Quantum Mechanics - Angular Momentum

[Tangent87]

Hi, I'm doing this question at the top of page 77 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2008/Part_2/list_II.pdf ). I am stuck on the last part where we have to verify perturbation theory with the exact result. Which means I think we need to find the eigenvalues of the simultaneous eigenstates for $$H=-\gamma(B_{1}J_1+B_{3}J_3)$$. I know that |j m> is the eigenstate for $$J_3$$ but this isn't an eigenstate of $$J_1$$ so what do we do?

I have found in my notes that the simultaneous eigenstates of $$J_1$$ are given by a unitary operator but I'm sure there must be an easier way as I find it unlikely this is how the question wants us to do it.

 

[gabbagabbahey]

Hi, I'm doing this question at the top of page 77 here (http://www.maths.cam.ac.uk/undergrad/pastpapers/2008/Part_2/list_II.pdf ). I am stuck on the last part where we have to verify perturbation theory with the exact result. Which means I think we need to find the eigenvalues of the simultaneous eigenstates for $$H=-\gamma(B_{1}J_1+B_{3}J_3)$$. I know that |j m> is the eigenstate for $$J_3$$ but this isn't an eigenstate of $$J_1$$ so what do we do?

I have found in my notes that the simultaneous eigenstates of $$J_1$$ are given by a unitary operator but I'm sure there must be an easier way as I find it unlikely this is how the question wants us to do it.

Re-read the question carefully...your unperturbed Hamiltonian is when $B_1=0$ (i.e. $H_0 = -\gamma B_3 J_3$ )...you shouldn't have much trouble finding the eigenvalues & eigenstates of that. Your Perturbed Hamiltonian is $H=-\gamma((B_{1}J_1+B_{3}J_3)$...if you rescale your energies as $$\tilde{H}=-\frac{H}{\gamma B_3}$$, your unperturbed and perturbed hamiltonians become [tex]\tilde{H}_0 = J_3[/itex] and $$\tilde{H}=\tilde{H}_0+\lambda J_1$$ where [tex]\lambda \equiv \frac{B_1}{B_3} \ll 1[/itex].[/tex][/tex]

 

[Tangent87]

Ok so as you say we can easily find the e'vals and e'states of $$\tilde{H}_0 = J_3$$ and apply the perturbation theory to $$\tilde{H}=\tilde{H}_0+\lambda J_1$$ but even after this rescaling surely we still have the problem of finding the e'states of $$J_1$$ when it comes to the exact result, unless I'm missing something?

 

[Dickfore]

@OP:

Your Hamiltonian may be written as:

$$H = -\gamma (\mathbf{B} \cdot \mathbf{J}), \; \mathbf{B} = \langle B_{1}, 0, B_{3} \rangle$$

i.e. it is proportional to the component of angular momentum along the $\mathbf{B}$ field. This component is quantized in the same manner as any component of the angular momentum, including $J_{3}$. With this information, can you write the exact eigenvalues of $H$?

 

[lpetrich]

The trick here is that the direction for angular-momentum quantization can be any direction. One can choose a direction where there magnetic field is parallel to it, where the magnetic field has zero perpendicular part.

So adjust the quantization direction accordingly, complete with relabeling the coordinates as appropriate. Make the new z axis along the magnetic field's direction.

 

[Tangent87]

@OP:

Your Hamiltonian may be written as:

$$H = -\gamma (\mathbf{B} \cdot \mathbf{J}), \; \mathbf{B} = \langle B_{1}, 0, B_{3} \rangle$$

i.e. it is proportional to the component of angular momentum along the $\mathbf{B}$ field. This component is quantized in the same manner as any component of the angular momentum, including $J_{3}$. With this information, can you write the exact eigenvalues of $H$?

Sorry I don't follow, $$J_3$$ is quantised like this right: $$J_{3}|j m>=m|j m>$$ (in units where hbar=1)? And we need to find a constant K basically such that
$$\mathbf{B} \cdot \mathbf{J}|j m> = K|j m>$$?

But how can we do this when we don't know the spin j, and we're pretending we don't know the components of $$\mathbf{B}$$?

 

[Tangent87]

The trick here is that the direction for angular-momentum quantization can be any direction. One can choose a direction where there magnetic field is parallel to it, where the magnetic field has zero perpendicular part.

So adjust the quantization direction accordingly, complete with relabeling the coordinates as appropriate. Make the new z axis along the magnetic field's direction.

This stuff about choosing your axis/directions to be other than what they've given us always confuses me, surely we can't change the magnetic field to be anything other than (B_1,0,B_3) because our answer will somehow involve B_1 and B_3?

And how do you know the direction for angular-momentum quantization can be any direction? Are the physical predictions of the quantum system along a different direction always guaranteed to be the same?

Cheers.

 

[lpetrich]

What you do is to change the coordinates so that the new z-axis is in the direction of the magnetic field:

J₁*B₁ + J₂*B₂ + J₃*B₃ = J₃' * sqrt(B₁² + B₂² + B₃²)

If you rotate the coordinates, the new J₃'s eigenstates become mixtures of the old J₃'s eigenstates:

|j,m,new> = R(j,m,m')|j,m',old>

 

[Tangent87]

What you do is to change the coordinates so that the new z-axis is in the direction of the magnetic field:

J₁*B₁ + J₂*B₂ + J₃*B₃ = J₃' * sqrt(B₁² + B₂² + B₃²)

If you rotate the coordinates, the new J₃'s eigenstates become mixtures of the old J₃'s eigenstates:

|j,m,new> = R(j,m,m')|j,m',old>

Why do they just become mixtures of the old J₃ eigenstates, how come the old J₁ eigenstates aren't relevant?

 

[lpetrich]

Because angular-momentum states cannot be eigenstates of two different components at the same time. That's because they don't commute. It's easy to prove that a state being an eigenstate of two different operators implies that they commute.

Let X be a state, A and B operators, and a and b their eigenvalues for that state:
A.X = a*X
B.X = b*X
A.B.X = a*b*X
B.A.X = b*a*X
(A.B - B.A).X = 0
[A,B].X = 0

 

[Tangent87]

Sorry, I understand what you're saying about choosing new coordinates so that the new z-axis is in the direction of the magnetic field but I don't really understand the actual specifics of how you can use that to solve the original problem.

 

[lpetrich]

The way it works is simple:

J.B = J_3,new*|B|

|B| = magnetic-field strength

 

[Tangent87]

Oh I see, and so J.B*|j m> = J3,new*|B|*m*|j m>?

Thanks a lot for your help

 

[lpetrich]

That's essentially it.

(J.B)|j,m,new> = (J_3,new*|B|)|j,m,new> = |B|*J_3,new|j,m,new> = m*|B|*|j,m,new>