Reflection operator for spin 1/2 particle

Question

How do I construct a matrix rep. of a reflection operator for spin 1/2 particle.

Reflection operator ($\hat R_z$ : reflection about x-y plane i.e, along z-direction) acts in following way:

$U^{\dagger}S_z U\rightarrow -S_z,\\ U^{\dagger}S_x U\rightarrow S_x,\\ U^{\dagger}S_y U\rightarrow S_y.$

I started with assumption: $U|\uparrow\rangle=c_1|\downarrow\rangle, U|\downarrow\rangle=c_2|\uparrow\rangle$, and tried to find $c_1,c_2$ but no solution satisfy all three constraints!

Answer 1

Careful, spin is a pseudo vector. Therefore the the reflection operator rather acts like ($P$ like parity, $R$ usually represents rotation operators): $$ \begin{align} P_z&: & S_x&\to-S_x & S_y&\to-S_y & S_z&\to S_z \end{align} $$ This can be implemented with a $\pi$ rotation about the $z$ axis, ie: $$ U=e^{i\pi S_z}=2S_z $$

In general, the spin $1/2$ representation is entirely determined by the spin operators that must satisfy the commutation relation ($\hbar=1$): $$ [S_i,S_j]=i\epsilon_{ijk}S_k $$ In fact, an automorphism can always be uniquely represented (up to a phase) by a rotation.

Note that in your case, the commutations switch signs, so such a $U$ is impossible to find. You were actually looking for the representation of reflection followed by time reversal $P_z T$. $T$ is defined by switching the sign of spin: $$ \begin{align} T&: & S_x&\to-S_x & S_y&\to-S_y & S_z&\to -S_z \end{align} $$ The latter has to be represented as an antiunitary transformation (conserves inner product, but antilinear). This explains why the commutation relations switch signs as well.

Hope this helps.