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.

  • 1
    $\begingroup$ Assuming everything else you did is correct, then yes, you can pull the constant factor out of the Tr so that the solution is that factor. See the properties of the Trace, en.wikipedia.org/wiki/Trace_(linear_algebra)#Properties $\endgroup$ Sep 16 '18 at 22:58
  • $\begingroup$ @N.Steinle, But am I correct in evaluating $\text{Tr}(|2p_x \rangle \langle 2p_x |)$? $\endgroup$
    – NoVa
    Sep 16 '18 at 23:05
  • 2
    $\begingroup$ The probability is $<2p_x| \rho |2p_x>$ $\endgroup$ Sep 17 '18 at 0:16
  • $\begingroup$ @RogerJBarlow So, I don't need to use the trace at all here? $\endgroup$
    – NoVa
    Sep 17 '18 at 0:18
  • $\begingroup$ The trace should yield the same, but you have miscalculated the 3rd line from the 2nd line. Do it right. $\endgroup$ Sep 19 '18 at 15:13

Long story short, the trace of X is : $$ \sum_k <{p_k}|X |p_k>$$ As you said the trace is the sum over the diagonal element and the element of a matrix in QM are $A_{ij} =<{p_i}|A|p_j>$.


Ismasou's answer was quite standard upvoted, but just to avoid confusion, you can drop all those fancy notation and just treat it in a simple angular momentum states with $L=1$ and $l=1,0,-1$. It's commonly known matrix and most time it was taught around when professor introducing Stern–Gerlach experiment. (https://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html) (Technically, one was spin, the other was orbital, but they really don't have much difference in terms of generalized angular momentum.) Then google Trace of density operator, the answer ought to be obvious.


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