Finding energy eigenvalue of a harmonic oscillator using a Hamiltonian

[JordanGo]

Homework Statement

Find the energy eigenvalue.

Homework Equations

H = (p^2)/2m + 1/2m(w^2)(x^2) + λ(x^2)

Hψ=Eψ

The Attempt at a Solution

So this is what I got so far:

((-h/2m)(∂^2/∂x^2)+(m(w^2)/2 - λ)(x^2))ψ=Eψ

I'm not sure if I should solve this using a differential equation method, or is there an easier trick?

Thank you!

 

[Simon Bridge]

I'm not sure if I should solve this using a differential equation method, or is there an easier trick?

... you should solve using the DE methods.

Since the potential is in x only, the partials become exact[1]:
$$\left ( \frac{-\hbar^2}{2m}\frac{d^2}{dx^2} + \zeta x^2 - E \right )\psi(x) = 0 : 2\zeta = m\omega^2+2\lambda$$... and it is really useful for these things to learn to use $\LaTeX$ :)

Note: You had:

((-h/2m)(∂^2/∂x^2)+(m(w^2)/2 - λ)(x^2))ψ=Eψ

I don't know why you didn't square your $\hbar$ or why you have a $-\lambda$ in there but the setup has a $+\lambda$.
Also, I'm guessing by "$h$" you meant $\hbar=h/2\pi$ since the momentum operator is $-i\hbar\frac{\partial}{\partial x}$.

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[1] if you are expected to do this in 3D, then you'll have to deal with the other two directions.

 

[JordanGo]

Ok, so now I tried to perform the differential equation, but then I get an equation in the form of:

Ψ=Ae^(ς-E)x+Be^(-(ς-E)x)

How can I find the energy eigenvalue from this equation?

 

[Simon Bridge]

You put it into the Schrodinger equation and solve for E.

But I am not sure that wavefunction is a solution... don't you associate Gaussians with harmonic oscillator potentials?

 

[JordanGo]

We haven't done any sort of Gaussian...

But I came up with another way to solve, can you see if this makes sense:

I can write the Hamiltonian as:

$H =\frac{p^2}{2m} + x^2 (\frac{m\varpi^2}{2} + \frac{\lambda}{\sqrt{2}})$

or

$H = \frac{p^2}{2m} + \frac{mω'^2x^2}{2}$

where

$ω'=\sqrt{ω^2+\frac{\sqrt{2}λ}{m}}$

then I can use

$E = (\hbar)ω'(n+\frac{1}{2})$

 

[Simon Bridge]

Sure - you can exploit existing solutions. What what stopping you from writing $m\omega^{\prime 2} = \zeta$ before? (and using the same trick?)

 

[JordanGo]

Good point! I can also do it that way, I'll try it out.

This problem involves a 2D harmonic oscillator (which the Hamiltonian was for x). To find the degeneracy of the first excited state, can I state that since ω_x=ω_y=ω, that n = n_x + n_y.

Then I can write n_y=n - _x, and set up the following sum:
$\sum(n-n_{x}) = \frac{1}{2}n(n+1)$
where the sum is from n_x=0 to n.

So, using the equation to solve for n=1 (first excited state), I get the degeneracy is equal to 1. Is this correct or am I missing something?

 

[Simon Bridge]

Have a look at:
https://www.physicsforums.com/showthread.php?t=82076

reality check conclusions by looking at them another way:
$E_n=(n_x+n_y+1)\hbar\omega$
$n=1$ corresponds to $n_x=n_y=0$ so there is only one state with that energy corresponding to wavefunction $\psi = C_{00}\psi_0(x)\psi_0(y)$ [*]

btw: do you know the form of $\psi(x)$?

[* if it is the same $\omega$ in both directions?]
https://www.physicsforums.com/showthread.php?t=586355