# Quantum mechanics: SHM expectation of $x^2$ time independent for one state but not superposition of 2 states?

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?

• 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 $\langle x \rangle$. Jan 12, 2016 at 20:38
• Or \left<x\right> = $\left<x\right>$ Jan 12, 2016 at 22:24

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.)
• Maybe I'm confused, but why is the expectation of $\frac{m \omega^2 x^2}{2}$ equal to that of $\frac{p^2}{2m}$? Jan 13, 2016 at 0:29
• 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.