# 纳米技术入门 Introduction to Nanotechnology PHYS131101

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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
$$\hat{\mathbf{p}}=\frac{\hbar}{i}\left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)=\frac{\hbar}{i} \boldsymbol{\nabla} .$$

## PHYS131101COURSE 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
$$1=\int_{\mathcal{V}} d \mathbf{r} n(\mathbf{r}, t)$$