Pertubation Theory - 3D harmonic oscillator

1. Apr 28, 2010

landpt

1. The problem statement, all variables and given/known data

A particle with mass m moves in the potential:

$$V(x,y,z) = \frac{1}{2} k(x^{2}+y^{2}+z^{2}+ \lambda x y z)$$

considering that lambda is low.

a) Calculate the ground state energy accordingly to Pertubations Theory of the second order.
b) Calculate the energies of the first three excited states accordingly to Pertubations Theory.

2. Relevant equations

Pertubation Theory:

$$E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}> = 0$$

$$E_{n}^{(2)} = \sum_{i\neq n}^{q} \frac{| <\psi_{n}^{(0)}|H^{1}|\psi_{i}^{(0)}> |^{2}}{E_{n}^{(0)}-E_{i}^{(0)}}$$

3. The attempt at a solution

a) This was easy, I guess. Not very problematic, except the correction of the second order. Let me resume what was my problem here:

We know that, accordingly to Pertubations Theory, we need to "fix" the ground state energy. Since we're going to do it second order, we'll have:

$$E_{0} = E_{0} ^{(0)} + E_{0} ^{(1)} + E_{0} ^{(2)}$$

The energies of the unpertubated state is calculated:

$$E_{n} ^{(0)} = (n_{x}+n_{y}+n_{z}+\frac{3}{2})h\omega$$

$$E_{0} ^{(0)} = \frac{3}{2}h\omega$$

However the pertubated states need to be calculated:

$$E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}>$$

We need to calculate the wave function. Since it's an harmonic oscillator, it will be for each coordinate:

$$u_{0}^{0}(x) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} x^{2}}$$

$$u_{0}^{0}(y) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} y^{2}}$$

$$u_{0}^{0}(z) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} z^{2}}$$

So the wave function will be:

$$\psi_{0}^{(0)}(x,y,z) = u_{0}^{0}(x).u_{0}^{0}(y).u_{0}^{0}(z) = (\frac{mw}{\pi h})^\frac{3}{4} e^{-\frac{mw}{h} (x^{2}+y^{2}+z^{2})}$$

Calculating and considering

$$H^{1} = \frac{1}{2}mw^{2}\lambda xyz$$

We'll have:

$$E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}> = 0$$

Which I think it's expected. However the second order is not that easy to calculate, we'll have an infinite series:

$$E_{n}^{(2)} = \sum_{i\neq n}^{q} \frac{| <\psi_{n}^{(0)}|H^{1}|\psi_{i}^{(0)}> |^{2}}{E_{n}^{(0)}-E_{i}^{(0)}}$$

I calculated each one, starting with
$$\psi_{1}^{(0)}$$
until
$$\psi_{3}^{(0)}$$
something like:

$$E_{0}^{(2)} \approx \frac{| <\psi_{0}^{(0)}|H^{1}|\psi_{1}^{(0)}> |^{2}}{E_{0}^{(0)}-E_{1}^{(0)}} + \frac{| <\psi_{0}^{(0)}|H^{1}|\psi_{2}^{(0)}> |^{2}}{E_{0}^{(0)}-E_{2}^{(0)}} + \frac{| <\psi_{0}^{(0)}|H^{1}|\psi_{3}^{(0)}> |^{2}}{E_{0}^{(0)}-E_{3}^{(0)}}$$

Didn't make more than 3 because it'd be very low that we could skip them... Is it okay to do it? I heard there was another method to do it, using partition energy, however i can't find that much information about it.

The real problem, however, is in b)... We're requested to calculate the energies of the first three states.
Since they share the same energy, we need to consider the degeneracy. So the first three states are:
$$\psi_{0,0,1}$$ and $$\psi_{0,1,0}$$ and $$\psi_{1,0,0}$$

Using the equation of the hermite polynomials (as for a)), I calculated for $$\psi_{0,0,1}$$:

$$u_{n}(x) = (\frac{mw}{\pi h})^\frac{1}{4} \frac{1}{\sqrt{2^{n}n!}} H_{n}(y) e^{-y^{2}/2}$$

where
$$y=\sqrt{mw/h}x$$

$$u_{0}(x) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} x^{2}}$$

$$u_{0}(y) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} y^{2}}$$

$$u_{1}(z) = (\frac{mw}{\pi h})^\frac{3}{4} (\frac{\sqrt{2}}{\pi ^{\frac{1}{4}}}) z e^{-\frac{mw}{h} z^{2}}$$

So it'll be:

$$\psi_{a} = \psi_{0,0,1} (x,y,z) = \frac{\sqrt{2} e^{-(\frac{mw}{h})(x^{2}+y^{2}+z^{2})} (\frac{mw}{h})^{\frac{5}{4}} z}{\pi^{\frac{3}{4}}}$$

Following the same procedures for the others we'll have:
$$\psi_{b} = \psi_{0,1,0} (x,y,z) = \frac{\sqrt{2} e^{-(\frac{mw}{h})(x^{2}+y^{2}+z^{2})} (\frac{mw}{h})^{\frac{5}{4}} y}{\pi^{\frac{3}{4}}}$$

$$\psi_{c} = \psi_{1,0,0} (x,y,z) = \frac{\sqrt{2} e^{-(\frac{mw}{h})(x^{2}+y^{2}+z^{2})} (\frac{mw}{h})^{\frac{5}{4}} x}{\pi^{\frac{3}{4}}}$$

Now we have to build a matrix $$W_{i,j} = <\psi_{i}^{0}|H^{1}|\psi_{j}^{0}>$$

So we have and calculate:

$$\psi_{aa} = <\psi_{a}^{0}|H^{1}|\psi_{a}^{0}> = 0 = \psi_{bb} = \psi_{cc}$$

$$\psi_{ab} = \psi_{ba} = <\psi_{b}^{0}|H^{1}|\psi_{a}^{0}> = 0$$

$$\psi_{ac} = \psi_{ca} = <\psi_{a}^{0}|H^{1}|\psi_{c}^{0}> = 0$$

Therefore the whole matrix is 0... and i can't calculate the energies of the degenerate states... What am I doing wrong in here?

I used Mathematica to solve some integrals in here (and to avoid handwritten mistakes). Any help or advice would be great. Thanks.

PS: I know it's not "h", but "h-bar", however i did write it in latex before. Shouldnt matter that much because thats not the problem.

Last edited: Apr 29, 2010
2. Apr 29, 2010

vela

Staff Emeritus
Are you familiar with solving the harmonic oscillator problem using annihilation and creation operators? It's quite easy to show that only a finite number of terms will contribute to the second-order correction.

You can show the same thing by considering the orthogonality of the Hermite polynomials. The argument's not as straightforward as with the operator approach, but it's not that difficult either.

3. Apr 29, 2010

vela

Staff Emeritus
I didn't see any mistakes in your work. What you're finding is that there is no first-order correction to the energies of those states. The perturbation doesn't lift the degeneracy to first order.

4. Apr 30, 2010

landpt

Yes, you're right... It should be zero (therefore there is no first order correction of those energies).

I found out today that there was a mistake in the equation of the potential - it should be only $$\lambda x y$$ instead $$\lambda x y z$$. Now there is actually a correction. Thanks.

About operators A+ and A-, i started reviewing it and now it's easier, indeed. Thanks, once again.