- #1
gasar8
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Homework Statement
[/B]
Particle is moving in 2D harmonic potential with Hamiltonian:
[tex] H_0 = \frac{1}{2m} (p_x^2+p_y^2)+ \frac{1}{2}m \omega^2 (x^2+4y^2) [/tex]
a) Find eigenvalues, eigenfunctions and degeneracy of ground, first and second excited state.
b) How does [tex] \Delta H = \lambda x^2y [/tex] split second excited state? (First order of perturbation)
Homework Equations
[tex] x= \frac{x_0}{ \sqrt{2} } (a_x+a_x^\dagger) \\
y= \frac{x_0}{ \sqrt{2} } (a_y+a_y^\dagger)\\
a |n> = \sqrt{n} |n-1>\\
a^\dagger |n> = \sqrt{n+1} |n+1>\\
a a^\dagger - a^\dagger a = 1 [/tex]
The Attempt at a Solution
a)
I was trying to solve Schroedinger's equation first with separating x,y, using [tex] H |\psi> = E |\psi>, |\psi>=X(x)Y(y), [/tex] but got stuck at:
[tex] -\frac{\hbar}{2m} \frac{X''}{X}+ \frac{1}{2}m \omega^2 x^2 = E [/tex],
[tex] -\frac{\hbar}{2m} \frac{Y''}{Y}+ 2 m \omega^2 y^2 = E [/tex].
Then, I tried by using anihillation and creation operators (for p and x), but gave up, because all the [tex](a_i+a_i^\dagger)^2[/tex] became too complicated for calculating the ground, first and second state.
b)
I am assuming that the correction is 0, because y in ΔH is even function and if we integrate it across all space it gives us 0, but I can't prove it.
Second excited state has got |11>, |20> and |02>, so I was calculating [tex]\Delta H |20>, \Delta H |02>,\Delta H |11>,[/tex]:
[tex]\lambda x^2 y |20> = \frac{\lambda \sqrt{2} x_0^3}{4} (a_x+a_x^\dagger)^2(a_y+a_y^\dagger)|20>[/tex]
using commutator, I got:
[tex] \lambda x^2 y |20> = \frac{\lambda \sqrt{2} x_0^3}{4} (a_x^2+a_x^{\dagger 2}+1+2 a^\dagger a)(a_y+a_y^\dagger)|20>[/tex].
Here I have got problems, because if I calculate [tex](a+a^\dagger)^2 = a^2+a a^\dagger + a^\dagger a + a^{\dagger 2}= \\ =a^2+a^{\dagger 2}+2 a a^\dagger-1 \\=a^2+a^{\dagger 2}+2 a^\dagger a+1[/tex]
and [tex] a^\dagger a [/tex]gives different result on |20> as [tex]a a^\dagger.[/tex]
Can anyone please help me?
Thank you very much.