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But my spin can't flip!

  1. Jun 16, 2005 #1
    I'm studying for a QM course test, and I was just doing a little thinking about the pauli matrices in a distracted moment. I came up with something, that to my knowledge isn't right, but I can't figure out where my argument goes 'boink'.

    The case is this:
    Imagine I have an electron polarised in the positive z-direction (call it |z_up> ). And can prepare this for instance with a Stern Gerlach apparatus.
    Now I perform a measurement on the spin in an orthogonal direction, say the x-direction. I either find |x_up> or |x_down>. Whatever the result, it can be expressed as a superposition of |z_up> and |z_down>, more specifically as one of the following 2 vectors: (1,1) or (1,-1) in the z-basis.

    Until this point, I think I got it right. Now comes what disturbs me:
    Measuring is mathematically represented by a Hermitian matrix, in this case the matrix S_x. But letting S_x act on (1,0) (wich is |z_up>) gives me (0,1), wich would say that my spin flipped in the z-direction.

    Now I think the problem is that I'm misusing/interpreting S_x...but I can't figure out what would be the correct procedure...
     
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  3. Jun 16, 2005 #2

    dextercioby

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    It makes no sense to speak of electrons (s=1/2) and infer spin eigen vectors like (1,1),(1,0) and (1,-1) which are not for a 1/2 spin particle.

    So reformulate your problem,unless you didn't mean the kets [itex] |1,1\rangle,|1,0\rangle,|1,-1\rangle [/itex],but something else.

    Daniel.
     
  4. Jun 16, 2005 #3
    Spin 1/2 particles 'live' in a 2D Hilbert space, wich is spanned by two basisvectors, wich I can choose to be the eigenspinor of Sz. I can label these vectors as (1,0) and (0,1), corresponding to spin up or down in the z direction...Right?

    (like for instance Griffiths page 156)
     
    Last edited: Jun 16, 2005
  5. Jun 16, 2005 #4

    dextercioby

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    You mean the columns.Sure you can.I was thinking about the kets and that's why i responded.

    Daniel.
     
  6. Jun 16, 2005 #5
    ah yes, I see the confusion. Sorry, the vectors I mentioned need to be transposed...But still, where does my reasoning go wrong?
     
  7. Jun 16, 2005 #6

    dextercioby

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    Okay.The system is initially in a pure state,an eigenstate of [itex]\hat{S}_{z} [/itex],that is represented by the ket [itex] |1/2,1/2\rangle [/itex].You make a measurement of the spin in the "x" direction and you want to determine the state vector after measurement.

    You'll have to apply the V-th postulate (so called von Neumann projection postulate).

    Daniel.
     
  8. Jun 16, 2005 #7

    Galileo

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    Measuring a quantity is not the same as multiplying the vector by the corresponding observable. It is not right to say that after the measurement the state is S_x|z_up> (this would be deterministic). The state afterwards will be the eigenstate of S_x corresponding to the measured eigenvalue.
     
  9. Jun 16, 2005 #8

    Gokul43201

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    If you are trying to find the Sx operator in the Sz basis, it makes more sense to consider the Sx;+ ans Sx;- beams coming out of a SG(x) apparatus and measure Sz on these beams. Since the probability of finding Sz in the up-state equals that for the down-state, you have :
    [tex]|\langle S_z;+|S_x;+ \rangle| = |\langle S_z;-|S_x;+ \rangle| = 1/ \sqrt{2} [/tex]


    [tex]|\langle S_z;+|S_x;- \rangle| = |\langle S_z;-|S_x;- \rangle| = 1/ \sqrt{2} [/tex]

    This lets you construct the Sx;+ and Sx;- kets (making sure they are orthogonal, since they are mutually exclusive) as:

    [tex]|S_x;+ \rangle = \frac{1}{\sqrt{2}} |S_z;+ \rangle + \frac{1}{\sqrt{2}} |S_z;- \rangle [/tex] and

    [tex]|S_x;- \rangle = \frac{1}{\sqrt{2}} |S_z;+ \rangle - \frac{1}{\sqrt{2}} |S_z;- \rangle [/tex]

    This (along with the eigenvalue equations for Sz) allows you to describe the Sx operator in the Sz basis:

    [tex]S_x = \frac {\hbar}{2}[(|S_z;+ \rangle \langle S_z;-|) + (|S_z;- \rangle \langle S_z;+|)] [/tex]

    True.

    No, the eigenvectors of Sz in its own basis are the (1,0) and (0,1) column vectors.

    Sx will only be Hermitian in its own basis, not in the eigenbasis of Sz.

    This is not the right way to do this. Sx is not Hermitian in the Sz basis. It is Hermitian in its own basis. So what you want to do is write the |z_up> vector in terms of |x_up> and |x_down> and then operate Sx on this.
     
