Riemann surface
matlab

Problems. Solution 1.1. Recall
$$\left{\alpha^j, \alpha^k\right}_{a b}=2 \delta^{j k} \delta_{a b}, \quad\left{\alpha^j, \beta\right}_{a b}=0 \quad\left(\beta^2\right){a b}=\delta{a b}$$
Suppose $\beta|\psi\rangle=\lambda|\psi\rangle$
$$\beta^2|\psi\rangle=\lambda^2|\psi\rangle=1|\psi\rangle=|\psi\rangle \lambda^2=1 \text { or } \lambda= \pm 1$$
Use hint. $\operatorname{Tr} \alpha_1^2 \beta=-\operatorname{Tr} \alpha_1 \beta \alpha_1=-\operatorname{Tr} \alpha_1^2 \beta \Longrightarrow \operatorname{Tr} \alpha_1^2 \beta=0$
$$\left{\alpha^1, \alpha^1\right}_{a b}=2\left(\alpha^1\right){a b}^2=2 \delta{a b}$$
so $\left(\alpha^1\right)^2=1 \Longrightarrow \operatorname{Tr} \beta=0$
Since $\operatorname{Tr} \beta=0$ and $\lambda_\beta= \pm 1$ and $\operatorname{Tr} \beta=\sum \lambda_\beta$, then $\beta$ must be even-dimensional for a set of \pm 1 ‘s to sum up and cancel each other.
Likewise,
$$\begin{gathered} \alpha_i|\psi\rangle=\lambda|\psi\rangle \ \alpha_i^2|\psi\rangle=\lambda \alpha_i|\psi\rangle=\lambda^2|\psi\rangle=|\psi\rangle \quad \Longrightarrow \lambda= \pm 1 \ \operatorname{Tr} \alpha_1^2 \alpha_j=-\operatorname{Tr} \alpha_1 \alpha_j \alpha_1=-\operatorname{Tr} \alpha_1^2 \alpha_j \Longrightarrow \operatorname{Tr} \alpha_1^2 \alpha_j=0 \text { so } \operatorname{Tr}\left(\alpha_j\right)=0 \end{gathered}$$
So then $\alpha^j$ even dimensional as well.

$$\begin{gathered} \int d^3 x V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=\sum{j=1}^n( \pm 1)^{j-1} \int d^3 x V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle= \ =\sum{j=1}^n( \pm 1)^{j-1} V_{\mathbf{x}j-\mathbf{y}} a{\mathbf{x}j}^{\dagger} a{\mathbf{y}}^{\dagger} a_{\mathbf{y}} \prod_{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle \end{gathered}$$ Now $$a{\mathbf{y}} \prod_{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle=\sum{k \neq j}^n \delta_{\mathbf{x}k-\mathbf{y}}(-1)^m \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle$$ So then $$\begin{gathered} \int d^3 x d^3 y V{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=\int d^3 y \sum{j=1}^n( \pm 1)^{j-1} V_{\mathbf{x}j}-\mathbf{y}{\mathbf{x}j}^{\dagger} a{\mathbf{y}}^{\dagger} \sum_{k \neq j}^n \delta_{\mathbf{x}k-y}(-1)^m \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle= \ =\sum{j=1}^n \sum_{k \neq j}^n V_{\mathbf{x}j-\mathbf{x}_k}( \pm 1)^{j-1+m} a{\mathbf{x}j}^{\dagger} a{\mathbf{x}k}^{\dagger} \prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}|0\rangle \end{gathered}$$ To bring $a{\mathbf{x}k}^{\dagger}$ back in its ordered place with $\prod{l \neq j, k}^n a_{\mathbf{x}l}^{\dagger}$, then $m$ (anti)commutations must occur (we really are repeating what we did before, exactly, again). So there’s another factor of $(-1)^m$. Likewise for $a{\mathbf{x}j}^{\dagger}$, to be back in its ordered place, another factor of $(-1)^{j-1}$ results. $$\Longrightarrow \sum{j=1}^n \sum_{k \neq j}^n V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle$$

Now $\sum_{j=1}^n \sum_{k \neq j}^n$ counts each unique pair twice so
$$\begin{gathered} \sum_{j=1}^n \sum_{k \neq j}^n V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle=2 \sum{j=1}^n \sum_{k=1}^{j-1} V\left(\mathbf{x}j-\mathbf{x}_k\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle \ \Longrightarrow \frac{1}{2} \int d^3 x d^3 y V(\mathbf{x}-\mathbf{y}) a{\mathbf{x}{\mathbf{y}}^{\dagger}}^{\dagger} a{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}} \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle= \ =\sum{j=1}^n \sum_{k=1}^{j-1} V\left(\mathbf{x}j-\mathbf{x}_k\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle \ \Longrightarrow i \hbar \frac{\partial}{\partial t}|\psi, t\rangle=H|\psi, t\rangle \text { is satisfied. } \end{gathered}$$

