# Ritz method in QM applied to a particular problem

1. Jun 4, 2013

### fluidistic

1. The problem statement, all variables and given/known data
Using my classnotes and what the professor did, it is not clear to me as how I must tackle the following problem.
I'm given the potential $V(x)= V_0 \left ( \frac{x^2}{2} -K \right ) e^{-a x^2}$ where a>0.
I must estimate via Ritz method the ground state energy. I must use a trial wave-function $\Psi \propto e^{-\alpha x^2}$.

2. Relevant equations
$E=\frac{\langle \Psi , \hat H \Psi \rangle }{||\Psi||^2}$.
I know that the energy value I should get is greater than the ground state energy, because the Ritz method gives an upper bound of the ground state energy.

3. The attempt at a solution
First confusion: in the formula $E=\frac{\langle \Psi , \hat H \Psi \rangle }{||\Psi||}$, is psi a linear combination of pure states psi's. i.e. $\Psi = \sum _{n=0} ^{\infty} c_n \psi$ (which I believe so) or is it just $\Psi =ce^{-\alpha x^2}$? To get us started on the exercise my professor used the second case, but I don't understand why.
So I've used the way my professor started us. I splitted the Hamiltonian operator into kinetic and potential energies operators. Yielding $\langle \Psi , V(\hat x ) \Psi \rangle =c^2V_0 \sqrt \pi \left [ \frac{1}{2(2\alpha +a)^{3/2}} -\frac{K}{(2\alpha +a )^{1/2}} \right ]$ (I think the units make sense). While $\langle \Psi , \frac{\hat{\vec p }}{2m} \Psi \rangle =\frac{c^2\hbar ^2 \sqrt{\pi \alpha}}{2^{3/2}m}$.
I found the constant c by normalizing: $\langle \Psi , \Psi \rangle =1 \Rightarrow c^2= \sqrt{\frac{2\alpha}{\pi}}$.

So that finally $\langle \Psi , \hat H \Psi \rangle = \sqrt{\frac{2\alpha}{\pi}} V_0 \sqrt \pi \left [ \frac{1}{2(2\alpha +a)^{3/2}} -\frac{K}{(2\alpha +a)^{1/2}} \right ] + \sqrt{\frac{2\alpha}{\pi}} \frac{\hbar ^2 \sqrt{\pi \alpha}}{2^{3/2}m}$. I know I can simplify slighlty this last expression but my doubt is too big to continue. Why am I working on a single psi that is not a linear combinations of others?!
To continue, I should derivate that expression with respect to alpha and equate to 0? As if I was looking for a minimum in $E(\alpha)$ so I would obtain the value for alpha that minimize E. Is this correct?

Edit: I've derived the expression E(alpha) with respect to alpha and equated to 0 but it doesn't seem an easy job to isolate alpha.

Last edited: Jun 4, 2013
2. Jun 5, 2013

### PhysicsGente

This is also called the variational principle. You guess the wavefunction and calculate an upper bound for the ground state energy $E_{0}$.

So suppose you know the Hamiltonian $H$ but you don't know the wavefunction $\psi$. You can say $\psi=\sum_{n}c_{n}\psi_{n}$, and $H\psi_{n}=E_{n}\psi_{n}$. If we assume your wavefunction is normalized and that the set of $\psi_{n}$'s is orthonormal, then $1=\langle\psi|\psi\rangle=\langle\sum_{m}c_{m}\psi_{m}|\sum_{n}c_{n} \psi_{n}\rangle=\sum_{m,n}c_{m}^{*}c_{n}\langle\psi_{m}|\psi_{n}\rangle=\sum_{n}|c_{n}|^{2}$.

Now, note that $\langle H\rangle = \langle\sum_{m}c_{m}\psi_{m}|H\sum_{n}c_{n}\psi_{n}\rangle=\sum_{m,n}c_{m}^{*}E_{n}c_{n}\langle\psi_{m}|\psi_{n}\rangle=\sum_{n}E_{n}|c_{n}|^{2}$. Since $E_{0}\leq E_{n}$, we can conclude that $\langle H\rangle \geq E_{0}\sum_{n}|c_{n}|^2=E_{0}$.

If you minimize the Hamiltonian, then you'll get $\langle H\rangle = E_{0}\sum_{n}|c_{n}|^2=E_{0}$. The formula that you are given follows and as you can see all you need to do is guess the wavefunction.

Also, remember what you do when you want to minimize something. GL

Last edited: Jun 5, 2013
3. Jun 5, 2013

### fluidistic

Thank you very much.
In my example my guess function isn't a linear combination of some "state functions". I've basically solved the problem and it involves apparently getting a numerical value for alpha (the parameter that I variate in my case) because the equation I get isn't solvable analytically.