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Schrodinger solution spin half particles

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    The evolution of a particular spin-half particle is given by the Hamiltonian [itex]\hat{H} = \omega\hat{S}_{z},[/itex] where [itex]\hat{S}_{z}[/itex] is the spin projection operator.
    a) Show that [itex]\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{-i\frac{\omega}{2}t}\\
    e^{i\frac{\omega}{2}t}
    \end{pmatrix} [/itex] is a solution to the Schrodinger equation.
    b) Calculate [itex]<\hat{S}_{x}>[/itex] as a function of time with respect to this state.

    We are told
    [itex]\hat{S}_{z} = \frac{\bar{h}}{2}\begin{pmatrix}
    1&0\\
    0&-1
    \end{pmatrix}, [/itex][itex]\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}
    0&1\\
    1&0
    \end{pmatrix}[/itex]

    2. Relevant equations
    det([itex]\hat{S}_{z} - λI)=0[/itex]


    3. The attempt at a solution
    This was a previous exam example - we went over it in class but I got a little bit lost. I know it has to do with eigenvalues and eigenvectors, and I can get up to a certain point but then I get stuck, and I've no clue about where to start for b, we didn't get time to do that in class.

    I get that det([itex]\hat{S}_{z} - λI)=0[/itex], and I know that ([itex]\hat{S}_{z} - λI) = \begin{pmatrix}
    \frac{\bar{h}}{2}-λ&0\\
    0&-\frac{\bar{h}}{2}-λ
    \end{pmatrix}.[/itex] Substituting this into det([itex]\hat{S}_{z} - λI)=0[/itex] gives [itex]\frac{-\bar{h}^{2}}{4}+λ^{2}=0.[/itex] Solving for λ gives [itex]λ=\frac{\bar{h}}{2}[/itex] or [itex]λ=-\frac{\bar{h}}{2}[/itex]. Then I substitute this back in, so that if [itex]λ=\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
    0&0\\
    0&-\bar{h}
    \end{pmatrix} Z=0[/itex] where Z is some vector. Also if [itex]λ=-\frac{\bar{h}}{2}[/itex] then [itex] \begin{pmatrix}
    \bar{h}&0\\
    0&0
    \end{pmatrix} Z=0[/itex] where Z is some vector. This is where I get stuck - I can't seem to solve for vector Z, which takes the form of \begin{pmatrix}a\\b\end{pmatrix}, without just ending up with zeros. Not sure how to find a meaningful value of Z. And even if I had one, not sure how to bring that around to prove [itex]\upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{-i\frac{\omega}{2}t}\\
    e^{i\frac{\omega}{2}t}
    \end{pmatrix} [/itex] is a solution to the Schrodinger equation.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 4, 2013 #2

    fzero

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    Let's take one more step and write, for ##\lambda = \hbar/2##,

    $$ \begin{pmatrix}
    0&0\\
    0&-\bar{h}
    \end{pmatrix} Z = \begin{pmatrix}
    0&0\\
    0&-\bar{h}
    \end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} =\begin{pmatrix}
    0&0\\
    0&-\bar{h} b
    \end{pmatrix} =0.$$

    This equation implies that ##b=0##, but ##a## is not constrained. So an eigenvector of ##\hat{S}_z## with eigenvalue ##\lambda = + \hbar## is a multiple of

    $$ \chi_+ = \begin{pmatrix}1\\0\end{pmatrix}.$$

    Similarly, an eigenvector with eigenvalue ##\lambda = + \hbar## is a multiple of (it'll help if you fill in the steps for yourself)

    $$ \chi_- = \begin{pmatrix}0\\1\end{pmatrix}.$$


    These unit vectors are probably familiar to you from the discussion in class. Eigenvectors of ##\hat{H}## are also multiples of these, but the corresponding eigenvalue also involves a factor of ##\omega##.

    The time-dependent Schrodinger equation (TDSE) is

    $$ \hat{H} \psi = i\hbar \frac{\partial \psi}{\partial t}.$$

    The idea is to write ##\psi ## as a linear combination of the eigenvectors of ##\hat{H}## that we found above:

    $$ \psi(t) = a(t) \chi_+ + b(t) \chi_- = \begin{pmatrix}
    a(t)\\
    b(t)
    \end{pmatrix}.$$

    When ##\hat{H}## acts on ##\chi_\pm##, we can substitute in the appropriate eigenvalue. The TDSE will then become a pair of independent equations for the functions ##a(t),b(t)##. You should see if you can make progress on the problem from this and post back if you're still having trouble,
     
  4. Nov 4, 2013 #3
    Okay so, from there I got to [itex]\hat{H} = ω\hat{S}_{z}[/itex] so then [itex]ω\hat{S}_{z} \begin{pmatrix}
    a(t)\\
    b(t)
    \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] so then
    [itex] \begin{pmatrix}
    \frac{ω \bar{h}}{2}&0\\
    0&-\frac{ω \bar{h}}{2}
    \end{pmatrix} \begin{pmatrix}
    a(t)\\
    b(t)
    \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex]
    From here I can multiply it out to [itex] \begin{pmatrix}
    \frac{iω}{2}a(t)\\
    -\frac{iω}{2}b(t)
    \end{pmatrix} = i\bar{h}\frac{d\psi}{dt}[/itex] and so I can solve for [itex]\psi = \begin{pmatrix}
    e^{\frac{iωt}{2}}\\
    e^{-\frac{iωt}{2}}
    \end{pmatrix} [/itex]

    This looks much closer to the final result, but I can't seem to work out a) why my signs appear to be around the wrong way and b) where the [itex]\frac{1}{\sqrt{2}}[/itex] comes into play. :S
     
  5. Nov 4, 2013 #4

    fzero

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    You've introduced an extra factor of ##i## in the next to last equation. It should be

    $$\begin{pmatrix}
    \frac{ω}{2}a(t)\\
    -\frac{ω}{2}b(t)
    \end{pmatrix} = i\hbar\frac{d\psi}{dt}$$

    The [itex]\frac{1}{\sqrt{2}}[/itex] was introduced to normalize the wavefunction.
     
