Anharmonic Oscillator - Energy Shift Calculation Using 1st Order Perturbation

[Hart]

Homework Statement

<br /> V(x) = \frac{1}{2}mw^{2}x^{2} + \lambdax^{4}<br /> <br />

Using first-order perturbation theory to calculate the energy shift of:

1. The ground state:

<br /> <br /> \psi_{0}(x) = (2\pi\sigma)^{\frac{-1}{4}}\exp(\frac{-x^{2}}{4\sigma})<br /> <br />

of the harmonic oscillator, where:

<br /> <br /> \sigma = \frac{\hbar}{2m\omega}<br /> <br />

Homework Equations

Within the stated question.

The Attempt at a Solution

If \lambda = 0 then the potential would correspond to a harmonic oscillator of classical frequency \omega with energy levels defined as:

E_{0n} = (n+\frac{1}{2})\hbar\omega

And with the ground state energy eigenfunction as:

U_{00} = (\frac{m\omega}{\pi\hbar})^{1/4}.exp(\frac{-m\omega}{2\hbar}x^{2})

The inclusion of the \lambdax^{4} term is what results in the anharmonic problem.

By substituting from the equation for the ground state energy eigenfunction above, into

E_{1n} = H_{nn}^{&#039;}

Get the result of:

E_{10} = ((\frac{m\omega}{\pi\hbar})^{1/2})\int_{\infty}^{\infty}\lambda x^{4}[exp(\frac{-m\omega}{\hbar}x^{2})]dx

Therefore:

E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda

Which I believe is the energy shift for the ground state.


I do need to do the same for the first excited state, so with a slightly different expression for \psi_{1}(x), but I think I could do that once I know how to do this.

 

[vela]

What's your question?

 

[masqau]

Yes E10 is the first order correction in ur perturbation equation.
its the expectation value.
similarly u can find out the higher order correction.
and ur perturb potential is x^4 not 4.

 

[Hart]

Sorry about the typo, the perturbation is indeed \lambda x^{4}.

 

[Hart]

Ok, so I've now got that expression for the first order correction to the ground state energy shift.. but where does \psi_{o}(x) factor into it?!

 

[masqau]

Ok, so I've now got that expression for the first order correction to the ground state energy shift.. but where does \psi_{o}(x) factor into it?!

$$E10=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)$$

 

[Hart]

$$E10=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)$$

E_{10}=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)

Sorry, I just don't see how what I need to do next at the moment :S

 

[masqau]

E_{10}=\int \psi_{0}^{*}(x)x^4\psi_{0}(x)

Sorry, I just don't see how what I need to do next at the moment :S

substitute the value of ground state wavefunction and integrate it.

thanks for making my equation nice...

 

[Hart]

\psi_{0}^{*}(x)\psi_{o}= <br /> <br /> ((2\pi\sigma)^{\frac{-1}{4}})^{2}(exp(\frac{-x^{2}}{4\sigma}))^{2} = (2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})

x^{4}\psi_{0}^{*}(x)\psi_{o}= x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})

\int x^{4}\psi_{0}^{*}(x)\psi_{o}= \int x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma}) = (2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma})

(2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma}) = <br /> <br /> <br /> \frac{(-2x)(2\pi\sigma)^{\frac{-1}{2}}}{2 \sigma}exp(\frac{-x^{2}}{2\sigma}) <br /> <br />

Hopefully this is along the right lines?

 

[vela]

Oh, so your "attempt at a solution" above actually wasn't your attempt at a solution, but part of the problem statement.

Everything up until you actually integrated looks fine. How did you come up with that result?

 

[Hart]

Apologies for any confusion with my initial post on this thread.

I did get that result, but that I believe is the purturbation?

So this integration I'm going through should give me the energy shift then?

 

[vela]

Yes, the integral you wrote down gives you the correction to the ground state energy, but you're not doing the actual integral correctly. What steps are you doing to try integrate that function? Your mistake is somewhere in those steps. (Also, don't forget the limits on the integral.)

 

[Hart]

\psi_{0}^{*}(x)\psi_{o}= <br /> <br /> ((2\pi\sigma)^{\frac{-1}{4}})^{2}(exp(\frac{-x^{2}}{4\sigma}))^{2} = (2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})

x^{4}\psi_{0}^{*}(x)\psi_{o}= x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma})

\int x^{4}\psi_{0}^{*}(x)\psi_{o}= \int x^{4}(2\pi\sigma)^{\frac{-1}{2}}exp(\frac{-x^{2}}{2\sigma}) = (2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma})

(2\pi\sigma)^{\frac{-1}{2}} \int x^{4}exp(\frac{-x^{2}}{2\sigma}) = <br /> <br /> <br /> \frac{(-2x)(2\pi\sigma)^{\frac{-1}{2}}}{2 \sigma}exp(\frac{-x^{2}}{2\sigma}) <br /> <br />

Hopefully this is along the right lines?

Yes, the integral you wrote down gives you the correction to the ground state energy, but you're not doing the actual integral correctly. What steps are you doing to try integrate that function? Your mistake is somewhere in those steps. (Also, don't forget the limits on the integral.)

Sorry, I don't see where in those integration steps is incorrect? Is that what you're hinting at?

