# Relation between Quantum Mechanics and Temperature

1. Sep 4, 2010

### leoneri

Hi, I am quite confused with few concepts and relations between quantum mechanics and statistical physics.

Here it goes:
1. In quantum mechanics, we can calculate the energy levels of a bound state (confining potential). So in short, we calculate all possible energy level that a system could have in any state of this given system.
2. In room temperature, statistics of particles can be approximated by Boltzmann statistics. And physical quantities can be calculated using partition function.
3. As temperature goes down, there will be distinction between fermions and bosons, where Fermi-Dirac and Bose-Einstein statistics must be used respectively, instead of Boltzmann statistics. Then, again, with partition function, we can calculate physical quantities that we are interested with.

So .. my question is, is there any connection between calculated energy from quantum mechanics with the partition function? Is the energy levels calculated with quantum mechanics will be like that forever in any condition? How about the relation with the change of temperature?

I feel like my understanding is in mess right now. The more book I read, the more confused I am .. Thanks in advance!

2. Sep 7, 2010

### xepma

Let's first focus on the idea of a single particle in a quantum mechanical system which is set at some temperature T. When a system is at some temperature T it means that its energy is not fixed. The particle can sit in each energy eigenstate, and the probability that it does so is given by $$e^{-E/(kT)}/Z$$ where E is the energy of that particular eigenstate. Z here is a normalization factor, also known as the partition function.

Now, in quantum mechanics the partion function is defined as the trace of an operator.

$$Z = Tr[ e^{-H/(kT)}]$$

with H being the Hamiltonian of the system. The trace runs over the Hilbert space you are working with. You can see the connection with the energy eigenstates appearing here: you can choose the basis of energy eigenstates for the Hilbert space and the trace will turn into a sum over the energy eigenstates.

$$Z = Tr[ e^{-H/(kT)}] = \sum_{i} \left\langle E_i|e^{-H/(kT)} | E_i \right\rangle = \sum_i e^{-E_i/(kT)}$$

This is nothing what you expect, since Z serves as a normalization factor. It also coincides with the definition you would use in a classical setup. The trace naturally incorporates degenerate states arising from spins for instance.

Now, the original definition of Z in terms of a trace can be employed to determine stuff like "the expectation value of an operator in a quantum mechanical system at temperature T". If you have some operator $$A$$ and you want to know its expecation value in such a system, you simply compute:

$$\langle A \rangle = \frac{1}{Z} Tr[ e^{-H/(kT)} A]$$

The factor of $$\frac{1}{Z}$$ serves as a normalization again. We can also define the following operator:

$$\rho = \frac{1}{Z} e^{-H/(kT)}$$

and the definition of an expectation value turns into:

$$\langle A \rangle =Tr[ \rho A]$$

The operator $$\rho$$ is called a density operator. It sort of generalizes the notion of a quantum state. A density operator is more general than used here, and in this specific case it's also known as a Gibbs state, meaning it does not change over time (i.e. it's an equilibrium state). It's not really a state in the usual sense though -- you implicitly performed an averaging over time to arrive at this state. Anyways, these technicalities should be looked up in a book.

Now, here I have only talked about temperature and not about particles nor their statistics. The are two seperate cases which I should mention. The first case is when you have a fixed number of particles which can be either fermions or bosons. In that case the above formalism actually still applies.

What changes when you go from one particle in a quantum mechanical system at temperature T to multiple particles? Well, the Hilbert space changes. Any state in the Hilbert space that describes fermions (bosons) has to be antisymmetrized (symmetrized). That means that any basis that you choose for this two-particle Hilbert space has to obey these statistics as well.

Specifically, the energy eigenstates you trace over in the definition of the partition function is no longer the trace over single particle energy eigenstates. Instead, you trace over the multiparticle energy eigenstates. The multiparticle energy eigenstates already incorporate the statistics of the particles! Simply put: the trace does not run over states which are not allowed by the statistics of the particle.

So if you have two fermions the trace will not run over states which contains two fermions in the same quantum state. It will only run over states which are proberly antisymmetrized. The same idea applies to bosons: the trace only runs over properly symmetrized states.

So suppose you have some Hamiltonian. This Hamiltonian has some energy spectrum, which follow form its energy eigenstates. Next, we add some particles -- either fermions or bosons. The combination of single particle energy eigenstates + particle number + particle species fixes the Hilbert space. After that we can introduce temperature through the introduction of the density operator. That's it.

Now what about stuff like Bose-Einstein and Fermi-Dirac distribution? Let me get back to that in another post.

3. Sep 7, 2010

### xepma

How can you obtain Bose-Einstein and Fermi-Dirac distributions through this setup?

For that you have to recall that these distributions arise in systems in which the particle number is not fixed. This is said to occur when a system is in contact with some particle reservoir and there are tons of way to realise this.

How is it incorporated in the scheme above? Long story short, the partition function (and therefore also the density operator) is now defined as:

$$Z = Tr[ e^{-(H - \mu N)/(kT)} ]$$

Here we have introduced a second operator N, which is the number operator. The number operator simply measures the number of particles that are in a given state. $$\mu$$ is the chemical potential of the system. This is ofcourse nothing but the quantum version of the grand canonical ensemble.

So the probability that the system is in a state with energy E_i and number of particles n_i is given by:

$$e^{-(E_i -\mu n_i)/(kT)}/Z$$

What has changed? Well, in this particular system the particle number is not a fixed quantity. That means that we are not dealing with just the n-particle Hilbert space, but rather with all Hilbert spaces associated with all possible particle numbers. This space is called the Fock space, and roughly it is defined as the direct sum of all n-particle Hilbert space (for all n).

$$\mathcal{F} = \mathcal{H}_1\oplus \mathcal{H}_2\oplus \cdots$$

The trace then runs over all states within this Fock space. And here we encounter the statistics of the particle species again: there are no states in the Fock space that violate the statistics of the particle. This is simply a result from the fact that the Fock space is build up from Hilbert spaces which respect the statistics of the particle species.

There is another way that things like Fermionic statistics is incorporated in this scheme. Namely, the number operator N will never be able to "count" more than one fermion in the same quantum state (spin counts as a seperate quantum number). Put differently, for fermions we know that the only eigenvalues of the number operator N are zero and 1 ! With a little bit of work and this information you can actually derive the Fermi-Dirac distribution from the above partition function and this information.

Alright, I'm ranting too much. If you got any questions, shoot.

4. Sep 11, 2010

### leoneri

xepma, thank you very much with your explanation. give me time to digest these valuable information.