Exchange operator

The exchange operator $\hat{P}_{ik}$ acts on a function resulting in the swapping of labels of any two identical particles, i.e.

$\hat{P}_{ik}\psi(q_1,\cdots,q_i,\cdots,q_k,\cdots,q_n)=\psi(q_1,\cdots,q_k,\cdots,q_i,\cdots,q_n)$

where the label $q_n$ refers to the $n$ -th particle.

If we apply $\hat{P}_{ik}$ twice on the function,

$\hat{P}_{ik}^{\; \; 2}f(\cdots,q_i,\cdots,q_k,\cdots)=\hat{P}_{ik}f(\cdots,q_k,\cdots,q_i,\cdots)=f(\cdots,q_i,\cdots,q_k,\cdots)$

Therefore, $\hat{P}_{ik}^{\; \; 2}=\hat{I}$ , where $\hat{I}$ is the identity operator.

Question

Show that the eigenvalues of $\hat{P}_{ik}$ are $\lambda=\pm1$ .

Answer

If $\psi$ is an eigenfunction of $\hat{P}_{ik}$ , then $\hat{P}_{ik}^{\; \; 2}\psi=\lambda\hat{P}_{ik}\psi=\lambda^{2}\psi$ . Since $\hat{P}_{ik}^{\; \; 2}=\hat{I}$ , we have $\psi=\lambda^{2}\psi$ . As an eigenfunction must be non-zero, $\lambda=\pm1$ .

Experiment data reveals that the wavefunction of a system of two identical fermions (particles with spin $s=\frac{1}{2},\frac{3}{2},\frac{5}{2},\cdots$ ) is antisymmetric with respect to label exchange (i.e. the eigenvalue is -1 when the exchange operator acts on the wavefunction), while the wavefunction of a system of two identical bosons ( $s=0,1,2,\cdots$ ) is symmetric with respect to label exchange (eigenvalue of +1). The antisymmetric property of fermion wavefunctions and the symmetric property of boson wavefunctions can be regarded as postulates of quantum mechanics.

For identical fermions, $\hat{P}_{ik}\psi(\cdots,q_i,\cdots,q_k,\cdots)=-\psi(\cdots,q_k,\cdots,q_i,\cdots)$ . Since the way we label identical particles cannot affect the state of the system, the commutation relation between $\hat{P}_{ik}$ and the Hamiltonian is

$\left [\hat{P}_{ik},\hat{H}\right ]\psi=\hat{P}_{ik}\hat{H}\psi-\hat{H}\hat{P}_{ik}\psi=E\hat{P}_{ik}\psi+\hat{H}\psi=-E\psi+E\psi=0$

This implies that, for a system of identical fermions, we can select a common complete set of eigenfunctions for $\hat{P}_{ik}$ and $\hat{H}$ , with the eigenfunctions being antisymmetric under label exchange.

Question

Show that $\hat{P}_{ik}$ commutes with $\hat{{S}^{2}}^{(T)}$ , $\hat{S}_z^{\; (T)}$ , $\hat{S}_1^{\; 2}$ , $\hat{S}_2^{\; 2}$ , $\hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2$ , $\hat{{L}^{2}}^{(T)}$ , $\hat{L}_z^{\; (T)}$ , $\hat{L}_{1z}$ , $\hat{S}_{1z}$ , $\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}$ , $\hat{J}^{2}$ and $\hat{J}_z$ .

Answer

$\left [ \hat{P}_{12},\hat{S}_1^{\; 2} \right ]\psi=s_1(s_1+1)\hbar^{2}\hat{P}_{12}\psi+\hat{S}_1^{\;2}\psi=-s_1(s_1+1)\hbar^{2}\psi+s_1(s_1+1)\hbar^{2}\psi=0$

Similarly, $\left [ \hat{P}_{12},\hat{S}_2^{\; 2} \right ]=0$ . For $\hat{{S}^{2}}^{(T)}$ and $\hat{S}_z^{\;(T)}$ , we have $\hat{{S}^{2}}^{(T)}\psi=S(S+1)\hbar^{2}\psi$ and $\hat{S}_z^{\;(T)}\psi=m_S\hbar\psi$ respectively. So $\left [\hat{P}_{12} ,\hat{{S}^{2}}^{(T)}\right ]=0$ and $\left [\hat{P}_{12} ,\hat{S}_z^{\;(T)}\right ]=0$ . Since $\hat{{S}^{2}}^{(T)}=\hat{S}_1^{\;2}+2\hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2+\hat{S}_2^{\;2}$ , we have $\left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}_1\cdot\hat{\boldsymbol{\mathit{S}}}_2\right ]=0$ . Similarly, $\hat{P}_{12}$ commutes with $\hat{{L}^{2}}^{(T)}$ , $\hat{L}_z^{\;(T)}$ , $\hat{L}_{1z}$ and $\hat{S}_{1z}$ .

Next, $\left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=\left [\hat{P}_{12},\left ( \hat{\boldsymbol{\mathit{S}}}_1+\hat{\boldsymbol{\mathit{S}}}_2\right )\cdot\left ( \hat{\boldsymbol{\mathit{L}}}_1+\hat{\boldsymbol{\mathit{L}}}_2\right )\right ]$ . Expanding this equation using eq76 and eq179, and noting that $\left [\hat{P}_{12},\hat{L}_{mk}\right ]=0$ and $\left [\hat{P}_{12},\hat{S}_{mk}\right ]=0$ , where $m=1,2$ and $k=x,y,z$ , we have $\left [\hat{P}_{12},\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=0$ . It follows that $\left [\hat{P}_{12},\hat{J}^{2}\right ]=\left [\hat{P}_{12},\hat{{L}^{2}}^{(T)}\right ]+\left [\hat{P}_{12},\hat{{S}^{2}}^{(T)}\right ]+\left [\hat{P}_{12},2\hat{\boldsymbol{\mathit{S}}}^{(T)}\cdot\hat{\boldsymbol{\mathit{L}}}^{(T)}\right ]=0$ and that $\left [\hat{P}_{12},\hat{J}_z\right ]=\left [\hat{P}_{12},\hat{L}_z^{\; (T)}\right ]+\left [\hat{P}_{12},\hat{S}_z^{\; (T)}\right ]=0$ .