QM: Issues with parity of spherical harmonics and Heisenberg

[renec112]

Hi physics forms! I'm practicing to for an Quantum mechanics exam, and i have a problem.

1. Homework Statement
I have two problems, but it's all related to the same main task. I have a state for the Hydrogen:
$\Psi = \frac{1}{\sqrt{2}}(\psi_{100} + i \psi_{211})$
where $\psi_{nlm}$.

Use Heisenberg uncertainty relation to find the lower value for the varians $\sigma_x$ of possible meassures for the electrons x-cordinate. HINT It can be used without proof that
$[L_z, x] = i \hbar y$
2. Equations
Heisenberg:
$\sigma_x \sigma_y \geq \hbar/2$

3. My try
Well here is where it gets awkward. I have a very hard time understanding what i can use the hint for.
I could try to isolate $\sigma_x \geq \frac{2}{\hbar \sigma_p}$

But given the hint is in a commutator i think i should use the generalized uncertainty principle.
$\sigma_A \sigma_B \geq |\frac{1}{2i} <[A,B]>|$
But then i should look at
$\sigma_x \sigma_{L_z}$
Which makes no sense to me.

I Would very much appreciate your comment, thank you

 

[VKint]

The trick here is that $\sigma_{L_x}$ is something we can actually compute directly in this state, using the definition $$\sigma_{L_x}^2 = \langle L_x^2 \rangle - \langle L_x \rangle^2.$$ Specifically, you should be able to compute $\langle L^2 \rangle$ without too much trouble, which gives you $\langle L_x^2 \rangle$ by symmetry in $x$ and $y$; that leaves $\langle L_x \rangle$, which can be computed without any work at all (why?).

Once you have the exact variance $\sigma_{L_x}$, you can use the generalized uncertainty relation to find a lower bound for $\sigma_x \sigma_{L_x}$ and solve the inequality for $\sigma_x$, as you suggested in your solution attempt. At some point, you'll have to compute the expected value $\langle y \rangle$, which could be kind of nasty, but otherwise this should be straightforward.

 

[renec112]

At some point, you'll have to compute the expected value $\langle y \rangle$, which could be kind of nasty, but otherwise this should be straightforward.

Me and my group had a laugh at this, we are just finished doing it and it was very nasty indeed :D
Thank you very much for the comment.
- You mean $<\sigma_{L_z}^2>$ and not $<\sigma_{L_x}^2>$ Right?
We have actually calculated $<L_z$ and $<L_z^2>$ from another task. We did it using:
$L_z f^m_l = \hbar m f^m_l$

I'm Curious why you suggest $<L^2>$? Would you use:
$L^2 f^m_l = \hbar^2 l(l+1) f^m_l$ a
and then exploit:
$L^2 = L_x^2 + L_y^2 +L_z^2$

Thanks for helping me out!
Btw, so we are just picking $L_z$ because it does not commute with $x$ ? it's just a bit random to me.

 

[VKint]

As you suggest, $\langle L^2 \rangle$ is just a way to get at $\langle L_x^2 \rangle$, which we need to know in order to compute $\sigma_{L_x}$.

I'm not sure what you mean by your first question--$\sigma_{L_x}$ can be calculated exactly, so we don't need to worry about "$\langle \sigma_{L_x} \rangle$."

As for $\langle L_z \rangle$, your calculation is correct, but this is actually not all that relevant to this question. In particular, $\langle L_x \rangle \neq \langle L_z \rangle$.

The reason we're choosing $L_x$ is twofold: (1) it doesn't commute with $x$, and (2) we're in a linear combination of angular momentum eigenstates, which means that $L_x$ should be (relatively) convenient to work with.

 

[renec112]

Okay i think i understand so we are calculating:
$\sigma_x \sigma_{L_x} \geq |\frac{1}{2i} <[x,L_x]>|$
and not:

$\sigma_x \sigma_{L_z} \geq |\frac{1}{2i} <[x,L_z]>|$?
Thank you for helping me.