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# 物理代写|量子力学代写Quantum mechanics代考|H-atom in an external electric field

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## 物理代写|量子力学代写Quantum mechanics代考|H-atom in an external electric field

We recall the Hamiltonian in the presence of an electromagnetic field
$$H=\frac{1}{2 m}\left(\mathbf{p}+\frac{e}{c} \mathbf{A}\right)^2-e A_0-\boldsymbol{\mu} \cdot \mathbf{B}+\mathbf{V}(r)$$

For a constant external electric field the gauge potential is given by $\mathbf{A}=0$ and $A_0=-\mathbf{E} \cdot \mathbf{r}$, so that $\mathbf{E}=-\nabla A_0$ is verified. We choose the electric field in the $z$ direction, hence $A_0=-E z=E r \cos \theta$. Therefore the Hamiltonian takes the form
$$H=H_0+e E r \cos \theta$$
where $H_0$ is the Hydrogen atom Hamiltonian. We compute the matrix elements in $\left|n l j m_j\right\rangle$ basis. Of course $H_0$ is diagonal as in the previous sections. The perturbation has the following matrix elements
$$\left\langle n l j m|e E r \cos \theta| n^{\prime} l^{\prime} j^{\prime} m^{\prime}\right\rangle=e E \delta_{m m^{\prime}}\left\langle n l|r| n^{\prime} l^{\prime}\right\rangle\left\langle l j m|\cos \theta| l^{\prime} j^{\prime} m\right\rangle$$
We will concentrate on a given level $n$, such as $n=n^{\prime}=2$ as an illustration. The orbital angular momentum $l, l^{\prime}$ must differ by one unit because the operator $\cos \theta$ or $\mathbf{r}$ has odd parity while the states have parity $(-1)^l$. Therefore we take the values $l=0$ and $l^{\prime}=1$. Thus we need to calculate $\left\langle 2 s_{1 / 2}|e E r \cos \theta| 2 p_{1 / 2,3 / 2}\right\rangle$
$$\left\langle 2 s_{1 / 2}, m|e E r \cos \theta| 2 p_{1 / 2,3 / 2}, m\right\rangle=e E\langle 2 s|r| 2 p\rangle\left\langle s_{1 / 2}, \pm \frac{1}{2}|\cos \theta| p_{1 / 2,3 / 2}, \pm \frac{1}{2}\right\rangle$$
First we compute
$$\langle 2 s|r| 2 p\rangle=\int_0^{\infty} d r r^3 R_{20}(r) R_{21}(r)=-3 \sqrt{3} \frac{a_0}{Z}$$

## 物理代写|量子力学代写Quantum mechanics代考|Interaction picture

To compute the time development operator for the general case $H^{\prime}(t)$, we will distinguish between different formalisms that deal with time dependent problems in quantum mechanics. These are called the Schrödinger picture, the interaction picture and the Heisenberg picture.

