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## 物理代写|量子场论代写Quantum field theory代考|Energy-momentum tensor

The field strengths allow to construct a tensor of second rank that comprises the mechanical properties energy and momentum of the electromagnetic field,
$$T^{\mu \nu}=F_\rho^\mu F^{\rho \nu}+\frac{1}{4} g^{\mu \nu}\left(F_{\rho \sigma} F^{\rho \sigma}\right) .$$
This energy-momentum tensor contains the energy density of the field,
$$T^{00}=\frac{1}{2}\left(\vec{E}^2+\vec{B}^2\right)$$
and the components of the Poynting vector $\vec{S}=\vec{E} \times \vec{B}$,
$$S^k=T^{0 k}$$
representing the momentum density. The purely spatial components $T^{k l}$, also known as Maxwell’s stress tensor, describe the momentum-current density of the electromagnetic field.

The force on a charge and current distribution $j^\mu$ is given by a generalization of Eq. (1.50) to continuous systems,
$$F^k=\int d^3 x(\rho \vec{E}+\vec{j} \times \vec{B})^k=\int d^3 x F^{k \nu} j_\nu \equiv \int d^3 x f^k(x)$$
with the force density $f^k$ that can be extended to a 4 -vector $f^\mu$ and written in a covariant way as follows,
$$f^\mu=F^{\mu \nu} j_\nu$$

## 物理代写|量子场论代写Quantum field theory代考|Relativistic quantum mechanics of spin-0 particles

The difficulty with the Hamiltonian as the square root of a differential operator can be circumvented by a twofold application of the operators on both sides of Eq. (2.3), yielding a differential equation of second order,
$$i \frac{\partial}{\partial t}\left(i \frac{\partial \phi}{\partial t}\right)=H^2 \phi \quad \Leftrightarrow \quad-\frac{\partial^2}{\partial t^2} \phi=\left(-\Delta+m^2\right) \phi .$$
It can be used to replace Eq. (2.3) as the quantum mechanical equation of motion for the wave function $\phi=\phi(t, \vec{x})$ of a particle state,
$$\left(\square+m^2\right) \phi=0$$
This differential equation, which is of second order also with respect to time, is known as the Klein-Gordon equation (although it was considered also by Schrödinger). Because of the invariant differential operator, it is a Lorentz-invariant equation when $\phi(x)$ is a scalar field under Lorentz transformations.

In the non-relativistic interpretation, $|\phi|^2=\rho$ is the probability density of the particle’s position in space with the normalization
$$\int d^3 x|\phi|^2=1$$
in accordance with the number of particles as a conserved quantity. In a relativistic version this interpretation cannot be maintained since the volume element $d^3 x$ is not invariant and thus the normalization would be dependent on the reference frame.

An analogy to the non-relativistic quantities, probability density $\rho$ and probability current $\vec{j}$, is found in terms of the 4 -current associated with the Klein-Gordon wave function $\phi$,
$$j^\mu=i\left[\phi^* \partial^\mu \phi-\left(\partial^\mu \phi^\right) \phi\right] .$$ This current is conserved, $$\partial_\mu j^\mu=0,$$ as can be easily derived from the Klein-Gordon equation and its Hermitian adjoint, \begin{aligned} &\phi^\left(\square+m^2\right) \phi-\phi\left(\square+m^2\right) \phi^=0 \ &=\partial_\mu\left(\phi^ \partial^\mu \phi-\phi \partial^\mu \phi^*\right) . \end{aligned}

# 量子场论代考

## 物理代写|量子场论代写Quantum field theory代考|Energy-momentum tensor

$$T^{\mu \nu}=F_\rho^\mu F^{\rho \nu}+\frac{1}{4} g^{\mu \nu}\left(F_{\rho \sigma} F^{\rho \sigma}\right) .$$

$$T^{00}=\frac{1}{2}\left(\vec{E}^2+\vec{B}^2\right)$$

$$S^k=T^{0 k}$$

$$F^k=\int d^3 x(\rho \vec{E}+\vec{j} \times \vec{B})^k=\int d^3 x F^{k \nu} j_\nu \equiv \int d^3 x f^k(x)$$

$$f^\mu=F^{\mu \nu} j_\nu$$

## 物理代写|量子场论代写Quantum field theory代考|Relativistic quantum mechanics of spin-0 particles

$$i \frac{\partial}{\partial t}\left(i \frac{\partial \phi}{\partial t}\right)=H^2 \phi \quad \Leftrightarrow \quad-\frac{\partial^2}{\partial t^2} \phi=\left(-\Delta+m^2\right) \phi .$$

$$\left(\square+m^2\right) \phi=0$$

$$\int d^3 x|\phi|^2=1$$

$$\partial_\mu j^\mu=0,$$

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