  6. Nov 4, 2013 #5
    Ah ok, the normalisation makes sense. Forgot about that. Also makes sense about the extra i - I stuffed up typing in the equation, it should have read:
    [itex] \begin{pmatrix}
    \frac{iω}{2}a(t)\\
    -\frac{iω}{2}b(t)\\
    \end{pmatrix} = \frac{\delta\psi}{\delta t} [/itex]
    Still struggling to work out how I get to:
    [itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{-\frac{iωt}{2}}\\
    e^{\frac{iωt}{2}}\\
    \end{pmatrix} [/itex]

    I seem to keep arriving at:
    [itex] \upsilon = \frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{\frac{iωt}{2}}\\
    e^{-\frac{iωt}{2}}\\
    \end{pmatrix} [/itex] instead? Does the change in signs occur at some point in the normalisation?
     
  7. Nov 4, 2013 #6

    fzero

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    When you moved the ##i## from one side of the equation to the other, you missed the minus sign coming from ##1/i = - i##. You should have

    [itex] \begin{pmatrix}
    -\frac{iω}{2}a(t)\\
    \frac{iω}{2}b(t)\\
    \end{pmatrix} = \frac{\delta\psi}{\delta t} [/itex]
     
  8. Nov 6, 2013 #7
    The first part all makes sense but now I'm stuck on part b) - calculate [itex]<\hat{S}_{x}>[/itex].
    [itex]\hat{S}_{x} = \frac{\bar{h}}{2}\begin{pmatrix}
    0&1\\
    1&0
    \end{pmatrix}
    [/itex]
    So the expectation value = [itex]\int^{\infty}_{-\infty} \psi*<\hat{S}_{x}>\psi dx[/itex] so I get to [itex]\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{i\frac{w}{2}t}\\
    e^{-i\frac{w}{2}t}
    \end{pmatrix}\begin{pmatrix}
    0&1\\
    1&0
    \end{pmatrix}\begin{pmatrix}
    e^{-i\frac{w}{2}t}\\
    e^{i\frac{w}{2}t}
    \end{pmatrix} [/itex].
    From here I get a bit stuck because if I multiply [itex]<\hat{S}_{x}>[/itex] by [itex] \psi [/itex] then I end up with [itex]\int^{\infty}_{-\infty} \frac{\bar{h}}{2}\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}\begin{pmatrix}
    e^{i\frac{w}{2}t}\\
    e^{-i\frac{w}{2}t}
    \end{pmatrix}\begin{pmatrix}
    e^{i\frac{w}{2}t}\\
    e^{-i\frac{w}{2}t}
    \end{pmatrix} [/itex] which comes out to
    [itex] \int^{\infty}_{-\infty}\frac{\bar{h}}{2}\psi*^{2}[/itex]. Is this the final answer? I feel like there should be a way to simplify, am I mixing up the calculations somewhere? If I can get to [itex] \int^{\infty}_{-\infty} \psi\psi*[/itex] then I know this cancels out to 1 since the solution is normalised.
     
  9. Nov 6, 2013 #8

    fzero

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    When the wavefunctions are spinors or vectors, instead of ##\psi^*## appearing in the inner product, we must have the adjoint wavefunction ##\psi^\dagger##. In addition to taking the complex conjugate, the adjoint also takes the transpose of the object: ##\psi^\dagger = (\psi^*)^T = (\psi^T)^*##. This is because the inner product ##\langle \psi | \chi \rangle## must be a c-number.
     
  10. Nov 6, 2013 #9
    Ok, so I've followed this through and I end up with [itex]\frac{1}{2}\frac{\bar{h}}{2}[/itex][itex]\int^{\infty}_{-\infty} \begin{pmatrix}
    e^{-i\frac{ω}{2}t}t&e^{i\frac{ω}{2}t}
    \end{pmatrix}\begin{pmatrix}
    e^{i\frac{ω}{2}t}\\
    e^{-i\frac{ω}{2}t}
    \end{pmatrix}[/itex] which then works out to be [itex]\frac{\bar{h}}{2}[/itex] ?
     
  11. Nov 6, 2013 #10

    fzero

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    In post #7, you had

    $$\psi^* = \begin{pmatrix}
    e^{i\frac{w}{2}t}\\
    e^{-i\frac{w}{2}t}
    \end{pmatrix}, $$

    which is correct. When we take the transpose of this, we have

    $$ \psi^\dagger = \begin{pmatrix}
    e^{i\frac{w}{2}t} &
    e^{-i\frac{w}{2}t}
    \end{pmatrix} .$$

    When you put this back in the expectation value, you should write the result in terms of trig functions to verify that the expectation value is real.
     
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