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Consider a system of five particles, inside a container where the allowed energy levels are nondegenerate and evenly spaced. For instance, the particles could be trapped in a one-dimensional harmonic oscillator potential. In this problem you will consider the allowed states for this system, depending on whether the particles are identical fermions, identical bosons, or distinguishable particles.

(a) Describe the ground state of this system, for each of these three cases.

(b) Suppose that the system has one unit of energy (above the ground state). Describe the allowed states of the system, for each of the three cases. How many possible system states are there in each case?

(c) Repeat part (b) for two units of energy and for three units of energy.

(d) Suppose that the temperature of this system is low, so that the total energy is low (though not necessarily zero). In what way will the behavior of the bosonic system differ from that of the system of distinguishable particles? Discuss.

Short Answer

Expert verified

(a) The ground state of this system is.

(b) The are five possible system states in each case.

(c) For two-unit of energy the graph is

and for three-unit of energy is

.

(d) The way that the behavior of the bosonic system differs from that of the system of distinguishable particles is discussed below.

Step by step solution

01

Part (a) Step 1: Given Information

We have to describe the ground state of the system, for each of these three cases.

02

Part (a) Step 2: Simplify

As bosons do not follow the Pauli exclusion principle, particles in the ground state on the same level are distinguishable, but if they are fermions, each one will occupy a level starting from the lowest level, resulting in something like this:

03

Part (b) Step 1: Given Information

We have to describe the allowed states of the system, for each of the three cases and find the possible system states in each case.

04

Part (b) Step 2: Simplify

Consider a particle that has been promoted to the second lowest level, one energy unit above the ground state. There is only one way to do this for identifiable particles or bosons, which is to promote one particle from the lowest level to the second lowest level. There are five ways to promote indistinguishable particles or fermions, and the particle is promoted from the fifth level to the sixth level, as represented graphically:

05

Part (c) Step 1: Given Information

We have to repeat part (b) for two units of energy and for three units of energy.

06

Part (c) Step 2: Simplify

Consider a particle promoted to the second lowest level with two energy units above the ground state. We can promote one particle up to two energy levels, leaving four particles in the first level, or we can promote two particles to the second lowest level. There is only one way to do this for either, which is to promote one particle from the lowest level to the second lowest level. There are ten ways to do the first arrangement (where the two particles are promoted to the second and third lowest levels above the last filled level) and five ways to do the second arrangement (where the last particle is promoted to the third level above the last filled level) for indistinguishable particles or fermions, as shown graphically:

Consider a particle promoted to the second lowest level with three energy units above the ground state. We can promote one particle up to three energy levels for distinguishable particles or bosons, leaving four particles in the first level, or promote one particle to the second lowest level and one to the third lowest level, or promote three particles to the second lowest level, but there is only one way to do this for each. There are ten ways to make the first arrangement, twenty ways to do the second arrangement, and five ways to do the third arrangement for indistinguishable particles or fermions, and these arrangements are depicted graphically as:

07

Part (d) Step 1: Given Information

We have to find the behavior of the bosonic system differ from that of the system of distinguishable particles.

08

Part (d) Step 2: Simplify

The Boltzmann factor is proportional to the likelihood that a system with a temperature of T, i.e.

Pe-E/kT

The energy of any system is the same, therefore this factor is the same, but when we factor in degeneracy (from the previous section, the degeneracy of the bosons is 3 and the degeneracy of the fermions is 35), we can see that at low temperatures, the fermions are more likely to be found than the bosons. The bosons are also likely to be discovered in the ground state.

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Most popular questions from this chapter

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is

(a) 1eVless than

(b) 0.01eVless than

(c) equal to

(d) 0.01eVgreater than

(e) 1eVgreater than

Problem 7.67. In the first achievement of Bose-Einstein condensation with atomic hydrogen, a gas of approximately 21010atoms was trapped and cooled until its peak density was1.81014atoms/cm3. Calculate the condensation temperature for this system, and compare to the measured value of50K.

If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate for T>Tc.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms dimensionless variables. So define t=T/Tc,c=/kTc,andx=/kTc. Express the integral that defines , equation N=0g()1e(-)/kT-1d, in terms of these variables, you should obtain the equation

2.315=0xdxe(x-c)/t-1

(b) According to given figure , the correct value of cwhen T=2Tc, is approximately -0.8. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary , holding Tfixed, to find the precise value of for T=2Tc. Repeat for values ofT/Tcranging from 1.2up to 3.0, in increments of 0.2. Plot a graph of as a function of temperature.

Consider any two internal states, s1 and s2, of an atom. Let s2 be the higher-energy state, so that Es2-Es1= for some positive constant. If the atom is currently in state s2, then there is a certain probability per unit time for it to spontaneously decay down to state s1, emitting a photon with energy e. This probability per unit time is called the Einstein A coefficient:

A = probability of spontaneous decay per unit time.

On the other hand, if the atom is currently in state s1 and we shine light on it with frequency f=/h, then there is a chance that it will absorb photon, jumping into state s2. The probability for this to occur is proportional not only to the amount of time elapsed but also to the intensity of the light, or more precisely, the energy density of the light per unit frequency, u(f). (This is the function which, when integrated over any frequency interval, gives the energy per unit volume within that frequency interval. For our atomic transition, all that matters is the value of u(f)atf=/h) The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coefficient:

B=probability of absorption per unit timeu(f)

Finally, it is also possible for the atom to make a stimulated transition from s2down to s1, again with a probability that is proportional to the intensity of light at frequency f. (Stimulated emission is the fundamental mechanism of the laser: Light Amplification by Stimulated Emission of Radiation.) Thus we define a third coefficient, B, that is analogous to B:

B'=probability of stimulated emission per unit timeu(f)

As Einstein showed in 1917, knowing any one of these three coefficients is as good as knowing them all.

(a) Imagine a collection of many of these atoms, such that N1 of them are in state s1 and N2 are in state s2. Write down a formula for dN1/dt in terms of A, B, B', N1, N2, and u(f).

(b) Einstein's trick is to imagine that these atoms are bathed in thermal radiation, so that u(f) is the Planck spectral function. At equilibrium, N1and N2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related by

B'=BandAB=8hf3c3

In Section 6.5 I derived the useful relation F=-kTln(Z)between the Helmholtz free energy and the ordinary partition function. Use analogous argument to prove that =-kTln(Z^), where Z^ is the grand partition function and is the grand free energy introduced in Problem 5.23.

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