Density matrix

  • Thread starter yukawa
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  • #1
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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
 

Answers and Replies

  • #2
tiny-tim
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Hi yukawa! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
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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.
 
  • #4
tiny-tim
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LaTeX

(by the way, how can i input a matrix in the post?)

Is this approch correct? I can't get the answer as shown in my notes.

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 [Broken]

And if you know the answer, always tell us!

it makes it much quicker for us to spot where you've gone wrong. :wink:
 
Last edited by a moderator:
  • #5
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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]
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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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:
  • #7
13
0
Oh! yes!
I got it.
Thank you very much.:smile:
 

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