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I'm currently enrolled in a statistical mechanics course and am a bit stuck on how to calculate the probabilities of a hydrogen atom in a given state. I'll post the exact question I'm working on and my work up until that point, and note exactly where I am confused. Any help from there would be appreciated.

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.

I'm stuck on part (c), which is:

Calculate the probability of measuring one hydrogen atom in the following states. $\big | 2p_x \big >$, $\big | 2p_y \big >$, $\big | 2p_z \big >$, $\big | 2p_{+1} \big >$, $\big | 2p_0 \big >$, $\big | 2p_{-1} \big >$.

I have found my density operator to be: $$ \rho = \frac{1}{4}\big | 2p_x \big >\big < 2p_x \big | + \frac{1}{4}\big | 2p_y \big >\big < 2p_y \big | +\frac{1}{4}\big | 2p_z \big >\big < 2p_z \big | +\frac{1}{4}\big | 2p_{+1} \big >\big < 2p_{+1} \big | $$

I've read that the probability of measuring a particle in a specific state is given by $\text{Tr}\big [ P \rho\big ] $, where $P$ is the projection operator. Following this, I attempted to calculate the probability of finding the hydrogen atom in the $\big | 2p_x \big >$ state. From here on, $p$ will denote probability. \begin{align} p_{ | 2p_x \rangle} &= \text{Tr} \big (|2p_x \rangle \langle 2p_x | \rho \big )\\ &=\text{Tr} \big ( | 2p_x \rangle \langle 2p_x | \big [ \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 | \big ] \big) \\ &= \text{Tr} \big (\big [\frac{1}{4}-\frac{1}{4\sqrt 2}\big ] |2p_x \rangle \langle 2p_x | \big ) \end{align}

I do not know how to evaluate the trace from here. I understand that the trace is simply the sum of the diagonal elements of a matrix, but I do not see how that applies here. I want to say that the solution is $\frac{1}{4}-\frac{1}{4\sqrt 2}$ since $\text{Tr} \big (|2p_x \rangle \langle 2p_x | \big ) = 1$ by completeness, but I am not confident in that answer.

Any guidance would be appreciated, thank you.

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You're right that $\operatorname{Tr}\left(|2p_x\rangle\langle2p_x|\right)=1$, but this isn't by completeness but by the cyclic invariance of the trace:

$$\operatorname{Tr}\left(|2p_x\rangle\langle2p_x|\right)=\operatorname{Tr}\left(\langle2p_x|2p_x\rangle\right)=1\;.$$

In this finite-dimensional case, you can think of the state vectors as vectors in $\mathbb R^3$, with $|2p_x\rangle$, $|2p_y\rangle$ and $|2p_z\rangle$ represented by the canonical basis. Then $|2p_x\rangle\langle2p_x|$ corresponds to

$$e_xe_x^\top=\pmatrix{1\\0\\0}\pmatrix{1&0&0}=\pmatrix{1&0&0\\0&0&0\\0&0&0}$$

and $\langle2p_x|2p_x\rangle$ to

$$e_x^\top e_x=\pmatrix{1&0&0}\pmatrix{1\\0\\0}=\pmatrix{1}\;;$$

the trace of both products is $1$. Completeness, on the other hand, is

$$ |2p_x\rangle\langle2p_x|+|2p_y\rangle\langle2p_y|+|2p_z\rangle\langle2p_z|=1\;, $$

where the right-hand side is the identity operator $1$, not the scalar $1$, corresponding to

$$e_xe_x^\top+e_ye_y^\top+e_ze_z^\top=\pmatrix{1&0&0\\0&0&0\\0&0&0}+\pmatrix{0&0&0\\0&1&0\\0&0&0}+\pmatrix{0&0&0\\0&0&0\\0&0&1}=\pmatrix{1&0&0\\0&1&0\\0&0&1}\;.$$

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