Verify that the probabilities shown in Table 9.1 for four distinguishable oscillators sharing energy 2δEagree with the exact probabilities given by equation (9-9).

Short Answer

Expert verified

The probabilities agree with the result from equation 1.

P(n)=(Mn+N2)!(Mn)!(N2)!/(M+N1)!M!(N1)!

Step by step solution

01

Concept of probability of finding a particle at an energy state

Probability of finding a particle at an energy state is:

P(n)=(Mn+N2)!(Mn)!(N2)!/(M+N1)!M!(N1)! ……. (1)

whereM, is the sum of nonnegative distinct quantum numbersniandNis the total number of particles.

02

Calculate the probability using equation (1)

Here, M=2,N=4, and the allowed values of nare n=0,1,2. Let us first consider n=0.

The probability is:

role="math" localid="1660022722768" P(n=0)=(20+42)!(20)!(42)!/(2+41)!2!(41)!=4!2!2!/5!2!3!=6/10=0.6

Forn=1, we get:

P(n=1)=(21+42)!(21)!(42)!/(2+41)!2!(41)!=3!1!2!/5!2!3!=3/10=0.3

Finally, for n=2, we get:

P(n=2)=(22+42)!(22)!(42)!/(2+41)!2!(41)!=2!0!2!/5!2!3!=1/10=0.1

We have thus verified that the probabilities agree with the result from equation 1.

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