BvU said:
Steady states for the harmonic oscillator are eigenfunctions of the Hamiltonian: ##H\phi(x) = E\phi(x)##. The solutions of the time-independen Schroedinger equation are distinct wavefunctions ##\phi_n(r)## with distinct energies ##E_n##. Their expectation value for ##\bar x## is zero because ##\phi_n^* \phi_n ## is symmetric and x is antisymmetric.
That is the same, I wanted to say with that, just graphical:
BvU said:
ok. here is my steady state: \psi(x,t) = \phi(x)e^{-iEt/\hbar}
\bar{x} = \int_{-\infty}^{+\infty} dx \phi^*(x)e^{iEt/\hbar} x \phi(x)e^{-iEt/\hbar} = \int_{-\infty}^{+\infty} |\phi(x)|^2 x
And this has to be zero. Because an integral is just the area between the graph of a function and the x-axis. and the "x" brings a "point reflection.
I make an example. I calculate the expectation value for the position for ## \psi_1 ## which is:
\psi_1(\zeta) = (\frac{m\omega}{\pi\hbar})^{1/4} \frac{1}{\sqrt{2^11!}} H_1(\zeta) e^{-0,5\zeta^2}<br />
= (\frac{m\omega}{\pi\hbar})^{1/4} \frac{1}{\sqrt{2}} 2\zeta e^{-0,5\zeta^2} = (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2}
(Is that right? because my skript says sth. different and I don`t find a proper source elsewere)
\rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} d\zeta (\frac{m\omega}{\pi\hbar})^{1/2} 2 \zeta^2 e^{-\zeta^2} \zeta = 0, because of the two made explanations. (the one part of this integral is symmetric, the other one antisymmetric...) and because of \zeta = \sqrt{\frac{m\omega}{\hbar}} x also the expectation value of x is zero.
BvU said:
What about e.g. ##\phi_1^* \phi_0 ## ?
\rightarrow \bar{\zeta} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} \cdot \zeta \cdot (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} = \int_{-\infty}^{+\infty} (\frac{m\omega}{\pi\hbar})^{1/2} e^{-\zeta^2} \sqrt{2} \zeta^2 = (\frac{m\omega}{\pi\hbar})^{1/2} \frac{1}{\sqrt{2}} \sqrt{\pi} The integral was solved with a gauss-integral... so not zero..
is there a rule for that?
just the integral ##\int \phi_1^* \phi_0 ## would be zero. The integral for Hermite polynomials says that.
BvU said:
Arbitrary general states ##\Phi(r)## can be written as ##\sum_n C_n\phi_n## with ##C_n## derived from ##\Phi(r)## at t= 0 and they develop in time as
##\Phi(r,t) = \sum_n C_n\phi_n e^{-{i\over \hbar} E_n t}## .
A simple case is ##\Phi(r) = {1\over \sqrt 2} \left ( \phi_0 + \phi_1 \right )## where the time development of ##\bar x## no longer vanishes. Doing the integrals with the Hermite polynomials is hard work (but worth it). Easier to use ladder operators, but you may not have seen those yet.
Check the first picture in
wikipedia
so the time development vanishes for ##\phi_n^* \phi_n ## and ##\phi_n^* \phi_m ## because they are steady states. ##\Phi(r) = {1\over \sqrt 2} \left ( \phi_0 + \phi_1 \right )## is also solution of schroedinger equation, but the time development doesn`t vanish.
so you mean I shall calculate this:::\bar{\zeta} = \int_{-\infty}^{+\infty} \frac{1}{\sqrt{2} } (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} e^{iE_0t/\hbar} + (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{iE_1t/\hbar} \cdot \zeta \cdot \frac{1}{sqrt{2}} (\frac{m\omega}{\pi\hbar})^{1/4} e^{-0,5\zeta^2} e^{-iE_0t/\hbar} + (\frac{m\omega}{\pi\hbar})^{1/4} \sqrt{2} \zeta e^{-0,5\zeta^2} e^{-iE_1t/\hbar} d\zeta?
So the answer to the question, how the expectation values for a linear harmonic oscillator are, is:
## \bar{x} ## is zero for ##\phi_n^* \phi_n ## and a constant for ##\phi_n^* \phi_m ##.
## \bar{p} ## is zero for ##\phi_n^* \phi_n ## and for ##\phi_n^* \phi_m ##
and ## \bar{x} ## + ## \bar{p} ## aren`t zero, when we use a combination of different states?
