How to Calculate Energy Corrections using Perturbation Theory

In summary, the conversation discusses the perturbation of a harmonic oscillator with the perturbation \hat{H}' = \alpha\hat{p}, and the calculation of first and second order energy corrections using ladder operators. It is found that the first order correction is 0 and the second order correction is E_n^2=m\alpha^2/2, independent of n. There is a discussion on the validity of this result and the use of perturbation theory. The conversation also touches on the classical Hamiltonian of a particle in an electromagnetic field and the relation between L_x and L_y for a mixed state of a hydrogen electron.
  • #1
broegger
257
0
Hi,

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

[tex]\hat{H}' = \alpha\hat{p}.[/tex]​

(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:

[tex]p=i\sqrt{\tfrac{\hslash m\omega}2}(a_+-a_-).[/tex]​

I find that the first order correction is 0, and that the second order correction is [tex]E_n^2=m\alpha^2/2[/tex], 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?
 
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  • #2
broegger said:
Hi,

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

[tex]\hat{H}' = \alpha\hat{p}.[/tex]​

(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:

[tex]p=i\sqrt{\tfrac{\hslash m\omega}2}(a_+-a_-).[/tex]​

I find that the first order correction is 0, and that the second order correction is [tex]E_n^2=m\alpha^2/2[/tex], 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?

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
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
broegger said:
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.

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
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?

[tex]E = \frac{q_1q_2}{4\pi\epsilon_0r^2}-\mu\cdot B[/tex]​
 
  • #6
broegger said:
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?

[tex]E = \frac{q_1q_2}{4\pi\epsilon_0r^2}-\mu\cdot B[/tex]​

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 : )
 
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  • #7
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
broegger said:
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?

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

Seratend
 
  • #9
Weeh, thanks for helping!
 
  • #10
Additionnal question for you: why does the pertubation theory give the same result in this case?

Seratend.
 
  • #11
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:
[tex]|\psi\rangle=\frac1{\sqrt2}(|210\rangle + |211\rangle)[/tex]​
(|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
broegger said:
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??

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

Seratend.
 
  • #13
broegger said:
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...

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

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

here is the rotation matrix for an angle phi:
Code:
(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. : )
 
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  • #16
seratend said:
I am not sure, but you should try to work a little : )
Hehe, you're right, I should...

seratend said:
here is the rotation matrix for an angle phi:
(exp(-i.phi/2) 0)
(0 exp(iphi/2))
Do you keep saying the state does not change?
Well no, I just thought "geometrically", you know :) I'll try working it out from the start again...
 
  • #17
broegger said:
Hehe, you're right, I should...
Well no, I just thought "geometrically", you know :) I'll try working it out from the start again...

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

Seratend.
 

Related to How to Calculate Energy Corrections using Perturbation Theory

What is Perturbation Theory?

Perturbation theory is a mathematical method used in physics and other sciences to approximate the solutions of complex problems by breaking them down into simpler, solvable parts. It is particularly useful in calculating energy corrections for quantum mechanical systems.

Why is Perturbation Theory used in calculating energy corrections?

Perturbation theory is used because it provides a systematic way of approximating the solutions of complex problems, making them easier to solve. In the case of energy corrections, perturbation theory allows us to calculate the small changes in energy caused by perturbations, or disturbances, in the system.

What are the steps involved in calculating energy corrections using Perturbation Theory?

The steps involved in calculating energy corrections using Perturbation Theory are:

  • First, we need to define the unperturbed system, which is the system without any disturbances.
  • Next, we introduce a perturbation to the system, which could be in the form of an external force or potential.
  • Then, we use the unperturbed system as a starting point and apply perturbation theory to calculate the perturbed energy levels.
  • Finally, we can use the perturbed energy levels to calculate the energy corrections caused by the perturbation.

What is the difference between first-order and higher-order perturbation theory?

First-order perturbation theory only considers the first-order effects of the perturbation on the energy levels, while higher-order perturbation theory takes into account higher-order effects. Higher-order perturbation theory is more accurate, but also more complex and time-consuming to calculate.

How can one determine the accuracy of the calculated energy corrections using Perturbation Theory?

The accuracy of the calculated energy corrections using Perturbation Theory can be determined by comparing the results to experimental data or more accurate theoretical calculations. In general, the accuracy improves as higher-order perturbation theory is used and as the perturbation becomes smaller.

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