Consider 7 particles in a simple harmonic oscillator potential Ei = n_ih omega (
ID: 1998175 • Letter: C
Question
Consider 7 particles in a simple harmonic oscillator potential Ei = n_ih omega (ni = 0.1, 2, ...) with a total energy of 7h omega. For the two cases Spin 3/2 fermions Spin 0 bosons Enumerate the microstates. Compute the probability that a measurement of a single particle will yield 3hw. A good way to list microstates is by the number of particle's in each state of ascending energy, e.g. (4, 1, 0, 2, 0, 0, 0, 0) corresponds to 4 particles of energy 0, 1 of hw and two of energy 3h omega. From such a list (assumed accurate), or a table with the same information, computing probabilities is straightforward.Explanation / Answer
b) For Bosons, finding the number of states is equivalent to finding partitions of 7.
7+0+0+0+0+0+0 DENOTES one particle in n=7 state and other particles in n=0 state.
So the microstates thus obtained are listed below.
1) 7+0+0+0+0+0+0
2) 6+1+0+0+0+0+0
3) 5+2+0+0+0+0+0
4) 4+3+0+0+0+0+0
5) 5+1+1+0+0+0+0 ***
6) 4+2+1+0+0+0+0 ***
7) 3+2+2+0+0+0+0 ***
8) 3+3+1+0+0+0+0 ***
9) 4+1+1+1+0+0+0 ***
10) 3+2+1+1+0+0+0 ***
11) 2+2+2+1+0+0+0 ***
12) 2+1+1+1+1+1+0
13) 1+1+1+1+1+1+1
14) 3+1+1+1+1+0+0 ***
15) 2+2+1+1+1+0+0 ***
The Probability of each microstate=1/15
Probability of each microstate having a particle of energy 3hw = n/7 where n is the number of particles in 3rd state
Total Probablity=1/15*1/7*(1+1+2+1+1)=6/105
a)For a 3/2 fermion each state cannot be occupied by more than 4 particles. Therefore only the states with *** given above are possible
The Probability of each microstate=1/9
Probability of each microstate having a particle of energy 3hw = n/7 where n is the number of particles in 3rd state
Total Probablity=1/9*1/7*(1+2+1+1)=5/54
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