Total angular momentum operator for a superposition

[DivGradCurl]

Hi all,

Quick quantum question. I understand the total angular momentum operation $$\hat{L}^2 \psi _{nlm} = \hbar\ell(\ell + 1) \psi _{nlm}$$ which means the total angular momentum is $$L = \sqrt{\hbar\ell(\ell + 1)}$$ But how about applying this to an arbitrary superposition of eigenstates such as this $$\psi = \frac{1}{\sqrt{2}} \psi _{310} + \frac{1}{\sqrt{2}} \psi _{420}$$ and trying to find the expected value of the total angular momentum? Here is my best guess: $$\hat{L}^2 \psi = \frac{1}{\sqrt{2}} \hbar \, 1(1+ 1) \psi _{310} + \frac{1}{\sqrt{2}} \hbar \, 2(2 + 1) \psi _{420} = \frac{2}{\sqrt{2}} \hbar \psi _{310} + \frac{6}{\sqrt{2}} \hbar \psi _{420}$$ so the expected value is $$\langle L ^2 \rangle = \left( \frac{2}{\sqrt{2}} \hbar \right)^2 + \left( \frac{6}{\sqrt{2}} \hbar \right)^2$$ which means the expected value of the total angular momentum is maybe $$\langle L \rangle = \sqrt{ \left( \frac{2}{\sqrt{2}} \hbar \right)^2 + \left( \frac{6}{\sqrt{2}} \hbar \right)^2}$$ Is this correct? Thank you

 

[Nugatory]

Just apply the Born rule. Your wave function is of the form $\psi=\alpha\psi_1+\beta\psi_2$ where $\psi_1$ and $\psi_2$ are eigenvectors of $L^2$ with eigenvalues $\lambda_1$ and $\lambda_2$, so the probability of getting the result $\lambda_1$ is $\alpha^2$ and the probability of getting the result $\lambda_2$ is $\beta^2$. That makes the expectation value $\alpha^2\lambda_1+\beta^2\lambda_2$.

Plug in your known values and you're done.

 

[DivGradCurl]

Yeah, I see your point. There is a simpler way to approach it. Looks like we're arriving at the same answer either way, which is good. Thanks

 

[DivGradCurl]

Let me bring an additional question, related to the original post with an added twist. If I were to change the wavefunction to, for example $$\psi = \frac{1}{\sqrt{2}} \psi_{100} + \frac{1}{\sqrt{2}} \psi_{310}$$ and wanted (1) the expectation value of the total angular momentum and (2) the probability of a single measurement yielding the value found in (1), is this correct...

(1) Finding the expectation value of the total angular momentum
$$\langle L \rangle = \sqrt{ \left( \frac{1}{\sqrt{2}} \right) ^2 \, 0 (0 +1) \hbar + \left( \frac{1}{\sqrt{2}} \right) ^2 \, 1 (1 +1) \hbar } = \sqrt{ \frac{1}{2} \, 2 \hbar} = \sqrt{\hbar}$$

(2) The probability of a single measurement yielding the value found in (1)

100% since the wavefunction collapses upon the measurement. Is that right? Thanks again! :)

 

[DivGradCurl]

I've just caught my own mistake. Every $$\hbar$$ in the above posts should be squared. Other than that, I'm assuming it's correct.

 

[kith]

(1): You have to be a bit careful with the operators. $\langle |\mathbf{L}| \rangle = \left\langle \sqrt{ \mathbf{L}^2 }\right\rangle$ is not the same as $\sqrt{\langle \mathbf{L}^2 \rangle}$ in general.

(2): How do you calculate the probabilities? Do you get different probabilities for the state of your initial post and if yes, why?

 

[blue_leaf77]

which means the expected value of the total angular momentum is maybe $$\langle L \rangle = \sqrt{ \left( \frac{2}{\sqrt{2}} \hbar \right)^2 + \left( \frac{6}{\sqrt{2}} \hbar \right)^2}$$ Is this correct? Thank you

In which context did you want or were asked to calculate the expectation value of the total angular momentum? Usually, the term "total angular momentum" is sufficiently referred to the square modulus of the angular momentum $|\mathbf{L}|^2$. Otherwise, angular momentum is a vector, it may be more relevant to calculate the expecation value of the vector angular momentum.

 

[Nugatory]

(2) The probability of a single measurement yielding the value found in (1)

100% since the wavefunction collapses upon the measurement. Is that right?

As long as you meant "subsequent" instead of "single", yes. First measurement applied to the superposed state collapses the wave function to the eigenfunction whose eigenvalue is what you measured; subsequent measurements give you that value with probability 100%.

(BTW, this only works because $L^2$ commutes with the Hamiltonian. Otherwise the post-measurement state, whose evolution is governed by the time-dependent Schrodinger equation, would evolve into something that is not a eigenfunction of $L^2$).

Edit: Oh... and also what BlueLeaf just said. You want to be talking about $L^2$ not $\vec{L}$

 

[DivGradCurl]

Hi,

On (1), thanks for bringing that point up. I guess I should express my solution as
$$\langle |\mathbf{L}| \rangle = \left\langle \sqrt{ \mathbf{L}^2 }\right\rangle$$
which is what it seems to be.
On (2), the probabilities are just the modulus squared of the coefficients, which I arbitrarily assumed as $$\frac{1}{\sqrt{2}}$$ Then, in the original post, the probability would be 1/2 for each eigenstate.

 

[kith]

On (2), the probabilities are just the modulus squared of the coefficients, which I arbitrarily assumed as $$\frac{1}{\sqrt{2}}$$ Then, in the original post, the probability would be 1/2 for each eigenstate.

Then why did you write about a probability of 1 in post #4? Did you intend to talk about subsequent measurements (like Nugatory suggested in post #8)?

 

[DivGradCurl]

Exactly, subsequent measurements and wavefunuction collapse, giving probability of 1. Thanks