Quantum Mechanics (finding the Hamiltonian of a quantum top)

[MelissaM]

Hello,

This was part of my midterm exam that i couldn't solve.
Any help is extremely appreciated.

Problem: The K.E. of a rotating top is given as L^2/2I where L is its angular momentum and I is its moment of inertia. Consider a charged top placed at a constant magnetic field. Assume that the magnetic momentum of the top is proportional to L, M=KL (K is a cst that can be derived from the classical distribution of the charge that is known). Write the Hamiltonian for the quantum top and find the energy eigenstates and energy eigenvalues of the quantum top.This is what i wrote during the exam:

T= L^2/2I and V= -M.B (where B is the magnetic field)
so H=L^2/2I -M.B = L^2/2I - KL.B

then i said since it's a rotating top let's suppose that B is along the z-axis which limits our L to L_z
then, H= L_z( L_z/2I - K.B_z)

and i stopped there.

Now looking calmly at the problem, i realized that i absolutely looked over the fact that it is charged. I think i need to be using A ( the electrostatic potential) along with B but still i have no idea how to start...


Please any help would be awesome!

Thank you.

 

[MelissaM]

maybe i should use : H= 1/2m (p-q.A)^2 + qφ but since V=0 ⇒ H= 1/2m × (p - q.A)^2

 

[paralleltransport]

maybe i should use : H= 1/2m (p-q.A)^2 + qφ but since V=0 ⇒ H= 1/2m × (p - q.A)^2

Hi melissa, I think you were on the right track? Because

maybe i should use : H= 1/2m (p-q.A)^2 + qφ but since V=0 ⇒ H= 1/2m × (p - q.A)^2

Hi melissa, if you start from your generic hamiltonian:

$$\hat{H} = \frac{(p-q A)^2}{ 2 m}$$

you'll just arrive at the generic Hamiltonian below:

$$\hat{H} \approx {p^2 \over 2m} + K B \cdot L + ... = {L^2 \over 2 I} + K B \cdot L$$

See http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec5_compressed.pdf

This is your original hamiltonian which is correct.

Suppose B is oriented a particular direction: Then rotate your basis so that B is in the z direction ( Because the kinetic term has no preferred direction it can be diagonalized in any direction of space). Then your eigenstates are now
$$|l_z \rangle$$

with energy following properties:

$$ L^2 | l_z \rangle = \hbar^2 l_z (l_z + 1) | l_z \rangle$$

I think you know how the rest looks like from here.

 

[MelissaM]

Hi melissa, I think you were on the right track? BecauseHi melissa, if you start from your generic hamiltonian:

$$\hat{H} = \frac{(p-q A)^2}{ 2 m}$$

you'll just arrive at the generic Hamiltonian below:

$$\hat{H} \approx {p^2 \over 2m} + K B \cdot L + ... = {L^2 \over 2 I} + K B \cdot L$$

See http://www.tcm.phy.cam.ac.uk/~bds10/aqp/lec5_compressed.pdf

This is your original hamiltonian which is correct.

Suppose B is oriented a particular direction: Then rotate your basis so that B is in the z direction ( Because the kinetic term has no preferred direction it can be diagonalized in any direction of space). Then your eigenstates are now
$$|l_z \rangle$$

with energy following properties:

$$ L^2 | l_z \rangle = \hbar^2 l_z (l_z + 1) | l_z \rangle$$

I think you know how the rest looks like from here.

Yes absolutely.
Thank you for the help!