Is the Hamiltonian H = p^2/2m + V_0r^2 Rotationally Invariant?

[malawi_glenn]

[SOLVED] rotationally invariant hamiltonian

Homework Statement

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.

Homework Equations

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

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!

 

[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.

 

[malawi_glenn]

As I wrote, I found it out...

 

[kdv]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

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

---

I got it!

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

 

[malawi_glenn]

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

 

[vincebs]

Then mark the thread down as SOLVED under the Thread Tools menu.

 

[kdv]

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

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.

 

[malawi_glenn]

Then mark the thread down as SOLVED under the Thread Tools menu.

[SOLVED] rotationally invariant hamiltonian

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1. Homework Statement

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?

 

[Gigi]

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

 

[malawi_glenn]

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

it is very very easy...

 

[Gigi]

Hi!

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

 

[malawi_glenn]

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

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

We don't post full solutions here, see forum rules.

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

 

[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

 

[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.

 

[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!

 

[malawi_glenn]

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!

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 =)

 

[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

 

[jambaugh]

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.

Not just roughly! It is exactly the condition. The action of the symmetry group on an operator is the exponential of applying the commutator:
$$g<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\mapsto e^{\theta [J,\cdot]}H = H + \frac{\theta}{1!}[J,H] + \frac{\theta^2}{2!}[J,[J,H]]+ ...$$
If the commutator is zero the effective action is $e^{0}=\mathbf{1}$, $g<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />\mapsto H$.