Imaginary and real parts of a ground state wave function

Question

Here're key-frames from a time evolution of a quantum harmonic oscillator in it's ground state (source):

So, I do understand that such a state may equally well describe grand variety of systems, but for now I want to focus on just 1 "particle on a spring" example, just to see if the following description will be all on itself valid:

Time	Particle's position	Particle's momentum (direction only)
$1., 9.$	$x_0$ (equilibrium point)	$x_1$ (maximum stretching)
$3.$	$x_1$	$x_{-1}$ (maximum compression)
$5.$	$x_0$	$x_{-1}$
$7.$	$x_{-1}$	$x_1$

Btw, I do understand also that:

• these $x$ points are arbitrary: we could've started at any other point than $x_0$ (with it's role as "equilibrium point" unchanged), and $x_1$ and $x_{-1}$ roles might've been swapped.
• measuring upon the system in the table would not have revealed a "particle" in it's $x$ position (with described momentum), but rather in any of possible outcomes.

I'm just interested in validity of meaning I've put in that table - say, as a time evolution of just 1 eigenstate.

Answer 1

In quantum mechanics, particles do not have well-defined positions or momenta. Instead, we can only talk about the probability distribution we would get if we measured them. These probability distributions are given by the Born rule, which says that the squared norm of the wavefunction (expressed in the basis of the variable being measured) is the probability density function for the measurement of that variable. For example, the probability of measuring the position to be in some interval $[a,b]$ is given by $\int_a^b\mathrm{d}x\,\lvert\psi(x)\rvert^2$.

When a state is an eigenstate of the Hamiltonian, such as the ground state, it is called a stationary state. This is because its time evolution is entirely given by a phase factor: $\psi(x,t)=e^{-iEt/\hbar}\psi(x,0)$, where $\psi(x,0)$ is an eigenfunction with energy eigenvalue $E$. Notably, when you take the squared norm, the time dependence vanishes, since $\lvert e^{-iEt/\hbar}\rvert^2=1$, meaning the probability distributions for measurements are the same at any time. Therefore, it does not make sense to think of a particle in a quantum harmonic oscillator as moving back and forth, analogous to a classical harmonic oscillator. In particular, the particle effectively does not change if it is in an energy eigenstate. However, if the particle is not in an energy eigenstate (instead being in a superposition of states with different energies), then the expectation values of the position and momentum will oscillate similar to the classical case.

Answer 2

As has already been said by Sandejo, the position and momentum do not have well-defined values for a quantum particle. However, it is possible to relate the behavior of the quantum particle with that of a classical particle by considering the expectation values of the position and momentum operators in the state of interest.

For the quantum harmonic oscillator, the expectation value of both position and momentum are zero at all times in an energy eigenstate, so from this point of view the motion of the quantum particle does not resemble at all the back-and-forth motion that you expect from a classical oscillator.

As also said by Sandejo, a superposition of energy eigenstates could lead to time-dependent position and momentum expectation values. At this point, it is interesting to ask the question of which quantum states would lead to a motion that most closely resembles the behaviour of the quantum harmonic oscillator. The answer is something called coherent states, typically labelled with $|\alpha\rangle$. They can be defined in a variety of ways (e.g. as the eigenstates of the lowering operator: $\hat{a}|\alpha\rangle=\alpha|\alpha\rangle$), and in the energy basis they take the following form:

$$ |\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle, $$

where $|n\rangle$ are the energy eigenstates. For these states, the expectation values of the position and momentum operators exactly follow the classical trajectories.