I'm taking my first course in graduate statistical mechanics and I'm struggling a bit with the math. I think I understand how to use dirac notation, at least the basic stuff for now, but I want to make sure I am evaluating these matrix elements correctly. The following is the problem I am working on.
Let's look at a density matrix for a quantum system which is well described by a basis with three states. An example is the electron of a hydrogen atom that is prepared with spin up and in a linear combination of the three $2p$ states. Two standard bases for the hydrogen atom are the Cartesian states given by $\big | 2p_x \big >$, $\big | 2p_y \big >$, $\big | 2p_z \big >$, and the good $L_z$ angular momentum states, $\big | 2p_{+1} \big >$, $\big | 2p_0 \big >$, $\big | 2p_{-1} \big >$. They are related by a unitary transformation: \begin{align} \big | 2p_{\pm 1} \big > &= \mp \frac{1}{\sqrt 2} [\big | 2p_x \big > \pm \big | 2p_y \big > ] \\ \big | 2p_0 \big > &= \big | 2p_z \big > \end{align} You can prepare a beam of hydrogen atoms so that with probability $1/4$ an atom is prepared in the $\big | 2p_x \big >$ state, with probability $1/4$ it is prepared in the $\big | 2p_y \big >$ state, with probability $1/4$ it is prepared in the $\big | 2p_z \big >$ state, and with probability $1/4$ it is prepared in the $\big | 2p_{+1} \big >$ state.
a) Calculate the matrix elements of the density operator $\rho$ for a hydrogen atom in this beam using the $\big | 2p_x \big >$, $\big | 2p_y \big >$, $\big | 2p_z \big >$ basis. That is, calculate the matrix $$ \begin{pmatrix} \big < 2p_x \big | \rho \big | 2p_x \big > & \big < 2p_x \big | \rho \big | 2p_y \big > & \big < 2p_x \big | \rho \big | 2p_z \big > \\ \big < 2p_y \big | \rho \big | 2p_x \big > & \big < 2p_y \big | \rho \big | 2p_y \big > & \big < 2p_y \big | \rho \big | 2p_z \big > \\ \big < 2p_z \big | \rho \big | 2p_x \big > & \big < 2p_z \big | \rho \big | 2p_y \big > & \big < 2p_z \big | \rho \big | 2p_z \big > \end{pmatrix} $$
This is my solution so far. I essentially just calculated the first matrix element. If I did this correct, then the rest will come easy. I just want to confirm I am doing it right. What is making me question my answer is when I evaluate terms like $\big < 2p_{+1} \big | 2p_x \big >$. I know that $\big < 2p_{+1} \big |$ is the complex conjugate of $\big | 2p_{+1} \big >$, but I'm not sure if that changes the calculation or not.
Here is my work:
We have that \begin{align} \rho = \rho_{op} (t) = \sum_n p_n \big | \psi_n \big > \big < \psi_n \big | \end{align}
Where $\rho$ is our thermal density matrix, $n$ is the state number, and $p_n$ is the probability of finding our particle in that state. Using the basis sets given to us in the problem, we find that \begin{align} \rho = \frac{1}{4} \big | 2p_x \big > \big < 2p_x \big | + \frac{1}{4} \big | 2p_y \big > \big < 2p_y \big | + \frac{1}{4} \big | 2p_z \big > \big < 2p_z \big | + \frac{1}{4} \big | 2p_{+1} \big > \big < 2p_{+1} \big | \end{align}
The process for calculating each matrix element in the density operator is fairly straightforward. I will show the calculation for the $\big < 2p_x \big | \rho \big | 2p_x \big >$ case and simply fill in the other elements.
So,
\begin{align} \big < 2p_x \big | \rho \big | 2p_x \big > = \frac{1}{4} \big < 2p_x \big | 2p_x \big > \big < 2p_x \big | 2p_x \big > + \frac{1}{4} \big < 2p_x \big | 2p_y \big > \big < 2p_y \big | 2p_x \big > + \frac{1}{4} \big < 2p_x \big | 2p_z \big > \big < 2p_z \big | 2p_x \big > + \frac{1}{4} \big < 2p_x \big | 2p_{+1} \big > \big < 2p_{+1} \big | 2p_x \big > \end{align}
With the orthogonal terms going to zero and the similar terms going to one, \begin{align} \big < 2p_x \big | \rho \big | 2p_x \big > = \frac{1}{4} + \frac{1}{4} \big < 2p_x \big | 2p_{+1} \big > \big < 2p_{+1} \big | 2p_x \big > \end{align} Now to expand the terms on the far right, \begin{align} \big < 2p_x \big | 2p_{+1} \big > &= -\frac{1}{\sqrt 2} \big < 2p_x \big | 2p_x \big > - \frac{1}{\sqrt 2} \big < 2p_x \big | 2p_y \big > \\ &= -\frac{1}{\sqrt 2} \end{align} \begin{align} \big < 2p_{+1} \big | 2p_x \big > &= -\frac{1}{\sqrt 2} \big < 2p_x \big | 2p_x \big > - \frac{1}{\sqrt 2} \big < 2p_y \big | 2p_x \big > \\ &= -\frac{1}{\sqrt 2} \end{align}
Therefore, \begin{align} \big < 2p_x \big | \rho \big | 2p_x \big > &= \frac{1}{4} + \frac{1}{4} \big < 2p_x \big | 2p_{+1} \big > \big < 2p_{+1} \big | 2p_x \big > \\ &= \frac{1}{4} + \bigg (\frac{1}{4}\bigg ) \bigg ( \frac{-1}{\sqrt 2}\bigg ) \bigg ( \frac{-1}{\sqrt 2}\bigg ) \\ &= \frac{3}{8} \end{align}
The other terms follow this same method, and I will show their answers in my homework. But, did I evaluate the first matrix element correctly?
EDIT:
Doing it the way I did above, I get the following answer.
$$ \begin{pmatrix} \big < 2p_x \big | \rho \big | 2p_x \big > & \big < 2p_x \big | \rho \big | 2p_y \big > & \big < 2p_x \big | \rho \big | 2p_z \big > \\ \big < 2p_y \big | \rho \big | 2p_x \big > & \big < 2p_y \big | \rho \big | 2p_y \big > & \big < 2p_y \big | \rho \big | 2p_z \big > \\ \big < 2p_z \big | \rho \big | 2p_x \big > & \big < 2p_z \big | \rho \big | 2p_y \big > & \big < 2p_z \big | \rho \big | 2p_z \big > \end{pmatrix} = \begin{pmatrix} \frac{3}{8} & \frac{3}{8} & 0 \\ \frac{3}{8} & \frac{3}{8} & 0 \\ 0 & 0 & 0 \end{pmatrix} $$
EDIT
I think I found the answer to my question. The density matrix is Hermitian, so $|a>^{\dagger}=<a|$ and $<a|^{\dagger} = |a>$. So, my solution should be correct.
Thanks!