# Perturbation theory

1. Feb 5, 2005

### JamesJames

For the harmonic oscillator $$V(x) = \frac{1}{2}kx^2$$, the allowed energies are $$E_n=(n+1/2)h \omega$$ where $$\omega = \sqrt{k/m}$$ is the classical frequency. Now suppose the spring constant increases slightly: k -> $$(1 + \epsilon)k$$. Calculate the first order perturbation in the energy.

This is 6.2 from Griffith' s book and after this question he gives the following hint although according to me, the hint is more confusing than the question without the hint:

Hint
What is H' here? It is not necessary- in fact it is not permitted - to calculate a single integral in doing this problem.

I understand what the formula looks like..it is
$$E_n^1 = <\psi_n^0 |H'| \psi_n^0>$$

but how can this be done without evaluating a single integral? Also what is H' ?

James

Last edited: Feb 5, 2005
2. Feb 5, 2005

### dextercioby

H' must the perturbation.Namely the small term added to the original hamiltonian and which shifts the energy levels and also the initial quantum states.U need to find that perturbation.

Daniel.

P.S.I think it should be linear term in "epsion"... :uhh:

3. Feb 5, 2005

### Galileo

The new potential is: $V'(x)=\frac{1}{2}(1+\epsilon)kx^2=\frac{1}{2}kx^2+\frac{1}{2}\epsilon kx^2$

Therefore the new hamiltonian is:

$$H=H_0+\frac{1}{2}\epsilon kx^2$$
where $H_0$ is the old hamiltonian.

EDIT: Whoops. Forgot the k.

Last edited: Feb 5, 2005
4. Feb 5, 2005

$$\frac{1}{2}\epsilon k x^{2} [/itex] if u drop the k (you have no reason to do that),it won't be that obvious that the perturbation is proportional (and much smaller due to the magnitude of the proportionality constant) to the unperturbed one. Daniel. 5. Feb 5, 2005 ### JamesJames Now just one second.....is H' going to affected? It was [tex]\sigma \hbar \omega x\hat$$

I don' t see how this is going to be affected by k changing.....$$\omega$$ contains the $$\epsilon$$ but $$\omega$$ itself does not change right? According to me H' must be same

James

6. Feb 5, 2005

### dextercioby

Hold on,that's the problem...Find H'...And then see whether it is any different (than what ?? )

Daniel.

7. Feb 5, 2005

### JamesJames

$$E_n^1 = <\phi_n^0 |\sigma \hbar\omega \hat x| \phi_n^0>$$ so according to me just the $$\omega$$ bit will be affected. I don' t see what else in the expression for $$E_n^1$$ could be affected. i.e. the results of the integrals won' t be affected

James

8. Feb 5, 2005

### dextercioby

From the expression u posted,should i understand that
$$\hat{H}'=\sigma\hbar\omega\hat{x}$$

?If so,then who's every symbol of the RHS??

Daniel.