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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 commentthis 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?

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?

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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Wavefunction of a system of particles

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?