# Putting the ket $|l,m_x \rangle$ in terms of the ket $|l, m_z \rangle$

Could someone guide me in my thought process of this problem? I don’t know if I’m thinking about it the right way.

The problem is the following:

I have a system which possible states are generated by a basis of four eigenvectors $|l, m_z \rangle$ of the angular momentum operators $\hat{L}^2$ and $\hat{L}_z$ where $l=0,1$ and $-l \leq m_z \leq l$.

As it says in the title, what I have to do next is put the common eigenstates between $\hat{L}^2$ and $\hat{L}_x$ $|l, m_x \rangle$ in terms of the kets $|l, m_z \rangle$

So, what I do know is that the possible states of my system are generated by the following basis $A := \{|0,0 \rangle, |1,-1 \rangle, |1,0 \rangle, |1,1 \rangle \}$. If I’m understanding the question correctly I have to find the coefficients of the following linear combination $$|l, m_x \rangle = c_1 |0,0 \rangle + c_2 |1,-1 \rangle + c_3 |1,0 \rangle + c_4 |1,1 \rangle$$ In order to find the coefficients I’ve thought of using the orthonormality between the spherical harmonics, meaning $\langle l’, m_z’ | l, m_z \rangle = \delta_{l, l’} \delta_{m_z, m_z’}$. However while only one coefficient will remain when I make the inner product, in the left side of the equation I will have the term $\langle l, m_z | l, m_x \rangle$ which I do not know what it yields.

Another thing I’ve tried is using $\hat{L}^2 |l, m_z \rangle = l(l+1) \hbar^2 |l, m_z \rangle$, $\hat{L}_z |l, m_z \rangle = m_z \hbar |l, m_z \rangle$ and $\hat{L}_x |l, m_x \rangle = m_x \hbar |l, m_x \rangle$, however I’m not able to get much further than that.

• I would just look at the spherical harmonics in cartesian coordinates: $1$, $x-iy$, $z$, $x+iy$ but I'm not sure that'll get credit on a homework problem.
– JEB
Aug 29, 2018 at 0:46
• What do you mean by getting credit?.
– user168027
Aug 29, 2018 at 0:51
• you've never had an instructor say "you solved the problem the wrong way"?
– JEB
Aug 29, 2018 at 0:59
• If the method I use is consistent, rigorous and I get the right answer using said method then there’s no problem.
– user168027
Aug 29, 2018 at 1:04

Here are some hints:

1. Because $L^2$ commutes with any $L_x, L_y$ or $L_z$, the action of $L^2$ cannot change the $\ell$ value of a state.
2. Physically, the eigenvalues of $L_x$ must be the same as the eigenvalues of $L_z$, since a system with spherical symmetry has no preferred quantization axis, i.e. the possible eigenvalues of the projection of angular momentum along any axis must be the same.

Thus, start with your $\vert \ell,m_x\rangle$ and express it as a linear combination of states $\vert \ell, m_z\rangle$ with the same $\ell$, pick an eigenvalue for $m_x$, and find the coefficients that make it work, in other words, restrict your $\vert \ell, m_x\rangle$ appropriately, use $L_x\vert \ell,m_x\rangle = m_x\vert \ell,m_x\rangle$ and find the $c$s that produce the eigenstate for the eigenvalue $m_x$.

• Thank you for the hint. So while $l$ stays the same, can $m_x$ and $m_z$ have different values? And because of $2.$ this means that $m_x$ must also take the values $-l \leq m_x \leq 1$, am I correct?
– user168027
Aug 29, 2018 at 1:13
• @R.Mor The possible values of $m_x$ are identical to those of $m_z$. Of course, as you have largely correctly guessed, a state $\vert \ell m_x\rangle$ is a linear combination of $\vert \ell m_z\rangle$ states, and eigenstates of $L_z$ are not eigenstates of $L_x$ (with one exception). Aug 29, 2018 at 1:15

In light of @ZeroTheHero's correct answer, I'll give mine. For short hand, I'll define $Z_l^m$ to be an eigenstate relative to the $x$-axis. I'm going to ignore normalization for brevity, and start with:

$$Y_0^0 \approx 1$$

$$Y_1^{\pm 1} \approx (x \pm iy)/\sqrt 2$$ $$Y_1^1 \approx z$$

Now the goal is to replace $(x, y, z)$ with $(y, z,x)$

By inspection:

$$Z_0^0 = 1= Y_0^0$$

$$Z_1^0 = x = Y_1^1 + Y_1^{-1}$$

Now to replace $x\pm iy$ we need $y \pm iz$, and

$$y \approx -i(Y_1^1 - Y_1^{-1})$$

$$z \approx Y_1^0$$

So plug (w/ normalization) that in and you're done. I hope you get credit for it.

• Your welcome. It's important to understand the operator methods, and representation theory results. But sometimes, it's useful to remember, these things are just functions on a sphere.
– JEB
Aug 29, 2018 at 1:40