# Singlet/Triplet state mixing question

Hey guys!
I am trying to show how singlet and triplet states mix via spin-orbit coupling, but I am having some trouble.
I've gotten most of the way through, but I am stuck on how an operator works. Sadly, I think my poor basis in quantum is going to show right here.

My starting point...

$\left(s_{1z}-s_{2z}\right)\left|0,0\right\rangle =\hbar \left|1,0\right\rangle$

The book I am looking at sort of skims past this and assumes that I have a clue what is going on here.

So I know
$\left|0,0\right\rangle =\frac{1}{\sqrt{2}}\left\{\left|-+\right\rangle -\left|+-\right\rangle \right\}$
$s_z=\frac{\hbar }{2}\left\{\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right\}$

and I am thinking that
$\left(s_{1z}-s_{2z}\right)\left|0,0\right\rangle = \frac{1}{\sqrt{2}}\left(s_{1z}\left\{\left|-+\right\rangle -\left|+-\right\rangle \right\}-s_{2z}\left\{\left|-+\right\rangle -\left|+-\right\rangle \right\}\right)$

After this it becomes a giant mess. Basically it hinges on my inability to evaluate things of the form
$\left|+\right\rangle \left\langle +|-+\right\rangle$

I am thinking that this is heavily reliant on some other basics as well. In determining the s_z spinor I am not sure how to evaluate something like
$\left\langle +|+\right\rangle \left\langle +|+\right\rangle - \left\langle +|-\right\rangle \left\langle -|+\right\rangle$
which is supposed to equal one and

$\left\langle +|+\right\rangle \left\langle +|-\right\rangle - \left\langle +|-\right\rangle \left\langle -|-\right\rangle$
which is supposed to equal zero

I was hoping that someone here would be able to help me figure out how
$\left(s_{1z}-s_{2z}\right)\left|0,0\right\rangle =\hbar \left|1,0\right\rangle$
and the above evaluations. Thanks a whole bunch!

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DrDu
You have two sub-systems. A convenient way to take care of this is the tensor product:
You write e.g. : $| ++>=|+>\otimes|+>$ and $s_{1z}=s_z \otimes 1$.
The action of the operator on the state then becomes:
$s_{1z}|++>=s_z|+>\otimes 1|+>=\frac{\hbar}{2}|++>$.
I hope that gives you an idea of how to proceed.

Yes! That definitely will help me on my way. Unfortunately, I am still having some difficulty.

This is the part that is giving me trouble.
$s_z|+\rangle ⊗ 1|+\rangle$

When I took a quantum class we never used the +/- notation, we were trained with α and β, but it seems that most modern books are using the +/-. Because of this, I am not very familiar with their mathematical meaning.

If you could be so kind as to help me solve: $s_z|+\rangle$?

Here's my current process:

If $s_z=\frac{\hbar }{2}\left[\left|+\right\rangle \left\langle +\right|-\left|-\right\rangle \left\langle -\right|\right]$
then $s_z\left|+\right\rangle =\frac{\hbar }{2}[|+\rangle \langle +|+\rangle - |-\rangle \langle -|+\rangle]$

This is where I am stuck (If I am even correct to this point... for all I know, I may not be clear on the proper way to handle the operator)
Does <+|+> = 1 and <-|+> = 0? because that would make sense for getting $\frac{\hbar }{2}|+\rangle$
(would this also mean that <-|-> = 1? or would it be -1?)

If so, how do I get the 0 and 1? If not, then how do I evaluate these correctly?

Thanks again!

DrDu
yes, + and - correspond one to one to alpha and beta.
They are defined to be the eigenstates of s_z.
The two states are orthogonal to each other (like any two eigenstates of a hermitian operator belonging to two different eigenvalues, namely that of s_z) whence <+|->=0 and <+|+>=1.

I think one alternative way of seeing things is to realize that you're working within a finite state space (4 dimensional) and that you can write down these operators explicitly. Take a look at this page:

http://electron6.phys.utk.edu/qm1/modules/m10/twospin.htm

I don't get what your question has to do with SO coupling though.

Thanks a bunch! After having a good think about it, it all makes perfect sense now and I now know exactly how the s_1z-s_2z operator works. I will post how it relates to spin orbit coupling and singlet triplet mixing a little later once I have written it out in an understandable manner.