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Spin-1 Particles Results of a Measurement

  1. Mar 11, 2013 #1
    1. The problem statement, all variables and given/known data

    Spin-1 Particles prepared in the state:

    $$ |\psi> = \frac{2}{\sqrt29} |1> + \frac{i 3}{\sqrt29}|0> - \frac{4}{\sqrt29} |-1> $$

    Where I'm guessing the ## |#> ## represents the spin state of -1,0 or 1.

    I'm looking to find the results of a measurements of the ## S_x ## component and the associated probabilities

    2. Relevant equations

    Also told that modulus of the spin vector always has the same value:

    $$ S^2 = S.S =s(s+1)\hbar ^2 = 2\hbar ^2 $$

    Although I don't see how that fits in.

    3. The attempt at a solution

    First normalise via:

    $$ <\psi|\psi> = \frac{4}{29} + \frac{i^2 9}{29} + \frac{16}{29} = \frac{11}{29} $$

    Therefore will need to multiply by ## 29/11 ##

    To get the probabilities

    $$ (\frac{2}{\sqrt29})^2 \times \frac{29}{11} = \frac{4}{11} $$

    $$ (\frac{i3}{\sqrt29})^2 \times \frac{29}{11} = \frac{-9}{11}$$

    $$ (\frac{-4}{\sqrt29})^2 \times \frac{29}{11} = \frac{16}{11} $$

    From there I'm a little stuck on how to actually get the results of a measurement.
     
    Last edited: Mar 11, 2013
  2. jcsd
  3. Mar 11, 2013 #2
    I do not remember exactly how you do this exercises (I did these things a couple of years ago and never used them again), but anyway first of all you have to understand in which direction is projected your state. Your wave function has |+1>, |0> and |-1>, meaning that they are the states of spin 1 projected along some direction (usually ##z## but you have to check).

    If they are really projected along ##z##, there should be a way to express the operator you want as a combination of ##S^2## and ##S_z##, and you DO know the result of these operators on these states.
     
  4. Mar 11, 2013 #3
    I can't see anything about which axis it is projected on, I thought at first that maybe each term represents projection on a different axis.
    I need to find results of the measurement for ##S_z## and ##S_x## though.

    If they are projected along the ##S_z## axis, I'm guessing the expression for ##S^2## is useful, but how do I use this in combination with the operator for ##S_z##?
     
  5. Mar 11, 2013 #4
    That you have to check where you found the text of the exercise... anyway usually these are the eigenstates projected along z axis (it is the one we usually use)

    That's exactly what I do not remember... but try check on your book, there should be as far as I remember some relation which permits to find out ##S_x## from the two operators you know...

    Sorry if I'm not able to be of more help
     
  6. Mar 11, 2013 #5

    vela

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    To be more precise, ##|m\rangle## denotes an eigenstate of ##\hat{S}_z##.

    The state is already normalized. To find the adjoint of ##|\psi\rangle##, you have to take the complex conjugate of the coefficients, so
    $$\langle\psi|\psi\rangle = \frac{4}{29} + \frac{(-3i)(3i)}{29} + \frac{16}{29} = \frac{29}{29} = 1.$$

    You should recognize these answers can not possibly be correct because you can't have a negative probability and because they don't sum to 1.

    To start, you need to familiarize yourself with two basic concepts in QM:

    1. How are measurements related to observables?
    2. If a system is in state ##|\psi\rangle##, how do you find the probability of finding it in state ##|\phi\rangle##?
     
  7. Mar 11, 2013 #6
    I thought that at first, but googling it I saw that there was a bit about Dirac saying they are possible, so I though it may be possible.

    To start, you need to familiarize yourself with two basic concepts in QM:

    Am I right in thinking that if you take a measurement or an observable the related operator acts on the system, which will push the system into a different state, this changing the observable?

    I think it might be this looking at my notes:

    $$ P_1(\phi_1) = \frac{|<\phi_1|\psi>|^2}{<\psi|\psi>} $$
     
  8. Mar 11, 2013 #7

    vela

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    No. First, an observable is the operator. Second, measurements have nothing to do with an operator acting on a state.

    That's right if ##|\phi_1\rangle## is normalized.
     
  9. Mar 11, 2013 #8
    Oh right, that makes a lot more sense thinking about it, considering you have the momentum operator and position operator.
    Have I got this completely wrong then, but I thought when you make a measurement you change the wavefunction, as in would is not change from ##\psi## to ##\phi##?

    Awesome, at least I got one part right, so once I have calculated ##phi## I can substitute this in, which would be

    $$ [(\phi|1> coefficient) \times \frac{2}{\sqrt29}] [ (\phi|0> coefficient)\times \frac{i3}{\sqrt29}] + [(\phi|-1> coefficient) \times \frac{-4}{\sqrt29}] ^2 $$

    All divided by 1 as ## <\psi|\psi> = 1 ##

    Edit: Wait, do I use the ##S_z## operator

    $$ S_z|s,m> =\hbar m|s,m> $$

    Similar to the ##S^2## operator given.

