# Homework Help: SHM and Quantum with gravity

1. Apr 18, 2015

### MaxwellsCat

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

2. Relevant 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

3. 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!

2. Apr 18, 2015

### Orodruin

Staff Emeritus
You are doing fine so far. What happens when you insert $x = \delta + g/\omega^2$ into your potential?

Are you familiar with how to describe translations using the momentum operator?

3. Apr 19, 2015

### MaxwellsCat

Duh, thanks...
$$V(δ) = \frac{1}{2}mω^2(δ+\frac{g}{ω^2})^2 + mg(δ + \frac{g}{ω^2}) = \frac{1}{2}mω^2(δ^2 + \frac{2δg}{ω^2} + \frac{g^2}{ω^4}) + mg(δ + \frac{g}{ω^2})$$
$$= \frac{1}{2}mω^2δ^2 + mgδ + \frac{1}{2}m\frac{g^2}{ω^2} + mgδ + \frac{mg^2}{ω^2}$$
and so
$$V(δ) = \frac{1}{2}mω^2δ^2 + 2mgδ + \frac{3}{2}\frac{mg^2}{ω^2}$$ for posterity.

Not as such, but wikipedia tells me that it's
$$\hat{T} = exp\bigg{[}\frac{-\imath x \cdot \hat{p}}{\hbar}\bigg{]} \approx 1-\frac{\imath x \cdot \hat{p}}{\hbar}$$ so I should just redefine $\hat{p}$ as $\imath \hbar (\nabla \hat{T})_{x = 0}$? Then
$$\hat{H} = -\frac{{\hbar}^2}{2m} \bigg{[}\nabla \bigg{(}1-\frac{\imath x \cdot \hat{p}}{\hbar} \bigg{)} \bigg{]}^2 + V$$ with my redefined V - I assume I can use the one that I have above? Further I'd assume that $\hat{p}$ is my original $-\imath \hbar \nabla$, right? On the right track there?

4. Apr 19, 2015

### MaxwellsCat

Due to a wonderful human who already took this class, it was suggested that I simplify $V(δ)$ by one more step:

$$V(δ) = \frac{1}{2} mω^2 \bigg{(}δ + \frac{2g}{ω^2} \bigg{)}^2 -\frac{mg^2}{ω^2}$$

Because then it's just a linear addition to $\hat{H}$ and you get energy eigenvalues that are the same $E_n = \hbar\,ω (n+\frac{1}{2}) + γ$ where γ is just a constant based on the constant offset of $\hat{H}$.

I'd still like to hear more about how to do this with the translations if someone is willing to help.

Also - I want to make clear that on my last post that 'duh' was supposed to be self-denigrating not insulting to the answerer!

Last edited: Apr 19, 2015