# Why is the energy of a quantum state, which is not a stationary state, the expected value of energy?

I know that expected energy $\langle E\rangle _{\text{av}}$ is the mean value of the whole series of measurements of energy of a state.

If the state is not a stationary state, one measurement would yield one value, other measurement of the same state would yield some other value & so on; expected energy is the average of those measurements.

But does this expected energy tell about the energy of the state before measurement? Meant to say is the expected energy the energy of the system being in that state?

According to me, expected energy is just a mean value of the distribution of the energy measurements; before measurement we don't know what energy the system was at as it was in superposition; isn't it?

But recently, I read the following excerpt of Feynman :

[...] we can calculate the average energy of an atomic state even without knowing its energy levels. All we need is the wave function. It’s an important law. We’ll tell you about one interesting application. Suppose you want to know the ground-state energy of some system—say the helium atom, but it’s too hard to solve Schrödinger’s equation for the wave function, because there are too many variables. Suppose, however, that you take a guess at the wave function—pick any function you like—and calculate the average energy. This energy will certainly be higher than the ground-state energy which is the lowest possible energy the atom can have. Now pick another function and calculate its average energy. If it is lower than your first choice you are getting closer to the true ground-state energy. If you keep on trying all sorts of artificial states you will be able to get lower and lower energies, which come closer and closer to the ground-state energy. [...] By varying the parameters to give the lowest possible energy, you are trying out a whole class of functions at once. Eventually you will find that it is harder and harder to get lower energies and you will begin to be convinced that you are fairly close to the lowest possible energy.

Now, what he is saying is to change the parameters in such a way so as to decrease the average energy to the ground energy as far as possible that is to decrease the average energy from the previous computed average energy.

But average energy that is the expected energy is just the mean value of those measurements of energies; we don't know the energy of the system if it is in linear combination.

So, why is the average energy or expected energy associated with a non-stationary state as if the system has that energy before the measurement?

In this page, it is shown that the non-stationary state has the expected energy associated with it viz:

Any trial function can formally be expanded as a linear combination of the exact eigenfunctions $\Psi_i$. Of course, in practice, we don't know the $\Psi_i$, since we're assuming that we're applying the variational method to a problem we can't solve analytically. Nevertheless, that doesn't prevent us from using the exact eigenfunctions in our proof, since they certainly exist and form a complete set, even if we don't happen to know them. So, the trial wavefunction can be written $$\Phi = \sum_i c_i \Psi_i, \tag{146}$$

and the approximate energy corresponding to this wavefunction is $$E[\Phi] = \frac{\int \Phi^* {\hat H} \Phi}{\int \Phi^* \Phi}. \tag{147}$$

Substituting the expansion over the exact wavefuntions, $$E[\Phi] = \frac{\sum_{ij} c_i^* c_j \int \Psi_i^* {\hat H} \Psi_j}{ \sum_{ij} c_i^* c_j \int \Psi_i^* \Psi_j}. \tag{148}$$

Since the functions $\Psi_j$ are the exact eigenfunctions of ${\hat H}$, we can use ${\hat H} \Psi_j = {\cal E}_j \Psi_j$ to obtain $$E[\Phi] = \frac{\sum_{ij} c_i^* c_j {\cal E}_j \int \Psi_i^* \Psi_j}{ \sum_{ij} c_i^* c_j \int \Psi_i^* \Psi_j}.\tag{149}$$

Now using the fact that eigenfunctions of a Hermitian operator form an orthonormal set (or can be made to do so), $$E[\Phi] = \frac{\sum_{i} c_i^* c_i {\cal E}_i}{ \sum_{i} c_i^* c_i}.\tag{150}$$

We now subtract the exact ground state energy ${\cal E}_0$ from both sides to obtain $$E[\Phi] - {\cal E}_0 = \frac{\sum_i c_i^* c_i ( {\cal E}_i - {\cal E}_0)}{ \sum_i c_i^* c_i}.\tag{151}$$

Since every term on the right-hand side is greater than or equal to zero, the left-hand side must also be greater than or equal to zero, or $$E[\Phi] \geq {\cal E}_0.\tag{152}$$

In other words, the energy of any approximate wavefunction is always greater than or equal to the exact ground state energy ${\cal E}_0$.

Notice, the last two statements & $152$; what is the energy of the wavefunction? The expectation value of energy $E(\Phi)$! That is the non-stationary state has the expected energy as its energy. But how can $E[\Phi]$ be the energy of the state? It is just the mean-value of the measurements of the energy; we don't know the energy of the state prior to the measurement; then how can we know its energy is the expectation of energy??

So, my question is

$\bullet$ What is the energy of a quantum-state that is in linear combination of exact eigenfunctions of Hamiltonian operator? Is it the expectation of energy? If so, how could the expected energy be the energy of a state that we don't know what energy it is in before measurement?? Why do the linked pages above constantly refer the expected energy as the energy of the state?

Pleas help me sought out this confusion.

• @user36790: I linked Ehrenfest's theorem after that. In particular, it shows that $\frac{\mathrm{d}}{\mathrm{d}t}\langle H \rangle = 0$ for time-independent Hamiltonians, i.e. the expectation value of energy is conserved in time. – ACuriousMind Oct 24 '15 at 14:08