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Calculating rms displacement

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    The question is as stated:
    "The ##H_2## molecule has oscillatory excitations. In classical physics the energy can be approximated to \begin{equation} E = \frac{p^2}{2m} + \frac{m \omega^2 x^2}{2} \end{equation}where m is the reduced mass. Quantum mechanics can be applied to this equation. Using as a starting point \begin{equation} \Psi \propto e^{\frac{-m \omega x^2 }{2 \hbar}} e^{- i E_{o} \frac{t}{ \hbar}} \end{equation}, show how to find ##<x^2>## for this state, from this formula and measured frequency 132 THz, calculate the rms displacement.
    2. Relevant equations
    Aside from the given ones, probably just the one for expectation value equation for ##<x^2>## :
    \begin{equation}
    <x^2> = \int_{- \infty}^{\infty} \Psi^{*}(x,t) x^2 \Psi(x,t) dx
    \end{equation}

    I guess the rms value is the the square root of that expectation value.

    3. The attempt at a solution
    So for ##\Psi##, I just made the equation given equal to a constant and the proportional part, and made that the wave equation:
    \begin{equation} \Psi (x,t) = k e^{\frac{-m \omega x^2}{2 \hbar}} e^{- i E_{o} \frac{t}{ \hbar}} \end{equation}

    Its complex conjugate:
    \begin{equation} \Psi (x,t) = k e^{\frac{-m \omega x^2}{2 \hbar}} e^{ i E_{o} \frac{t}{ \hbar}} \end{equation}

    Then, I sat up the integral:

    \begin{equation}
    <x^2> = k^2 \int_{- \infty}^{\infty} e^{\frac{-m \omega x^2}{2 \hbar}} e^{ i E_{o} \frac{t}{ \hbar}} x^2 e^{\frac{-m \omega x^2 }{2 \hbar}} e^{- i E_{o} \frac{t}{ \hbar}} dx
    \end{equation}

    The complex parts just cancel out and become 1, and the real parts can be combined, so the integral simplifies to:
    \begin{equation}
    <x^2> = k^2 \int_{- \infty}^{\infty} x^2 e^{\frac{-m \omega x^2 }{ \hbar}} dx \end{equation}

    I wasn't sure if what I did was right, since this is a non-trivial integral, but some other examples also had them, so I kept on going. According to a source:
    \begin{equation} \int_{- \infty}^{\infty} x^2 e^{-ax^2} = \frac{ \sqrt{\pi}}{2 a^{\frac{3}{2}}}
    \end{equation}

    Using that, I just substituted ##a## for ##\frac{m \omega}{\hbar}##, and got:
    \begin{equation}
    <x^2> = k^2 \frac{ \sqrt{\pi}}{2 \frac{m \omega}{\hbar}^{\frac{3}{2}}}
    \end{equation}

    After this, I guess I will have to find the value of the constant, for that, I just set up a regular expectation value:
    \begin{equation} <x^2> = k^2 \int_{- \infty}^{\infty} e^{\frac{-m \omega x^2 }{ \hbar}} dx \end{equation}

    After looking up the value of the general integral, which is ##\sqrt{\frac{\pi}{a}}## where ##a = \frac{m \omega}{\hbar}##, I get:
    \begin{equation} k^2 = \sqrt{\frac{2 m \omega}{h}} \end{equation}

    After some plugging in and simplifications, I get the answer:
    \begin{equation}
    <x^2> = \frac{\hbar}{\sqrt{2} m \omega}
    \end{equation}

    What I wanted to know is if my process and the formulae I used were correct. I am still somewhat rusty on expectation values so I actually wasn't sure on whether my whole process is the correct approach. I am also not exactly sure on how to get the rms displacement, do I really just square the expectation value or is there more to the calculation than that? I'm also not sure if the first equation given in the question has any relevance in the calculations. Thanks in advance.
     
  2. jcsd
  3. Nov 13, 2014 #2
    I'd say to recheck your k (constant value), since it only requires a normalisation i.e.

    [tex]\int_{-infinity}^{+infinity}\Psi ^{*}\Psi dx = 1[/tex]

    seems correct other than that.
     
  4. Nov 13, 2014 #3

    DrClaude

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    Staff: Mentor

    Correct.

    Gaussian integrals are everywhere, better get used to them!

    I strongly suggest you keep it in terms of ##\hbar##, it will help below.

    You have a slight error here. Redo the algebra (see my suggestion above).

    Everything is fine, although I think is often better to start by normalizing the wave function. (Maybe it is just a question of taste.)

    I guess you mean to take the square root. The root-mean-square of any observale is defined as ##\sqrt{\langle \hat{A}^2 \rangle}##.

    I guess it is mostly to provide context.
     
  5. Nov 13, 2014 #4

    DrClaude

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    Staff: Mentor

    The ##k## value he gets is correct.
     
  6. Nov 13, 2014 #5
    Would my error just be that the denominator has a 2 and not its square root? I realized I didn't notice that the 2 in the original expression was not under a square root.
     
  7. Nov 14, 2014 #6

    DrClaude

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    Staff: Mentor

    Yes, the only problem was the superfluous square root.
     
  8. Nov 14, 2014 #7

    Orodruin

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    This problem can also be solved without actually computing the integral or normalizing the wave function (i.e., finding k). I can think of two different paths, one involving annihilation and creation operators (which the OP may or may not be familiar with) and another using partial integration.
     
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