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# 物理代写|粒子物理代写Particle Physics代考|Lagrangian, Hamiltonian and Green functions

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## 物理代写|粒子物理代写Particle Physics代考|Lagrangian, Hamiltonian and Green functions

The Lagrangian formulation of the Dirac equation presents some subtleties, because the latter is a first order differential equation. Recall the results we obtained for the case of a complex scalar field. The Lagrangian density (7.27) depends on the fields $\phi$ and $\phi^*$ as well as their first derivatives. This is consistent with the fact that the equations of motion are second order differential equations and we must assign initial values for the fields and their first derivatives. When we vary with respect to the field $\phi$ in order to obtain the Euler-Lagrange equations, we must also take into account the variation of $\partial \phi$. For the Dirac equation, however, which is a first order equation, we can vary only with respect to $\psi$ and $\bar{\psi}$, not to their derivatives. This implies in turn that the standard way to obtain the Hamiltonian through a Legendre transformation should be reformulated. In this section we want to present the rules which will make it possible for us to use the Lagrangian and Hamiltonian formalisms, without attempting a mathematically rigorous justification.

We choose the Lagrangian density corresponding to the Dirac equation in the form
$$\mathcal{L}{\mathrm{D}}=\frac{\mathrm{i}}{2}\left(\bar{\psi} \gamma^\mu \partial\mu \psi-\partial_\mu \bar{\psi} \gamma^\mu \psi\right)-m \bar{\psi} \psi$$
This is justified by the fact that we obtain the Dirac equations for $\psi$ and $\bar{\psi}$ as a consequence of the stationarity requirement of the action $S=\int \mathcal{L}{\mathrm{D}} d^4 x$ under independent variations of $\bar{\psi}$ and $\psi$, respectively. Because of the linear dependence of $\mathcal{L}{\mathrm{D}}$ on $\psi$ or $\bar{\psi}$, the action has neither a minimum nor a maximum. Thus, the overall sign of the action can be chosen at will. Provided that the field vanishes at infinity, we can rewrite the action as
$$S=\int \mathrm{d}^4 x(\mathrm{i} \bar{\psi} \not \partial \psi-m \bar{\psi} \psi)$$

## 物理代写|粒子物理代写Particle Physics代考|The plane wave solutions

We have shown that the solutions of the Dirac equation are solutions of the KleinGordon equation $\left(\square+m^2\right) \psi=0$ as well. Consequently, a plane wave solution $\psi(x) \sim$ $\exp (-\mathrm{i} k \cdot x)$ of the Dirac equation has to satisfy the condition $k^2=k_0^2-\boldsymbol{k}^2=m^2$, which means that its energy $k_0$ can have either sign. We shall be interested in the full set of plane wave solutions of both positive and negative energies, since only their union forms a basis. We fix the zero component of the wave vector $k^\mu$ to $k_0=+\sqrt{\boldsymbol{k}^2+m^2} \equiv E_k$. Then, we denote the positive energy solution of wave vector $\boldsymbol{k}$ by
$$\psi^{(+)}(x)=\mathrm{e}^{-\mathrm{i} k \cdot x} u(\boldsymbol{k})$$
and the negative energy one by
$$\psi^{(-)}(x)=\mathrm{e}^{\mathrm{i} k \cdot x} v(\boldsymbol{k})$$
where $u$ and $v$ are four-component spinors, whose components are labelled $u_r$ and $v_r$, $r=\mathbf{1}, \ldots, 4$
From $\left(\mathrm{i} \gamma^\mu \partial_\mu-m\right) \psi^{( \pm)}(x)=0$, we obtain
$$(\not k-m) u(\boldsymbol{k})=0 \text { and }(\not k+m) v(\boldsymbol{k})=0$$
Let us choose the $\gamma$ matrices in the standard representation. For $k=0$, the equations (7.72) simplify to
$$\left(\gamma^0-1\right) u(\mathbf{0})=0 \quad \text { and }\left(\gamma^0+1\right) v(\mathbf{0})=0$$
and lead to $u_3=u_4=v_1=v_2=0$. A possible basis of the solutions is
$$\hat{u}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 1 \ 0 \ 0 \ 0 \end{array}\right), \quad \hat{u}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 1 \ 0 \ 0 \end{array}\right), \quad \hat{v}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 1 \ 0 \end{array}\right), \quad \hat{v}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 0 \ 1 \end{array}\right)$$

## 物理代写|粒子物理代写Particle Physics代考|Lagrangian, Hamiltonian and Green functions

$$\mathcal{L D}=\frac{\mathrm{i}}{2}\left(\bar{\psi} \gamma^\mu \partial \mu \psi-\partial_\mu \bar{\psi} \gamma^\mu \psi\right)-m \bar{\psi} \psi$$

$$S=\int \mathrm{d}^4 x(\mathrm{i} \bar{\psi} \partial \partial-m \bar{\psi} \psi)$$

## 物理代写|粒子物理代写Particle Physics代考|The plane wave solutions

$$\psi^{(+)}(x)=\mathrm{e}^{-\mathrm{i} k \cdot x} u(\boldsymbol{k})$$

$$\psi^{(-)}(x)=\mathrm{e}^{\mathrm{i} k \cdot x} v(\boldsymbol{k})$$

$$(k-m) u(\boldsymbol{k})=0 \text { and }(k+m) v(\boldsymbol{k})=0$$

$$\left(\gamma^0-1\right) u(\mathbf{0})=0 \quad \text { and }\left(\gamma^0+1\right) v(\mathbf{0})=0$$

$$\hat{u}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 1 \ 0 \ 0 \ 0 \end{array}\right), \quad \hat{u}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 1 \ 0 \ 0 \end{array}\right), \quad \hat{v}^{(1)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 1 \ 0 \end{array}\right), \quad \hat{v}^{(2)}(m, \mathbf{0})=\left(\begin{array}{l} 0 \ 0 \ 0 \ 1 \end{array}\right)$$

## MATLAB代写

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