# 物理代写|量子力学代写quantum mechanics代考|PHYSICS3544

## 物理代写|量子力学代写quantum mechanics代考|Boost from the rest frame

Now we can demonstrate how a boost transforming a particle at rest into a moving particle leads directly to the Dirac equation, which thus appears as a consequence of Lorentz symmetry. Consider a special boost from the rest frame of the particle to a moving inertial frame where the particle has a 4 -momentum $p^\mu$,
\begin{aligned} &\text { rest frame } K \quad \longrightarrow \text { moving frame } K^{\prime} \ &\text { 4-momentum }(m, \overrightarrow{0}) \longrightarrow(E, \vec{p})=\left(p^0, \vec{p}\right) \end{aligned}

$K^{\prime}$ moves relatively to $K$ into the direction opposite to the momentum, i.e. the boost axis $\vec{n}_b$ is given by
$$\vec{n}_b=-\vec{n}, \quad \vec{n}=\frac{\vec{p}}{|\vec{p}|}$$
and the rapidity is determined by
$$\cosh \phi=\frac{E}{m} .$$
The 2-dimensional representation matrices thus read as follows,
\begin{aligned} &D=e^{-\frac{\phi}{2} \vec{n} \cdot \vec{\sigma}}=\mathbb{1} \cosh \frac{\phi}{2}-(\vec{n} \cdot \vec{\sigma}) \sinh \frac{\phi}{2} \ &\bar{D}=e^{+\frac{\phi}{2} \vec{n} \cdot \vec{\sigma}}=\mathbb{1} \cosh \frac{\phi}{2}+(\vec{n} \cdot \vec{\sigma}) \sinh \frac{\phi}{2} \end{aligned}

## 物理代写|量子力学代写quantum mechanics代考|Covariance of the Dirac equation

A Lorentz transformation $\Lambda$ without inversion is represented in the 4-dimensional spinor space by the matrices
$$\Lambda \longrightarrow\left(\begin{array}{cc} D & 0 \ 0 & \bar{D} \end{array}\right)=: S(\Lambda)$$ in the chiral representation. Thus, a 4-component spinor $\psi(x)$ transforms according to
$$\psi(x) \longrightarrow \psi^{\prime}\left(x^{\prime}\right)=S(\Lambda) \psi(x) \quad \text { with } x^{\prime \mu}=\Lambda_\nu^\mu x^\nu .$$
For the following discussion we introduce a compact notation for the Pauli and Dirac matrices by defining
$$\sigma^0=\mathbb{1}, \quad \sigma^\mu=\left(\sigma^0, \vec{\sigma}\right), \quad \bar{\sigma}^\mu=\left(\sigma^0,-\vec{\sigma}\right) .$$
With this notation the $\gamma$ matrices in the chiral representation read as follows,
$$\gamma^\mu=\left(\begin{array}{cc} 0 & \sigma^\mu \ \bar{\sigma}^\mu & 0 \end{array}\right)$$
For the matrices $D$ und $\bar{D}$ the following relations hold ${ }^2$
$$D^{\dagger}=\bar{D}^{-1}, \quad \bar{D}^{\dagger}=D^{-1}$$
as can easily be seen from the expressions (2.111) and (2.112).
This implies the fundamental properties
\begin{aligned} &\bar{D}^{\dagger} \sigma^\mu \bar{D}=\Lambda_\nu^\mu \sigma^\nu \ &D^{\dagger} \bar{\sigma}^\mu D=\Lambda_\nu^\mu \bar{\sigma}^\nu \end{aligned}
In case of rotations these relations are reduced to those in Eq. (2.105).

## 物理代写|量子力学代写quantum mechanics代考|Boost from the rest frame

rest frame $K \quad \longrightarrow$ moving frame $K^{\prime} \quad$ 4-momentum $(m, \overrightarrow{0}) \longrightarrow(E, \vec{p})=\left(p^0, \vec{p}\right)$
$K^{\prime}$ 相对移动 $K$ 进入与动量相反的方向，即助推轴 $\vec{n}_b$ 是 (谁) 给的
$$\vec{n}_b=-\vec{n}, \quad \vec{n}=\frac{\vec{p}}{|\vec{p}|}$$

$$\cosh \phi=\frac{E}{m} .$$

$$D=e^{-\frac{\phi}{2} \vec{n} \cdot \vec{\sigma}}=1 \cosh \frac{\phi}{2}-(\vec{n} \cdot \vec{\sigma}) \sinh \frac{\phi}{2} \quad \bar{D}=e^{+\frac{\phi}{2} \vec{n} \cdot \vec{\sigma}}=1 \cosh \frac{\phi}{2}+(\vec{n} \cdot \vec{\sigma}) \sinh \frac{\phi}{2}$$

## 物理代写|量子力学代写quantum mechanics代考|Covariance of the Dirac equation

$$\Lambda \longrightarrow\left(\begin{array}{llll} D & 0 & 0 & \bar{D} \end{array}\right)=: S(\Lambda)$$

$$\psi(x) \longrightarrow \psi^{\prime}\left(x^{\prime}\right)=S(\Lambda) \psi(x) \quad \text { with } x^{\prime \mu}=\Lambda_\nu^\mu x^\nu .$$

$$\sigma^0=1, \quad \sigma^\mu=\left(\sigma^0, \vec{\sigma}\right), \quad \bar{\sigma}^\mu=\left(\sigma^0,-\vec{\sigma}\right) .$$

$$D^{\dagger}=\bar{D}^{-1}, \quad \bar{D}^{\dagger}=D^{-1}$$

$$\bar{D}^{\dagger} \sigma^\mu \bar{D}=\Lambda_\nu^\mu \sigma^\nu \quad D^{\dagger} \bar{\sigma}^\mu D=\Lambda_\nu^\mu \bar{\sigma}^\nu$$

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