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Density matrix

  1. Aug 29, 2008 #1
    How to obtain the density matrix of the following system at thermal equilibrium?

    Given:

    Hamiltonian H :(in 4x4 matrix form)
    Hij = the i-th row and j-th column element of H
    H11 = (1+c)/2
    H22 = -(1+c)/2
    H23 = 1-c
    H32 = 1-c
    H33 = -(1+c)/2
    H44 = (1+c)/2
    where c is a parameter and all other elements are zero
     
  2. jcsd
  3. Aug 30, 2008 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi yukawa! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Aug 31, 2008 #3
    I tried to find it by using this formula:

    [tex]\rho = \frac{e^{-\beta H}}{Z} [/tex]

    where [tex]\beta = \frac{1}{kT}[/tex] and [tex]\ Z = tr (e^{-\beta H}) [/tex]

    I was stuck at finding [tex]\ e^{-\beta H} [/tex].

    The following is what i tried in order to find [tex]\ e^{-\beta H} [/tex].

    i) finding the eigenvectors and eigenvalues of H and write H in form of XDX-1 (where X is the matrix formed by the eigenvectors of H and D is a diagoanl matrix with elements equal to the eigenvalues of H) :

    explicitly,
    X11 = X44= 1
    X22 = X23 = X33 = [tex]1/\sqrt{2}[/tex]
    X32 = -[tex]1/\sqrt{2}[/tex]
    all other elements are zero.
    and
    D11 = D44 = (1+c)/2
    D22 = (-3+c)/2
    D33 = (1-3c)/2
    all other elements are zero.
    (by the way, how can i input a matrix in the post?)

    ii) then i calculate [tex]\ e^{-\beta H} [/tex] by:
    [tex]\ e^{-\beta H} = X e^{-\beta D} X^{-1} [/tex] and [tex]\ e^{-\beta D}[/tex] is just taking the exponential of the digonal elements of [tex] -\beta D[/tex].

    Is this approch correct? I can't get the answer as shown in my notes.
     
  5. Aug 31, 2008 #4

    tiny-tim

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    Homework Helper

    LaTeX

    Hi yukawa! :smile:

    I've just woken up … :zzz: not in functioning mode yet … :confused:

    For the LaTeX for matrices tables and long equations, see http://www.physics.udel.edu/~dubois/lshort2e/node56.html#SECTION00850000000000000000

    And if you know the answer, always tell us!

    it makes it much quicker for us to spot where you've gone wrong. :wink:
     
  6. Sep 1, 2008 #5
    the eigenvectors(left) and eigenvalues(right) of the Hamiltonian are:

    [tex]\left|\uparrow\uparrow\right\rangle : (1+c)/2[/tex]

    [tex]\left|\downarrow\downarrow\right\rangle : (1+c)/2
    [/tex]

    [tex]\frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right\rangle - \left|\downarrow\uparrow\right\rangle ) : (-3+c)/2[/tex]

    [tex]\frac{1}{\sqrt{2}}(\left|\uparrow\downarrow\right\rangle + \left|\downarrow\uparrow\right\rangle ) : (1-3c)/2[/tex]

    Write H = XDX-1,
    [tex]\begin{displaymath}
    \mathbf{H} =
    \left(\begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}\right)

    \left(\begin{array}{cccc}
    \frac{1+c}{2} & 0 & 0 & 0 \\
    0 & \frac{-3+c}{2} & 0 & 0 \\
    0 & 0 & \frac{1-3c}{2} & 0 \\
    0 & 0 & 0 & \frac{1+c}{2} \\
    \end{array}\right)

    \left(\begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
    0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}\right)

    \end{displaymath}

    [/tex]

    Therefore, take k = 1,
    [tex]\begin{displaymath}
    \mathbf{e^{-\beta H}} =
    \left(\begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}\right)

    \left(\begin{array}{cccc}
    e^{-\frac{1+c}{2T}} & 0 & 0 & 0 \\
    0 & e^{-\frac{-3+c}{2T}} & 0 & 0 \\
    0 & 0 & e^{-\frac{1-3c}{2T} }& 0 \\
    0 & 0 & 0 & e^{-\frac{1+c}{2T}} \\
    \end{array}\right)

    \left(\begin{array}{cccc}
    1 & 0 & 0 & 0 \\
    0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
    0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
    0 & 0 & 0 & 1 \\
    \end{array}\right) =

    \left(\begin{array}{cccc}
    e^{-\frac{1+c}{2T}} & 0 & 0 & 0 \\
    0 & \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) & \frac{1}{2}(-e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) & 0 \\
    0 & \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) & \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) & 0 \\
    0 & 0 & 0 & e^{-\frac{1+c}{2T}} \\
    \end{array}\right)



    \end{displaymath}

    [/tex]

    However, the ans. in the notes is:

    [tex]\begin{displaymath}
    \mathbf{e^{-\beta H}} =
    \left(\begin{array}{cccc}
    e^{-\frac{1+c}{T}} & 0 & 0 & 0 \\
    0 & cosh\frac{1-c}{T} & -sinh\frac{1-c}{T} & 0 \\
    0 & -sinh\frac{1-c}{T} & cosh\frac{1-c}{T} & 0 \\
    0 & 0 & 0 & e^{-\frac{1+c}{T}} \\
    \end{array}\right)
    \end{displaymath}

    [/tex]
     
  7. Sep 1, 2008 #6

    tiny-tim

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    Homework Helper

    Hi yukawa! :smile:

    Best ever LaTeX!! :biggrin:

    Now I know the answer, it's easy to see where you've gone wrong …

    you haven't!

    you're just out by a factor of e-(1+c)/2T! :wink:
     
    Last edited: Sep 1, 2008
  8. Sep 2, 2008 #7
    Oh! yes!
    I got it.
    Thank you very much.:smile:
     
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