这是一份nottingham诺丁汉大学PHYS3002作业代写的成功案例

$$
i \hbar \frac{\partial}{\partial t} \int d^{3} r u_{k}^{}(\mathbf{r}) \Psi(\mathbf{r}, t)=\int d^{3} r u_{k}^{}(\mathbf{r})\left[H^{\prime} \Psi^{\prime}(\mathbf{r}, t)\right]
$$
on the right,
$$
\begin{aligned}
i \hbar \frac{\partial}{\partial t} \psi_{k^{k}}(t) &=\int d^{3} r u_{\mathbf{k}^{}}(\mathbf{r}) H^{\prime} \sum_{\mathbf{k}^{\prime}} \psi_{\mathbf{k}^{\prime}}(t) u_{k^{\prime}} \ &=\sum_{k^{k}} \psi_{k^{\prime}}(t)\left[\int d^{3} r u_{k}^{}(\mathbf{r}) H^{\prime} u_{\mathbf{k}^{\prime}}(\mathbf{r})\right]
\end{aligned}
$$ can be rewritten as
$$
i \hbar \frac{d}{d t} \psi_{k}(t)=\sum_{k^{\prime}} H_{k_{k}}^{\prime}(t) \psi_{k}(t)
$$
where the matrix elements $H_{\mathrm{k}, \mathrm{k}}^{\prime}$ are defined by
$$
H_{k, k^{\prime}}^{\prime}(t)=\int d^{3} r u_{k}^{*}(\mathbf{r}) H^{\prime} u_{k^{\prime}}(\mathbf{r})
$$

PHYS3002 COURSE NOTES ：

$$
\mathbf{V} \cdot \mathbf{V}=\sum_{1} V_{i}^{*} V_{i}
$$
There is an analogous result for state vectors as well.
$$
\begin{aligned}
\int d^{3} r \Psi * \Psi &=\int d^{3} r\left[\sum_{r} \psi_{i} u_{i}(\mathbf{r})\right]^{} \sum_{i} \psi_{j} u_{j}(\mathbf{r}) \ &=\sum_{i} \psi_{i}^{} \psi_{j} \int d^{3} r u_{i}^{}(\mathbf{r}){j}(\mathbf{r}) \end{aligned} $$ Using eq. $(2,16)$, $$ \int d^{3} r \psi r \psi \psi=\sum{1} \psi_{i}^{} \psi_{i}
$$