Riemann surface
matlab
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量子场的正则形式：研究量子场的正则量子化方法，如场算符、场正则对易关系等。
相对论性量子场论：介绍相对论性量子场论的基本原理，如洛仑兹不变性、自由场等。
量子电动力学：研究电磁场与带电粒子的相互作用，如量子电动力学的基本框架和计算方法。
量子色动力学：探索强相互作用的量子场论，如夸克和胶子的相互作用和强子的产生与衰变。
量子场论中的计算方法：介绍量子场论中的计算技巧和近似方法，如费曼图、重整化等。
量子场论在凝聚态物理中的应用：探索量子场论在凝聚态物理领域中的应用，如拓扑相、量子液体等。
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Show that the superposition of plane waves
$$
\psi_\sigma(\vec{x}, t)=\int \frac{d^3 k}{(2 \pi)^{\frac{3}{2}}} e^{i \vec{k} \cdot \vec{x}-i\left(\frac{h^2 \vec{k}^2}{2 m}\right) t / \hbar} a_\sigma(\vec{k})
$$
is a solution of the “free wave equation” which is obtained by setting $\lambda=0$ in equation (1.1).
Solution:
We operate the wave operator, $\left(i \hbar \frac{\partial}{\partial t}+\frac{\hbar^2}{2 m} \vec{\nabla}^2\right)$, on the proposed solution (1.6) to find
$$
\begin{aligned}
& \left(i \hbar \frac{\partial}{\partial t}+\frac{\hbar^2}{2 m} \vec{\nabla}^2\right) \psi_\sigma(\vec{x}, t) \
& =\left(i \hbar \frac{\partial}{\partial t}+\frac{\hbar^2}{2 m} \vec{\nabla}^2\right) \int \frac{d^3 k}{(2 \pi)^{\frac{3}{2}}} e^{i \vec{k} \cdot \vec{x}-i\left(\frac{\hbar^2 \vec{k}^2}{2 m}\right) t / \hbar} a_\sigma(\vec{k}) \
& =\int \frac{d^3 k}{(2 \pi)^{\frac{3}{2}}}\left(i \hbar \frac{\partial}{\partial t}+\frac{\hbar^2}{2 m} \vec{\nabla}^2\right) e^{i \vec{k} \cdot \vec{x}-i\left(\frac{h^2 \vec{k}^2}{2 m}\right) t / \hbar} a_\sigma(\vec{k}) \
& =\int \frac{d^3 k}{(2 \pi)^{\frac{3}{2}}}\left(\frac{\hbar^2 \vec{k}^2}{2 m}-\frac{\hbar^2}{2 m} \vec{k}^2\right) e^{i \vec{k} \cdot \vec{x}-i\left(\frac{\hbar^2 \vec{k}^2}{2 m}\right) t / \hbar} a_\sigma(\vec{k})=0
\end{aligned}
$$
where we have assumed that the integral is sufficiently convergent that we can interchange the order of integration and differentiation. There is always the possibility of taking the matrix of this operator in a state such that $<\vartheta\left|a_\sigma(\vec{k})\right| \vartheta^{\prime}>$ is a square integrable function of $\vec{k}$. Then the interchange of differentiation and integration is valid. We have also used the identity
$$
\frac{\partial}{\partial k_a} \exp (\chi(\vec{k}))=\frac{\partial \chi(\vec{k})}{\partial k_a} \exp (\chi(\vec{k}))
$$
Show that, in the $\lambda=0$ non-interacting case, the multi-particle state which we found in equation (1.13) is an eigenstate of the Hamiltonian
$$
\begin{aligned}
& H a^{\dagger \sigma_1}\left(\vec{k}1\right) a^{\dagger \sigma_2}\left(\vec{k}_2\right) a^{\dagger \sigma_3}\left(\vec{k}_3\right) \ldots a^{\dagger \sigma_N}\left(\vec{k}_N\right) \mid 0> \ & =\left[\sum{i=1}^N \frac{\hbar^2 \vec{k}_i^2}{2 m} N\right] a^{\dagger \sigma_1}\left(\vec{k}_1\right) a^{\dagger \sigma_2}\left(\vec{k}_2\right) a^{\dagger \sigma_3}\left(\vec{k}_3\right) \ldots a^{\dagger \sigma_N}\left(\vec{k}_N\right) \mid 0>
\end{aligned}
$$
Show that, since $\mid 0>$ has the smallest possible eigenvalue of the hermitian operator $\mathcal{N}$, operating on it with the lowering operators for particle number should give the zero vector, in other words,
$$
\psi_\sigma(\vec{x}, t) \mid 0>=0 . \quad \forall \vec{x}, t
$$
and, as a consequence,
$$
\mathcal{N}|0>=0, H| 0>=0, \vec{P}|0>=0, \vec{L}| 0>=0, \vec{\Sigma} \mid 0>=0
$$
Solution:
Assume that $\psi_\sigma(\vec{x}, t) \mid 0>\neq 0$, that is, it is not the zero vector in the Hilbert space. Then, if $\mathcal{N}\left|0>=N_{\text {min }}\right| 0>$, by the above algebra we have $\mathcal{N}\left[\psi_\sigma(\vec{x}, t) \mid 0>\right]=\left(N_{\min }-1\right)\left[\psi_\sigma(\vec{x}, t) \mid 0>\right]$ which is a contradiction to the assumption that $\mid 0>$ is the eigenstate corresponding to the smallest eigenvalue of $\mathcal{N}$. The only other possibility is that $\psi_\sigma(\vec{x}, t) \mid 0>=0$. Then all operators which have a $\psi$ on the right annihilate the state $\mid 0>$.
The empty vacuum thus has no particles, no energy, no momentum, no angular momentum and no spin. This should not be a surprise to us.
The states with one particle are created from the empty vacuum by operating the raising operator , $\psi^{\dagger \sigma}(\vec{x}, t) \mid 0>$
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