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Electron Spin State and Values

  1. Dec 4, 2007 #1
    [SOLVED] Electron Spin State and Values

    1. The problem statement, all variables and given/known data
    An electron is in the spin state:

    [tex]X = A\begin{pmatrix} 1-2i \\ 2 \end{pmatrix}[/tex]

    (a) Determine the constant A by normalizing [tex]X[/tex]

    (b) If you measured [tex]S_{z}[/tex] on this electron, what values could you get, and what is the probability of each? What is the expectation value of [tex]S_{z}[/tex]?


    2. Relevant equations
    [tex]S_{z} = \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}[/tex]

    [tex]\left\langle S_{z}\right\rangle = \left\langle X | S_{z}X\right\rangle[/tex]


    3. The attempt at a solution
    Part a):
    [tex]A^{2}\left[ |1-2i|^{2} + |2|^{2}\right] = 1 \Rightarrow[/tex]
    [tex]A^{2}\left[ 1-4i+4+4\right] = 1 \Rightarrow[/tex]
    [tex]A^{2}\left[ 9-4i\right] = 1 \Rightarrow[/tex]
    [tex]A^{2} = \frac{1}{9-4i} \Rightarrow[/tex]
    [tex]A = \sqrt{\frac{1}{9-4i}} \Rightarrow[/tex]

    Part b):
    For this part I'm a bit confused (thus the posting). I'm not sure what they mean when they talk about measuring [tex]S_{a}[/tex] on the electron. Are they just saying, if you measured the z-component of the spin of the electron? And if so would I have something like:

    [tex]\frac{\hbar}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix}1-2i \\ 2 \end{pmatrix} \Rightarrow[/tex]
    [tex]\frac{\hbar}{2}\begin{pmatrix} 1-2i \\ -2 \end{pmatrix}[/tex]

    And where do I go from there? And as far as the probabilities I may be able to get that if I know the normalization constant. And I think I can get the expectation value from the normalization constant as well.
     
  2. jcsd
  3. Dec 5, 2007 #2
    Any thoughts on normalizing this? I tried another way and got [tex]\frac{1}{3}[/tex] and another way and I got [tex]\frac{16\hbar - 2\hbar i}{18}[/tex]. And neither of those seem to work. I thought that 1/3 would work; but when I try to find the expectation value I get [tex]\frac{\hbar}{2} \frac{13}{3}[/tex].
     
  4. Dec 5, 2007 #3
    Ok, I have from my teacher that [tex]A = \frac{1}{3}[/tex], and that [tex]\left\langle S_{x}\right\rangle = \frac{2\hbar}{9}[/tex].

    I can normalize it and get [tex]A = \frac{1}{3}[/tex]; but I can't get the correct expectation value. Here is what I am doing:

    [tex]\left\langle X | S_{x}X\right\rangle = A^{2}\frac{h}{2}[(1+2i), 2]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (1-2i) \\ 2 \end{pmatrix} \Rightarrow[/tex]

    [tex]A^{2}\frac{h}{2}[(1+2i), 2][2 + (1-2i)] \Rightarrow[/tex]

    [tex]A^{2}\frac{h}{2}(2 + 4i + 4 + 1 + 4 + 2 -4i) = \frac{13\hbar }{18}[/tex]

    So what am I doing wrong? I try including the the constants during the calculations and it comes out the same:

    [tex]\left\langle X | S_{x}X\right\rangle = \begin{pmatrix}\frac{(1+2i)}{3} & \frac{2}{3} \end{pmatrix} \begin{pmatrix} 0 & \frac{h}{2} \\ \frac{h}{2} & 0 \end{pmatrix} \begin{pmatrix} (1-2i)/3 \\ 2/3 \end{pmatrix} \Rightarrow[/tex]

    [tex]\frac{h}{2}\begin{pmatrix} \frac{(1+2i)}{3} & \frac{2}{3} \end{pmatrix} \begin{pmatrix} \frac{2}{3} & \frac{(1-2i)}{3} \end{pmatrix} \Rightarrow[/tex]

    [tex]\frac{h}{2}\frac{(2 + 4i + 4 + 1 + 4 + 2 -4i)}{18} = \frac{13\hbar }{18}[/tex]

    So, it still doesn't work. Any thoughts?
     
  5. Dec 5, 2007 #4
    A matrix times a vector is a vector, so you want:

    [tex]\left\langle X | S_{x}X\right\rangle = A^{2}\frac{\hbar}{2}[(1+2i), 2]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} (1-2i) \\ 2 \end{pmatrix} \Rightarrow[/tex]

    [tex]A^{2}\frac{\hbar}{2}[(1+2i), 2][2, (1-2i)] \Rightarrow[/tex]

    Now take the dot product of the two vectors:

    [tex]A^{2}\frac{\hbar}{2}(2 + 4i + 2 - 4i) = \frac{2\hbar }{9}[/tex]
     
  6. Dec 5, 2007 #5
    Ah... Ok. Great! Thanks 2Tesla. That helps a lot. That makes sense now. Ok, I can see why I wasn't getting it. Thanks again.
     
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