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.

 

[paranormal]

Dear student,

Thank you for sharing your thought process and attempts at solving this problem. It seems like you are on the right track by considering the Lx operator and its eigenvalues. However, it is important to note that the Lx operator is not responsible for changing basis, but rather for measuring the angular momentum in the x direction.

To find the column vector in the z-basis that represents the initial orbital angular momentum state, we can use the eigenvalue equation for the Lz operator, which is responsible for measuring the angular momentum in the z direction. This equation is given by Lz|l,ml> = ml|l,ml>, where ml is the z component of the orbital angular momentum.

Since we are given |l,ml>x=|1,1>x, we can see that ml=1. Therefore, the corresponding column vector in the z-basis would be [0, 0, 1]. This means that there is a 100% probability of measuring the z component of the orbital angular momentum to be ml=1.

I hope this helps clarify the steps needed to solve this problem. It is important to carefully consider which operator is responsible for measuring the desired component of the angular momentum, and to use the corresponding eigenvalue equation.

Best,