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Sakurai problem 1.12

  1. Oct 21, 2009 #1
    1. The problem statement, all variables and given/known data

    A spin ½ system is known to be in an eigenstate of [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex] with eigenvalue [itex]\frac{\hbar}{2}[/itex] , where [tex]\hat{\textbf{n}}[/tex] is a unit vector lying in the xy-plane that makes an angle γ with the positive z-axis.

    a. Suppose [tex]S_x[/tex] is measured. What is the probability of getting [itex]+\frac{\hbar}{2}[/itex]?


    3. The attempt at a solution

    [tex]|\hat{\textbf{n}},+\rangle = \cos\left(\frac{\beta}{2}\right)|+\rangle + \sin\left(\frac{\beta}{2}\right)|-\rangle[/tex]

    [tex]\langle S_x,+|\hat{\textbf{n}},+\rangle = \left(\frac{\langle +|+\langle -|}{\sqrt{2}}\right)\left(\cos\left(\frac{\gamma}{2}\right)|+\rangle + \sin\left(\frac{\gamma}{2}\right)|-\rangle\right)=\frac{1}{\sqrt{2}}\left(\cos\left(\frac{\gamma}{2}\right)+\sin\left(\frac{\gamma}{2}\right)\right)[/tex]

    The probability of getting [itex]\frac{\hbar}{2}[/itex] is

    [tex]P=|{\langle S_x,+|\hat{\textbf{n}},+\rangle|^2=\frac{1+\sin\gamma}{2}[/tex]

    I don't really understand this solution.

    Why are they using [tex]\langle S_x,+|\hat{\textbf{n}},+\rangle[/tex] instead of [tex]\langle S_x|\hat{\textbf{n}}\rangle[/tex]?

    And how does [itex]\frac{\hbar}{2}[/itex] come into play here? What if we were looking for the probability of getting something other than [itex]\frac{\hbar}{2}[/itex], like [itex]\frac{\hbar}{4}[/itex] for example? How would that change it?
     
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  3. Oct 21, 2009 #2

    gabbagabbahey

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    What are the eigenvalues and eigenstates of [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] ?
     
  4. Oct 21, 2009 #3
    According to the problem statement:

    A spin ½ system is known to be in an eigenstate of [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex] with eigenvalue [itex]\frac{\hbar}{2}[/itex] .

    So I guess [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex] is the eigenstate, and [itex]\frac{\hbar}{2}[/itex] is its eigenvalue.
     
  5. Oct 21, 2009 #4

    gabbagabbahey

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    No, [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] is an operator, not an eigenstate. What are its eigenvalues and eigenstates?
     
  6. Oct 21, 2009 #5
    I don't know.
     
  7. Oct 21, 2009 #6

    gabbagabbahey

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    You do understand that [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] is the component of the total spin operator along [itex]\mathbf{\hat{n}}[/itex] right?

    What are the eigenvalues and eigenvector of any component of the spin operator?
     
  8. Oct 21, 2009 #7
    Yes. But isn't it a state?

    [tex]S_x=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)[/tex]
    [tex]S_y=\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)[/tex]
    [tex]S_z=\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)[/tex]
     
  9. Oct 21, 2009 #8
    Is this where you're going?

    [tex]\textbf{S}\cdot\hat{\textbf{n}}=\left(S_x\hat{\textbf{x}}+S_y\hat{\textbf{y}}+S_z\hat{\textbf{z}}\right)\cdot\left(\sin{\theta}cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}\right)=S_x\cos{\phi}+S_y\sin{\phi}[/tex]

    Because [tex]\theta=\frac{\pi}{2}[/tex]

    So that means

    [tex]\textbf{S}\cdot\hat{\textbf{n}}=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}+\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)\sin{\phi}[/tex]
     
  10. Oct 21, 2009 #9

    gabbagabbahey

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    Is [itex]S_x[/itex] a state?



    In the [itex]S_z[/itex] eigenbasis, yes.

    More generally, in the [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] eigenbasis, the eigenvalues and corresponding eigenstates of [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] are [itex]\pm\frac{\hbar}{2}[/itex] and [itex]|\mathbf{\hat{n}},\pm\rangle[/itex]... Have you really not seen this before?
     
