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I'm doing an assignment for my quantum class at the moment and I'm having trouble figuring out how to act a Spin operator on a two-particle state - specifically in finding the eigenvalues - I've spent several hours going through examples but they tend to not be too enlightening and It'd be appreciated if someone could explain if a) what I'm doing is correct b) how operators work over tensor products as well.

My operator is $S_z = S_{1z}+ S_{2z}$ Where $S_{iz}|\pm\rangle= \pm\frac{\hbar}{2}|\pm\rangle $. I wish to find eigenvalues of $S_z$ of the following state: $|+-\rangle - |-+\rangle $.

Knowing $S_z = S_{1z}\otimes \mathbb{1} + \mathbb{1}\otimes S_{2z}$ I do the following:

$(S_{1z}+S_{2z})(|+-\rangle - |-+\rangle) = S_{1z} (|+-\rangle - |-+\rangle) +S_{2z}(|+-\rangle - |-+\rangle)$

$=\left(\frac{\hbar}{2}-\frac{-\hbar}{2}\right)(|+-\rangle + |-+\rangle)+\left(\frac{-\hbar}{2}-\frac{\hbar}{2}\right)(|+-\rangle + |-+\rangle)$

$=\hbar (|+-\rangle - |-+\rangle) -\hbar (|+-\rangle + |-+\rangle)$

$=(\hbar-\hbar) (|+-\rangle - |-+\rangle)$

$=0$

I'm confused here now for several reasons. Do I now factorise getting an eigenvalue of 0? When I say $S_{1z}+S_{2z}$ what does "+" mean? Is it regular addition or is it a direct sum or something? Have I evaluated this correctly? Most of the examples I find are a bit simple and not too enlightening.

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$\newcommand{\ket}[1]{\left| #1 \right>}$ $\newcommand{\o}{\mathbf 1}$

Physicists are lazy people we all are! When you see something like $S_{1z}+S_{2z}$ you should really think of the following:

$$S_{1z}+S_{2z} \equiv S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z}$$

Since you get tired of writing it over and over you just shorten it by an addition and remember how to act on the states. That is basically what you are doing wrong in your calculation. Remember that

$$\ket {+-} \equiv \ket + \otimes \ket -$$

Which means the following:

\begin{align} \left(S_{1z}+S_{2z}\right) \ket{+-} &= \big( S_{1z} \otimes \mathbf 1+ \mathbf 1 \otimes S_{2z} \big) \big( \ket + \otimes \ket - \big) \tag{1} \\ &= S_{1z}\ket + \otimes \ket - + \ket + \otimes S_{2z}\ket -\\ &=\frac{\hbar}{2}\big( \ket + \otimes \ket- - \ket + \otimes \ket -\big) =0 \\ \end{align}

Notice how long the equation ($1$) is, which is, as I said before, usually shortened at the cost of ambiguity.

I think you can work out your question by yourself after this explanation.

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  • $\begingroup$ Thanks for the reply! Two questions: 1) $$\ket {+-} \equiv \ket + \otimes \o + \o \otimes \ket -$$ In this here - is the "+" a regular plus? $ $ 2) Going from the second last line to the last line - how come you can factorise $\frac{\hbar}{2}$ from the first term but not the second? Thanks again! $\endgroup$ – AXidenT May 14 '15 at 17:02
  • $\begingroup$ That is my friend a genuine plus sign:) Notice the sign change in the last step... Shoot! I should edit that there shouldn't be any $S_{2z}$ there $\endgroup$ – Gonenc May 14 '15 at 17:05
  • $\begingroup$ Ah ok this makes sense now thank you very much! :) $\endgroup$ – AXidenT May 14 '15 at 17:07
  • $\begingroup$ Actually one more question sorry - in your example, how would you reduce this down to the simplified notation? With the negative introduced there it doesn't seem possible? ie You would have $|+\rangle\otimes \mathbb{1}+(-)\mathbb{1}\otimes|-\rangle\neq |+-\rangle$. In fact I'm not sure what that would equal? $\endgroup$ – AXidenT May 14 '15 at 17:32
  • $\begingroup$ @AXidenT I'm now 100% sure that it is correct. I should apologize again for misleading you. $\endgroup$ – Gonenc May 14 '15 at 17:42

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