Finding the Density Matrix of a 4x4 System at Thermal Equilibrium

[yukawa]

How to obtain the density matrix of the following system at thermal equilibrium?

Given:

Hamiltonian H :(in 4x4 matrix form)
H_ij = 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!

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

 

[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]

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!

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

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.

 

[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<br />

\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}<br /> \mathbf{H} =<br /> \left(\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; -\frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 \\<br /> \end{array}\right)<br /> <br /> \left(\begin{array}{cccc}<br /> \frac{1+c}{2} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{-3+c}{2} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; \frac{1-3c}{2} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; \frac{1+c}{2} \\<br /> \end{array}\right)<br /> <br /> \left(\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 \\<br /> \end{array}\right) <br /> <br /> \end{displaymath}<br /> <br />

Therefore, take k = 1,
\begin{displaymath}<br /> \mathbf{e^{-\beta H}} =<br /> \left(\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; -\frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 \\<br /> \end{array}\right)<br /> <br /> \left(\begin{array}{cccc}<br /> e^{-\frac{1+c}{2T}} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; e^{-\frac{-3+c}{2T}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; e^{-\frac{1-3c}{2T} }&amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; e^{-\frac{1+c}{2T}} \\<br /> \end{array}\right)<br /> <br /> \left(\begin{array}{cccc}<br /> 1 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; -\frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; \frac{1}{\sqrt{2}} &amp; \frac{1}{\sqrt{2}} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 \\<br /> \end{array}\right) =<br /> <br /> \left(\begin{array}{cccc}<br /> e^{-\frac{1+c}{2T}} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) &amp; \frac{1}{2}(-e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) &amp; 0 \\<br /> 0 &amp; \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) &amp; \frac{1}{2}(e^{-\frac{-3+c}{2T}}+e^{-\frac{1-3c}{2T}}) &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; e^{-\frac{1+c}{2T}} \\<br /> \end{array}\right)<br /> <br /> <br /> <br /> \end{displaymath}<br /> <br />

However, the ans. in the notes is:

\begin{displaymath}<br /> \mathbf{e^{-\beta H}} =<br /> \left(\begin{array}{cccc}<br /> e^{-\frac{1+c}{T}} &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; cosh\frac{1-c}{T} &amp; -sinh\frac{1-c}{T} &amp; 0 \\<br /> 0 &amp; -sinh\frac{1-c}{T} &amp; cosh\frac{1-c}{T} &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; e^{-\frac{1+c}{T}} \\<br /> \end{array}\right)<br /> \end{displaymath}<br /> <br />

 

[tiny-tim]

Hi yukawa!

Best ever LaTeX!

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!

 

[yukawa]

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