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A three-dimensional volume $V$ contains a certain number $N$ of electrons and they can't escape the volume $V$. Assume for simplicity that the potential $\mathcal{V}(\mathbf{r})$ is zero in all the volume. Each electron is in a stationary state, so with a well-defined energy value. The $k$-th electron has a wavefunction $\psi_k(\mathbf{r}, t)$ and an energy $E_k$, and obviously $k = 1, 2, \ldots, N$.

So,

$$i \hbar \displaystyle \frac{\partial}{\partial t} \psi_k(\mathbf{r}, t) = E_k \psi_k(\mathbf{r}, t)$$

The average value of the energy of an electron in such a system has to be computed. I read about simple problems like the quantum well with a single electron, but never about multiple particles all in the same volume.

Is it legitimate to consider a global wavefunction

$$\psi (\mathbf{r}, t) = \sum_{k = 1}^N \psi_k(\mathbf{r}, t)$$

and then compute the average energy as

$$E_{av} = \displaystyle \frac{\displaystyle \int_V \psi^* (\mathbf{r}, t) i \hbar \displaystyle \frac{\partial}{\partial t} \psi (\mathbf{r}, t) dV}{N} = \displaystyle \frac{\displaystyle \int_V \psi^* (\mathbf{r}, t) i \hbar \displaystyle \frac{\partial}{\partial t} \left[ \sum_{k = 1}^N \psi_k(\mathbf{r}, t) \right] dV}{N}$$

?

The result here would be

$$E_{av} = \displaystyle \frac{\displaystyle \int_V \psi^* (\mathbf{r}, t) \left[ \sum_{k = 1}^N E_k \psi_k(\mathbf{r}, t) \right] dV}{N} = \displaystyle \frac{\displaystyle \int_V \sum_{n = 1}^N \psi^*_n(\mathbf{r}, t) \left[ \sum_{k = 1}^N E_k \psi_k(\mathbf{r}, t) \right] dV}{N}$$

and knowing that $\psi^*_n(\mathbf{r}, t) \psi_k(\mathbf{r}, t) = 1$ if $n = k$ and $0$ for $n \neq k$ (these wavefunctions are orthonormal functions) trivially gives

$$E_{av} = \displaystyle \frac{\displaystyle \sum_{k = 1}^N E_k}{N}$$

Or another approach is required for such a problem?

For example, according to this comment, stating that «in a system with two electrons each individual electron would have its own wave function» «is wrong since they instead have a joint wave function». Is this the case also for this system of $N$ electrons?

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BowPark
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1 Answers1

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The way you combine quantum systems is not by summing their wavefunctions, but by taking the tensor product, see this question.

In particular, if the systems you are composing are just single electrons described by square-integrable wavefunctions in $L^2(\mathbb{R}^3,\mathrm{d}x)$, then a system of $N$ electrons is described by a square-integrable function of all the $3N$ coordinates, i.e. an element in $L^2(\mathbb{R}^{3N},\mathrm{d}x) = \bigoplus_{i=1}^N L^2(\mathbb{R}^3,\mathrm{d}x)$. You have "individual wavefunctions" for the electrons if and only if the joint wavefunction $\psi(x_1,\dots,x_{3N})$ factors into the product of such individual wavefunctions, i.e. $$ \psi(x_1,\dots,x_{3N}) = \psi(x_1,x_2,x_3)\cdot\psi(x_4,x_5,x_6)\cdot\dots\cdot\psi(x_{3N-2},x_{3N-1},x_{3N})$$

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