Statistical operator of hydrogen atom

[sunrah]

Homework Statement

Individual hydrogen atoms have been prepared in the energy state n = 2. However, nothing is known about the remaining quantum numbers. Fine structure and all corrections can be ignored.

What is the micro-canonical statistical operator.

Homework Equations

$\hat{\rho_{mc}} = \frac{1}{\Omega (E)}\delta(E-\hat{H})$

$\hat{H} = -\frac{\hbar^{2}}{2m_{p}}\Delta_{p} -\frac{\hbar^{2}}{2m_{e}}\Delta_{e} - \frac{e^{2}}{4\pi \epsilon_{0}}\frac{1}{\left|r_{p} - r_{e}\right|}$

$\Omega (E) = Tr( \delta(E-\hat{H}) )$

The Attempt at a Solution

I don't understand this. If the atoms are all in state n = 2 then with have a system in a pure state: therefore the probability $p_{i} = \frac{1}{\Omega (E)} = 1$ and $\rho = |2\rangle \langle 2|$.

is that so?

 

[TSny]

All of the atoms have the same energy determined by the quantum number n = 2. But that doesn't mean that all the atoms are in the same quantum state. Consider the "remaining quantum numbers" and list all the possible quantum states that have this energy.

 

[sunrah]

Consider the "remaining quantum numbers" and list all the possible quantum states that have this energy.

n = 2
l = 0, 1
m_l = -1, 0, 1
m_s = -1/2, 1/2

so if the generalised state vector is $|n,l,m_{l},m_{s}\rangle$ then

$\hat{\rho} = \frac{1}{P} \cdot \sum\limits_{i = 0}^{1} \left(\sum\limits_{j = 0}^{2} \left(\sum\limits_{k = 0}^{1}|2,i,j-1,k - \frac{1}{2}\rangle\langle k - \frac{1}{2}, j-1,i,2| \right)\right)$

where $\frac{1}{P}$ is the probability associated with each state, which I assume to be all equal, e.g. normal distribution (I think 1/12 but not sure )

 

[TSny]

When $l = 0$ what are the possible values of $m_l$?

Otherwise, I think you're on the right track.