Simple question about Schrodinger equation (time independent)

For a quantum mechanical description of a system (like a small molecule) we can write:

$$\langle\psi|\hat {H}|\psi\rangle = \overline E$$

Question:

Is that energy the same as zero Kelvin energy obtained by statistical mechanics (using $E_n$ energies and partition functions)?

The average $\langle E\rangle$ you have on the right is not to be understood in the sense of statistical mechanics (and thus as a temperature).

The wavefunction $\psi$ can be a linear combination of states $\psi_n$ of definite energy of your molecule. The $\psi_n$’s are solutions of the Schrodinger equation. Thus, \begin{align} \psi(x,t)&=\sum_i a_i \psi_i(x,t)\, ,\qquad \sum_i \vert a_i\vert^2=1\, , \tag{1} \\ \hat H\psi_i(x,t)&=E_i\psi_i(x,t)\, . \end{align} With this wavefunction the average energy $$\langle E\rangle = \sum_i \vert a_i\vert^2 E_i \tag{2}$$ is a weighted average of the possible energies of your molecule, with the modulus square $\vert a_i\vert^2$ of the complex amplitude $a_i$ functioning as statistical weight. $\psi(x,t)$ given in (1) is an example of a pure state.

For instance, if you have a hydrogen wavefunction of the form $$\psi(r)=a_1\psi_{100}(r)+a_2\psi_{200}(r)$$ then the average energy of this system is $$\vert a_1\vert^2\times (-13.6) + \vert a_2\vert^2 \times (-13.6/4). \tag{3}$$ Measuring the energy, you will sometimes get the value $-13.6$ (this outcome will occur $\vert a_1\vert^2$ of the time) and sometimes get the value $-13.6/4$ (this outcome will occur $\vert a_2\vert^2$ of the time). The average energy is exactly given by (3).

In quantum statistical mechanics one introduces the concept of a mixted state. For mixed states one cannot define a wavefunction for the system as in (1). The system is described using a matrix that represents a statistical mixture of wavefunction that is not of the form given in (1); the average energy for this statistical mixture is also not of the form (2) since the latter comes from (1), which does not exist for mixted states.

• good answer. That equation doesn't seems time independent..anyway, not a problem. I understand nothing in the last paragraph :(. But those energies can still be used in statistical mechanics at any T? – HernanProust Oct 1 '17 at 10:19
• You can always set $t=0$ to make it time-independent. In stat mech. you need the mixted states (or density matrix) formulation and this introduces another type of averaging. It is this second type of averaging that is the averaging of classical statistical mechanics and which you can relate to temperature. You just can’t do this - i.e. think of (2) as an average in the sense of statistical mechanics - with pure states like $\psi(x,t)$. This is a deep and subtle point usually covered in advanced courses. You will get in trouble thinking of $\langle E\rangle$ as related to temperature. – ZeroTheHero Oct 1 '17 at 10:23
• Oh..I am a poor chemistry student(hate you physicists!). Although I understand a bit about density matrix, it was related to coeficients of wavef, if I am not mistaken. So there's no correlation between T=zero in stat mech mean energy and mean energy of schrod eq ? I am a bit confused also about what is E in sch eq (it not even seems something "real", or mesurable). Promiss not to bother again. – HernanProust Oct 1 '17 at 10:35
• @HernanProust Added some text to the answer to clarify your last point. – ZeroTheHero Oct 1 '17 at 10:42

Statistical mechanics does not really apply to the kind of system your equation refers to: your equation is good for a pure quantum state $\psi$. In statistical mechanical terms, the system's microstate is exactly and fully specified by $\psi$, and you can't really therefore talk about a system's temperature when it is in a pure state.

Recall that pure states can be non ground states, or in a superposition containing non ground states. So there is, in general, no relationship definable with a "$0{\rm K}$ energy".

A classical mixture of pure quantum states can describe a statistical mechanical system, can therefore be assigned a temperature and is described by a density matrix $\rho$. In such a case, your equation becomes replaced by:

$$\langle E\rangle = \mathrm{tr}(\rho\,\hat{H})$$

Perhaps a kind of converse to your question makes more sense and yields a more concrete answer: as we cool a statistical mechanical system down, and if the ground state is nondegenerate, the system becomes "forced" into a state where all particles / ensemble members are in the ground state. If, further, the ground state is nondegenerate (i.e. there is only one ground quantum energy eigenstate), then the $0{\rm K}$ state is a multiparticle pure quantum state and this state is the ground energy eigenstate. User Joshphysics analyzes this behavior in more detail in his answer here.

• $\psi$ can be the wavefunction of a molecule, for example? I am talking about those systems – HernanProust Oct 1 '17 at 9:36
• @HernanProust indeed. It can be the quantum state of a system comprising any number of particles. But it's a pure state: It defines the future (and past) system state completely by unitary evolution and there are no unspecified degrees of freedom. Statistical mechanical notions apply to classical probabilistic mixtures of such states and temperature is a statistical parameter of the probability distributions describing the mixture. – Selene Routley Oct 1 '17 at 23:42