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Density matrix and averages

  1. Sep 11, 2011 #1
    I'm really excited to get this as a homework problem. I have wanted to feel good about this formalism is quantum mechanics for a while now but my own stupidity has been getting in the way... With this homework problem hopefully I can move on to a new level.

    1. The problem statement, all variables and given/known data
    The most general observable is a density matrix. Generally it is a non-negative self-adjoint operator with trace 1. It has the general form
    [tex]
    \rho=\sum_{n}p_n |n\rangle\langle n|
    [/tex]

    where [itex]p_n[/itex] is a classical probability distribution ([itex]\sum_{n} p_n=1,\; 0\leq p_n \leq 1[/itex]) and [itex]|n\rangle\langle n|[/itex] are projection operators that are not necessarily orthogonal. [itex]\rho[/itex] represents a classical statistical ensemble of quantum states where the state [itex]|n\rangle[/itex] appears with probability [itex]p_n[/itex]. The ensemble average of an operator [itex]O[/itex] is an ensemble of states described by a density matrix [itex]\rho[/itex] is
    [tex]
    \langle O \rangle_{\rho}=\mathbf{Tr}(O\rho )
    [/tex]
    Physically this is the average of a number of measurements of [itex]O[/itex] in a classical probability distribution of different states. Consider a polarized beam of protons where 30% of the protons have spin up in the x-direction and 70% have spin down in the z direction. Find the density matrix for this ensemble and compute the ensemble average of [itex]s_z[/itex] in this ensemble of protons.


    2. Relevant equations
    [tex]
    \mathbb{I}=\sum_{n}|n\rangle\langle n|
    [/tex]

    3. The attempt at a solution
    I set up the density matrix like this
    [tex]
    \rho=\frac{3}{10}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}|\downarrow_{z}\rangle \langle \downarrow_{z} |
    [/tex]
    and with
    [tex]
    s_z=\frac{\hbar}{2}\begin{pmatrix}
    1 & 0 \\
    0 & -1
    \end{pmatrix}
    [/tex]
    Then
    [tex]
    \langle s_z\rangle_{\rho}=\mathbf{Tr}\left[\frac{3}{10}\frac{\hbar}{2}\begin{pmatrix}
    1 & 0 \\
    0 & -1
    \end{pmatrix}|\uparrow_{x}\rangle \langle \uparrow_{x} |+\frac{7}{10}\frac{\hbar}{2}\begin{pmatrix}
    1 & 0 \\
    0 & -1
    \end{pmatrix}|\downarrow_{z}\rangle \langle \downarrow_{z} |\right]
    [/tex]
    Now I need help with how to compute the above...

    May I have some help?

    Thanks
     
  2. jcsd
  3. Sep 11, 2011 #2

    vela

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    Try finding the matrix representing the density operator with respect to the Sz eigenbasis.
     
  4. Sep 11, 2011 #3
    Well, here what I think I know...
    [tex]
    \mathbb{I}=|\uparrow_{z}\rangle\langle \uparrow_{z}|+|\downarrow_{z}\rangle \langle \downarrow_{z} |
    [/tex]
    so
    [tex]
    \begin{align}
    |\uparrow_{x}\rangle &=|\uparrow_{z}\rangle\langle \uparrow_{z}|\uparrow_{x}\rangle+|\downarrow_{z} \rangle \langle \downarrow_{z}|\uparrow_{x}\rangle \\
    &= \begin{pmatrix} 1 \\ 0 \end{pmatrix}\frac{1}{\sqrt{2}}+\begin{pmatrix}0 \\ 1 \end{pmatrix}\frac{1}{\sqrt{2}} \\
    &= \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1 \end{pmatrix}
    \end{align}
    [/tex]
    so
    [tex]
    |\uparrow_{x}\rangle\langle \uparrow_{x}|=\frac{1}{2}\begin{pmatrix}1 \\ 1\end{pmatrix}\begin{pmatrix}1 \\ 1 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}
    [/tex]

    Does that look right?
     
  5. Sep 11, 2011 #4

    vela

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    Minor correction:
    [tex]\lvert \uparrow_{x} \rangle \langle \uparrow_{x}\rvert = \frac{1}{2} \begin{pmatrix}1 \\ 1\end{pmatrix} \begin{pmatrix}1 & 1 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}[/tex]
     
  6. Sep 11, 2011 #5
    Thank you,

    Then
    [tex]
    \begin{align}
    \rho &=\frac{3}{20}\begin{pmatrix}1 & 1 \\ 1 & 1 \end{pmatrix}+\frac{7}{10}\begin{pmatrix}0 & 0 \\ 0 & 1 \end{pmatrix} \\
    &=\frac{1}{20}\begin{pmatrix} 3 & 3 \\ 3 & 17 \end{pmatrix}
    \end{align}
    [/tex]
    Then
    [tex]
    \begin{align}
    s_z \cdot \rho &=\frac{\hbar}{2}\frac{1}{20}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix}3 & 3 \\ 3 & 17 \end{pmatrix} \\
    &=\frac{\hbar}{40}\begin{pmatrix} 3 & 3 \\ -3 & -17 \end{pmatrix}
    \end{align}
    [/tex]
    So
    [tex]
    \mathbf{Tr}(s_z\cdot\rho)=-\frac{7\hbar}{20}
    [/tex]

    Does this look good?
     
  7. Sep 11, 2011 #6

    vela

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    Yes, looks good.
     
  8. Sep 11, 2011 #7
    Thanks for your help.
     
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