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It is sometimes useful to find the average energy of a certain number [itex]N[/itex] of particles contained in a box of volume [itex]V[/itex].

In order to find this quantity, the total energy is required and then divided by [itex]N[/itex]. The result is

[itex]E_{average} = \displaystyle \frac{1}{N} \sum_{n = 1}^{N} \left| a_n \right|^2 E_n[/itex]

where [itex]\left| a_n \right|^2[/itex] is the probability, for the [itex]n[/itex]-th particle, of having energy [itex]E_n[/itex].

If the wavefunction related to the [itex]n[/itex]-th particle is [itex]\psi_n (\mathbf{r}, t)[/itex], a total wavefunction can be built:

[itex]\psi (\mathbf{r}, t) = \displaystyle \sum_{n = 1}^N a_n \psi_n (\mathbf{r}, t)[/itex]

and the above result is obtained by computing the energy of this (global) wavefunction, divided by [itex]N[/itex]:

[itex]E_{average} = \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \mathcal{H} \psi (\mathbf{r}, t) dV = \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \mathcal{H} \left( \displaystyle \sum_{n = 1}^N a_n \psi_n (\mathbf{r}, t) \right) dV =[/itex]

[itex]= \displaystyle \frac{1}{N} \int_{V} \psi^* (\mathbf{r}, t) \left( \displaystyle \sum_{n = 1}^N a_n E_n \psi_n (\mathbf{r}, t) \right) dV = \displaystyle \frac{1}{N} \int_{V} \left( \sum_{m = 1}^N a_m^* \psi_m^* (\mathbf{r}, t) \right) \left( \displaystyle \sum_{n = 1}^N a_n E_n \psi_n (\mathbf{r}, t) \right) dV[/itex]

(and then the orthonormality of the [itex]\psi_n[/itex] must be used).

[itex]E_n[/itex] is the energy related to the [itex]n[/itex]-th wavefunction, that is the eigenvalue of the hamiltonian

[itex]\mathcal{H} \psi_n (\mathbf{r}, t) = E_n \psi_n (\mathbf{r}, t)[/itex]

But if the single [itex]\psi_n (\mathbf{r}, t)[/itex] must satisfy by itself the normalization condition

[itex]\displaystyle \int_{V} \left| \psi_n (\mathbf{r}, t) \right|^2 dV = 1[/itex]

why an [itex]a_n[/itex] has to be used? I've not found this computation on the web, but just in some notes.

If the total number of particles is [itex]N[/itex] and the [itex]n[/itex]-th particlemusthave its own wavefunction

[itex]\psi_n (\mathbf{r}, t)[/itex], how can we consider a probability that the particle will have this wavefunction?

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# I Energy of a number of particles

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