# 物理代写|量子力学代写quantum mechanics代考|PHYS7076

## 物理代写|量子力学代写quantum mechanics代考|Static magnetic field

A static magnetic field is described by the 4-potential
$$A^0=0, \quad \vec{A}=\vec{A}(\vec{x})$$

In the non-relativistic approximation for $|\vec{p}| \ll m,|E-m| \ll m$ and with the ansatz in the Dirac representation
$$\psi=e^{-i m t}\left(\begin{array}{l} \varphi \ \chi \end{array}\right) \quad \text { where } \chi=\mathcal{O}\left(\frac{|\vec{p}|}{m}\right) \cdot \varphi,$$
an expansion in the leading terms yields a 2-component equation for $\varphi$,
$$i \partial_0 \varphi=\left[\frac{1}{2 m}(-i \nabla-e \vec{A})^2-\frac{e}{2 m} \vec{\sigma} \cdot \vec{B}\right] \varphi$$
which is recognized as the Pauli equation. The spin part within the brackets,
$$\frac{e}{2 m} \vec{\sigma} \cdot \vec{B}=\frac{e}{2 m} g \vec{S} \cdot \vec{B},$$
describes the interaction of the electron magnetic moment with the magnetic field. The $g$-factor is predicted as $g=2$; the small deviation from the experimental value, of the order $10^{-3}$, is another quantum effect of the electromagnetic field.

The derivation of the Pauli equation in the non-relativistic limit is briefly explained. Starting from Eq. (2.136), the ansatz (2.140) leads to coupled equations for the 2 -component spinors $\varphi$ and $\chi$,
\begin{aligned} &i \partial_0 \varphi=\sigma \cdot(\vec{P}-e \vec{A}) \chi, \ &i \partial_0 \chi=\sigma \cdot(\vec{P}-e \vec{A}) \varphi-2 m \chi, \end{aligned}
yielding $\chi$ as follows,
$$\chi=\frac{1}{2 m} \sigma \cdot(\vec{P}-e \vec{A}) \varphi-\frac{1}{2 m} i \partial_0 \chi \approx \frac{1}{2 m} \sigma \cdot(\vec{P}-e \vec{A}) \varphi$$
since the second term is of the order $\mathcal{O}\left(|\vec{p}|^2 / \mathrm{m}^2\right)$ and does not contribute to the expansion in the leading terms.

## 物理代写|量子力学代写quantum mechanics代考|Free Electromagnetic Field

The basis for the description of the classical electromagnetic field has already been given in Sec. 1.6. In the 4-dimensional formulation the vector potential $A^\mu(x)$ in Lorentz gauge $\partial_\mu A^\mu=0$ fulfills Maxwell’s equations in terms of an inhomogeneous wave equation
$$\square A^\mu=j^\mu$$
with the source given by the 4-current $j^\mu$. In QED the current is formed by the charged leptons and quarks.

As a first item we consider the free field with $j^\mu=0$, which is a pure radiation field. By a gauge transformation, the solutions $A^\mu$ of the free wave equation can always be cast into the form
$$\left(A^\mu\right)=(0, \vec{A}) \quad \text { with } \partial_\mu A^\mu=\nabla \cdot \vec{A}=0$$
denoted as the radiation gauge. A complete set of solutions is given by the system of plane waves,
$$\epsilon_\lambda^\mu e^{\pm i k x}$$
with the wave vector $\left(k^\mu\right)=\left(k^0, \vec{k}\right)$ complying with
$$k^0=|\vec{k}|,$$ and the transverse polarization vectors
$$\left(\epsilon_\lambda^\mu\right)=\left(0, \vec{\epsilon}\lambda\right), \quad \lambda=1,2,$$ which are orthogonal to the wave vector, $$\epsilon\lambda \cdot k=\epsilon_\lambda^\mu \cdot k_\mu=0$$

## 物理代写|量子力学代写quantum mechanics代考|Static magnetic field

$$A^0=0, \quad \vec{A}=\vec{A}(\vec{x})$$

$$\psi=e^{-i m t}(\varphi \chi) \quad \text { where } \chi=\mathcal{O}\left(\frac{|\vec{p}|}{m}\right) \cdot \varphi,$$

$$i \partial_0 \varphi=\left[\frac{1}{2 m}(-i \nabla-e \vec{A})^2-\frac{e}{2 m} \vec{\sigma} \cdot \vec{B}\right] \varphi$$

$$\frac{e}{2 m} \vec{\sigma} \cdot \vec{B}=\frac{e}{2 m} g \vec{S} \cdot \vec{B},$$

$$i \partial_0 \varphi=\sigma \cdot(\vec{P}-e \vec{A}) \chi, \quad i \partial_0 \chi=\sigma \cdot(\vec{P}-e \vec{A}) \varphi-2 m \chi,$$

$$\chi=\frac{1}{2 m} \sigma \cdot(\vec{P}-e \vec{A}) \varphi-\frac{1}{2 m} i \partial_0 \chi \approx \frac{1}{2 m} \sigma \cdot(\vec{P}-e \vec{A}) \varphi$$

## 物理代写|量子力学代写quantum mechanics代考|Free Electromagnetic Field

$$\square A^\mu=j^\mu$$

$$\left(A^\mu\right)=(0, \vec{A}) \quad \text { with } \partial_\mu A^\mu=\nabla \cdot \vec{A}=0$$

$$\epsilon_\lambda^\mu e^{\pm i k x}$$

$$k^0=|\vec{k}|,$$

$$\left(\epsilon_\lambda^\mu\right)=(0, \vec{\epsilon} \lambda), \quad \lambda=1,2,$$

$$\epsilon \lambda \cdot k=\epsilon_\lambda^\mu \cdot k_\mu=0$$

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