So, I am working on the following equation:
$$\frac{|A^2|}{|B^2|+|A^2|} = \frac{e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}}{Tr[e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}]} \tag{1}$$
Which I am told is expressible in the form:
$$|A^2| = C|B^2| \tag{2}$$
Where C is a value that is a function of $\omega$ and $k_b T$ (both constants).
With some basic rearranging I have obtained:
$$|A^2| = \frac{e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}}{Tr[e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}}] - e^{\frac{-\hbar\omega\lvert e \rangle \langle e \rvert}{k_B T}} }|B^2| \tag{3}$$
But one issue that I have with my solution is the $Tr[e^{...}]$ term; I cannot further simplify it. I am familiar with computing the trace of a function involving kets and bras, but no so much when taking the trace of an exponential to the power of bras and kets.
My question is: how does one compute such a trace? I have attempted to use the Macluaurin expansion of $e$ and compute the trace of each individual term but didn't have much luck. Thank you for your time.
For clarification $\langle e\vert e\rangle=1$ and $\langle e\vert k\rangle=0$ for $k\ne e$.
What you have written is a little vague but let's go with some reasonable assumptions: $\langle e\vert e\rangle=1$ and $\langle e\vert k\rangle=0$ for $k\ne e$. Then your operator \begin{align} \hat E=\vert e\rangle\langle e\vert \end{align} is diagonal and idempotent: $\hat E^2=\hat E$, with eigenvalue $1$) (or at there there is a basis where it's diagonal). Thus \begin{align} e^{\alpha \hat E}= \sum_n \frac{\alpha^n}{n!} \hat E^n = \sum_n \frac{\alpha^n}{n!} \hat E = e^{\alpha}\vert e\rangle\langle e\vert. \end{align} You can easily deal with the rest.
Note that your denominator in (3) must be handled with care (or is downright incorrect) unless you have a way of defining "division by an operator". It could be defined by the power series of $(1+\hat A)^{-1}$ and you won't have to worry about ordering since $\hat E$ will commute with powers of $\hat E$ in the numerator. You better start with (1) which is free of this issue.