Pauli Spin Matrices - Lowering Operator - Eigenstates

1. Jan 15, 2017

ChrisJ

This is not part of my coursework but a question from a past paper (that we don't have solutions to).

1. The problem statement, all variables and given/known data

Construct the matrix $\sigma_{-} = \sigma_{x} - i\sigma_{y}$ and show that the states resulting from $\sigma_{-}$ acting on the eigenstates of $\sigma_{z}$ are also eigenstates of $\sigma_{z}$ and comment on your result.

2. Relevant equations
pauli spin matrices

3. The attempt at a solution
I need more help with the commenting on the result and the actual physics rather than the maths here,

I constructed the matrix $\sigma_{-} = \left( \begin{array}{ccc} 0 & 0 \\ 2 & 0 \end{array} \right)$

and in previous bit of question found the eigenstates of $\sigma_{z}$ to be $\left( \begin{array}{ccc} 1 \\ 0 \end{array}\right)$ and $\left( \begin{array}{cc} 0 \\ 1 \end{array}\right)$ respectively.

So therefore $\sigma_{-} \left( \begin{array}{ccc} 1 \\ 0 \end{array}\right) =2 \left( \begin{array}{ccc} 0 \\ 1 \end{array}\right)$

and also $\sigma_{-} \left( \begin{array}{ccc} 0 \\ 1 \end{array}\right) = \left( \begin{array}{ccc} 0 \\ 0 \end{array}\right)$

I am pretty sure the math is correct as I ran the math past a few people who agreed but I cant see how/why that that shows they are also eigenstates of $\sigma_{z}$. I can maybe see it mathematically with the first result, as that is explicitly an eigenstate, but the zero matrix result, I am not sure how in words I can say that it is. And what it means physically. As I said this isnt part of any coursework, just a question from a past exam paper, any help/advice is much appreciated.

2. Jan 15, 2017

blue_leaf77

An eigenvector is defined to be a non-zero vector, so the statement that the resulting vector is an eigenstate of $\sigma_z$ as mentioned in the question shall only apply to cases when the resulting vector is non-zero.
As for the comment, I think it asks you about how $\sigma_-$ changes the eigenvector of $\sigma_z$, for example how the eigenvalue changes.

3. Jan 15, 2017

ChrisJ

Ah ok, that zero vector/matrix was throwing me. Ok so its just as simple as saying that it lowers the eigenvalue by one? A previous part of the question was to find the eigenvalue of $\sigma_z$ and was found to be 3 and 3.

4. Jan 15, 2017

blue_leaf77

At least that's the most notable property of $\sigma_-$.

5. Jan 15, 2017

vela

Staff Emeritus
3 and 3?

6. Jan 15, 2017

ChrisJ

Sorry, worded that wrong. Meant to say that I found the eigenvalues of the two eigenstates to be 3 and 3.

7. Jan 15, 2017

vela

Staff Emeritus
That's what I thought you meant, but you should have found them to be 1 and -1.

8. Jan 15, 2017

blue_leaf77

Not sure why I missed that, but that's clearly not true.

9. Jan 16, 2017

ChrisJ

My bad! I did the previous bit of the question a couple weeks ago, sorry my fault. It seems the 3 and 3 I had in my head was not about that, it was to find the eigenvalues of $\sigma^2=\sigma_x^2+\sigma_y^2+\sigma_z^2$ Sorry for the confusion. I just remembered in my head that I found eigenvalues and they were 3, so when I went back to do the rest of the multiple part question, I had those 3's in my head.

10. Jan 16, 2017

ChrisJ

But now that you pointed out the mix up. My original question about commenting on the result $\sigma_{-}$ acting on $\sigma_z$? as when the lowering operator acts on $\sigma_z$ it results in an eigenstate of $\sigma_z$ but with an eigenvalue of 2, but its original eigenvalues were 1 and -1?

11. Jan 16, 2017

blue_leaf77

No, the constant 2 in $\sigma_- |+\rangle = 2 |-\rangle$ is not the eigenvalue - it should be obvious that this equation is not an eigenvalue equation. What $\sigma_-$ does is that it changes the eigenstate of $\sigma_z$ to another eigenstate whose eigenvalue is one unit smaller than that of the initial eigenstate. Lowering operator, for more general angular momentum operator, obeys the condition $L_z L_- |n\rangle = (n-1)\hbar L_- |n\rangle$ where $|n\rangle$ is an eigenstate of $L_z$.