2
$\begingroup$

enter image description here

my answers for the first bits

$$\langle H\rangle =n\hbar\omega$$ $$\langle x\rangle =\sqrt\frac{\hbar n}{2m\omega}\cos(\omega t)$$ $$\langle p\rangle =-\sqrt\frac{\hbar m\omega n}{2}\sin(\omega t)$$

I got that the amplitude for classical oscillator is twice as big, which I dont understand why that would be?

isn't $$\langle x^2\rangle =\langle\psi (t)|x^2|\psi (t)\rangle ~?$$

I tried doing the last part by Ehrenfest's theorem but it didn't seem to work, especially for the superposition of 2 states part, any help?

$\endgroup$
2
  • 1
    $\begingroup$ For future reference < and > are marked up by MathJax (and LaTeX) as operators and therefore have space on both sides like this $<x>$. For bra-ket notation you should use \langle and \rangle to obtain $\langle x \rangle$. $\endgroup$ Jan 12, 2016 at 20:38
  • 1
    $\begingroup$ Or \left<x\right> = $\left<x\right>$ $\endgroup$ Jan 12, 2016 at 22:24

1 Answer 1

2
$\begingroup$

Let $\left|n\right\rangle$ denote the $n$th excitation beyond the ground state $\left|0\right\rangle$ so $$\left\langle n\right|\frac{m\omega^2}{2}x^2\left|n\right\rangle=\left\langle n\right|\frac{p^2}{2m}\left|n\right\rangle=\frac{2n+1}{4}\hbar\omega$$ is time-independent. A superposition $\left|\psi\right\rangle=a_i\sum_i \left|i\right\rangle$ with $\sum_i \left|a_i\right|^2=1$ satisfies $$\left\langle\psi\right|x^2\left|\psi\right\rangle=\sum_i \left|a_i\right|^2\left\langle i \right|x^2\left| i \right\rangle + 2\sum_{i<j}\text{Re} \left\langle i \right|x^2\left| j \right\rangle.$$ The time-dependence (if any) comes from diagonal terms. The annihilation operator $a=\frac{m\omega x + ip}{\sqrt{2m\hbar\omega}}$ satisfies $x\propto a+a^\dagger$ and $a a^\dagger + a^\dagger a = \frac{2H}{\hbar\omega}$, so $x^2 \propto a^2 + a^{\dagger 2}+\frac{2H}{\hbar\omega}$. Thus any time-dependence of an off-diagonal term occurs only if $i=j\pm 2$. This explains why a superposition of only two consecutive eigenstates gives a time-independent mean of $x^2$. (If your calculations indicate otherwise, you made a mistake.)

$\endgroup$
2
  • $\begingroup$ Maybe I'm confused, but why is the expectation of $\frac{m \omega^2 x^2}{2}$ equal to that of $\frac{p^2}{2m}$? $\endgroup$
    – Prahar
    Jan 13, 2016 at 0:29
  • $\begingroup$ It's well-known that, in energy eigenstates, those operators have equal means. (Do not assume this holds in superpositions of energy eigenstates.) I used this fact to accelerate an explanation of what's going on here. If you want to prove it, write each operator in terms of $a,\,a^\dagger$. You'll find the same coefficient of $aa^\dagger+a^\dagger a$ in each case, while the $a^2,\,a^{\dagger 2}$ terms have expectation 0 in energy eigenstates. Roughly, the reason this happens is because the exchange $p\leftrightarrow m\omega x$ preserves the Hamiltonian. $\endgroup$
    – J.G.
    Jan 13, 2016 at 8:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.