Degeneracy in quantum statistics

[arneet]

degeneracy,this word appears in my textbook many times,but i could not understand what it means in quantum statistics.also in my textbook it is said in bose-einstein statistics that " the deviation from perfect gas behaviour exhibited by bose-einstein gas is called gas degeneracy".but i can't understand it how degeneracy is related to deviation from ideal behaviour of gas.

 

[sa1988]

It's the amount of times one particular state is satisfied.

It's actually easier to use the standard quantum physics example, so I'll use that:

Imagine a situation where some amount of energy is described by

$E = k(n_x^2 + n_y^2 + n_z^2)$

where n_x, n_y, n_z are positive integers.

For different values of each n, you can sometimes get the same E value.

For example, for n_x=1, n_y=1, n_z=1 , we have E=3k

Now try some others:

n_x=1, n_y=1, n_z=2 ... E=6k
n_x=1, n_y=2, n_z=1 ... E=6k
n_x=2, n_y=1, n_z=1 ... E=6k

This clearly shows that there are three totally different situations which give the same result for E.

This is the degeneracy. The E=6k level has a degeneracy of 3.

The same can be applied to any statistical system.

The degeneracy of any particular state is essentially the number of different ways that same state can be achieved.

Hope that helps!

EDIT: Just seen you ask about how it relates to Bose-Einstein statistics and 'gas degeneracy'. I'm not sure of that one, to be honest!

 

[vanhees71]

That's degeneracy of eigenvalues in linear algebra (or quantum mechanics which is applied linear algebra on the separable Hilbert space).

In statistical physics degeneracy refers to the deviations of a many-body system in thermal equilibrium from the classical statistics. The classical statistics holds, if the occupation per state is small compared to 1. If you have large occupation numbers per state, the Fermi-Dirac or Bose-Einstein statistics must be used instead of the Maxwell-Boltzmann statistics of classical statistics.

 

[arneet]

It's the amount of times one particular state is satisfied.

It's actually easier to use the standard quantum physics example, so I'll use that:

Imagine a situation where some amount of energy is described by

$E = k(n_x^2 + n_y^2 + n_z^2)$

where n_x, n_y, n_z are positive integers.

For different values of each n, you can sometimes get the same E value.

For example, for n_x=1, n_y=1, n_z=1 , we have E=3k

Now try some others:

n_x=1, n_y=1, n_z=2 ... E=6k
n_x=1, n_y=2, n_z=1 ... E=6k
n_x=2, n_y=1, n_z=1 ... E=6k

This clearly shows that there are three totally different situations which give the same result for E.

This is the degeneracy. The E=6k level has a degeneracy of 3.

The same can be applied to any statistical system.

The degeneracy of any particular state is essentially the number of different ways that same state can be achieved.

Hope that helps!

EDIT: Just seen you ask about how it relates to Bose-Einstein statistics and 'gas degeneracy'. I'm not sure of that one, to be honest!

thank you,it was helpful.

 

[arneet]

That's degeneracy of eigenvalues in linear algebra (or quantum mechanics which is applied linear algebra on the separable Hilbert space).

In statistical physics degeneracy refers to the deviations of a many-body system in thermal equilibrium from the classical statistics. The classical statistics holds, if the occupation per state is small compared to 1. If you have large occupation numbers per state, the Fermi-Dirac or Bose-Einstein statistics must be used instead of the Maxwell-Boltzmann statistics of classical statistics.

sir,it was helpful.thank you.it would be great if you can explain a bit more about degeneracy particularly in statistical physics.