# Perturbation theory

1. Jun 1, 2005

### broegger

Hi,

I'm new to this subject, so bear with me. We consider the harmonic oscillator with a pertubation:

$$\hat{H}' = \alpha\hat{p}.$$​

(What kind of a perturbation is that anyway, it's not a disturbance in the potential, what does it correspond to physically.)

Now I have to calculate the first and second order energy corrections. I express p by the ladder operators:

$$p=i\sqrt{\tfrac{\hslash m\omega}2}(a_+-a_-).$$​

I find that the first order correction is 0, and that the second order correction is $$E_n^2=m\alpha^2/2$$, that is, the energy shift is INDEPENDENT of n (i.e. it is the same for all excited states of the oscillator), but this can't be true, can it?

Last edited: Jun 1, 2005
2. Jun 2, 2005

### seratend

Hint 1: H_harm= k1^2.p^2+k2^2.q^2
If you add this potential (alpha.p) what do you obtain?

Hint 2: what is the interaction of a spinless charged particle with a magnetic field?

Seratend.

3. Jun 2, 2005

### broegger

Well, it could be solved analytically, but I don't know how. I'm not particularly familiar with solving such differential equations. I've tried solving it numerically in Matlab, and this confirms my result, but it's not quite satisfying.

I don't know much about magnetic interactions in QM, haven't touched the subject in our intro course, unfortunately.

4. Jun 2, 2005

### seratend

Hint 3: a^2+ka= (a+k/2)^2 - k^2

Hint 4: do you known the classical hamiltonian of a particle with an electromagnetic field?

Seratend.

5. Jun 2, 2005

### broegger

I've tried completing the square, but I don't know how to go on.

The energy of a classical particle in an electromagnetic field is just the sum of the energy associated with the electric and magnetic field, respectively, right?

$$E = \frac{q_1q_2}{4\pi\epsilon_0r^2}-\mu\cdot B$$​

6. Jun 2, 2005

### seratend

Ok,

First, Let't take the harmonic oscillator hamiltonian:
H= p^2+k.q^2
You know the solutions of such an hamiltonian. Therefore I suppose you know the solution of H'=(p+a)^2+k.q^2, don't you?

(k>0, I have choosen the units to got simple equations, you are free to choose your own units)

If I had the interaction hamiltonian, I get:

choosing alpha= 2a

H=p^2+k.q^2+2a.p= (p+a)^2 + k.q^2 -a^2

We just recover another harmonic oscillator with p'=p+a:
[p,q]=[p+a,q]=-ihbar.

The constant -a^2 does not change the spectrum (when you write the hamiltonian it is defined up to a constant).
Or if you prefer a more mathematical way:
the commutator [p'^2 + k.q^2 -a^2, p'^2 + k.q^2]=0 => we have the same eigenvalues. Or just solve the equation with an additional constant (not very difficult)

Second: Classical hamiltonian: H= (P - qA)^2/2m + eV. Where A and V are the electromagnetic field potentials.
Now, develop the square and assume A^2 is very very small, what do you obtain?

Seratend.

Edit: correction of minor typo errors + others : )

Last edited: Jun 2, 2005
7. Jun 2, 2005

### broegger

I don't know if I'm doing it right, but I get the same result as perturbation theory - E = (n+1/2)hw - ma^2/2 - (that's a good sign, isn't it , so perturbation theory yields the exact result in this case?

8. Jun 2, 2005

### seratend

For the sign, I think it's ok.
bravo!

Seratend

9. Jun 2, 2005

### broegger

Weeh, thanks for helping!

10. Jun 2, 2005

### seratend

Additionnal question for you: why does the pertubation theory give the same result in this case?

Seratend.

11. Jun 2, 2005

### broegger

Hmm, I think it's because we were able to factor the Hamiltonian perfectly, but I don't know, I was wondering the same thing...

I also have a quick question for you (I asked this in a another thread, but I'm still thinking about it). Suppose you have a hydrogen electron in the mixed state:
$$|\psi\rangle=\frac1{\sqrt2}(|210\rangle + |211\rangle)$$​
(|nlm> denotes an eigenstate of H, L^2 and L_z - n is the main quantum number and l and m is the quantum numbers associated with L^2 and L_z, respectively).

Now, I find that L_x and L_y has different expectation values (h/sqrt(2) and 0, respectively), but how can this be, when there is nothing to distinguish the x- and y-directions in this case??

12. Jun 2, 2005

### seratend

Apply a rotation (z axis) to the state, does it change?

Seratend.

13. Jun 2, 2005

### seratend

Think on the functions: f(x) and f(x+h)~f(x)+hf'(x). When do you have the equality?
Think now on the expansion of the hamiltonian, with the pertubation theory. It just allows more possibilities.

Seratend.

EDIT change = into ~to avoid confusion : )

14. Jun 2, 2005

### broegger

No, I don't think so. It's symmetric with respect to x and y, that's the problem. I might be wrong, though :)

15. Jun 2, 2005

### seratend

I am not sure, but you should try to work a little : )

here is the rotation matrix for an angle phi:
Code (Text):
(exp(-i.phi/2)      0)
(0      exp(iphi/2))
Do you keep saying the state does not change?

Seratend.

EDIT: corrected a missing i in the roation matrix. : )

Last edited: Jun 2, 2005
16. Jun 2, 2005

### broegger

Hehe, you're right, I should...

Well no, I just thought "geometrically", you know :) I'll try working it out from the start again...

17. Jun 2, 2005

### seratend

Good. Progression in the knowledge requires somtimes some efforts : )

Seratend.