The Schrödinger picture is the one we discussed above. In this case operators such as $\mathbf{r}, \mathbf{p}$ or their functions $O(\mathbf{r}, \mathbf{p})$ are all taken at time zero in the Hamiltonian formalism. The time dependence is all in the states and is governed by the Schrödinger equation
$$i \hbar \partial_t|\psi, t\rangle=H|\psi, t\rangle$$
The Heisenberg picture is the opposite, where operators such as $\mathbf{r}(t), \mathbf{p}(t)$ or their functions $O(\mathbf{r}(t), \mathbf{p}(t))$ are taken as dynamical observables that develop in time while the states are all time independent. To distinguish the Heisenberg states from the Schrödinger states we will append an extra $H$ as a subscript. When the Hamiltonian is independent of time the Heisenberg states are related to the Schrödinger states, which have no subscript, as
$$|\psi, t\rangle=e^{-i t H / \hbar}|\psi\rangle_H \leftrightarrow|\psi\rangle_H=e^{i t H / \hbar}|\psi, t\rangle$$
while the Heisenberg operators $O_H$ are related to the Schrödinger operators $O$ by
$$O_H(t)=e^{i t H / \hbar} O e^{-i t H / \hbar} .$$
The interaction picture is somewhere in between the Heisenberg and Schrödinger pictures. It is defined by
$$|\psi, t\rangle=e^{-i t H_0 / \hbar}|\psi, t\rangle_I \leftrightarrow|\psi, t\rangle_I=e^{i t H_0 / \hbar}|\psi, t\rangle$$
where, compared to the Heisenberg picture, $e^{i t H_0 / \hbar}$ appears instead of $e^{i t H / \hbar}$. The equation of motion for the interaction picture state $|\psi, t\rangle_I$ is derived by computing its time derivative as follows
\begin{aligned} & i \hbar \partial_t|\psi, t\rangle_I=i \hbar \partial_t\left(e^{i t H_0 / \hbar}|\psi, t\rangle\right)=e^{i t H_0 / \hbar}\left(i \hbar \partial_t|\psi, t\rangle\right)-H_0 e^{i t H_0 / \hbar}|\psi, t\rangle \ & =e^{i H_0 t / \hbar}\left(H_0+H^{\prime}(t)\right)|\psi, t\rangle-H_0|\psi, t\rangle_I=\left(e^{i t H_0 / \hbar} H^{\prime}(t) e^{-i t H_0 / \hbar}\right)|\psi, t\rangle_I \end{aligned}

## 物理代写|量子力学代写Quantum mechanics代考|H-atom in an external electric field

$$H=\frac{1}{2 m}\left(\mathbf{p}+\frac{e}{c} \mathbf{A}\right)^2-e A_0-\boldsymbol{\mu} \cdot \mathbf{B}+\mathbf{V}(r)$$

$$H=H_0+e E r \cos \theta$$

$$\left\langle n l j m|e E r \cos \theta| n^{\prime} l^{\prime} j^{\prime} m^{\prime}\right\rangle=e E \delta_{m m^{\prime}}\left\langle n l|r| n^{\prime} l^{\prime}\right\rangle\left\langle l j m|\cos \theta| l^{\prime} j^{\prime} m\right\rangle$$

$$\left\langle 2 s_{1 / 2}, m|e E r \cos \theta| 2 p_{1 / 2,3 / 2}, m\right\rangle=e E\langle 2 s|r| 2 p\rangle\left\langle s_{1 / 2}, \pm \frac{1}{2}|\cos \theta| p_{1 / 2,3 / 2}, \pm \frac{1}{2}\right\rangle$$

$$\langle 2 s|r| 2 p\rangle=\int_0^{\infty} d r r^3 R_{20}(r) R_{21}(r)=-3 \sqrt{3} \frac{a_0}{Z}$$

## 物理代写|量子力学代写Quantum mechanics代考|Interaction picture

$$i \hbar \partial_t|\psi, t\rangle=H|\psi, t\rangle$$

$$|\psi, t\rangle=e^{-i t H / \hbar}|\psi\rangle_H \leftrightarrow|\psi\rangle_H=e^{i t H / \hbar}|\psi, t\rangle$$

$$O_H(t)=e^{i t H / \hbar} O e^{-i t H / \hbar}$$

$$|\psi, t\rangle=e^{-i t H_0 / \hbar}|\psi, t\rangle_I \leftrightarrow|\psi, t\rangle_I=e^{i t H_0 / \hbar}|\psi, t\rangle$$

$$i \hbar \partial_t|\psi, t\rangle_I=i \hbar \partial_t\left(e^{i t H_0 / \hbar}|\psi, t\rangle\right)=e^{i t H_0 / \hbar}\left(i \hbar \partial_t|\psi, t\rangle\right)-H_0 e^{i t H_0 / \hbar}|\psi, t\rangle \quad=e^{i H_0 t / \hbar}\left(H_0+H^{\prime}(t)\right)|\psi, t\rangle-H_0|\psi, t\rangle_I=\left(e^{i t H_0 / \hbar} H^{\prime}(t) e^{-i t H_0 / \hbar}\right) \mid$$

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