# Rotationally invariant hamiltonian

1. Feb 9, 2008

### malawi_glenn

[SOLVED] rotationally invariant hamiltonian

1. The problem statement, all variables and given/known data

Show that the Hamiltonian $$H = p^2/2m+V_0r^2$$ corresponding to a particle of mass m and
with $$V_0$$ constant is
a) rotationally invariant.

2. Relevant equations

Rotation operator: $$U_R(\phi ) = \exp (-i \phi \vec{J} / \hbar )$$, where $$\vec{J}$$ is the angular momentum operator.

3. The attempt at a solution

I think I should show that [U,H] = 0 ?
Or is it [J,H] = 0 ?

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I got it!

Last edited: Feb 9, 2008
2. Feb 9, 2008

### joelperr

A Hamiltonian is considered to be rotationally invariant if, after a rotation, the system still obeys Schrödinger's equation.
Equivalently, you may show that $$[U,H] = 0$$, which should be rather easy since your Hamiltonian is time-independent.

3. Feb 9, 2008

### malawi_glenn

As I wrote, I found it out...

4. Feb 9, 2008

### kdv

As you probably realized by now, proving that [H,J] = 0 implies that [U,H] = 0.

5. Feb 9, 2008

### malawi_glenn

But I said I got it... (2nd time in this post)

6. Feb 10, 2008

### vincebs

7. Feb 10, 2008

### kdv

I know. I did not know how you had solved it and I thought that you might not have used the fact that [J,anything] =0 implies [U, anything] =0. And even if you had used that, I thought that this point could be useful to someone else reading the thread. That's all.

8. Feb 11, 2008

### malawi_glenn

[SOLVED] rotationally invariant hamiltonian

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1. The problem statement, all variables and given/known data

https://www.physicsforums.com/attachment.php?attachmentid=12586&stc=1&d=1202722914

I already did, maybe its been a bug so that change did not shown up to everyone?

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9. Mar 2, 2008

### Gigi

Is it possible I can have a look at the actual calculation? It looks like a nice problem.

10. Mar 2, 2008

### malawi_glenn

it is very very easy...

11. Mar 2, 2008

### Gigi

Hi!

I really like this problem. Can I see the actual calculation? Thanks

12. Mar 2, 2008

### malawi_glenn

Sorry, it is quite lenghty, but trivial calculations..

We dont post full solutions here, see forum rules.

If you want to solve it, I can help you, but not give you the solution.

13. Mar 2, 2008

### Gigi

Yes, sure. I can try to solve it and sent it to you.
I just want to understand rotational invariance better.

I have read the theory regarding rotationally invariant Hamiltionians, i.e. that in the case I have a rotation about the z axis by an angle b, then my Hamiltionian is invariant if H(r, theta, fi)=H(r, theta, fi+b).

If H is invariant, then [Lz, H]=0 and Lz is a constant of motion.
Now [J, H] in the case that my Hamiltonian is invariant, must be equal to zero.
Thus I think this is what you are doing. Trying to show that. Yes?

I can see U is a function of J, but U is the rotation right?

Why would you need to show [U, H]=0? Wouldnt [J, H]= 0 be enough?
Couldnt one prove that H is rotationally invariant even without knowing U?

Hope you do not find my questions too trivial :)

Many thanks

14. Mar 2, 2008

### malawi_glenn

yes [J, H]= 0 is enough,

if a operator commutes with the generator of a specific symmetry, then the operator is symmetric with respect to that symmetry.

roughly speaking.

Last edited: Mar 2, 2008
15. Mar 2, 2008

### Gigi

Is it true that under a rotation SU(2) the Hamiltonian remains invariant?

Nevertheless under a Lorentz group rotation it doesn't? Thus the need for the Klein Gordon and Dirac equations?

Thanks!

16. Mar 3, 2008

### malawi_glenn

for a rotation of SU(2) you must see if H commutes with that gruops generator. And the same holds for Lorentz group, but I am not 100% sure, since my knowledge in relativistic QM at the moment is not so good, but will become in the future =)

17. Feb 10, 2012

### Miglior

Another way is by proving the expression of the hamiltonian looks the same after changing the coordinate x,y,z to x' y' z' by rotation by an angle theta

18. Feb 10, 2012

### jambaugh

Not just roughly! It is exactly the condition. The action of the symmetry group on an operator is the exponential of applying the commutator:
$$g\mapsto e^{\theta [J,\cdot]}H = H + \frac{\theta}{1!}[J,H] + \frac{\theta^2}{2!}[J,[J,H]]+ ...$$
If the commutator is zero the effective action is $e^{0}=\mathbf{1}$, $g\mapsto H$.