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## 物理代写|量子场论代写Quantum field theory代考|Quantum field theoretical formulation

The states of both particles and antiparticles are combined in a common Hilbert space. The eigenstates for momentum and energy of a particle or antiparticle with the eigenvalues
$$\vec{p} \text { and } p^0=\sqrt{\vec{p}^2+m^2}$$
are described by ket vectors in the conventional quantum mechanical language,
Furthermore, there is a zero-particle state,
$$\text { vacuum }|0\rangle$$
with the normalization
$$\langle 0 \mid 0\rangle=1 .$$
The momentum eigenstates as part of a continuous spectrum can only be normalized in terms of a $\delta$-function,
\begin{aligned} &\left\langle\pm, p \mid \pm, p^{\prime}\right\rangle=2 p^0 \delta^3\left(\vec{p}-\vec{p}^{\prime}\right) \ &\left\langle\pm, p \mid \mp, p^{\prime}\right\rangle=0 \end{aligned}
This differs from the standard normalization in quantum mechanics by a factor $2 p^0$. The choice of $\mathrm{Eq}$. (2.14) has the advantage of being invariant under Lorentz transformations,
$$\int \frac{d^3 p}{2 p^0}\left\langle\pm, p \mid \pm, p^{\prime}\right\rangle=1$$
with the invariant volume element $d^3 p /\left(2 p^0\right)$. The last-mentioned property follows from the invariance of the 4-dimensional volume element $d^4 p$ according to
$$\begin{array}{r} \int d^4 p \delta\left(p^2-m^2\right) \theta\left(p^0\right)=\int d^3 p \underbrace{\int p^0 \delta\left(p^{02}-\vec{p}^2-m^2\right) \theta\left(p^0\right)}_{=\frac{1}{2 p^0} \quad \text { with } p^0=\sqrt{\vec{p}^2+m^2}} . \end{array}$$

## 物理代写|量子场论代写Quantum field theory代考|Current and charge

With the field operator $\phi$ the current (2.7) becomes a current operator
$$j^\mu=i\left[\phi^{\dagger} \partial^\mu \phi-\left(\partial^\mu \phi^{\dagger}\right) \phi\right]$$
with a conserved charge following from the continuity equation (2.8),
$$Q=\int d^3 x j^0 .$$
By insertion of the Fourier expansion for $\phi$ in $j^0$ one can express the charge operator in terms of the number operators $N_{+}(p)$ and $N_{-}(p)$ for particles and antiparticles as follows,
$$Q=\int \frac{d^3 p}{2 p^0}\left[a^{\dagger}(p) a(p)-b^{\dagger}(p) b(p)\right] \equiv \int \frac{d^3 p}{2 p^0}\left[N_{+}(p)-N_{-}(p)\right] .$$
One can easily verify by means of Eq. (2.17) that the one-particle states are eigenstates of $Q$ with eigenvalues $\pm 1$ :
$$Q|+, p\rangle=+|+, p\rangle, \quad Q|-, p\rangle=-|-, p\rangle .$$
A special case is the real (self-adjoint) field with $\phi(x)=\phi^{\dagger}(x)$, which describes neutral particles of charge zero where particles and antiparticles are identical.

# 量子场论代考

## 物理代写|量子场论代写Quantum field theory代考|Quantum field theoretical formulation

$$\vec{p} \text { and } p^0=\sqrt{\vec{p}^2+m^2}$$

vacuum $|0\rangle$

$$\langle 0 \mid 0\rangle=1 .$$

$$\left\langle\pm, p \mid \pm, p^{\prime}\right\rangle=2 p^0 \delta^3\left(\vec{p}-\vec{p}^{\prime}\right) \quad\left\langle\pm, p \mid \mp, p^{\prime}\right\rangle=0$$

$$\int \frac{d^3 p}{2 p^0}\left\langle\pm, p \mid \pm, p^{\prime}\right\rangle=1$$

$$\int d^4 p \delta\left(p^2-m^2\right) \theta\left(p^0\right)=\int d^3 p \underbrace{\int p^0 \delta\left(p^{02}-\vec{p}^2-m^2\right) \theta\left(p^0\right)}_{=\frac{1}{2^0} \quad \text { with } p^0=\sqrt{p^2+m^2}} .$$

## 物理代写|量子场论代写Quantum field theory代考|Current and charge

$$j^\mu=i\left[\phi^{\dagger} \partial^\mu \phi-\left(\partial^\mu \phi^{\dagger}\right) \phi\right]$$

$$Q=\int d^3 x j^0 .$$

$$Q=\int \frac{d^3 p}{2 p^0}\left[a^{\dagger}(p) a(p)-b^{\dagger}(p) b(p)\right] \equiv \int \frac{d^3 p}{2 p^0}\left[N_{+}(p)-N_{-}(p)\right] .$$

$$Q|+, p\rangle=+|+, p\rangle, \quad Q|-, p\rangle=-|-, p\rangle .$$

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