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:

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

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

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

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

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

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

However, the ans. in the notes is:

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

 

[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.