# States of a System | Thermal Physics Lecture Notes

Since the quantum state is time independent, we can use the Schrodinger equation to determine the energy of the system. Thus each quantum state has a definite energy. States with identical energy are said to belong to the same energy level. The multiplicity (or degeneracy) of an energy level is the number of quantum states with the same energy. Whereas in quantum mechanics it is the energy levels that are the important consideration, in thermal physics we are most concerned with the number of quantum states in a certain energy level.

Example:

What is the multiplicity of the hydrogen atom?

Without going into the details of the calculation, which is relatively involved, application of the Schrodinger equation to the hydrogen atom shows that the energy levels are given by

(2.1)

where enis the energy associated with the nth level,  is the reduced mass (with m being the electron’s mass and M being the mass of the nucleus), c is the speed of light, Z is the atomic number of the atom, and ais fine structure constant (a-1= 137.036).   In obtaining this result, we have neglected the fact that the proton in the nucleus also has a spin of ½.  In addition to this result, the Schrodinger equation shows that there are three “quantum numbers” associated with the atom.  These numbers, denoted n, l, and m, denote the “radial quantum number”, the “total angular momentum quantum number”, and the “z component of the angular momentum quantum number”.  The radial quantum number is directly related to the energy, as seen in (2.1), and satisfies .  The total angular momentum quantum number is related to the total angular momentum of the electron, and is associated with the azimuthal angle.  It must satisfy .  The z component of the angular momentum quantum number is associated with the zenith angle, and is related to that component of the angular momentum pointing along this axis.  It must satisfy .

The fact that the hydrogen atom’s wave function requires three separate quantum numbers to uniquely specify it leads to the result that we want.  Notice that the energy level only explicitly depends on the radial quantum number.  That means that the other two quantum numbers, l and m, can take on any value in their allowed ranges without changing the energy.  Thus, for a specific energy level, the total number of states is the sum of the allowed values for l and m.  From the results of the Schrodinger equation, we see that each l value has (2l+1) possible m values, and there are n possible l values.  So the total number of states is

Since the electron has two possible spin states for any energy state, the final degeneracy is given by 2n2.  For example, for the first three energy levels the multiplicity is

n

Multiplicity

1

2

2

8

3

16

Example:

What is the multiplicity of a particle in a box whose sides have a length of L?

For an infinite potential well we saw that the energy levels were given by

.

The particle in a box problem can be solved by inspection once we realize that the box is simply three infinite wells at right angles to each other.  The net result of this is that each direction has it’s own quantum number, so the energy levels of the particle in a box becomes

where nx, ny and nz each independently run from 1 to infinity.  For example, for the lowest six energy states, the multiplicity is

multiplicity

12+12+12 = 3

1

22+12+12 = 6

3

22+22+12 = 9

3

32+12+12 = 11

3

22+22+22 = 12

1

32+22+12 = 14

6

The important thing to notice in both of these examples is that the energy of the system is the total energy of the particle, kinetic and potential.  If the system consists of more than one particle, the total energy of the system is the total energy of all the particles, including the energy involved in interactions between the particles.  Another thing to notice from these examples is that we can easily reach states which have large multiplicities.  In addition, we will usually be dealing with systems that consist of a large number of particles, each of which can be treated as being independent of each other.  Thus we need to start talking about the statistical properties of the system, rather than properties of individual particles in the system.  In order to describe the statistical properties of a system of N particles, it is essential to know the possible values of the energy es(N), where eis the energy of the quantum state s of the N particle system.  As a first example of this approach, let’s consider a Binary Model System.

## Binary Model System

The Binary Model System is the simplest thermodynamic system.  It uses the Fundamental Assumption of Thermal Physics: A system is equally likely to be in any accessible quantum state.  How do we describe the Binary Model System?  Classically, we know that for a magnetic field B and a magnetic moment m, the energy is .  Because there are two possible spin orientations, quantum mechanics tells us that the magnetic moment can be either positive or negative, so the energy becomes U = ±mB.  Consider a lattice of N magnetic moments

How many different orientations are possible?  Since each magnetic moment can have two orientations and there are N magnetic moments, there are a total of 2N possible orientations.  Thus the probability of the system being in any one specific orientation is

.                                      (2.2)

Using the arrow notation, we see that every unique state of the system can be symbolically denoted as

.                              (2.3)

This symbolic notation is called the generating function: it generates all the possible states of the system.  The total magnetic moment of the system is given by M, which can vary from Nm to –Nm.  The possible magnetic moments are then

M = Nm, (N-2)m, (N-4)m, …, -(N-2)m, –Nm.                         (2.4)

We see that there are a total of N+1 possible values for the magnetic moment.  If N is very large, we see that there are a lot more possible states than there are magnetic moment values.  So, since the energy of the system depends on the magnetic moment, we see that, for large N, we expect to have a large multiplicity for any energy level.

Now let’s consider the total magnetic moment for the system.  We see that it depends on the number of magnetic moments pointing up minus those pointing down.  However, it does not depend on which magnetic moments are pointing in any specific direction.  So, let Nup be the number of magnetic moments pointing up, and Ndown be the number of magnetic moments pointing down.  Then Nup = ½N + s and Ndown = ½Ns, where s is called the spin excess and is defined by

NupNdown = 2s.                                                (2.5)

How many different ways can we get a specific spin excess, s?  Since we are no longer interested in a specific orientation, we can drop the subscripts in (2.3) and write it as

.                                                      (2.6)

This can be expanded as a binomial series to yield

.                       (2.7)

## Multiplicity Function

Each piece of the summation represents a collection of states with the same spin excess, and the coefficient in the summation is the number of states with a specific spin excess of 2s and a total magnetic moment of 2sm.  We define this coefficient to be the multiplicity function

.                                      (2.8)

Since the multiplicity function is the number of states having the same spin excess s, and we have seen that the total magnetic moment, and thus the energy, depends on the spin excess, we see that g(N,s) is the total number of states in a specific energy level.

Example:

What are the values of the multiplicity function for N = 10?

Using (2.8), we see that for N = 10

s

g(10,s)

5

1

4

10

3

45

2

120

1

210

0

252

-1

210

-2

120

-3

45

-4

10

-5

1

A graph of this looks like

As N gets big, this curve approaches a gaussian curve. To see this, use Stirling’s approximation:

.                                       (2.9)

Taking the natural logarithm of this yields

.                       (2.10)

Applying this to g(N,s)

.             (2.11)

In order to proceed, expand ln(Nup) as

(2.12)

and ln(Ndown) as

.                               (2.13)

Note that we used the approximation

in both of these expansions.  Substituting the (2.12) and (2.13) into (2.11) yields

which for large N can be simplified to

.                          (2.14)

Taking the exponential to recover g(N,s),

.                                 (2.15)

Notice that when .  Numerically, if N ~ 1022, 2s/N ~10-11.

## Mean Spin Excess Average

In order to determine the average of the spin excess, we need to determine the probability function.  The binomial distribution (2.8) has the sum

,                                    (2.16)

so the probability function becomes

.                                               (2.17)

We can use this to calculate <s2> and <(2s)2>.  From the definition of averages, we have that

.

Changing variables to , this becomes

.                                    (2.18)

and so

.                                               (2.19)

Physically, the quantity <(2s)2> is known as the mean square spin excess.  The root mean square spin excess is then

.

With this result we can find the fractional fluctuation in the spin excess.  By the definition of the fractional fluctuation we get

,                                                (2.20)

so we see that as N gets large the central peak of the distribution function becomes sharper.  This is important because the sharpness of the distribution function is related to the stability of the solution.

Share.