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Probability of measuring a specific value in a spin-1/2 system

  1. Aug 12, 2014 #1

    mtd

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    1. The problem statement, all variables and given/known data
    A spin-1/2 system in the state [itex] \left|ψ\right\rangle = \left|0.5, z\right\rangle[/itex] of the [itex]S_{z}[/itex] spin operator has eigenvalue [itex]s = +\hbar/2[/itex]. Find the expectation values of the [itex]S_{z}[/itex] and [itex]S_{x}[/itex] operators.

    2. Relevant equations
    [itex]\left\langle S_{x,z}\right\rangle = \left\langle ψ \right| S_{x,z}\left|ψ\right\rangle[/itex]

    3. The attempt at a solution
    Multiplied out above equations to find [itex]\hbar z/2[/itex] and [itex]\hbar (0.25 - z^{2})/2[/itex] for the x and z directions, respectively. I assume [itex]z[/itex] is just "some variable" - is it safe to normalize the eigenstate and set z equal to root 0.75?



    1. The problem statement, all variables and given/known data
    Find the probability of measuring [itex]\hbar /2[/itex] in a measurement of [itex]S_{x}[/itex] in the same system.

    2. Relevant equations
    The probability of measuring the eigenvalue [itex]a_{n}[/itex] in a measurement of the observable [itex]A[/itex] is [itex]P \left( a_{n} \right) = \left| \left\langle b_{n} |ψ \right\rangle \right| ^{2}[/itex] where [itex]\left|b_{n}\right\rangle[/itex] is the normalised eigenvector of [itex]A[/itex] corresponding the the eigenvalue

    3. The attempt at a solution
    I believe this should just be the eigenvalue squared i.e. [itex]\hbar ^{2}/4[/itex], but I'm not sure if or why this is the case.
     
    Last edited: Aug 12, 2014
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  3. Aug 12, 2014 #2

    vanhees71

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    [itex]S_x[/itex] and [itex]S_z[/itex] are the operators, representing the spin-x and -z components. Usually you write them with help of the Pauli spin matrices as
    [tex]S_j=\frac{\hbar}{2} \sigma_j, \quad j \in \{x,y,z\}.[/tex]

    In the 2nd problem just calculate what you've written down and not guess some eigenvalue. Think also about the question, how can a probability be a dimensionful quantity as your result suggests?
     
  4. Aug 12, 2014 #3

    mtd

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    I used the Pauli matrices to get my solution for the first question. For the second question I found the eigenvector for [itex]S_{x}[/itex] to be [itex](1,1)[/itex] and the probability of measuring [itex]\hbar /2[/itex] to be [itex](z+0.5)^{2}[/itex]. However, if the eigenstate of the system is normalised and [itex]z = \sqrt{3/4}[/itex] then I must be incorrect.

    Thank you

    edit: Using the other eigenvector [itex](1,−1)[/itex] yielded [itex](z−0.5)^{2} = 0.134[/itex] - my first choice of eigenvector corresponded to [itex]-\hbar /2[/itex] rather than [itex]\hbar /2[/itex]. I think I've solved the question, but why is the probability incorrect when I use the negative eigenvalue? I believe it is a valid value of [itex]s_{x}[/itex]
     
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4

    vanhees71

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    I don't understand what [itex]z[/itex] should mean. In the formula you quote, there is no variable [itex]z[/itex].

    You have [itex]|1/2,x \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} [/itex] (I've normalized the vector properly for you). Your system is prepared in the spin state [itex]|\psi \rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}[/itex]. Now you should be able to calculate the probability to measure [itex]\hbar/2[/itex] for [itex]\sigma_x[/itex]!
     
  6. Aug 12, 2014 #5

    mtd

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    Strange, the system state should be displayed like this: ebc6814477.png

    edit: Okay, I think I understand why the system state should be [itex]\begin{pmatrix} 1 \\ 0 \end{pmatrix}[/itex], i.e. [itex]z = 0[/itex] - by using the [itex]S_{z}[/itex] operator on the given eigenstate you can see the z has to be zero.

    Using this information I find expectation values of 0 and [itex]\hbar /2[/itex] for x and z respectively. However with [itex]z = 0[/itex] I find that probability of measuring [itex]\hbar /2[/itex] for [itex]S_{x}[/itex] is 1 which is inconsistent with the expectation value.
     
    Last edited: Aug 12, 2014
  7. Aug 13, 2014 #6

    vanhees71

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    Argh! Sorry, I've overlooked this. I understood the question such that [itex]|\psi \rangle[/itex] should be the eigenstate of [itex]\hat{s}_z[/itex] with eigenvalue [itex]\hbar/2[/itex]. Then, in the usual convention to choose the Pauli matrices with respect to the [itex]\hat{s}_z[/itex] eigenbasis, [itex]z=0[/itex].

    For the probability you have to normalize the eigenvector of [itex]\hat{\sigma}_x[/itex] to 1!
     
  8. Aug 13, 2014 #7

    mtd

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    Aha, got it (P = 0.5). Must remember to normalise eigenkets. Thanks!
     
  9. Aug 13, 2014 #8

    vanhees71

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