Also recall the commutation and anticommutation rules:
\begin{aligned} {\left[a(\mathbf{x}), a\left(\mathbf{x}^{\prime}\right)\right] } & =0 \ {\left[a^{\dagger}(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right] } & =0 \ {\left[a(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right] } & =\delta^3(\mathbf{x}-\mathbf{x}) \end{aligned}
\begin{aligned} \left{a(\mathbf{x}), a\left(\mathbf{x}^{\prime}\right)\right} & =0 \ \left{a^{\dagger}(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right} & =0 \ \left{a(\mathbf{x}), a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right} & =\delta^3\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \end{aligned}
Using some shorthand notation,
$$H|\psi, t\rangle=\left(\int d^3 x a_{\mathbf{x}}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla^2+U_{\mathbf{x}}\right) a_{\mathbf{x}}+\frac{1}{2} \int d^3 x d^3 y V_{\mathbf{x}-\mathbf{y}} a_{\mathbf{x}}^{\dagger} a_{\mathbf{y}}^{\dagger} a_{\mathbf{y}} a_{\mathbf{x}}\right) \cdot \int d^3 x_1 \ldots d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger} \mid 0$$ Now $$a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}=\left(( \pm 1) a{\mathbf{x}1}^{\dagger} a{\mathbf{x}}+\delta_{\mathbf{x}-\mathbf{x}1}\right) a{\mathbf{x}2}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}=( \pm 1) a{\mathbf{x}1}^{\dagger}\left(( \pm 1) a{\mathbf{x}2}^{\dagger} a{\mathbf{x}}+\delta_{\mathbf{x}-\mathbf{x}2}\right) a{\mathbf{x}3}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}+\delta{\mathbf{x}-\mathbf{x}1} a{\mathbf{x}2}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}$$ If this term is applied to $|0\rangle$, note $a_x|0\rangle=0$. $$a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger} \ldots a{\mathbf{x}n}^{\dagger}|0\rangle=\sum{j=1}^n( \pm 1)^{j-1} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle$$ For the kinetic and self-potential terms of $H|\psi, t\rangle$, $$\int d^3 x a{\mathbf{x}}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla^2+U_{\mathbf{x}}\right) \int d^3 x_1 \ldots d^3 x_n \psi\left(\mathbf{x}1, \ldots, \mathbf{x}_n, t\right) \sum{j=1}^n( \pm 1)^{j-1} \delta_{\mathbf{x}-\mathbf{x}j} \prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}|0\rangle$$ Evaluate $\int d^3 x$, in particular, with $\delta{\mathbf{x}-\mathbf{x}j}$ $$\sum{j=1}^n a_{\mathbf{x}j}^{\dagger}\left(\frac{-\hbar^2}{2 m} \nabla_j^2+U{\mathbf{x}j}\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots, \mathbf{x}_n, t\right)( \pm 1)^{j-1} \prod{k \neq j}^n a_{x_k}^{\dagger}|0\rangle$$
To (anti)commmute $a_{\mathbf{x}j}^{\dagger}$ over to its ordered place in $\prod{k \neq j}^n a_{\mathbf{x}k}^{\dagger}( \pm 1)^{j-1}, j-1$ (anti)commutations must take place. $$\Longrightarrow \sum{j=1}^n\left(-\frac{\hbar^2}{2 m} \nabla_j^2+U_{\mathbf{x}j}\right) \int d^3 x_1, \ldots, d^3 x_n \psi\left(\mathbf{x}_1, \ldots, \mathbf{x}_n, t\right) a{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}n}^{\dagger}|0\rangle$$ Now, on $a{\mathbf{x}} a_{\mathbf{x}1}^{\dagger}, \ldots, a{\mathbf{x}_n}^{\dagger}|0\rangle$.

Problem 2.2. Verify that eq. (2.14) follows from $U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)$.

Solution 2.2. $\Lambda^{-1} \Lambda^{\prime} \Lambda=\Lambda^{-1}\left(1+\delta \omega^{\prime}\right) \Lambda=1+\Lambda^{-1} \delta \omega^{\prime} \Lambda$
$$\begin{gathered} U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)=1+\frac{i}{2}\left(\Lambda^{-1} \delta \omega^{\prime} \Lambda\right){\rho \sigma} M^{\rho \sigma}=1+\frac{i}{2}\left(\Lambda^{-1}\right)\rho^a\left(\delta \omega^{\prime}\right){a b}(\Lambda)\sigma^b M^{\rho \sigma} \ U\left(\Lambda^{\prime}\right)=U\left(1+\delta \omega^{\prime}\right)=1+\frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu} \ U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U(\Lambda)^{-1} U(\Lambda)+\frac{i}{2} U(\Lambda)^{-1} \delta \omega_{\mu \nu} M^{\mu \nu} U(\Lambda) \end{gathered}$$
Using $U(\Lambda)^{-1} U\left(\Lambda^{\prime}\right) U(\Lambda)=U\left(\Lambda^{-1} \Lambda^{\prime} \Lambda\right)$,
$$\Longrightarrow U\left(\Lambda^{-1}\right) \delta \omega_{\mu \nu} M^{\mu \nu} U(\Lambda)=\delta \omega_{\mu \nu} U\left(\Lambda^{-1}\right) M^{\mu \nu} U(\Lambda)=\delta \omega_{\mu \nu}\left(\Lambda^{-1}\right)\rho^\mu(\Lambda)\sigma^\nu M^{\rho \sigma}$$
Note that $\delta \omega_{\mu \nu}$ is just an entry in a rank-2 tensor, a number, and for arbitrary Lorentz transformation, each a uniquelly determined quantity $\delta \omega_{\mu \nu}$. (6 of them for antisymmetric $\delta \omega_{\mu \nu}$ ).
$$\Longrightarrow U\left(\Lambda^{-1}\right) M^{\mu \nu} U(\Lambda)=\left(\Lambda^{-1}\right)\rho^\mu(\Lambda)\sigma^\nu M^{\rho \sigma}$$

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