Breathing of a wave packet in the harmonic oscillator

The possible results are $E_0=\hbar\omega/2$ and $E_2=5\hbar\omega/2$, each with probability $1/2$. The expectation value of the energy is

$$\langle\hat H\rangle=\frac12\cdot\frac{\hbar\omega}{2}+\frac12\cdot\frac{5\hbar\omega}{2}=\frac32\hbar\omega,$$

and it is constant. Each eigenstate evolves with its phase:

$$\ket{\psi(t)}=\frac{1}{\sqrt2}\Bigl(e^{-i\omega t/2}\ket0+e^{-5i\omega t/2}\ket2\Bigr)=\frac{e^{-i\omega t/2}}{\sqrt2}\Bigl(\ket0+e^{-2i\omega t}\ket2\Bigr).$$

The operators $\hat X$ and $\hat P$ are combinations of $a$ and $a^\dagger$, which connect a state $\ket n$ only to the states $\ket{n\pm1}$. The matrix elements $\bra0\hat X\ket0$, $\bra2\hat X\ket2$ and $\bra0\hat X\ket2$ therefore all vanish, since $\ket0$ and $\ket2$ differ by two quanta. The same holds for $\hat P$. Thus $\langle\hat X\rangle=\langle\hat P\rangle=0$ at all times. This is consistent with Ehrenfest's theorem: for the oscillator, the expectation values obey exactly the classical equations $\frac{d\langle\hat X\rangle}{dt}=\frac{\langle\hat P\rangle}{m}$ and $\frac{d\langle\hat P\rangle}{dt}=-m\omega^2\langle\hat X\rangle$; with the initial conditions $\langle\hat X\rangle(0)=\langle\hat P\rangle(0)=0$, their solution is identically zero. The centre of the wave packet remains stationary at the bottom of the well.

From the hint, $\hat X^2=\frac{\ell^2}{2}\bigl(a^2+(a^\dagger)^2+2N+\mathbf{1}\bigr)$. Let us calculate its relevant matrix elements. The diagonal terms involve only $2N+\mathbf{1}$:

$$\bra0\hat X^2\ket0=\frac{\ell^2}{2},\qquad\bra2\hat X^2\ket2=\frac{\ell^2}{2}(2\times2+1)=\frac{5\ell^2}{2}.$$

The off-diagonal term comes from $a^2$: since $a^2\ket2=a\sqrt2\ket1=\sqrt2\ket0$, we have $\bra0\hat X^2\ket2=\frac{\ell^2}{2}\sqrt2=\frac{\ell^2}{\sqrt2}$, and $\bra2\hat X^2\ket0$ is its complex conjugate, equal to the same real number. For the state $\ket{\psi(t)}$, with coefficients $c_0=\frac{1}{\sqrt2}$ and $c_2=\frac{1}{\sqrt2}e^{-2i\omega t}$ up to the global phase,

$$\langle\hat X^2\rangle(t)=|c_0|^2\frac{\ell^2}{2}+|c_2|^2\frac{5\ell^2}{2}+2\operatorname{Re}\Bigl(c_0^*c_2\frac{\ell^2}{\sqrt2}\Bigr)=\frac32\ell^2+\frac{\ell^2}{\sqrt2}\cos(2\omega t).$$

For $\hat P^2=-\frac{\hbar^2}{2\ell^2}\bigl(a^2+(a^\dagger)^2-2N-\mathbf{1}\bigr)$, the same calculations give, with the sign of the $a^2$ and $(a^\dagger)^2$ terms reversed:

$$\langle\hat P^2\rangle(t)=\frac{\hbar^2}{\ell^2}\Bigl(\frac32-\frac{1}{\sqrt2}\cos(2\omega t)\Bigr).$$

Finally, using $\frac{\hbar^2}{2m\ell^2}=\frac12m\omega^2\ell^2=\frac{\hbar\omega}{2}$,

$$\langle\hat H\rangle=\frac{\langle\hat P^2\rangle}{2m}+\frac12m\omega^2\langle\hat X^2\rangle=\frac{\hbar\omega}{2}\Bigl(\frac32-\frac{\cos2\omega t}{\sqrt2}+\frac32+\frac{\cos2\omega t}{\sqrt2}\Bigr)=\frac32\hbar\omega,$$

which is indeed constant.

Since the expectation values vanish, $\langle\hat X^2\rangle$ and $\langle\hat P^2\rangle$ are the variances. The wave packet remains centred at $x=0$, but its width oscillates at the angular frequency $2\omega$: it periodically broadens and contracts, as though it were breathing. When its position width is maximal, its momentum width is minimal, and conversely: the energy periodically changes between potential and kinetic forms, while its total remains constant. The angular frequency $2\omega$ is the Bohr frequency $(E_2-E_0)/\hbar$, the only one present in the state. It is twice the classical frequency, which can be understood as follows: a cloud of classical particles oscillating about the origin with symmetric phases broadens and contracts twice per period. The product of the standard deviations is

$$\sigma_X\sigma_P=\hbar\sqrt{\Bigl(\frac32+\frac{\cos2\omega t}{\sqrt2}\Bigr)\Bigl(\frac32-\frac{\cos2\omega t}{\sqrt2}\Bigr)}=\hbar\sqrt{\frac94-\frac{\cos^22\omega t}{2}}\geq\hbar\sqrt{\frac74}\simeq1{,}32\,\hbar,$$

which is always greater than $\hbar/2$.

We have

$$\psi(x,0)=\frac{1}{\sqrt2}\Bigl(1+\frac{2\xi^2-1}{\sqrt2}\Bigr)\phi_0(x)=\frac{\sqrt2-1+2\xi^2}{2}\,\phi_0(x).$$

The density is proportional to $(\sqrt2-1+2\xi^2)^2e^{-\xi^2}$. It is even, and its maximum lies away from the origin. Differentiating the logarithm of this function with respect to $\xi$, $\frac{8\xi}{\sqrt2-1+2\xi^2}-2\xi$, shows that the non-zero extrema satisfy $2\xi^2=4-(\sqrt2-1)=5-\sqrt2$, so that $|\xi|\simeq1{,}34$. The particle is therefore most likely to be detected near $x\simeq\pm1{,}34\,\ell$, which is consistent with the fact that at $t=0$, $\langle\hat X^2\rangle$ takes its maximum value $\bigl(\frac32+\frac{1}{\sqrt2}\bigr)\ell^2$.