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# 物理代写|粒子物理代写Particle Physics代考|C P T symmetry

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## 物理代写|粒子物理代写Particle Physics代考|C P T symmetry

The Dirac equation is by construction invariant under space inversions, or parity transformations $(\mathcal{P})$. We just showed the existence of a second symmetry transformation, that of charge conjugation $(\mathcal{C})$. We wish to investigate here the consequences of time reversal $\mathcal{T}$, with $(\mathcal{T} x)^\mu=(-t, \boldsymbol{x})$. We can easily check that $\gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$, which is an anti-unitary operation, also satisfies the Dirac equation. It is interesting to consider the product of these three symmetries, namely the product of ${ }^5$
$$\begin{gathered} \mathcal{P}: \Psi(x) \rightarrow \mathrm{i} \gamma^0 \Psi(\mathcal{P} x) \ \mathcal{C}: \Psi(x) \rightarrow \Psi^C(x)=C \gamma^0 \Psi^(x) \end{gathered}$$ and $$\mathcal{T}: \Psi(x) \rightarrow \gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$$ where $(\mathcal{P} x)^\mu=(t,-\boldsymbol{x})$ denotes space inversion. We can verify that the product $C P T$ of the three operators taken in any order is an invariance of the theory. Indeed, start with \begin{aligned} P C T \Psi(x) & =\mathrm{i} \gamma^0\left(C \gamma^0\left[\gamma^0 \gamma^5 C \gamma^0 \Psi^(\mathcal{P} \mathcal{T} x)\right]^*\right) \ & =\mathrm{i} \gamma^0\left(C \gamma^0\left(\gamma^0 \gamma^5 C^{-1} \gamma^0 \Psi(-x)\right)\right) \ & =\mathrm{i} \gamma^5 \Psi(-x) \end{aligned}
and ask the following question: Given a wave function $\Psi(x)$, which satisfies the Dirac equation in an external electromagnetic field $A_\mu(x)$, can we find the equation satisfied by its $C P T$-transformed wave function $\mathrm{i} \gamma^5 \Psi(-x)$ ? The answer is simple:
\begin{aligned} 0 & =\left[\mathrm{i} \not \partial_x-e \not A(x)-m\right] \Psi(x) \ & =\mathrm{i} \gamma^5\left[\mathrm{i} \not \partial_x-e \not A(x)-m\right] \Psi(x) \ & =\left[-\mathrm{i} \not \partial_x+e \not A(x)-m\right] \mathrm{i} \gamma^5 \Psi(x) \ & =\left[\mathrm{i} \not \partial_x+e \not A(-x)-m\right] C P T \Psi(x) \end{aligned}
where, in the last step, we have performed the change of variable $x^\mu \rightarrow-x^\mu$.

## 物理代写|粒子物理代写Particle Physics代考|Chirality

In section 7.3.6 we introduced the so-called “standard” representation, which separates the components of a Dirac spinor that vanish in the $c \rightarrow \infty$ non-relativistic limit. At the other end, at the ultra-relativistic limit where the mass is negligible, we expect, at least in the absence of any external potential, the Dirac equation to split into a pair of Weyl equations. We shall introduce a formal way to demonstrate this fact.

We define two orthogonal projectors $P_L$ and $P_R, L$ and $R$ standing for “left” and “right” respectively, by
$$P_L=\frac{1+\gamma^5}{2}, \quad P_R=\frac{1-\gamma^5}{2}$$
They are Hermitian operators and satisfy the projection, orthogonality and completeness relations: $P_{L, R}^2=P_{L, R}, P_L P_R=P_R P_L=0$ and $P_L+P_R=1$. With the help of these projectors we define the “left” and “right” components of a general Dirac spinor $\Psi(x)$ by
$$\Psi_{L, R}(x)=P_{L, R} \Psi(x), \quad \Psi(x)=\Psi_L(x)+\Psi_R(x)$$
In the Weyl basis we used in equation (7.38), in which $\gamma^5$ is diagonal, $P_L$ and $P_R$ project onto the $\xi$ and $\eta$ spinors, respectively. Using $\Psi_L$ and $\Psi_R$, the free Dirac action (7.61) becomes
$$S=\int \mathrm{d}^4 x\left[\bar{\Psi}_L \mathrm{i} \not \partial \Psi_L+\bar{\Psi}_R \mathrm{i} \not \partial \Psi_R-m\left(\bar{\Psi}_L \Psi_R+\bar{\Psi}_R \Psi_L\right)\right]$$

## 物理代写|粒子物理代写Particle Physics代考|C P T symmetry

$$\left.\mathcal{P}: \Psi(x) \rightarrow \mathrm{i} \gamma^0 \Psi(\mathcal{P} x) \mathcal{C}: \Psi(x) \rightarrow \Psi^C(x)=C \gamma^0 \Psi^{(} x\right)$$

$$\mathcal{T}: \Psi(x) \rightarrow \gamma^0 \gamma^5 \Psi^C(\mathcal{T} x)$$

$$\left.P C T \Psi(x)=\mathrm{i} \gamma^0\left(C \gamma^0\left[\gamma^0 \gamma^5 C \gamma^0 \Psi^{(\mathcal{P} T} x\right)\right]^*\right) \quad=\mathrm{i} \gamma^0\left(C \gamma^0\left(\gamma^0 \gamma^5 C^{-1} \gamma^0 \Psi(-x)\right)\right)=\mathrm{i} \gamma^5 \Psi(-x)$$

$$0=\left[\mathrm{i}, \partial_x-e A(x)-m\right] \Psi(x) \quad=\mathrm{i} \gamma^5\left[\mathrm{i} \partial \partial_x-e A(x)-m\right] \Psi(x)=\left[-\mathrm{i} \partial \partial_x+e A(x)-m\right] \mathrm{i} \gamma^5 \Psi(x) \quad=\left[\mathrm{i}, \partial_x+e A(-x)-m\right] C P T \Psi(x)$$

## 物理代写|粒子物理代写Particle Physics代考|Chirality

$$P_L=\frac{1+\gamma^5}{2}, \quad P_R=\frac{1-\gamma^5}{2}$$

$$\Psi_{L, R}(x)=P_{L, R} \Psi(x), \quad \Psi(x)=\Psi_L(x)+\Psi_R(x)$$

$$S=\int \mathrm{d}^4 x\left[\bar{\Psi}_L \mathrm{i} \not \partial \Psi_L+\bar{\Psi}_R \mathrm{i} \not \partial \Psi_R-m\left(\bar{\Psi}_L \Psi_R+\bar{\Psi}_R \Psi_L\right)\right]$$

## MATLAB代写

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