A system has non-degenerate energy levels with energy

[Eats Dirt]

Homework Statement

A system has non-degenerate energy levels with energy$$\epsilon=(n+1/2)\hbar\omega$$ where h-bar*omega=1.4*10^-23J and n positive integer zero what is the probability that it is in n=1 state with a heat bath of temperature 1K

Homework Equations

$$Z=\exp^\frac{-E_i}{k_b T} \\ p_r=\frac{\exp^\frac{-E_i}{k_b T}}{\sum^N_j \exp^\frac{-E_j}{k_b T}}$$

The Attempt at a Solution

I'm not really sure what to do now, I don't know how to sum the total number of states to get the fraction of states in the n=1 state

 

[mfb]

The sum is a sum over q^i for some q<1, this has an analytic result. You can just plug in all values and calculate the result.

 

[Eats Dirt]

Ok I think I might have gotten it, to deal with the infinite sum use a geometric series,

$$\sum_0^\inf e^\frac{-(n+\frac{1}{2})}{k_b T} \\ =e^\frac{-\hbar\omega}{2k_b T}\sum_0^\inf e^\frac{-n}{k_b T}\\ =\frac{e^\frac{-\hbar\omega}{2k_b T}}{1-e^\frac{-\hbar\omega}{k_b T}$$

then evaluate using the pr as stated before.

Also I don't know why my LaTeX is not displaying correctly.

 

[mfb]

Some error, probably with brackets.
Yes the approach is good.

 

[HalfLostAndSearching]

I would first clarify the question and make sure I understand the context and variables involved. It appears that the system in question has energy levels given by \epsilon=(n+1/2)\hbar\omega, where n is a positive integer and h-bar*omega is a constant. The question also mentions a heat bath with a temperature of 1K.

To find the probability that the system is in the n=1 state, we can use the Boltzmann distribution, which states that the probability of a system being in a particular state with energy E_i is proportional to the Boltzmann factor, \exp^\frac{-E_i}{k_b T}, where k_b is the Boltzmann constant and T is the temperature.

In this case, the probability of the system being in the n=1 state is given by p_1=\frac{\exp^\frac{-\epsilon_1}{k_b T}}{\sum^N_j \exp^\frac{-\epsilon_j}{k_b T}}, where \epsilon_1 is the energy of the n=1 state and the sum in the denominator is taken over all possible states of the system.

To find the total number of states, we can use the formula for the partition function, Z=\sum^N_j \exp^\frac{-\epsilon_j}{k_b T}, which sums over all states of the system. We can simplify this by noting that the energy levels are evenly spaced, with \Delta\epsilon=\hbar\omega. Therefore, the partition function can be written as Z=\sum^N_j \exp^\frac{-(j+1/2)\hbar\omega}{k_b T}, where j ranges from 0 to N-1.

To find the value of N, we can use the fact that the highest energy state of the system is given by \epsilon_N=(N+1/2)\hbar\omega. We can set this equal to the energy of the system, \epsilon=(n+1/2)\hbar\omega, and solve for N. This gives N=n.

Substituting this value of N into the partition function, we get Z=\sum^N_j \exp^\frac{-(j+1/2)\hbar\omega}{k_b T}=\sum^N_j \exp^\frac{-j\hbar\omega}{k_b T}\exp^\frac{-\