues of that operator, which are [itex]\pm 1[/itex]. Apparently, it's completely random which value (plus or minus 1) she actually gets. It's also apparently random which of [itex]\pm 1[/itex] Bob gets. But despite their being random, they are correlated. (If they weren't correlated then the expectation value of the product of Bob's result and Alice's result would be zero.) So the question is how to explain the correlation through local mechanisms. Suppose that they both decide to measure spins along axis [itex]\vec{a}[/itex]. Suppose further that Alice gets her result first, and it's +1. Then at that point, she knows for certain that Bob will measure -1. That seems to be a fact about Bob and his particle that is not reflected by any variable local to Bob. So what guarantees that Bob will get -1?

    Pointing out that [itex]\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle = -1[/itex] in this case (the case in which [itex]\vec{b} = \vec{a}[/itex]) doesn't seem like an explanation for the fact, it's just a mathematical way of stating the fact.

    The derivation of the expectation value

    Since [itex]|\Psi\rangle = \frac{1}{\sqrt{2}} (|u\rangle |d\rangle - |d\rangle |u\rangle)[/itex] and [itex]\vec{a} \cdot \vec{\sigma_1}[/itex] only affects the first component of the state, and [itex]\vec{b} \cdot \vec{\sigma_2}[/itex] only affects the second component of the state, we can write:

    [itex]\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle[/itex]
    [itex]= \frac{1}{2} [\langle u |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|d\rangle [/itex]
    [itex]- \langle u |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle d|(\vec{b} \cdot \vec{\sigma_2})|u\rangle [/itex]
    [itex]- \langle d |(\vec{a} \cdot\vec{\sigma_1}) |u\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|d\rangle [/itex]
    [itex]+ \langle d |(\vec{a} \cdot\vec{\sigma_1}) |d\rangle \langle u|(\vec{b} \cdot \vec{\sigma_2})|u\rangle][/itex]

    So now we can use the representation of the Pauli matrices to derive:

    [itex]\langle u | \vec{a} \cdot \vec{\sigma} |u\rangle = a_z[/itex]
    [itex]\langle d | \vec{a} \cdot \vec{\sigma} |u\rangle = (a_x + i a_y)[/itex]
    [itex]\langle u | \vec{a} \cdot \vec{\sigma} |d\rangle = (a_x - i a_y)[/itex]
    [itex]\langle d | \vec{a} \cdot \vec{\sigma} |d\rangle = -a_z[/itex]
    [itex]\langle u | \vec{b} \cdot \vec{\sigma} |u\rangle = b_z[/itex]
    [itex]\langle d | \vec{b} \cdot \vec{\sigma} |u\rangle = (b_x + i b_y)[/itex]
    [itex]\langle u | \vec{b} \cdot \vec{\sigma} |d\rangle = (b_x - i b_y)[/itex]
    [itex]\langle d | \vec{b} \cdot \vec{\sigma} |d\rangle = -b_z[/itex]

    So:
    [itex]\langle \Psi |(\vec{a} \cdot\vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2})|\Psi\rangle[/itex]
    [itex]= \frac{1}{2} [- a_z b_z - (a_x - i a_y) (b_x + i b_y) - (a_x + i a_y) (b_x - i b_y) - a_z b_z][/itex]
    [itex]= - a_z b_z - a_x b_x - a_y b_y[/itex]
     
  6. Mar 9, 2017 #5

    PeterDonis

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    The physical significance is that the dot product of this vector with a vector ##\hat{a}## that points along some axis, gives you an operator (i.e., a matrix) that represents measuring the spin about that axis. So, as stevendaryl said, ##\hat{a} \cdot \boldsymbol{\sigma_1}## is the operator that represents Alice measuring the spin of her particle (hence the ##1## subscript on the ##\boldsymbol{\sigma}##) about axis ##\hat{a}##. And similarly for Bob, with the subscript ##2## and the axis ##\hat{b}##.
     
  7. Mar 11, 2017 #6

    N88

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    Thanks for this. Please, what name do you give to "this vector" -- "the dot product of this vector with a vector ##\hat{a}##".

    I want to be clear about the terminology that I use in replying to stevendaryl.
     
  8. Mar 11, 2017 #7

    stevendaryl

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    You can call it "the component of spin in the [itex]\hat{a}[/itex] direction".
     
  9. Mar 11, 2017 #8

    N88

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    I'd like to be clear about this. With respect to the vector that I asked about, PeterDonis (Post #2) wrote: The symbol σ refers to a "vector" of matrices, the Pauli matrices.

    So I thought it was the dot-product of σ with a unit-vector [itex]\hat{a}[/itex] that gave "the component of spin in the [itex]\hat{a}[/itex] direction.

    So my new question is this: Is σ called "the Pauli-vector"?

    So when it is described as "a vector of matrices" with a dot-product, this is a slightly misleading for beginners unless they understand that it is a special element of a vector space (maybe)?

    Because when I searched for "Pauli vector" I found this: https://www.physicsforums.com/threads/pauli-vector.635079/

    So, hoping I have not misunderstood, I am thinking that dextercioby's answer is helpful: we are working with "a shorthand notation which is useful, but can be misleading".

