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

[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?

 

[olgranpappy]

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?

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".

 

[KFC]

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".

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.

 

[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.

 

[KFC]

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.

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

 

[Dr Transport]

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

Exactly...

 

[Avodyne]

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

 

[KFC]

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

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!?