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Approx. Solution To Quantum Harmonic Oscillator for |x| large enough

  1. Apr 4, 2014 #1
    Hi folks!


    \Psi(x) = Ax^ne^{-m \omega x^2 / 2 \hbar}

    is an approximate solution to the harmonic oscillator in one dimension

    -\frac{\hbar ^2}{2m} \frac{d^2\psi}{dx^2} + \frac{1}{2}m \omega ^2 x^2 \psi = E \psi

    for sufficiently large values of |x|. I thought this would be a simple matter of just plugging in the approximate solution into the harmonic oscillator equation and erase terms where large values of |x| reduces the term to 1 or 0.

    However, this turned out to be harder than expected. The first thing I am wondering is whether my approach is correct.

    Any help would be appreciated!

  2. jcsd
  3. Apr 4, 2014 #2


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    Why not just use an exact eigenfunction of the 1D oscillator and take the limit of large ##|x|##? An exact eigenfunction contains a Hermite polynomial ##H_n(x)## and for large ##|x|## we have ##H_n(x) \sim x^n##.
  4. Apr 5, 2014 #3
    My background in mathematics is not very broad, and I have not ever worked with Hermite polynomials. Would you care to show how that limit develops? My calculus experience is quite limited, unfortunately.

    H_n(x) = (-1)^ne^{x^2} \frac{d^n}{dx^n} e^{-x^2}

    I found this on wikipedia, but it seems there are other definitions of the polynomial as well.
  5. Apr 5, 2014 #4


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    Anders, I imagine six people will follow up with clever methods for solving the harmonic oscillator exactly, but I wanted to answer your question directly, namely how do you derive an approximate solution for large x. Approximation methods are important! And they often do not get the treatment they deserve.

    First, for convenience, write the equation in dimensionless form,

    -ψ'' + x2ψ = Eψ

    For large x, we ask which of the three terms we can say will dominate the others. Clearly x2ψ >> Eψ, so we can discard Eψ. We are left with two terms, and that's the least number of terms you can retain and still have a nontrivial equation! So the first approximate equation is

    -ψ'' + x2ψ = 0

    Now orders of magnitude come into play. The familiar ones are powers xn, but exceeding all of these are exponentials exp(ax), and even larger than that is exp(ax2). So the first approximate solution is ψ = C exp(x2/2). Or C x exp(x2/2), or C xn exp(x2/2), these all work equally well. Or in fact C f(x) exp(x2/2) where f(x) is any function that's polynomially bounded (no larger than xn as |x| → ∞, for some n.)

    What we do is factor out the exponential behavior by substituting ψ(x) = f(x) exp(x2/2), and put this back in the original exact equation. This gives us an exact equation for f:

    - f'' + 2xf' + f = E f

    and we must now start over, trying to find an approximation to this equation. This time a power of x is good enough, say f = xn. For any power of x, the first term is negligible compared to the other three, and with f = xn the (second) approximate equation is

    2xf' + f = E f

    giving the condition E = 2n + 1, which is the sequence of energy levels for the quantum oscillator.

    (Someone will say, "Aha, but how do you know that n must be an integer?" That's because the same solution must be valid for both x > 0 and x < 0, and on general principles a bound state wavefunction in one dimension must be real.)
    Last edited: Apr 5, 2014
  6. Apr 6, 2014 #5


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    An approach to figuring out the asymptotic behavior of the wave function is to write it in terms of a new function [itex]W(x)[/itex] as follows:

    [itex]\Psi(x) = e^{i W(x)/\hbar}[/itex]

    If we let [itex]p(x) = \dfrac{d W}{dx}[/itex], then the Schrodinger's equation
    [itex]-i \hbar^2/(2 m) \dfrac{d^2 \Psi}{dx^2} + V(x) \Psi = E \Psi[/itex]


    [itex]p^2/(2m) - i \hbar/(2m) \dfrac{dp}{dx} + V(x) = E[/itex]

    At this point, you assume that [itex]p[/itex] can be written as a power series in [itex](1/x)[/itex]:

    [itex]p = \sum_j p_j x^{\alpha - j}[/itex]

    Then you can solve for the coefficients and the leading power [itex]x^\alpha[/itex]

    For the harmonic oscillator, with [itex]V(x) = 1/2 k x^2[/itex], this leads to the result:

    [itex]p = i \sqrt{m k} x - i \hbar n/x + ...[/itex]

    where [itex]...[/itex] is terms of order [itex]1/x^2[/itex] and [itex]n[/itex] is related to the energy [itex]E[/itex] through

    [itex]E = (n+1/2) \hbar \sqrt{\dfrac{k}{m}}[/itex]

    Going back to [itex]\Psi[/itex], this gives the asymptotic form of
    [itex]\Psi = C x^n e^{- \sqrt{m k} x^2/\hbar }[/itex]
    (if I haven't made a mistake).
  7. Apr 8, 2014 #6
    Thanks, I appreciate your help. Approximation is new to me, but I I followed your reasoning most of the way. I have a few questions, if you do not mind:

    \psi '' + x^2 \psi = 0

    I understand how you came to this equation, but I did not follow your reasoning after. We have to consider orders of magnitude? This can be solved exactly, yes? Where do your orders of magnitude come in? In the approximate solutions are finding?

    I followed the separation of variables, and I was able to get the same expression as you, but again I am confused about the talk of orders of magnitude.
  8. Apr 8, 2014 #7


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    It's immediately clear that the approximate equation has (to the same order of accuracy) the approximate solution
    [tex]\psi(x)=A \exp(-x^2/2)+B \exp(+x^2/2),[/tex]
    [tex]\psi''(x)=x^2[1+\mathcal{O}(1/x^2)] \psi(x).[/tex]
    The next step is to write, using the fact that [itex]\psi[/itex] must be square integrable,
    [tex]\psi(x)=\tilde{\psi}(x) \exp(-x^2/2)[/tex]
    for the exact eigenvalue problem and then make a power-series ansatz for [itex]\tilde{\psi}[/itex], which then leads to the conclusion that the power series must in fact be a polynom in order to give a square-integrable function. This leads to a condition for the eigenvalues of the energy and also gives the solution of the time-dependent Schrödinger equation. The polynomials are the Hermite polynomials (up to normalization and a phase factor).
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