Question

2. Using the defi tions of the spin operators, sy and s, found in P10 ? show the following relationship (Hint Operate both sides ofthe equation on a spin wavefunction ?.)

P10.6 :

Please explain/clarify variable meanings too!

Answer 1

Expand the commutator on left hand side of the equation.

$[\hat{S_{x}},\hat{S_{y}}] = \hat{S_{x}}\hat{S_{y}} - \hat{S_{y}}\hat{S_{x}}$

State $\alpha$ is the spin up eigen state of $\hat{S_{z}}$ , the z component of the Spin operator. $\hat{S_{x}}$ and $\hat{S_{y}}$ are the x and y components respectively of the spin operators, all of which acts on the spin up and spin down states $\left ( \beta \right )$ as given in P10.6 except for $\hat{S_{y}}$ acting on $\beta$ . The correct expression is given below. Note this correction.

$\hat{S_{y}}\beta = - i$\hbar$\alpha /2$  

Act the commutator on the state $\alpha$

$[\hat{S}_{x},\hat{S}_{y}] \alpha = \hat{S}_{x}(\hat{S}_{y}\alpha) - \hat{S}_{y}(\hat{S}_{x}\alpha)$

$\Rightarrow [\hat{S}_{x},\hat{S}_{y}] \alpha =\hat{S}_{x}(i$\hbar$\beta)/2 - \hat{S}_{y}($\hbar$\beta)/2$

$\Rightarrow [\hat{S}_{x},\hat{S}_{y}] \alpha =i$\hbar$(\hat{S}_{x}\beta)/2 - $\hbar$ (\hat{S}_{y}\beta)/2$

$\Rightarrow [\hat{S}_{x},\hat{S}_{y}] \alpha =i$\hbar$($\hbar$ \alpha/2)/2 - $\hbar$(-i$\hbar$\alpha/2)/2$

$\Rightarrow [\hat{S}_{x},\hat{S}_{y}] \alpha =i$\hbar$^{2} \alpha/4 + i$\hbar$^{2}\alpha/4 = i$\hbar$^{2}\alpha/2$

Now note the action of $\hat{S_{z}}$ on $\alpha$

$$\hbar$\alpha /2 = \hat{S_{z}}\alpha$

Substitute this result in the evaluated answer for the commutator and the desired result is obtained

$[\hat{S}_{x},\hat{S}_{y}] \alpha = i$\hbar$\hat{S}_{z}$