- #1
sam400
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Homework Statement
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.
Homework 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.
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.