    Where s = principal spin quantum number, and I'm not sure on m, maybe the mass?
     
    Last edited: Mar 11, 2013
  10. Mar 11, 2013 #9

    vela

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    I'm not sure what you're getting at here.

    Yes, it's possible the state will change, which gets back to the topic of measurements in QM.

    1. You have some observable that corresponds to the measurement you want to take. How do you know what the possible outcomes of the measurement are?
    2. Say you make a measurement and obtain a certain result. How are the initial state and the new state related?
     
  11. Mar 11, 2013 #10
    Well I thought that the operator would act on the system, meaning I would apply ##S_z## to ##\psi##, but now I'm thinking that is wrong.
     
  12. Mar 11, 2013 #11

    vela

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    Yes, that's wrong. Don't you have a textbook you can consult? The concepts of observables and taking measurements are a few of the first topics typically covered in quantum mechanics.
     
  13. Mar 11, 2013 #12
    Going through my lecture notes from the start of the year looking for the relevant stuff.

    I think this is where I might be getting my idea of the operator acting from
    "3. In classical mechanics it is always assumed that the value of an observable may
    be measured without acting the state of the system. In quantum mechanics,
    the act of measuring an observable usually causes an irreversible change in the
    state of the system. One consequence of this is that for an arbitrary pair of
    observables, it matters fundamentally in which order they are measured."

    I think this statement may be more useful
    "Every observable is represented by a linear hermitian operator whose eigenvalues are
    the possible values of the observable and whose eigenfunctions are the wavefunctions
    of the states in which the observable has a de finite value. When an observable is
    measured, the state of the system changes discontinuously so that its wavefunction
    becomes the eigenfunction corresponding to the value measured"

    I may be interpreting this extremely wrong, but do I want the eigenvalue instead?
     
  14. Mar 11, 2013 #13

    vela

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    This just means that observing will cause the state to change. It doesn't mean applying the operator to the state.

    Yes. The eigenvalues of the operator are the possible outcomes. So if you know the eigenvalues of ##S_z##, you know the possible outcomes of a measurement of the z-component of the spin.

    The equation ##\hat{S}_z | s, m\rangle = m\hbar | s, m\rangle## tells you that ##|s, m\rangle## is an eigenstate of ##\hat{S}_z## with eigenvalue ##m\hbar##. Usually the value of s is understood, so it's omitted, and you write simply ##|m\rangle##. That's the notation being used in this problem.

    The probability of measuring the spin to be ##m\hbar## is the probability of finding the system in state ##|m\rangle##. You should be able to calculate this now. For ##\hat{S}_z##, the calculations are straightforward because ##|\psi\rangle## was given in terms of the eigenstates of ##\hat{S}_z##. For ##\hat{S}_x##, you have a bit more work to do.
     
  15. Mar 11, 2013 #14
    Is ## m ## just an constant then?

    Does that mean the actual measured value is just ##m\hbar## then, where the m changed depending on which spin state it is?
     
  16. Mar 11, 2013 #15

    vela

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    Yup.
     
  17. Mar 12, 2013 #16
    Is the relationship needed for to get the ##S_x##, the formula to get the ##S_x## eigenstates in the terms of the ##S_z## basis?
    I thought it would have to do with the relationship given, but I'm struggling to find anything.

    Also am I wrong in thinking ##m = +1,-1, 0##?
     
    Last edited: Mar 12, 2013
  18. Mar 12, 2013 #17

    vela

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    You can express the eigenstates of ##\hat{S}_x## in terms of the ##\hat{S}_z## basis if you want. That's a straightforward way of doing it because you don't have to deal with a change of basis.

    Why don't you start by finding the matrix that represents ##\hat{S}_x## in this basis?
     
  19. Mar 12, 2013 #18
    $$\frac{\hbar}{\sqrt2} \left(\begin{array}{cc} 0&1&0\\1&0&1\\0&1&0\end{array}\right) $$

    Mod note: fixed your LaTeX for you.
     
  20. Mar 12, 2013 #19

    vela

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    Good. Now find the eigenvalues and eigenvectors of this matrix.
     
  21. Mar 12, 2013 #20
    Haha, thanks for that was having a little trouble with that code.

    Hmm, I think I might have been really stupid here.
    Previously I calculated the eigenvalues and eigenvectors for the three spin operators ##S_x , S_y , S_z##, which I presumed was unrelated, but thinking about it they probably are related.

    So if I have the eigenvalues of ## S_z ## and ## S_x ## I already have the possible outcomes?
     
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