When is the temperature relevant for a quantum many body system?

Question

Let me illustrate my question with an example. Suppose I want to consider a system of $N$ identical bosons. The Hamiltonian of the system is typically of the form: $$H_{N} = \sum_{j=1}^{N}(-\Delta_{x_{j}}) + \sum_{1 \le i < j \le N}V(x_{i}-x_{j})$$ for some interaction potential $V$. Such systems are usually said to be in the zero temperature setting (e.g. I have seen such terminology in many books on many-body quantum mechanics) as opposed to systems with positive temperature where the equilibrium state is given some density matrix.

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve Schrödinger's equation $H_{N}\psi_{N} = E\psi_{N}$ we would have a lot of possible excited states given a full range of different energies. Thus, I assume that what is meant by zero temperature systems is actually low temperature, not zero.

But if the temperature is low, this means that the average energy should be also low, right? But there is no restriction on our systems to have a low average $\langle H_{N}\rangle$. Why is temperature not relevant then? And when is it, so we must consider density matrices instead of pure states?

Answer 1

The question in the title, the "when" question, is basically impossible to answer. I mean, it is only when your level of precision required for whatever experiment it is you are doing, happens to disagree with the zero-temperature prediction from quantum theory. Nobody knows how to define a more precise answer. Maybe they might have a better answer, but it is going to be some vague notion.

Instead, I think it is more interesting to tackle what you say:

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve Schrödinger's equation $H_N\psi_N=E\psi_N$ we would have a lot of possible excited states given a full range of different energies. Thus, I assume that what is meant by zero temperature systems is actually low temperature, not zero.

You are extremely wrong here. You can solve for the potential-as-in-not-actualised excited states without filling them up, staying strictly in the ground state.

Or you can pick a better definition of temperature. In the Gibbs entropy definition, being in any exact energy eigenstate alone means that $T=0$, so even if that is an excited state, that is $T=0$.

Or in the case of a two-level system, being in the excited state means that $T\to-0$, and yes, you read it correctly, that is negative temperature. To define this temperature, you have to relax the requirement of equilibrium, but has already been shown to be experimentally useful to define a temperature that behaves this way.

In particular, when in many-body physics we assert that a scheme is only for $T=0$, we literally mean it, and we definitely do not mean that it is suitable for low temperature. Low temperature is always an expansion on top of the $T=0$ behaviour, and we always have to insert that extra work to get there, and so it is conversationally convenient to make this distinction.

Answer 2

I will write my answer by addressing some of your statements in your question, and elaborate a bit around them.

You start by saying:

The Hamiltonian of the system is typically of the form (expression here) for some interaction potential $V$. Such systems are usually said to be in the zero temperature setting.

The Hamiltonian operator of a system does not depend on the temperature of the state the system is in. So this first statement is imprecise. I agree that many sources claim that the ground state of a given Hamiltonian "is" the state at zero temperature, but this is as you say only true in a certain context I will detail below.

Generally, the Schrödinger equation is an initial value problem, meaning that if we know the state of a system at time $t=0$ then we can find its state at future times $t>0$. In particular, if the system starts out in some excited state, it will stay that way if it does not interact with its environment.

Now if your system interacts with an environment that can be modelled as a *heat bath at temperature $T$ then it will attain equilibrium in a mixed state, where the average properties are given in terms of the partition function, involving both the ground state and excited states, and temperature.

What happens if we lower the temperature of the heat bath towards $T=0$? It is in this case we can make the claim you state: If the system is at T=0, all particles should be at the ground state and no excited state is possible. But this claim is not universal, it applies on this specific context, the one I mentioned before.

Finally you say:

But if the temperature is low, this means that the average energy should be also low, right? But there is no restriction on our systems to have a low average $\langle H_N \rangle$.

If the temperature of the heat bath with which your system is in equilibrium with is low, then this is precisely the restriction. Otherwise, it does not apply.

Answer 3

If the system is at $T=0$, all particles should be at the ground state and no excited state is possible. However, this is clearly not true for an arbitrary quantum system, given that if we solve Schrödinger's equation $H_{N}\psi_{N} = E\psi_{N}$ we would have a lot of possible excited states given a full range of different energies. Thus, I assume that what is meant by zero temperature systems is actually low temperature, not zero.

One has to distinguish the (eigen)states of the Hamiltonian, which we obtain by solving $H_{N}\psi_{N} = E\psi_{N}$, and the actual state that the particle occupies. Forgetting for a moment about statistical physics, the particle will be either in one of these eigenstates or in their superposition: $\Psi = \sum_{N}c_N\psi_N$.

In reality, if we start with a system in an excited state, there is high probability that it "relaxes" to a lower energy state via its interaction with the environment, e.g., when colliding with other particles in a gas or via emitting photons. The inverse process (getting excited to a state of higher energy) is also possible, but is usually less probable, as we usually assume the environment in thermodynamic equilibrium (referred to as bath or reservoir in statistical physics.) On eventually expects that the particle will be found in state $N$ with probability proportional to Boltzmann factor $p_N\propto e^{-\frac{E_N}{k_BT}}$. Note that the interaction with the environment destroys the coherence of the wave function, and we now have to describe it by a density matrix: $$ \hat{\rho}=Z^{-1}\sum_N e^{-\frac{E_N}{k_BT}}|N\rangle\langle N|$$

If the temperature of the environment is very low (lower than the spacing between the ground state and the first excited state), we expect that all the terms in the sum are negligeable in comparison to that corresponding to the lowest energy state, hence the particle will be found in this state.