Variational Principle: Find Best Bound State for 1D Harmonic Oscillator

[cepheid]

Homework Statement

Find the best bound state on E_gs for the one-dimensional harmonic oscillator using a trial wave function of the form

\psi(x) = \frac{A}{x^2 + b^2}

where A is determined by normalization and b is an adjustable parameter.

Homework Equations

The variational principle

\langle \psi |H| \psi \rangle \geq E_{gs}

The Attempt at a Solution

This problem has some really nasty integrals. E.g., just to normalize the wavefunction, you need to calculate:

\int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2}

I was able to find most of them using residue theory. It was a last resort...the only method I could think of! I had to dig through my old complex analysis notes to remind myself of the technique using a semicircular contour of radius r and then letting r go to infinity. The integral of f(z) over the arc is zero, leaving the integral on the real axis, which is the part I want to evaluate. Anyway, I'm actually pretty sure of my answers, which are as follows

The best value of b is:

b^2 = \frac{\hbar}{m\omega \sqrt{2}}

\langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs}

My question is, in a subsequent problem, Griffiths asks us to generalize this to a trial wavefunction of the form

\psi(x) = \frac{A}{(x^2 + b^2)^n}

Huh? Did I miss something?! I had enough trouble with the integrals the first time around! I don't think contour integration is going to help me. Am I missing some more obvious method for evaluating these integrals?

 

[pam]

That first integral is more easily done with the substitution x=tan y, and some trig.

 

[cepheid]

hmm...you're right that does work rather nicely. Not too sure about the general case though.

 

[kdv]

Homework Statement

Find the best bound state on E_gs for the one-dimensional harmonic oscillator using a trial wave function of the form

\psi(x) = \frac{A}{x^2 + b^2}

where A is determined by normalization and b is an adjustable parameter.

Homework Equations

The variational principle

\langle \psi |H| \psi \rangle \geq E_{gs}

The Attempt at a Solution

This problem has some really nasty integrals. E.g., just to normalize the wavefunction, you need to calculate:

\int_{-\infty}^{\infty} \frac{dx}{(x^2 + b^2)^2}

I was able to find most of them using residue theory. It was a last resort...the only method I could think of! I had to dig through my old complex analysis notes to remind myself of the technique using a semicircular contour of radius r and then letting r go to infinity. The integral of f(z) over the arc is zero, leaving the integral on the real axis, which is the part I want to evaluate. Anyway, I'm actually pretty sure of my answers, which are as follows

The best value of b is:

b^2 = \frac{\hbar}{m\omega \sqrt{2}}

\langle \psi |H| \psi \rangle = \frac{\hbar \omega}{\sqrt{2}} \geq E_{gs}

My question is, in a subsequent problem, Griffiths asks us to generalize this to a trial wavefunction of the form

\psi(x) = \frac{A}{(x^2 + b^2)^n}

Huh? Did I miss something?! I had enough trouble with the integrals the first time around! I don't think contour integration is going to help me. Am I missing some more obvious method for evaluating these integrals?

note that

\frac{1}{(x^2+b^2)^3} = -\frac{1}{2} \frac{d}{db^2} ( \frac{1}{(x^2+b^2)^2})

You can generalize to an arbitrary n. Therefore, if you know how to integrate with 1/(b^2+x^2), you can integrate any power n. Your result will contain the nth derivative which you can probably find explicitly (it depends, what is the result of the integral with 1/(x^2+b^2) ?)

 

[niqolas619]

Not to grave dig here too bad, but I don't think the OP came to the right solution and I didn't want someone else to be confused.

The ground state energy of a harmonic oscillator is known to be

E_{gs}=\frac{1}{2} \hbar \omega &gt;<br /> \frac{1}{\sqrt{2}} \hbar \omega

The variational principle can only find a lower bound on the energy, meaning that the approximation must be greater than or equal to the actual energy of the system.

Working out the problem myself, I got a little different value of b. I believe this is where the OP went wrong.