The symmetrized wave functions for a pair of non-interacting bosons/fermions are given by [ Psi_{alpha}left(mathbf{r}_{1}, mathbf{r}_{2} ight)=frac{1}{sqrt{2}
Question:
The symmetrized wave functions for a pair of non-interacting bosons/fermions are given by
\[
\Psi_{\alpha}\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right)=\frac{1}{\sqrt{2} V}\left(e^{i \mathbf{k}_{1} \cdot \mathbf{r}_{1}} e^{i \mathbf{k}_{2} \cdot \mathbf{r}_{2}} \pm e^{i \mathbf{k}_{1} \cdot \mathbf{r}_{2}} e^{i \mathbf{k}_{2} \cdot \mathbf{r}_{1}}\right)
\]
The probability density operator \(\hat{W}_{2}\) of the pair is then given through the matrix elements
\[
\begin{aligned}
\left\langle 1^{\prime}, 2^{\prime}\left|\hat{W}_{2}\right| 1,2\rightangle= & 2 \lambda^{6} \sum_{\alpha} \Psi_{\alpha}\left(1^{\prime}, 2^{\prime}\right) \Psi_{\alpha}^{*}(1,2) e^{-\beta E_{\alpha}} \\
= & \frac{\lambda^{6}}{V^{2}} \sum_{\alpha}\left(e^{i \mathbf{k}_{1} \cdot \mathbf{r}_{1}^{\prime}} e^{i \mathbf{k}_{2} \cdot \mathbf{r}_{2}^{\prime}} \pm e^{i \mathbf{k}_{1} \cdot \mathbf{r}_{2}^{\prime}} e^{i \mathbf{k}_{2} \cdot \mathbf{r}_{1}^{\prime}}\right) \times \\
& =\frac{\lambda^{6}}{2 V^{2}} \sum_{\mathrm{k}_{1}} \sum_{\mathrm{k}_{2}}\left[\begin{array}{l}
e^{i \mathbf{k}_{1} \cdot\left(\mathbf{r}_{1}^{\prime}-\mathbf{r}_{1}\right)} e^{i \mathbf{k}_{2} \cdot\left(\mathbf{r}_{2}^{\prime}-\mathbf{r}_{2}\right)}+e^{i \mathbf{k}_{1} \cdot\left(\mathbf{r}_{2}^{\prime}-\mathbf{r}_{2}\right)} e^{i \mathbf{k}_{2} \cdot\left(\mathbf{r}_{1}^{\prime}-\mathbf{r}_{1}\right)} \pm \\
\left.e^{i \mathbf{k}_{1} \cdot\left(\mathbf{r}_{2}^{\prime}-\mathbf{r}_{1}\right)} e^{i \mathbf{k}_{2} \cdot\left(\mathbf{r}_{1}^{\prime}-\mathbf{r}_{2}\right)} \pm e^{i \mathbf{k}_{1} \cdot\left(\mathbf{r}_{1}^{\prime}-\mathbf{r}_{2}\right)} e^{i \mathbf{k}_{2} \cdot\left(\mathbf{r}_{2}^{\prime}-\mathbf{r}_{1}\right)}\right]
\end{array}\right] \\
& =\frac{1}{2}\left[\left\langle 1^{\prime}\left|\hat{W}_{1}\right| 1\rightangle\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 2\rightangle+\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 2\rightangle\left\langle 1^{\prime}\left|\hat{W}_{1}\right| 1\rightangle \pm\right. \\
& \left.\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 1\rightangle\left\langle 1^{\prime}\left|\hat{W}_{1}\right| 2\rightangle \pm\left\langle 1^{\prime}\left|\hat{W}_{1}\right| 2\rightangle\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 1\rightangle\right] \\
= & \left\langle 1^{\prime}\left|\hat{W}_{1}\right| 1\rightangle\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 2\rightangle \pm\left\langle 2^{\prime}\left|\hat{W}_{1}\right| 1\rightangle\left\langle 1^{\prime}\left|\hat{W}_{1}\right| 2\rightangle .
\end{aligned}
\]
Comparing this with eqn. (10.6.18), we obtain the desired result.
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