QM: Issues with parity of spherical harmonics and Heisenberg

  • #1
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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 uncertanty 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
 

Answers and Replies

  • #2
139
12
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.
 
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  • #3
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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.
 
  • #4
139
12
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.
 
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  • #5
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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.
 

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