    PS: I'm happy to work with the shorthand, but I want to be clear for my response to your nicely expanded work above. Thanks
     
  10. Mar 11, 2017 #9

    stevendaryl

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    Yes. It's a vector with components [itex]\sigma_x, \sigma_y, \sigma_z[/itex], where each of those is a 2x2 matrix.

    Yes, except it's an operator, not a value. In the same way that [itex]p_x = -i \hbar \frac{\partial}{\partial x}[/itex] is the component of momentum in the x-direction, but it's an operator, not a value.

    I can't see how it's misleading to call it a vector of matrices. It's not an element of a vector space, it is a vector of matrices in the sense of a 3-vector with an x-component, a y-component and a z-component. But those components are matrices, instead of numbers.
     
  11. Mar 12, 2017 #10

    N88

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    Elsewhere, as I recall, you asked about my local-realistic views. Above you wondered about the sense of "local" that I use.

    I'm attaching a 1-page PDF extract from an essay that I'm working on. Equations marked (X.) are not relevant to our current discussion. The QM results are mainstream: whether I have derived them properly, under local-realism, remains the question. Also: Do they answer the correlation difficulties that you mention above?

    With my thanks again for the detail that you provide, I look forward to your comments.
     

    Attached Files:

  12. Mar 13, 2017 #11

    stevendaryl

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    No, I don't see how it answers the problems. You seem to be giving a local realistic model of the type that Bell proved impossible. As I said, the fact about Pauli matrices, that [itex]\langle \Psi |(\vec{a} \cdot \vec{\sigma_1})(\vec{b} \cdot \vec{\sigma_2}) \Psi \rangle = - \vec{a} \cdot \vec{b}[/itex] does not explain EPR correlations, it is simply a mathematically succinct statement of those correlations.

    You say in that paper that

    That is the starting point of a possible local-realistic model of the type that Bell proved is impossible. What you're saying is that
    • Alice's particle has an intrinsic spin [itex]\vec{\sigma}[/itex] pointing in some direction.
    • The particle interacts with the electromagnetic field of Alice's detector (which has orientation [itex]\hat{a}[/itex]).
    • As a result of the interaction, the particle's spin afterward is either spin-up or spin-down relative to [itex]\hat{a}[/itex].
    Before we move on to Bob, let's ask a follow-up question about this: Given [itex]\vec{\sigma}[/itex] and [itex]\hat{a}[/itex], is it nondeterministic, or deterministic whether the particle ends up being measured to have spin-up or spin-down relative to [itex]\hat{a}[/itex]?
    • If it's deterministic, then you're proposing a deterministic local hidden-variable theory, and Bell proved that no such theory produces the EPR correlations.
    • If it's nondeterministic, it's provable that no nondeterministic theory can reproduce the perfect correlations of EPR.
    The fact about products of Pauli spin matrices seems completely irrelevant. Or at least, I don't see the relevance. Once again, let's consider the case in which Alice has measured spin along axis [itex]\hat{a}[/itex] and gotten result [itex]+1[/itex] (or [itex]+\frac{\hbar}{2}[/itex]). Bob has not yet measured his particle, but he also plans to measure spin along the same axis, [itex]\hat{a}[/itex]. Alice can confidently predict that Bob will get result [itex]-1[/itex]. That's a fact to be explained.

    If you want to say that it's explained by saying that Bob's particle has an intrinsic spin vector [itex]\vec{\sigma_2}[/itex], fine. How does that explain that Bob will definitely get result [itex]-1[/itex]? In keeping with what you wrote (quoted above), you want to say that the interaction of Bob's particle with the electromagnetic field of Bob's detector results in the outcome [itex]-1[/itex]. And, since Alice's prediction is certain, that means that it is certain that the combination of Bob's detector setting, [itex]\vec{a}[/itex], and Bob's particle's intrinsic spin vector, [itex]\vec{\sigma_2}[/itex] results in -1. How is that different from saying that there is a function, [itex]B(\hat{b}, \vec{\sigma})[/itex] giving [itex]\pm 1[/itex] depending on Bob's setting [itex]\hat{b}[/itex] and the particle's spin [itex]\vec{\sigma}[/itex]?

    It seems to me that you don't have a different idea of a local realistic theory. You have the same idea that Bell had, and which Bell proved is impossible.
     
  13. Mar 13, 2017 #12

    N88

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    Thanks again for the detail. I have highlighted four key points but I start my reply with this (which I find very helpful):
    Continuing:
    To embrace both options (because that's the way I think here), my preferred answer is to say that it is "law-like": specifically, under EPRB, like Malus' Law.

    By law-like I mean this: if a fair robot is tossing a fair Head/Tail coin under experiment Z, I can call: P(H|Z) = P(T|Z) = [itex]\tfrac{1}{2}[/itex]. So my answer boils down to: "It doesn't matter, so I don't care: for I can handle both, as we'll see." (But I suspect -- if push comes to shove -- most (thinking like me), would see the case of EPRB being under determinism.)

