(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A particle with mass m moves in the potential:

[tex]V(x,y,z) = \frac{1}{2} k(x^{2}+y^{2}+z^{2}+ \lambda x y z) [/tex]

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:

[tex] E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}> = 0 [/tex]

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

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:

[tex] E_{0} = E_{0} ^{(0)} + E_{0} ^{(1)} + E_{0} ^{(2)} [/tex]

The energies of the unpertubated state is calculated:

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

[tex] E_{0} ^{(0)} = \frac{3}{2}h\omega [/tex]

However the pertubated states need to be calculated:

[tex] E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}> [/tex]

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

[tex]u_{0}^{0}(x) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} x^{2}} [/tex]

[tex]u_{0}^{0}(y) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} y^{2}} [/tex]

[tex]u_{0}^{0}(z) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} z^{2}} [/tex]

So the wave function will be:

[tex]\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})}[/tex]

Calculating and considering

[tex]H^{1} = \frac{1}{2}mw^{2}\lambda xyz[/tex]

We'll have:

[tex] E_{0} ^{(1)} = <\psi_{(0)}^{0}|H^{1}|\psi_{0}^{(0)}> = 0 [/tex]

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

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

I calculated each one, starting with

[tex]\psi_{1}^{(0)}[/tex]

until

[tex]\psi_{3}^{(0)}[/tex]

something like:

[tex] 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)}}[/tex]

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:

[tex] \psi_{0,0,1} [/tex] and [tex]\psi_{0,1,0}[/tex] and [tex]\psi_{1,0,0}[/tex]

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

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

where

[tex]y=\sqrt{mw/h}x[/tex]

[tex]u_{0}(x) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} x^{2}} [/tex]

[tex]u_{0}(y) = (\frac{mw}{\pi h})^\frac{1}{4} e^{-\frac{mw}{h} y^{2}} [/tex]

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

So it'll be:

[tex] \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}}} [/tex]

Following the same procedures for the others we'll have:

[tex] \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}}} [/tex]

[tex] \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}}} [/tex]

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

So we have and calculate:

[tex] \psi_{aa} = <\psi_{a}^{0}|H^{1}|\psi_{a}^{0}> = 0 = \psi_{bb} = \psi_{cc} [/tex]

[tex] \psi_{ab} = \psi_{ba} = <\psi_{b}^{0}|H^{1}|\psi_{a}^{0}> = 0 [/tex]

[tex] \psi_{ac} = \psi_{ca} = <\psi_{a}^{0}|H^{1}|\psi_{c}^{0}> = 0 [/tex]

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Pertubation Theory - 3D harmonic oscillator

**Physics Forums | Science Articles, Homework Help, Discussion**