Question

Calculate the probability that a harmonic oscillator en = (n + 1/2) nw is a . state with n an odd number if the oscillator is in contact with a heat bath / at temperature T

Answer 1

Answer

Energy is given

$E_n = (n + 1/2) \hbar w$

According to concept of physics

the probability of finding the system in a state corresponding to any odd value of

$P(n~\text{is odd}) = (1-x)(x + x^3 + \ldots + x^{2k-1} + \ldots) = \frac{(1-x)x}{1-x^2} = \frac{x}{1+x}$

Where $x = e^{-\beta\hbar\omega}$

To give some confidence that this is indeed the right answer, we can check some limits as below

A) At T=0K we expect that oscillator to be in its ground state, and therefore n

n cannot be odd. T=0

T=0 corresponds to β=∞

β=∞, and therefore x=0

x=0, which indeed gives P=0

P=0 in our formula.

B) As T→∞, 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→∞ corresponds to x→1, for which our formula gives P→1/2.