Riemann surface
matlab

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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