- #1

MaxwellsCat

- 12

- 2

## Homework Statement

1. Harmonic Oscillator on Earth Gravity :

In class, we solve the Harmonic Oscillator Problem, with a potential

$$ V(x) = \frac{m ω^2 x^2}{2} \quad (1)$$

with ##ω = \frac{k}{m}## being the classical frequency. Now, assume that x is a vertical direction and that we place the harmonic oscillator close to the Earth surface. Now, if x grows upwards, the potential will be

$$V(x) = \frac{m ω^2 x^2}{2} + mgx + C \quad (2)$$

with ##g = 9.8\frac{m}{s^2}## and C an arbitrary (and irrelevant) constant.

a. First think about the classical problem. The equilibrium point is no longer at ##x = 0##, but a displaced point where the tension and gravity forces are equilibrated. Find that point and rewrite the potential in terms of a new variable representing departures from the equilibrium point. What would be the motion of a classical particle under the potential given in Eq. (2) ?

b. Now think about the quantum problem. Without gravity, the energy eigenvalues are given by ##E_{n}^{h.o.} = \frac{\hbar ω (2n + 1)}{2}## and the corresponding wave functions ##ψ_{n}^{h.o.}(x)## can be written in terms of odd and even Hermite polynomials and a Gaussian function of x (Here h.o. means harmonic oscillator). Using these results, derive the new energy eigenvalues ##E_n## and eigen- functions ##ψ_n## in the presence of gravity, Eq. (2). Hint : Can you make a similar redefinition of the coordinates as you did in the classical case ?

## Homework Equations

$$ \hat{H}\,|ψ_n\!> = E_n\,|ψ_n\!> $$

$$ H_{ij} = <\!ψ_i|\,\hat{H}\,|ψ_j\!>$$

$$ \bigg{[}\frac{-{\hbar}^2}{2m} {\partial_x}^2 + \frac{m ω^2 x^2}{2} + mgx\bigg{]} ψ = E_n ψ $$

$$ ψ(x) = f_n(x)\exp{\frac{-ξ^2}{2}} $$

$$ ξ = \sqrt{\frac{m ω}{\hbar}} x $$

$$f_n(x) = \sqrt{\frac{m ω}{\hbar π}} \frac{1}{\sqrt{2^n n!}} H_n(ξ)$$ where ##H_n(ξ)## are the Hermite Polynomials

## The Attempt at a Solution

For the first part I have tried this:

$$V(x) = \frac{1}{2} m ω^2 x^2 + mgx \implies -\partial_x V = -mω^2x - mg $$

For equilibrium, ##F = 0## and so

$$0 = -mω^2x - mg \implies x = \frac{-g}{ω^2} $$

##mg## is constant so define

$$ δ = x + \frac{-g}{ω^2}$$

where x is the old equilibrium position and δ the new.

Am I on the right track or is this just completely wrong? This should be trivial but for some reason I'm having a lot of trouble with it.

For the second part, I've got and idea that perhaps I can use ##ψ_0## and then define the others in terms of the raising and lowering operators, but the issue comes in actually finding the ##ψ_n## because I'm not sure that I can redefine the coordinates like I did in the classical case, because I don't know if I can/have to redefine ##\hat{p}## in that case. If I don't have to, then it should be the relatively simple case of just applying our good pal Frobenius to the Schrödinger equation and crank it out, but I don't know if that's valid. It should be possible, and I think you should just end up with the same ##ψ## as above, but I'm not sure. I've tried it that way and it obviously works out, you get the same solutions for ##ψ## just in a different variable, but the problem is that ##E_n## also work out to be the same, but there should be an offset term, right? Or can you just say that the offset is your ##mgx## and go? I also thought that if I knew how to write ##\hat{H}## as a matrix that would help immensely because finding eigenvectors is easy, but sadly, I don't - at least not without ##ψ_n##. This last idea I have high hopes for, if I could find out how to actually write the darn matrix!