Is Quantum Angular Momentum Rotationally Invariant?

In summary: Thanks so much!In summary, the homework statement is that the rotational invariance of the quantum angular momentum under the rotation generated by the intrinic spin s is shown to be true. The equation for this invariance is found to be H,J = 0 and/or [H,J^2] = 0. The attempt at a solution is to think that the second part, the radial operator commutes with the angular momentas. This is correct, as is the assumption that spin has differential form. The final statement is that the homework will be covered in differential geometry/GR/supersymmetry/string theory/etc etc. Thanks so much for your help!
  • #1
malawi_glenn
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[SOLVED] Quantum Angular momentum Question

Homework Statement



Show that [tex] \dfrac{{\vec{p}} ^2 }{2m} = \dfrac{{\vec{l} }^2}{2mr^2} - \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] is rotational invariant under the rotation generated by: [tex] \vec{j} = \vec{l} + \vec{s} [/tex] , s is intrinic spin.

Homework Equations



[H,J] = 0 and/or [H,J^2] = 0


The Attempt at a Solution



I think that the second part, the radial operator [tex] \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] commutes with the angular momentas, since it is just a function of radial coordinate, whereas angular momenta depends on the angles (direction) Is that correct?
 
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  • #2
malawi_glenn said:

Homework Statement



Show that [tex] \dfrac{{\vec{p}} ^2 }{2m} = \dfrac{{\vec{l} }^2}{2mr^2} - \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] is rotational invariant under the rotation generated by: [tex] \vec{j} = \vec{l} + \vec{s} [/tex] , s is intrinic spin.

Homework Equations



[H,J] = 0 and/or [H,J^2] = 0


The Attempt at a Solution



I think that the second part, the radial operator [tex] \dfrac{\hbar ^2}{2mr^2}\dfrac{\partial}{\partial r} (r^2 \dfrac{\partial}{\partial r}) [/tex] commutes with the angular momentas, since it is just a function of radial coordinate, whereas angular momenta depends on the angles (direction) Is that correct?

Yes, any function of r i strivially invariant under rotation since [tex] [f(r), L_i] = 0 [/tex] because the [tex] L_i [/tex] depend only on the angles, as you say.
 
  • #3
Just as I thought then, Didn't want to search or work out the differential form of J before I was sure :) Thanx!
 
  • #4
malawi_glenn said:
Just as I thought then, Didn't want to search or work out the differential form of J before I was sure :) Thanx!

You're welcome.

They are given in spherical coordinates toward the middle of the page at xbeams.chem.yale.edu/~batista/vvv/node16.html
 
  • #5
Thanx!

Was wondering if you know if spin also have differential form? and j (total angular mom.). Or if you must work em out using group theory?
 
  • #6
malawi_glenn said:
Thanx!

Was wondering if you know if spin also have differential form? and j (total angular mom.). Or if you must work em out using group theory?

You have to work using group theory. The differential form approach only produces the integer angular momentum representations. It's only by working with the abstract formalism of commutator and operators that one can generate all the spin representation including the half integer ones. For the spatial angular momentum calculations, one has the choice of working with explicit spatial wavefunctions or with matrices and column vectors, etc. For spin, one must work with the matrix representations.
 
  • #7
Ok I got it :)

So how can I argue that f(r) and s commutes? same as with L, that s only depends on direction?
 
  • #8
malawi_glenn said:
Ok I got it :)

So how can I argue that f(r) and s commutes? same as with L, that s only depends on direction?

They commute but the reason is that s acts on a totally different space, so they commute trivially. The total Hilbert space is a direct product of the spin space and the Hilbert space of spatial wavefunctions. Any operator acting in one of the space commutes with any operator acting in the other space.
 
  • #9
yeah, of course.. I have done angular momentum in QM now in 6h.. maybe shall go and cook some food ;) this one was so obviuos, I should be ashamed..

Thanx again :) Next time I'll help you
 
  • #10
malawi_glenn said:
yeah, of course.. I have done angular momentum in QM now in 6h.. maybe shall go and cook some food ;) this one was so obviuos, I should be ashamed..

Thanx again :) Next time I'll help you

I am glad I could help!
I know the feeling of not seeing simple things after having worked for several hours.
I am sure you'll help me at some point with something in differential geometry/GR/supersymmetry/string theory/etc etc :-)
 

Related to Is Quantum Angular Momentum Rotationally Invariant?

1. What is quantum angular momentum?

Quantum angular momentum is a physical quantity that describes the rotational motion of a particle or system of particles in the framework of quantum mechanics. It is a fundamental property of particles and is related to their intrinsic spin.

2. How is quantum angular momentum measured?

Quantum angular momentum is measured in units of Planck's constant (h-bar). It can be measured using various experimental techniques, such as electron spin resonance or nuclear magnetic resonance.

3. What is the significance of quantum angular momentum?

Quantum angular momentum plays a crucial role in many areas of physics, including atomic and molecular physics, condensed matter physics, and particle physics. It also has applications in technology, such as in magnetic resonance imaging (MRI) and quantum computing.

4. How does quantum angular momentum differ from classical angular momentum?

Classical angular momentum is a continuous quantity and can take on any value, while quantum angular momentum is quantized and can only take on discrete values. Additionally, classical angular momentum follows the laws of classical mechanics, while quantum angular momentum follows the laws of quantum mechanics.

5. Can quantum angular momentum be changed?

Quantum angular momentum is a conserved quantity, meaning it cannot be created or destroyed. However, it can be transferred between particles or systems through interactions, such as collisions or electromagnetic forces.

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