这是一份leeds利兹大学PHYS131101作业代写的成功案例

This is often written in the abbreviated form,
$$
i \hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t)=\left[-\frac{\hbar^{2}}{2 m} \nabla^{2}+V(\mathbf{r})\right] \psi(\mathbf{r}, t),
$$
and the momentum operator $\hat{p}$ itself as
$$
\hat{\mathbf{p}}=\frac{\hbar}{i}\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)=\frac{\hbar}{i} \boldsymbol{\nabla} .
$$

PHYS131101 COURSE NOTES ：

$$
n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\mathbf{E}(\mathbf{r}, t)|^{2} .
$$
Naturally, the total number of photons $N(t)$ at time $t$ is found by integrating over the entire volume $\mathcal{V}$,
$$
N(t)=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t) .
$$
Returning to the quantum wavefunction $\psi(\mathbf{r}, t)$ of a single particle we postulate in analogy withthat the intensity $|\psi(\mathbf{r}, t)|^{2}$ of the wavefunction is related to the particle density $n$ and write
$$
n(\mathbf{r}, t) \propto I(\mathbf{r}, t) \propto|\psi(\mathbf{r}, t)|^{2} .
$$
But since we have only one particle the analogy reduces to
$$
1=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t)
$$