Expectation values for angular momentum

[Albereo]

Consider a quantum system with angular momentum 1, in a state represented by the vector

\Psi=\frac{1}{\sqrt{26}}[1, 4, -3]​

Find the expectation values <L_{z}> and <L_{x}>


I'm reviewing my quantum mechanics; I had a pretty horrible course on it during undergrad. I feel like this should be fairly simple, but I'm just not sure how to start. I think I can say that the state above is an eigenfunction of L^{2} with eigenvalue h^{2}l(l+1), and similarly for L_{z} with eigenvalue hm, but is this on the right track? How do I proceed from there?

 

[vela]

Consider a quantum system with angular momentum 1, in a state represented by the vector

\Psi=\frac{1}{\sqrt{26}}[1, 4, -3]​

Find the expectation values <L_{z}> and <L_{x}>


I'm reviewing my quantum mechanics; I had a pretty horrible course on it during undergrad. I feel like this should be fairly simple, but I'm just not sure how to start. I think I can say that the state above is an eigenfunction of L^{2} with eigenvalue h^{2}l(l+1), and similarly for L_{z} with eigenvalue hm, but is this on the right track? How do I proceed from there?

You need to get the basics down first before you can solve the problem. A convenient basis to work with when dealing with angular momentum consist of those states which are simultaneous eigenstates of $\hat{L}^2$ and $\hat{L}_z$. You can label these states using the quantum numbers $l$ and $m_l$. For a given $l$, $m_l$ can assume the values $l, l-1, \ldots, -(l-1), -l$. So the first couple of eigenstates correspond to:

\begin{align*}
l &= 0, m_l = 0 \\
\\
l &= 1 \ \
\begin{cases}
m_l = 1 \\
m_l = 0 \\
m_l = -1
\end{cases}
\\\\
l &= 2 \ \
\begin{cases}
m_l = 2 \\
m_l = 1 \\
m_l = 0 \\
m_l = -1 \\
m_l = -2
\end{cases}
\end{align*}
In light of this information, what linear combination of eigenstates is given in your problem?

 

[Albereo]

You need to get the basics down first before you can solve the problem. A convenient basis to work with when dealing with angular momentum consist of those states which are simultaneous eigenstates of $\hat{L}^2$ and $\hat{L}_z$. You can label these states using the quantum numbers $l$ and $m_l$. For a given $l$, $m_l$ can assume the values $l, l-1, \ldots, -(l-1), -l$. So the first couple of eigenstates correspond to:

\begin{align*}
l &= 0, m_l = 0 \\
\\
l &= 1 \ \
\begin{cases}
m_l = 1 \\
m_l = 0 \\
m_l = -1
\end{cases}
\\\\
l &= 2 \ \
\begin{cases}
m_l = 2 \\
m_l = 1 \\
m_l = 0 \\
m_l = -1 \\
m_l = -2
\end{cases}
\end{align*}
In light of this information, what linear combination of eigenstates is given in your problem?

I've got a linear combination of the states |$l$ , $m$_{l}> = |1, -1>, |1, 0>, and |0, 1>. Do I now need to determine the coefficients, or do I proceed some other way?

 

[vela]

|0, 1>? That's not an allowed combination.

 

[Albereo]

Oops, I meant |1, 1>.

 

[vela]

OK, good. You're given the coefficients. How would you express the state $\Psi$ in Dirac notation?

Once you have that, I'd try applying $\hat{L}_z$ to that state. What result do you get?

 

[Albereo]

Okay, the Dirac notation is definitely one spot where I'm a bit fuzzy. But I think it would be something like:

\Psi=\frac{1}{\sqrt{26}}|1, -1> + \frac{4}{\sqrt{26}}|1, 0> - \frac{3}{\sqrt{26}}|1, 1>​

But is the ordering of states important here? (If this is correct)?

 

[vela]

Yeah, you have the right idea. I'd write $\vert \Psi \rangle$ on the lefthand side, though. The order is indeed important. You should look up what convention your class is using to get it right.

 

[Albereo]

Thanks for all your help so far. This isn't coursework, so perhaps I'll look up the conventions later and just pick my ordering for the moment.

So with that written for the state, to find the expectation values, I'd have to do

<$L$_{z}> = <\Psi|\frac{h}{i}\frac{\partial}{\partial \varphi}|\Psi&gt;​

and so on, where the |$l$, $m$> is shorthand for the spherical harmonics?

I can calculate those with the aid of a spherical harmonics table, but I wonder if I'm missing a quicker way to do it? I only ask because this comes from an old qualifying exam, and with time and memory constraints this would seem a bit unreasonable.

 

[vela]

No, you're making it more complicated than needed. $\lvert l \ m \rangle$ is an eigenstate of $\hat{L}_z$, which means that $\hat{L}_z \lvert l \ m \rangle = m\hbar \lvert l \ m \rangle$.

If you were working in the $\theta$, $\phi$ basis, then you'd have
$$\langle \theta\ \phi\ \vert\ l\ m \rangle = Y_{lm}(\theta,\phi)$$ and
$$\hat{L}_z \rightarrow \frac{\hbar}{i}\frac{\partial}{\partial \phi}.$$ In this representation, the eigenvalue equation would be
$$ \frac{\hbar}{i}\frac{\partial}{\partial \phi} Y_{lm} = m\hbar Y_{lm},$$ which you should easily be able to verify, but going to this basis doesn't buy you anything in this problem. It's just needless work. Stick with the abstract notation for now.

 

[Albereo]

Ah, got it!

So then the expectation value for the z-component is straightforward, as well as the one for $L$^{2}.

But I can't do the same thing for $L$_{x} because the state isn't an eigenfunction of that operator. Do I then turn to the commutation relations?

 

[vela]

Try looking up the matrix representation of $\hat{L}_x$ and see if you can figure it out from there.