Density matrix

[yukawa]

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

[tiny-tim]

Hi yukawa! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

[yukawa]

I tried to find it by using this formula:

\rho = \frac{e^{-\beta H}}{Z}

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

I was stuck at finding \ e^{-\beta H} .

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

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 = 1/\sqrt{2}
X32 = -1/\sqrt{2}
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 \ e^{-\beta H} by:
\ e^{-\beta H} = X e^{-\beta D} X^{-1} and \ e^{-\beta D} is just taking the exponential of the digonal elements of -\beta D.

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

[tiny-tim]

(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

And if you know the answer, always tell us! …

it makes it much quicker for us to spot where you've gone wrong. :wink:

[yukawa]

the eigenvectors(left) and eigenvalues(right) of the Hamiltonian are:

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

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


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

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

Write H = XDX-1,
\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}



Therefore, take k = 1,
\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}



However, the ans. in the notes is:

\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}

[tiny-tim]

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:

[yukawa]

Oh! yes!
I got it.
Thank you very much.:smile: