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SHM and Quantum with gravity

  1. Apr 18, 2015 #1
    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. jcsd
  3. Apr 18, 2015 #2

    Orodruin

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    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?
     
  4. Apr 19, 2015 #3
    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?
     
  5. Apr 19, 2015 #4
    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
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