# How do I calculate the probability that the oscillator is in a certain state using partition function?

So let's say I have a single ($N=1$) quantum harmonic oscillator and the energy is determined by $E_n = (n + 1/2) \cdot \hbar \omega$ (where $n$ is the quantum number and $n$ = $0, 1, 2, \ldots$)

What's the probability that the oscillator is in the state labeled $n$ at temperature $T$?

So according to my calculation, $Z$, the partition function, is $Z = 1 / (1 - x)$ where $x = e^{-\beta \hbar \omega} \Rightarrow P = x ^ n (1 - x)$.

Is it correct? Also, how do I go about calculating the probability of finding the oscillator in a state with odd quantum number?

• Hi abcXYZ, and welcome to Physics.SE! If you want to, you can use LaTeX code in your question to make the formulas more readable: \$x^2 \$ becomes $x^2$ – jdm Dec 7 '12 at 11:51

Your calculation looks fine to me (technically, your partition function should have an extra factor of $e^{-\frac 12 \beta\hbar\omega}$, but this is unimportant, as it cancels in all observables).

Edit: As in the comment by abcXYZ, the probability of finding the system in a state corresponding to any odd value of $n$ is $$P(n~\text{is odd}) = (1-x)(x + x^3 + \ldots + x^{2k-1} + \ldots) = \frac{(1-x)x}{1-x^2} = \frac{x}{1+x}$$ where $x = e^{-\beta\hbar\omega}$. To give some confidence that this is indeed the right answer, we can check some limits:

• At $T=0$, we expect that oscillator to be in its ground state, and therefore $n$ cannot be odd. $T=0$ corresponds to $\beta = \infty$, and therefore $x=0$, which indeed gives $P=0$ in our formula.
• As $T \to \infty$, we expect the oscillator to spread out over all energy eigenstates, and therefore the probabilities of $n$ being odd or even to be equal. And indeed, $T \to \infty$ corresponds to $x \to 1$, for which our formula gives $P \to \frac 12$.

It's a good habit to do these sorts of simple checks on any formulae you derive.

• I was thinking, for the probability of finding the oscillator in states with odd values of n, the probability would just probably be P_odd = P1 + P3 + P5 + .... = (1 - x)(x + x^3 + x^5 + ... ) = x / (1 + x) – AbcXYZ Dec 7 '12 at 13:24
• But I'm not entirely sure. – AbcXYZ Dec 7 '12 at 13:25
• Ah, I see. I thought you meant a particular, odd, value of $n$. You are right; I'll add this to my answer. – Rhys Dec 7 '12 at 14:49

Given your partition function, you can use a heuristic approach like in the other answer, but if you want to calculate it, then you want to construct the density operator $$\rho$$ for your system and then find probabilities via

$$\mathcal{P}_n = Tr(\rho P_n)$$

where $$P_n$$ is the projection operator of the $$n^{th}$$ pure state.

This is a very similar question I recently answered. Here's a nice intro to quantum stat phys which details the properties of $$\rho$$. The last page of this article shows how to analytically derive the partition function.