    Last edited: Jun 16, 2005
  10. Jun 16, 2005 #9

    Galileo

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    I think wizzart was referring to the eigenvectors of S_x in the Sz-eigenbasis.

    I`m not so sure about that one. I'd check it, but I`m leaving. Will come back to it later...
     
  11. Jun 16, 2005 #10

    Gokul43201

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    That doesn't make much sense then. I've written out the eigenvectors of Sx, namely |Sx;+> and |Sx;-> in terms of the eigenkets of Sz (in the previos post).


    Do check it. Sx is not diagonal in the Sz basis. Neither will be Sy, S+ or S-.
     
  12. Jun 16, 2005 #11

    dextercioby

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    Yes,Gokul is right in everything he said.A very elegant (à la Sakurai,i might say) way of deriving the "x" spin operator.:approve:

    Daniel.
     
  13. Jun 16, 2005 #12
    thanks all...je comprend :)
    Like I said, just something I thought up while in the middle of stressful exam preps...just couldn't find where I was going wrong. Turns out I mixed measurement and expectation value, to put it in a short way.
     
  14. Jun 16, 2005 #13

    Galileo

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    Right, but you said Hermitian :wink:
     
  15. Jun 16, 2005 #14

    Here are some basic facts (that don't need checking):

    On observable is always represented by a Hermition Operator. The Hermitian nature of the operator does NOT care for which basis you expand the operator in (for your own convenience). In terms of matrices Hermitian means - take an element of the matrix- tranpose and complex conjugate it to get the element that corresponds to reflecting it across the diagonal. Clearly, no where does this require the matrix to be diagonal.

    So, Sx and Sz both are Hermitian in both the Sx and Sz bases.

    Indeed, Hermicity is a representation independent attribute - an attribute that is true even if you dont represent the operator by matrices - let alone matrices is a particular basis.

    S+ and S- are constructs for computational convenience. They do NOT correspond to observables are hence NOT hermitian.

    More facts follow.

    Adi.
     
  16. Jun 16, 2005 #15

    Gokul43201

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    Yes, that was a giant blunder in my previous two posts :redface:. I meant to say "diagonal", when I said "hermitian".

    Let me try again : What I should have said was that you can not operate Sx on the eigenkets of Sz and expect to see eigenvalues of Sx.

    I should have said that Sx is not diagonal in the Sz basis.
     
    Last edited: Jun 16, 2005
  17. Jun 16, 2005 #16
    Resolution:

    1) Observables are indeed represented by Hermitian operators.

    2) The operation of Sx on Iz up> (excuse the notation) can be studied in ANY basis of your choice.

    a)Let's do what you did first - use the Sz basis. Then Sx is the Pauli_x matrix multiplied by hbar/2 and Sz is the column matrix with components (1, 0). Operating Sx on Iz up> INDEED produces the spin flipped state (0, 1), i.e.
    Iz down > multiplied by hbar/2. So the answer to your question is: You are right - the spin does flip.

    b)To understand why this answer is no big surprise and to interpret this equation correctly, let's go to the Sx basis. Now in this basis

    Iz up> = 1/(sqrt 2)[ISx up> + ISx down>]

    Operating Sx on this simply pulls out the eigenvalues from both states, so you get:

    1/(sqrt 2)[ hbar/2 ISx up> - hbar/2 ISx down> ]
    = hbar/2 (Iz down>

    as before.

    Now here is the interpretation:
    You operated Sx on a LC of it's eigenstates with the expected result that you got another LC of it's eigenstates, where the coeffecients got suitably modified by the eigenvalues. It just turns out that this LC is an eigenvector of the Sz operator. No big deal.

    So you were right all along.

    Adi.
     
    Last edited: Jun 16, 2005
  18. Jun 16, 2005 #17
    It makes perfectly good sense to represent the basis states of the 2D Hilbert space of spin half particles by ANY two linearly independent 2-D vectors (or dual vectors).

    The representation of the vectors can be perfectly legitimately made in terms of EITHER kets (column vectors) or bras (row vectors). Once again, the whole of Quantum mechanics can be done by representing states by EITHER kets OR bras. It doesn't matter. It is Conventional (not necessary) to represent states by kets (column vectors).
     
  19. Jun 16, 2005 #18
    How is Sz represented by this ket?

    What is the V-th postulate (so called von Neumann projection postulate)?
    Why is it necessary here?
     
  20. Jun 16, 2005 #19
    The operation of measurement is represented by the operation of a projection operator on the state.
     
  21. Jun 16, 2005 #20
    The alzebra is right, but pray how does it answer the question one way or the other? Does the spin flip? Or does it not?
     
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