1. The problem statement, all variables and given/known data Can someone please give me some hints how to solve this problem. Show that expected value for the kinetic energy is the same as the expected value for the potential energy for a harmonic oscillator in gound state. 2. Relevant equations how to start with it? 3. The attempt at a solution 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
One way would be to calculate the expectation values <T> and <V> separately and show that they are equal. You know what the ground state wavefunction for the harmonic oscillator is, so you have to do a couple of integrals.
hello! I have been trying to write my calculations by using 'latex' but then i get a problem, it only shows me the very first part of my solutions when i want to add more it does not work at all.....do you know what might be a problem?
Sorry, I don't know what could be wrong. Why don't you start slow and write just a couple of simple equations for practice. You may wish to follow the link below to see how latex is used to write multiple equations. https://www.physicsforums.com/library.php?do=view_item&itemid=58
[tex]\psi_{0}= (\frac{\alpha}{\pi})^{\frac{1}{4}} e^{\frac{-y^2}{2}}[/tex] [tex]y= \sqrt{\frac{m\omega}{\hbar}}x\Rightarrow y=\sqrt{\alpha}x[/tex] [tex]\alpha= \frac{m\omega}{\hbar}[/tex] [tex]<|x^2|>=\int_{-\infty}^{\infty}dxx^2|{\psi_{0}}^2|=\sqrt{\frac{m\omega}{\pi \hbar}}\int_{-\infty}^{\infty}dxx^2e^{\frac{-m \omega x^2}{\hbar}}=I[/tex] [tex]\int_{-\infty}^{\infty}dxx^2e^{-\alpha x^2}=\frac{1}{2\alpha}\sqrt{\frac{\pi}{\alpha}}[/tex] [tex]I= \frac{1 \hbar}{2m \omega}[/tex] for [tex]<|p^2|>=\frac{m \hbar \omega}{2}[/tex] [tex]<|E_{k}| >= \frac{1}{2m}|<|p^2>|= \frac{\hbar \omega}{4}[/tex] [tex]<|E_{p}|> = \frac{m\omega^2}{2}<|x^2|>= \frac{\hbar \omega}{4}[/tex] can i calculate it this way? I have problems with finding formulas for the expected value for kinetic and potential energy....
[tex] <|p^2|>=-\hbar^2 \int_{-\infty}^{\infty}dxe^{\frac{-m \omega x^2}{2\hbar}}\frac{\partial ^2}{\partial x^2}e^{\frac{-m \omega x^2}{2\hbar}}\sqrt{\frac{m\omega}{\pi \hbar}}= \hbar m \omega -m^2 \omega^2<|x^2|>=\frac{m \hbar \omega}{2}[/tex]