Change of Basis With Orbital Angular Momentum

[tristan3214]

Homework Statement

We have the initial orbital angular momentum state in the x basis as |l,m_l>_x=|1,1>_x. We are asked to find the column vector in the z-basis that represents the initial orbital angular momentum of the above state. It then says "hint: use an eigenvalue equation".

Homework Equations

I feel like this is a simple change of basis from the x basis to the z basis with the L_x operator. However, I am not convinced this is quite the steps I need to take.

The Attempt at a Solution

To start I applied the L_x operator (the 3x3 operator that I think is responsible for changing basis) it is equal to half the angular momentum ladder operators added with each other. This only gives me one column vector where there should be 3, I think, since there should be some probability with each value of the quantum number m possible for the z component.

Another attempt was to try to find the eigenvalues of the L_x operator and to apply the 3 eigenvalues found as the probability to each vector but this didn't pan out so well. So here I am wondering how I might approach this problem in a different way. Any idea is appreciated.

 

[stevendaryl]

Homework Statement

We have the initial orbital angular momentum state in the x basis as |l,m_l>_x=|1,1>_x. We are asked to find the column vector in the z-basis that represents the initial orbital angular momentum of the above state. It then says "hint: use an eigenvalue equation".

Homework Equations

I feel like this is a simple change of basis from the x basis to the z basis with the L_x operator. However, I am not convinced this is quite the steps I need to take.

The Attempt at a Solution

To start I applied the L_x operator (the 3x3 operator that I think is responsible for changing basis) it is equal to half the angular momentum ladder operators added with each other. This only gives me one column vector where there should be 3, I think, since there should be some probability with each value of the quantum number m possible for the z component.

Another attempt was to try to find the eigenvalues of the L_x operator and to apply the 3 eigenvalues found as the probability to each vector but this didn't pan out so well. So here I am wondering how I might approach this problem in a different way. Any idea is appreciated.

You're on the right track. In terms of raising and lowering operators,
L_x = \frac{1}{2}(L_{+} + L_{-})

The effect of the raising and lowering operators on eigenstates of L_z are:
L_{+} |\mathcal{l}, m \rangle = \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m+1)} |\mathcal{l}, m+1 \rangle
L_{-} |\mathcal{l}, m \rangle = \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m-1)} |\mathcal{l}, m-1 \rangle

So the effect of L_x on eigenstates of L_z is:
L_x |\mathcal{l}, m \rangle
= \frac{1}{2} (\sqrt{\mathcal{l}(\mathcal{l}+1) - m(m+1)} |\mathcal{l}, m+1 \rangle + \sqrt{\mathcal{l}(\mathcal{l}+1) - m(m-1)} |\mathcal{l}, m-1 \rangle)

In the case l=1, we have 3 equations:

1. L_{x} |1, -1\rangle = \frac{1}{\sqrt{2}}|1,0\rangle
2. L_{x} |1,0\rangle = \frac{1}{\sqrt{2}}(|1,1\rangle + |1,-1\rangle)
3. L_{x} |1,1\rangle =\frac{1}{\sqrt{2}}|1,0\rangle

So if you assume that |\psi\rangle is an eigenstate of L_x with eigenvalue +1, then write:
|\psi\rangle = \alpha |1,-1\rangle + \beta |1,0\rangle + \gamma |1,1\rangle] and see what \alpha, \beta and \gamma have to be so that L_{x} |\psi\rangle = +1 |\psi\rangle

 

[stevendaryl]

In terms of matrices, if you represent |1, -1\rangle, |1,0\rangle, and |1,1\rangle[/itex] as
\left(\begin{array} \\ 0 \\ 0 \\ 1 \end{array} \right), \left(\begin{array} \\ 0 \\ 1 \\ 0 \end{array} \right) and \left(\begin{array} \\ 1 \\ 0 \\ 0 \end{array} \right)

then L_x can be represented as:

\frac{1}{\sqrt{2}} \left(\begin{array} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 &1& 0 \end{array} \right)

Then the problem becomes a matrix problem:

\frac{1}{\sqrt{2}} \left(\begin{array} \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 &1& 0 \end{array} \right) \left(\begin{array} \\ \alpha \\ \beta \\ \gamma \end{array} \right)= \left(\begin{array} \\ \alpha \\ \beta \\ \gamma \end{array} \right)

 

[tristan3214]

Thanks that makes a lot of sense that you would act on an unknown matrix to then equal the eigenvalue times that same unknown matrix. Then we can go and expand it out into a linear combination.