# Quantum Mechanics - question about spin

1. Oct 22, 2005

### touqra

Given $$S=\frac{1}{2}\hbar{\sigma}$$ where $$\sigma = \left(\left(\begin{array}{cc}0&1\\1&0\end{array}\right),\left(\begin{array}{cc}0&-i\\i&0\end{array}\right),\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\right)$$, show that
$$|+> = \left(\begin{array}{cc}1\\0\end{array}\right)$$ and $$|-> = \left(\begin{array}{cc}0\\1\end{array}\right)$$ are the eigenfunctions for $$S_z$$ . Obtain the matrix representation for $$S_y$$ and $$S_x$$ in the basis $$(|+>,|->)$$ .

Last edited: Oct 22, 2005
2. Oct 23, 2005

### George Jones

Staff Emeritus
$$S_{z} = \frac{1}{2}\hbar \sigma_z$$

What happens when $$S_{z}$$ operates on $$\left| + \right>$$?

How is a matrix representation (with respect to a basis) of an operator found? You've basically written down the final answer.

Regards,
George

3. Oct 23, 2005

### Gokul43201

Staff Emeritus
The first part is just an eigenvalue problem. From the matrix for Sz, find the eigenvalues (and eigenvectors) from the characteristic (and eigenvalue) equation.

4. Oct 23, 2005

### George Jones

Staff Emeritus
The is the comprehensive way to do things, which is probably a good idea, but it not strictly necessary to do this in order to show that |+> and |-> are eignevectors of S_z.

Regards,
George

5. Oct 23, 2005

### Gokul43201

Staff Emeritus
Oh, they've already given you the eigenvectors ! Yes, that would be unnecessary then.

6. Oct 23, 2005

### touqra

The original question was exactly the same as I've written here. Initially when I look at the question, I was dumbfounded.
The first part of the question, I would agree with Gokul43201 on finding the characteristic equation first.
How about the second part of the question? How do you find the matrix representation? I really have no idea.

Last edited: Oct 23, 2005
7. Oct 23, 2005

### Gokul43201

Staff Emeritus
I'm a little puzzled too...
As George said above, you can do it this way, if you wish, but it's not necessary. All you have to do is operate Sz on |+> and |-> and show what happens.

This is the weird part. What are you supposed to start from ? The matrix representation is what you're actually given.

The matrix form comes from the outer-product representation of those operators (by writing their own eigenkets as suitable linear combinations of the eigenkets of Sz).

Is this from a textbook ? If so, which one ?

8. Oct 24, 2005

### touqra

It's a past year exam question. I wonder how the students did for this question. IMO, if the question was phrased in another way, like, prove that the matrix representation for ... is .... Then, it should be ok.

9. Oct 24, 2005

### George Jones

Staff Emeritus
If $$A$$ is a linear operator on an n-dimensional vector space $$V$$, then the matrix representation of $$A$$ with respect to the basis $$\{v_{1}, \dots , v_{n} \}$$ of $$V$$ is given by

$$A v_{i} = \sum_{j} A_{ji} v_{j}.$$.

In this case, $$V = \mathbb{R}^{2}$$, $$v_{1} = |+> = \left(\begin{array}{cc}1\\0\end{array}\right)$$ and $$v_{2} = |-> = \left(\begin{array}{cc}0\\1\end{array}\right)$$.

The case where $$v_{1} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc}1\\1\end{array}\right)$$ and $$v_{2} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc}-1\\1\end{array}\right)$$ is more interesting.

Regards,
George

Last edited: Oct 24, 2005