I am calculating the partition function of a quantum harmonic oscillator and I am getting a surprising result. I am probably wrong at some point of the derivation, but I can't find out where.
I have;
$$Z = \mathrm{Tr} \left[e^{-\beta H}\right]\\
Z = \mathrm{Tr} \left[e^{-\beta a \left(P^2+X^2\right)}\right]$$
Then I invoke the Zassenhaus formula (variant of BCH formula):
$$Z = \mathrm{Tr} \left[e^{-\beta a P^2}e^{-\beta a X^2}e^{\beta \frac{a^2}{2} \left[X^2,P^2\right]}\right]$$
If I am not mistaken, $$\left[X^2,P^2\right] = -2\hbar^2$$ so,
$$Z = \mathrm{Tr} \left[e^{-\beta a P^2}e^{-\beta a X^2}e^{-\beta a^2 \hbar^2}\right]$$
The factor $e^{-\beta a^2 \hbar^2}$ does not seem to play an important role, so I factor it out and ignore it. I do the trace in the $x$ basis:
$$Z = \int \left\langle x \right| e^{-\beta a P^2}e^{-\beta a X^2} \left| x \right\rangle\mathrm{d}x\\
Z = \int e^{-\beta a x^2} \left\langle x \right| e^{-\beta a P^2} \left| x \right\rangle\mathrm{d}x$$
Inserting the closure relation for $p$, I get:
$$Z = \iint e^{-\beta a x^2} \left\langle x | p \right\rangle e^{-\beta a p^2} \left\langle p | x \right\rangle\mathrm{d}x {d}p$$
Since $\left\langle x | p \right\rangle \sim e^{ipx}$, I get the classical result:
$$Z = \iint e^{-\beta a x^2} e^{-\beta a p^2} \mathrm{d}x {d}p$$
Where is the mistake?
Best Answer
Your commutator is wrong. The correct formula is $$[X^2,P^2]=2i\hbar( XP+PX)$$ As such you need to include more terms in the Zassenhaus formula, as higher order commutators don't vanish.
You get the classical result because you're precisely ignoring terms $\mathcal{O}(\hbar)$.