 

[vela]

I'm referring to the last line you wrote. The first three lines have nothing to do with integration. They're just algebra. In the last line, you apparently tried to integrate the function, and something went wrong. (If you differentiate the RHS, you'll find you don't recover the integrand.)

 

[Hart]

\int_{-\infty}^{\infty} exp \left( \frac{-x^{2}}{2\sigma} \right) = \left(\frac{-1}{x^{2}}\right)\left(-2x\right)exp \left( \frac{-x^{2}}{2\sigma} \right) = \left(\frac{2}{x}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)<br /> <br />

Therefore:

\int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = \left(\left(2x^{3}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)\right) - \int_{-\infty}^{\infty}\left(\left(8x^{2}\right)exp\left(\frac{-x^{2}}{2\sigma}\right)\right)<br />

Back on track?

 

[vela]

Nope. I have no idea what you did in the first line. I'm kind of guessing you tried to use

\int e^{ax}dx=\frac{1}{a}e^{ax}+c

which you can't do because the exponent in your integral is of the form ax^2 not just ax and that makes all the difference in the world. Moreover, if that's what you did, I don't see where the factor of (-2x) comes from or what happened to the factor of 2\sigma.

You need to show or explain what steps you're taking rather than just posting your results. When all you write down is the integral and an incorrect answer, it's impossible to see where your calculations are going astray.

 

[Hart]

I did try to use that rule, but that was incorrect method then. I don't know how else to compute the integral :|

 

[vela]

Have you learned about the gamma function? If not, integrating by parts is probably the most appropriate way for you.

 

[Hart]

No I havn't learned about that function, hence why I used integration by parts method.

Just calculated the integral again:

\int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = 3\sqrt{2\pi}\sigma^{3/2}

 

[vela]

That's almost right. The numerical part of your answer is right.

Whenever you do these complicated calculations, you should perform some sanity checks to see if you might have made a mistake. One of the quickest and most useful is to see if your answer has the right units. In this case, the LHS has units of length^5 (4 from the x^4 plus 1 from the dx) while the RHS has units of length^3 (\sigma has units of length^2). So look for places in your calculations where you might have dropped factors of \sigma.

 

[Hart]

\int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = 3\sqrt{2\pi}\sigma^{5/2}

 

[Hart]

<br /> <br /> \left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\int_{-\infty}^{\infty} x^{4}exp\left(\frac{-x^{2}}{2\sigma}\right) = \left(\left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\right)\left(3\sqrt{2\pi}\sigma^{3/2}\right) = 6 \sqrt{\pi}\sigma^{2}

.. hopefully?

 

[vela]

<br /> \left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\int_{-\infty}^{\infty} x^{4}exp\left(\frac{-x^{2}}{2\sigma}\right) = \left(\left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\right)\left(3\sqrt{2\pi}\sigma^{3/2}\right) = 6 \sqrt{\pi}\sigma^{2}

.. hopefully?

Nope. You made numerous errors besides using what you should already know is the incorrect value for the integral.

\int_{-\infty}^{\infty} x^{4}exp(\frac{-x^{2}}{2\sigma}) = 3\sqrt{2\pi}\sigma^{5/2}

This is correct.

 

[Hart]

I'll try again..

\left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\int_{-\infty}^{\infty} x^{4}exp\left(\frac{-x^{2}}{2\sigma}\right) = \left(\left(2\pi\sigma\right)^-\left(\frac{1}{2}\right)\right)\left(3\sqrt{2\pi}\sigma^{5/2}\right) = 3(2\pi)\sigma^{(\frac{-1}{2})+(\frac{5}{2})} = 6 {\pi}\sigma^{2}

 

[vela]

No. Try again.

Fixed your LaTeX:

\left(2\pi\sigma\right)^{-1/2}}\int_{-\infty}^{\infty} x^{4}\exp\left(\frac{-x^{2}}{2\sigma}\right) = \left(2\pi\sigma\right)^{-1/2}} (3\sqrt{2\pi}\sigma^{5/2}) = \cdots

 

[Hart]

.. = 3\sigma^{2}

 

[vela]

Yup!

 

[Hart]

Yay!

So just to check, that is the result for the energy shift of the ground state?

i.e. E_{10}^{&#039;} = 3\sigma

If so, where does this factor in (defined at the start):

E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda

I seem to have two values for the same thing :(

 

[vela]

Yay!

So just to check, that is the result for the energy shift of the ground state?

i.e. E_{10}^{&#039;} = 3\sigma

No. Go back and look at what you were trying to calculate. The result you got earlier was just part of the calculation. Also, I assume you had a typo here. If not, you're doing something I don't understand again.

If so, where does this factor in (defined at the start):

E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda

I seem to have two values for the same thing :(

What is \sigma defined as?

I think at this point it would help you to go back over this problem and write up what you have so far cleanly. Get rid of the false starts and the dead ends. Whittle it down the solution to its essentials. Think about how you'd explain how to solve this problem to someone else.

 

[Hart]

Just substituted:

\sigma = \frac{\hbar}{2m\omega}

Into:

E_{10} = 3\sigma

Which becomes the previously defined statement:

E_{10} = \frac{3\hbar^{2}}{4m^{2}w^{2}}\lambda