  11. Oct 21, 2009 #10
    No. But the problem statement says it's in an eigenstate of [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex]. What does that mean?


    Not that I can recall.
     
  12. Oct 21, 2009 #11

    gabbagabbahey

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    It means that ther system is in some state [itex]|\psi\rangle[/itex] such that [itex]\textbf{S}\cdot\hat{\textbf{n}}|\psi\rangle=\frac{\hbar}{2}|\psi\rangle[/itex]....this is the definition of eigenstate.




    Do you understand it though?
     
  13. Oct 21, 2009 #12
    Is this what you're saying:

    [tex]
    \textbf{S}\cdot\hat{\textbf{n}}=\textbf{S}|\hat{\textbf{n}}\pm\rangle=\frac{\hbar}{2}|\hat{\textbf{n}}\pm\rangle
    [/tex]

    ?
     
  14. Oct 21, 2009 #13

    gabbagabbahey

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    No. Again, [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex] is an operator, not an eigenstate. [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex] is an operator just like [itex]S_z\equiv\textbf{S}\cdot\hat{\textbf{z}}[/itex] is an operator.
     
  15. Oct 22, 2009 #14
    What, then, is it operating on?
     
  16. Oct 22, 2009 #15

    gabbagabbahey

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    In this problem, it isn't acting on any state at all. But, like every other operator it has eigenvalues and eigenstates. The system in this problem is known to be in one of those eigenstates; specifically, the eigenstate with corresponding eigenvalue of [itex]+\frac{\hbar}{2}[/itex].

    I should have caught this mistake last night; but in the [itex]S_z[/itex] eigenbasis,

    [tex]S_z=\frac{\hbar}{2}\left(|+\rangle\langle+| - |-\rangle\langle-|\right)\neq\frac{\hbar}{2}\left(|+\rangle - |-\rangle\right)[/tex]

    The first expression is an operator (as it should be), while the second is just a linear combination/superposition of two eigenstates.

    Similarly, in the [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex] eigenbasis, you have

    [tex]\textbf{S}\cdot\mathbf{\hat{n}}=\frac{\hbar}{2}|\mathbf{\hat{n}},+\rangle\langle\mathbf{\hat{n}},+|-\frac{\hbar}{2}|\mathbf{\hat{n}},-\rangle\langle\mathbf{\hat{n}},-|[/tex]

    The extra label [itex]\mathbf{\hat{n}}[/itex] inside the Bras and Kets is just there to make it clear that they are the eigenstates of [itex]\textbf{S}\cdot\mathbf{\hat{n}}[/itex], in its eigenbasis.
     
  17. Oct 23, 2009 #16
    I find out what is [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex]

    I get the following:

    [tex]\textbf{S}\cdot\hat{\textbf{n}}=\left(S_x\hat{\textbf{x}}+S_y\hat{\textbf{y}}+S_z\hat{\textbf{z}}\right)\cdot\left(\sin{\theta}cos{\phi}\hat{\textbf{x}}+\sin{\theta}\sin{\phi}\hat{\textbf{y}}+\cos{\theta}\hat{\textbf{z}}\right)=S_x\cos{\phi}+S_y\sin{\phi}[/tex]

    [tex]S_x=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)[/tex]
    [tex]S_y=\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)[/tex]

    So

    [tex]\textbf{S}\cdot\hat{\textbf{n}}=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}+\frac{i\hbar}{2}\left(|-\rangle\langle +|-|+\rangle\langle -|\right)\sin{\phi}[/tex]
    [tex]=\frac{\hbar}{2}\left(|+\rangle\langle -| + |-\rangle\langle +|\right)\cos{\phi}-i\left(|+\rangle\langle -|-|-\rangle\langle +|\right)\sin{\phi}[/tex]


    Since the system is in an eigenstate of [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex] with eigenvalue of [itex]\frac{\hbar}{2}[/itex] it has to satisfy this equation:

    [tex]\textbf{S}\cdot\hat{\textbf{n}}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle=\frac{\hbar}{2}|\textbf{S}\cdot\hat{\textbf{n}};+\rangle[/tex]

    I understand everything up to here. I want to know why it has to satisfy that last equation. I know the definition of the eigenvalue. I want to know why [tex]|\textbf{S}\cdot\hat{\textbf{n}};+\rangle[/tex] is the eigenvector.

