1
$\begingroup$

$$\begin{align*}S_z &= \frac{\hbar}{2} \left(\left|+\right>\left<+\right| - \left|-\right>\left<-\right|\right)\\ S_y &= i\frac{\hbar}{2} \left(\left|-\right>\left<+\right| - \left|+\right>\left<-\right|\right)\\ S_x &= \frac{\hbar}{2} \left(\left|+\right>\left<-\right| + \left|-\right>\left<+\right|\right) \end{align*} $$

I keep reading that this is the way the spin operators for $\frac{1}{2}$-spin systems are 'defined'. I suspect the reason to define this way is very obvious and that's why explanations behind the definitions are omitted, but I can't figure it out.

What are the reasons for the order of the $\pm$ kets/bras, and what determines whether you add them ($S_x$) or subtract ($S_z$, $S_y$)?

$\endgroup$
4
$\begingroup$

The way you've written them, those are the spin operators in the $\hat{S}_z$ eigenbasis for a spin-1/2 particle. The two $\hat{S}_z$ eigenstates are spin up (written as $|+\rangle$ or $\uparrow$) and spin down ($|-\rangle$ or $\downarrow$), which can be written as $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$ in the $\hat{S}_z$ eigenbasis. When constructing the operators, you want to end up with $$ \hat{S}_z |+\rangle = \frac{\hbar}{2} |+\rangle, \quad \hat{S}_z |-\rangle = -\frac{\hbar}{2} |-\rangle. $$ Since $\hat{S}_z$ is diagonalized in its eigenbasis with eigenvalues of $\pm {\hbar}/{2}$, we know its form is $$ \hat{S}_z = \frac{\hbar}{2} \pmatrix{1&0\\0&-1} $$ which can also be written as $$ \hat{S}_z = \frac{\hbar}{2}\big(|+\rangle \langle +|- |-\rangle \langle -| \big) $$ You can get the other two operators by rotating the $\hat{S}_z$ operator to the $x$ or $y$ axes, or by constructing the eigenvectors of $\hat{S}_x$ and $\hat{S}_y$ in the $\hat{S}_z$ basis, from the commutation relations $[\hat{S}_i,\hat{S}_j] = i \varepsilon_{ijk}\hat{S}_k$, or from the ladder operators $\hat{S}_\pm = \hat{S}_x \pm i\hat{S}_y$. These operators shift the eigenvalue and eigenket by one unit of $\hbar$, $$ \hat{S}_\pm |\mp\rangle = \hbar \,|\pm\rangle, \quad \hat{S}_\pm |\pm\rangle = 0. $$ This yields the form of the ladder operators $$ \hat{S}_+ = \hbar \pmatrix{0&1\\0&0},\quad \hat{S}_- = \hbar \pmatrix{0&0\\1&0}. $$ Using the expression for $\hat{S}_\pm$ in terms of $\hat{S}_x$ and $\hat{S}_y$, you end up with the forms in your post. You can also check that they follow the commutation relations (which they do).

$\endgroup$
  • 1
    $\begingroup$ can you expand on transition between last two equations? $\endgroup$ – aaaaa says reinstate Monica Mar 10 '15 at 2:09
  • 3
    $\begingroup$ $\langle +|$ is the transpose of $|+\rangle$, so when you multiply $|+ \rangle \langle +|$, it's the same as multiplying $\pmatrix{1\\0} \pmatrix{1&0}$, which yields the 2-by-2 matrix $\pmatrix{1&0\\0&0}$. Similarly, when you multiply $| -\rangle \langle -|$ you get $\pmatrix{0&0\\0&1}$. Subtracting the two yields the $S_z$ operator. $\endgroup$ – Akano Mar 10 '15 at 2:19
  • $\begingroup$ Why do you want to end up with those values when you operate $\hat{S}_z$ on $\left|\pm\right>$? Does $\hat{S}_z \left|+\right> = \frac{\hbar}{2} \left|+\right>$, for example, just mean that when you measure number of times the $\hat{S}_z$ (spin in the $z$ axis) is spin up you're going to find that it will be so $\frac{\hbar}{2}$ amount of the time, proportionally? You'll get spin down equally as frequently, so $\frac{\hbar}{2}$ occurs for $\left|-\right>$ too, now negative so that overall they = $0$? (Why must they = $0$ accumulatively?) I'm going in circles trying to understand this, sorry. $\endgroup$ – perilousGourd Mar 10 '15 at 3:50
  • $\begingroup$ I realize the following is a question that might take a lot of working to answer, so feel free to just point me in the direction of another source. How do you construct $\hat{S}_x$ and $\hat{S}_y$ in the $z$ basis? $\endgroup$ – perilousGourd Mar 10 '15 at 3:50
  • $\begingroup$ You've already helped me a fair bit, so thank you, and sorry for my overly basic follow-up questions. $\endgroup$ – perilousGourd Mar 10 '15 at 3:51
2
$\begingroup$

These are mainly conventions. Conventionally, the kets $|+\rangle$ and $|-\rangle$ are taken to be eigenkets of the z-spin operator with, respectively, z-spin of $+\hbar/2$ and -$\hbar/2$.

S_x and S_y are chosen such that they obey the canonical commutation relations for angular momenta $$ [S_i,S_j]=i\epsilon_{ijk}S_k $$ E.g., $$ [S_x,S_y]=iS_z $$ and so on.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.