    Thus, in responding to the dilemma that you offer (if push comes to shove), I choose this option: "If it's deterministic, then you're proposing a deterministic local hidden-variable theory, and Bell proved that no such theory produces the EPR correlations."

    BUT Bell 1964:(14a) ≠ Bell 1964:(14b) under my theory. So I join Peres (textbook, p. 162) and many others in saying: "Bell's theorem is not a property of quantum theory." [EDIT: in my opinion, EPR correlations are not at stake here. Some EPR ideas might be. But, as I see it, those are not relevant to our discussion.]

    As in that 1-page extract, I invoke my generalisation of Malus' Law to quantum systems under EPRB. (NB: In conventional quantum theory, some (eg, Aspect) discuss collapse and Malus' Law in the same context. I have no collapse -- so I interpret it in QT as good old Bayesian updating.

    Given Malus' Law, am I missing something? At the moment, I see no difference. However, to be clear re my view: Neither Alice's nor Bob's pristine particles need have the measured spin-direction prior to each particle-detector interaction. (Many senior physicists support this; eg, Kochen (2015).)

    As I see it, the difference arises in this way: We start with identical equations -- Bell 1964:(1)-(3) -- but proceed with a very different assumption. For, under Bell's assumption (as identified after his (14b) Bell has (14a) = (14b). I make no such assumption.

    HTH, with my thanks again -- N88
     
    Last edited: Mar 13, 2017
  14. Mar 13, 2017 #13

    stevendaryl

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    Once again, if Alice and Bob both agree to perform measurements along axis [itex]\hat{a}[/itex], and Alice measures spin-up, then she knows for certain that Bob will measure spin-down. So the combination of Bob's setting and the intrinsic spin of Bob's particle is guaranteed to give the result -1. That's a deterministic model. It's provably false, according to Bell's theorem.
     
  15. Mar 13, 2017 #14

    N88

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    1. Is my law-like model provably false? It agrees with QM. (I put that push-shove response in so that I could learn about my private passion to avoid it.) [EDIT: I understand that determinism does NOT mean that Alice's result determines Bob's result; and vice-vera.]

    2. Since Bell and I start with the same equations -- but proceed with different assumptions -- can you help me by being more specific, please?

    3. For example: where does the math change and what is the physical significance of that change, if it is not at our disagreement re (14a) = (14b)? Is it not (then) an experimental fact that: "N88 is correct, that unless you assume that the averaging process C gives approximately the same result as the theoretical averages D, then you can't prove Bell's inequality, and in fact, you have a much weaker inequality: …"?

    That is, is this next not true? C delivers QM results but does not deliver D; QM delivers C results but does not deliver D.
     
  16. Mar 13, 2017 #15

    stevendaryl

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    I don't see that you have a local model. Can you explain how it works in the situation I'm talking about?
     
  17. Mar 13, 2017 #16

    N88

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    I take it that you are happy for me to discuss my model in the context of intrinsic spin only? It's easier; so that's what I'll do.

    (For generality, I prefer to discuss it in the context of total-spin using Bell's λ.)
     
  18. Mar 13, 2017 #17

    N88

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    As I see it:
    Given conservation of intrinsic spin under EPRB: [itex]\boldsymbol{\sigma_1}+\boldsymbol{\sigma_2}=0.[/itex] (1)
    Alice's result is given by [itex]\hat{a}\circ\boldsymbol{\sigma_1}=+1[/itex]. (2)
    Bob's result is given by [itex]\hat{a}\circ\boldsymbol{\sigma_2}=-\hat{a}\circ\boldsymbol{\sigma_1}=-1[/itex]. (3)
    QED?
     
  19. Mar 13, 2017 #18

    stevendaryl

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    If [itex]\vec{\sigma}[/itex] were an actual vector, yes. But if [itex]\vec{\sigma}[/itex] were an actual vector, then there would be a nonzero probability of Alice choosing [itex]\hat{a}[/itex] such that [itex]\hat{a} \cdot \vec{\sigma} = 0[/itex]. So it's not an actual vector. It's an operator. Since it's an operator, it's not literally true that [itex]\hat{a} \cdot \vec{\sigma} = +1[/itex]. The left-hand side is an operator, while the right-hand side is a number.
     
  20. Mar 13, 2017 #19

    stevendaryl

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    The issue is that there is no such thing as a vector [itex]\vec{\sigma}[/itex] such that for any choice of a measurement axis [itex]\hat{a}[/itex],

    [itex]\vec{\sigma} \cdot \hat{a} = \pm 1[/itex]
     
  21. Mar 13, 2017 #20

    N88

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    Not at all, in what I have written. As allowed by Bell under EPRB (via his λ), [itex]\hat{a}[/itex] and [itex]\vec{\sigma}[/itex] are both vectors. The operator is [itex]\circ[/itex]; ie, the spin-product as introduced and defined in my 1-page PDF.

    I do not use the dot-product. I use the spin-product -- AKA the \circ-product and the [itex]\circ[/itex]-product -- as defined above. Its output is definitely [itex]\pm 1[/itex] (and nothing else).
     
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