    Or to put it another way, why isn't [tex]|\textbf{S}\cdot\hat{\textbf{n}};-\rangle[/tex] the eigenvector? Or why isn't [tex]|\textbf{S}\cdot\hat{\textbf{n}}\rangle[/tex] the eigenvector?
     
  18. Oct 23, 2009 #17
    Also, if we skip to the end of the problem, put another way, the probability of getting [itex]\frac{\hbar}{2}[/itex] when [tex]S_x[/tex] is measured is given by

    [tex]|\langle S_x;+|\textbf{S}\cdot\hat{\textbf{n}};+\rangle|^2[/tex]

    Why is it that instead of this:

    [tex]|\langle S_x|\textbf{S}\cdot\hat{\textbf{n}}\rangle|^2[/tex]
    ?
     
  19. Oct 24, 2009 #18

    gabbagabbahey

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    My copy of Sakurai has [itex]\mathbf{\hat{n}}[/itex] lying in the [itex]xz[/itex]-plane, making an angle [itex]\gamma[/itex] with the postive [itex]z[/itex]-axis

    [tex]\implies\mathbf{\hat{n}}=\sin\gamma\mathbf{\hat{x}}+\cos\gamma\mathbf{\hat{z}}[/tex]


    [itex]|\textbf{S}\cdot\hat{\textbf{n}};+\rangle[/itex] is defined as the eigenstate of [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex], with corresponding eigenvalue of [itex]\frac{\hbar}{2}[/itex] (In its eigenbasis!), and [itex]|\textbf{S}\cdot\hat{\textbf{n}};-\rangle[/itex] is defined as the eigenstate of [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex], with corresponding eigenvalue of [itex]-\frac{\hbar}{2}[/itex]

    So, if the system is known to be in an eigenstate of [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex] with corresponding eigenvalue [itex]\frac{\hbar}{2}[/itex], then it must be in the state [itex]|\textbf{S}\cdot\hat{\textbf{n}};+\rangle[/itex]. (If it were instead known to be in an eigenstate of [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex] with corresponding eigenvalue [itex]-\frac{\hbar}{2}[/itex], then it would be in the state [itex]|\textbf{S}\cdot\hat{\textbf{n}};-\rangle[/itex])

    Also, writing [itex]|\textbf{S}\cdot\hat{\textbf{n}}\rangle[/itex] makes absolutely no sense. As I said earlier, [itex]\textbf{S}\cdot\hat{\textbf{n}}[/itex] is an operator, not a state; so writing it inside a Ket like this makes no sense.

    Well, the initial state of the system is [itex]|\psi_i\rangle=|\textbf{S}\cdot\hat{\textbf{n}};+\rangle[/itex]. If [itex]S_x[/itex] is measured, and the result is [itex]\frac{\hbar}{2}[/itex], what will the final state[itex]|\psi_f\rangle[/itex] of the system be? What is the probability of this outcome? (If you can't immediately answer these question, you need to re-read section 1.4 of Sakurai!)
     
  20. Oct 24, 2009 #19
    While this is generally true, Problem 12 follows from Problem 9, so the angles above should be [tex]\gamma/2[/tex]
     
  21. Oct 24, 2009 #20

    gabbagabbahey

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    Ermm... no, they should still just be [itex]\gamma[/itex]...you need only draw a picture to convince yourself of this.

    The [itex]\gamma/2[/itex] present in the equation for the eigenstates [itex]|\textbf{S}\cdot\hat{\textbf{n}};\pm\rangle[/itex] results from a straightforward solving of the eigenvalue equation using [itex]\hat{\textbf{n}}=\sin\gamma\hat{\textbf{x}}+\cos\gamma\hat{\textbf{z}}[/itex].
     
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