Density matrix in the canonical ensemble

[mdk31]

Homework Statement

We have a quantum rotor in two dimensions with a Hamiltonian given by $$\hat{H}=-\dfrac{\hbar^2}{2I}\dfrac{d^2}{d\theta^2}$$. Write an expression for the density matrix $$\rho_ {\theta' \theta}=\langle \theta' | \hat{\rho} | \theta \rangle$$

Homework Equations

$$\hat{H}=-\dfrac{\hbar^2}{2I}\dfrac{d^2}{d\theta^2}$$
$$\rho_ {\theta' \theta}=\langle \theta' | \hat{\rho} | \theta \rangle$$

The Attempt at a Solution

In the canonical ensemble, I know that $$\hat{\rho}$$ is given by:
$$\hat{\rho} =\dfrac{1}{Z} e^{-\beta \hat{H}}$$ where Z is the partition function. But this is about as far as I can get. Any assistance towards a solution would be greatly appreciated.

 

[mark726]

Hi there,

Great question! The density matrix for a quantum rotor in two dimensions can be written as:

\rho_ {\theta' \theta}=\langle \theta' | \hat{\rho} | \theta \rangle = \dfrac{1}{Z} \langle \theta' | e^{-\beta \hat{H}} | \theta \rangle

Now, we can use the fact that \hat{H} is diagonal in the basis of eigenstates of the operator \hat{\theta} (position operator), which we can denote as | \theta_n \rangle with corresponding eigenvalues \theta_n. This means that we can write \hat{\rho} as:

\hat{\rho} = \sum_n p_n | \theta_n \rangle \langle \theta_n |

where p_n is the probability of the rotor being in the state | \theta_n \rangle. Substituting this into the expression for \rho_ {\theta' \theta}, we get:

\rho_ {\theta' \theta}=\langle \theta' | \hat{\rho} | \theta \rangle = \dfrac{1}{Z} \sum_n p_n \langle \theta' | e^{-\beta \hat{H}} | \theta_n \rangle \langle \theta_n | \theta \rangle

Since the Hamiltonian is diagonal in the basis of eigenstates of \hat{\theta}, we can simplify this expression to:

\rho_ {\theta' \theta}=\dfrac{1}{Z} \sum_n p_n e^{-\beta \hat{H}(\theta_n)} \langle \theta' | \theta_n \rangle \langle \theta_n | \theta \rangle

Now, using the fact that \hat{H} is given by \hat{H}=-\dfrac{\hbar^2}{2I}\dfrac{d^2}{d\theta^2}, we can write the eigenvalue equation for the operator \hat{H} as:

\hat{H} | \theta_n \rangle = E_n | \theta_n \rangle

where E_n = \dfrac{\hbar^2 \theta_n^2}{2I} is the corresponding eigenvalue. Substituting this into the expression for \rho_ {\theta' \theta}, we get:

\rho_ {\theta' \theta}=\dfrac