# What does it mean to find eigenvectors of some operator in some basis?

1. Oct 26, 2008

### KFC

I am studying QM by myself. I got a quite confusing problem which annoying me for a certain time. Well, this question is about the angular momentum opeator Lx, Ly and Lz. The matrix form for these operatore are given, so by solving the corresponding secular equation, it is easy to find the corresponding eigenvalues and eigenvectors. Here, Lz is given in a diagonalized form, so one can directly obtain the eigenvalues by reading the diagonal elements. Let's say the eigenvalues for Lz is a, b and c. And the corresponding eigenvector are

$$\left( \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right), \qquad \left( \begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right),\qquad \left( \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)$$

But I got a question "To find the eigenvectors of Lx in the Lz basis", I have no idea what does it mean? Since we already know the form of Lx operator, what can't we just solve the secular equation to find the eigenvectors? What is the significance of written the eigenvectors of Lx in Lz basis?

2. Oct 26, 2008

### olgranpappy

The matrices you are given are already in the Lz basis. Everything you are doing is "in the Lz basis". When you find the eigenvectors of Lx you will be finding those eigenvectors "in the Lz basis".

3. Oct 26, 2008

### KFC

Why? How do you tell that? Still confusing ...

What about this, if it gives me Lx, Ly and Lz and all of them are not diagonalized matrices. This time, I need to find the eigenvectors of Ly in Lx basis, how can I do that? Sorry, I am not majoring in physics. But I do interest in it so I am learning by myself.

Last edited: Oct 26, 2008
4. Oct 26, 2008

### Dr Transport

You find the eigen-values and eigen-vectors for the $$L_x$$ and $$L_y$$ then see how they can be written in terms of the $$L_z$$ eigen-vectors.

5. Oct 26, 2008

### KFC

Oh, I got it. That is, we use the usual way to find the eigenvectors and then EXPAND it with those of Lz, is that right?

Thanks

6. Oct 26, 2008

### Dr Transport

Exactly.....

7. Oct 26, 2008

### Avodyne

Equivalently, an operator A is represented by a diagonal matrix in the "A basis".

8. Oct 27, 2008

### KFC

So, it means if I solve the secular equation for Lx to find the corresponding eigenvalues and eigenvectors, I actually diagonalize it with a matrix composing of Lx's eigenvectors?

If so, can I use Lz's eigenvectors to form a matrix (says S), and then use the similarity transformation on Lx (with S) to diagonalize Lx? I don't how if this works!?