# Superposition and stationary states

1. Sep 27, 2015

### DiracPool

I was watching some Steve Spicklemire QM videos and had a question/check my knowledge..

When we measure a the state of a system, say a particle in a box or a quantum harmonic oscillator (QSHO), we "collapse" the superposition of the system and end up with one eigenstate and one eigenvalue. This collapsed state may reflect the position of a particle or it's energy or momentum at the time we measured it.

Before the measurement though, the system exists as a superposition of several eigenstates which reflect a myriad of energies, positions and momentums of the particle, the participation of each in the total wave function is given by a factor, c1, c2, c3, etc. Is this correct?

Two questions:

1) In a given superposed state, are ALL possible energies, positions, etc. featured in the general wave function, with the more extreme eigenstates only minimally represented or "weighted" proportionally in the vector space? Or is it that there are typically only a few eigenstates involved in the superposition. If the latter, how is it determined what eigenstates are included?

2) It seems from Steve's videos of the particle in a box and QSHO animations that the time evolution of the superposition state is necessary for motion to exist in the system. That is, it is wholly the fact that the QSHO exists as a superposition of these many eigenstates that yields the behavior of, say, a particle oscillating back and forth. If not for the superposition, then there's no oscillation or dynamic behavior in the system. So measurement kills the dynamic behavior of the system and superposition allows the system to "move" or evolve? Is this how this works?

FF to 13:15

2. Sep 27, 2015

### Staff: Mentor

It depends on your system and where it comes from.
Take a ground state atom in a laser field tuned to a specific resonance, and you'll get a superposition between two states with negligible contributions from other states.
Take a ground state atom in a very intense laser field not tuned to a resonance and you'll get something much more complex.
Only if you measure energy eigenstates. If you measure momentum, for example, you still have motion afterwards (and you even know how much).

3. Sep 28, 2015

### DiracPool

How does that work? Doesn't a stationary state mean by definition that system is not moving? Aren't the only thing doing any moving are the omega phasors rotating in time? How do you get momentum here? However, I'm not really even looking at exceptional cases here, although they'd also be interesting to know. I'm just asking, in general, does a quantum system typically have to exist as a superposition of eigenstates in order for it to exhibit dynamic properties, such as a ball bouncing, a ballistic trajectory of a bullet, atoms oscillating in a molecule, etc. It just seems from the video I posted above that this would have to be the case..

4. Sep 28, 2015

### Staff: Mentor

No, it means that the state is an eigenstate of the Hamiltonian (or something that commutes with the Hamiltonian, which comes down to the same thing). These states do not change (except for a phase) as they evolve over time according to the Schrodinger equation. One such state is a state of constant momentum, equivalent to a classical particle moving at a constant speed. Of course the (expectation value of the) position of the particle will be changing in this stationary state, not surprising as the position operator does not commute with the Hamiltonian.

There's no such thing as a state that is or is not a superposition. You pick a basis, and in that basis some states will be superpositions and others won't; pick a different basis and some of the superpositions in the old basis won't be in the new basis, and vice versa.

Last edited: Sep 29, 2015
5. Sep 29, 2015

### vanhees71

This is a bit misleading. First of all you are right that a pure stationary state is described by a statistical operator
$$\hat{R}(t)=|u_E(t) \rangle \langle u_E(t) |$$
with a normalized energy eigenvector, which in the Schrödinger picture of time evolution evolves like
$$|u_E(t) \rangle=\exp(-\mathrm{i} \hat{H} t) |u_E(0) \rangle=\exp(-\mathrm{i} E t) |u_E(0) \rangle,$$
i.e., the representing vector changes by a phase factor. The state doesn't change at all, i.e.,
$$\hat{R}(t)=\hat{R}(0)=\text{const}.$$
That's why it's a stationary state, which is defined to be constant in time (in the Schrödinger picture of time evolution).

Consequently also the expectation values of observables that are not explicitly time dependent are constant in time. Indeed, in the Schrödinger picture such an observable is represented by a self-adjoint operator that doesn't depend on time at all, i.e.,
$$\hat{A}(t)=\hat{A}(0)=\text{const}$$
and consequently
$$\langle A(t) \rangle=\mathrm{Tr} [\hat{R}(t) \hat{A}(t)] = \mathrm{Tr} [\hat{R}(0) \hat{A}(0)]=\text{const}.$$
Of course, that does not mean that a particle in a stationary state always has 0 momentum.

Now to the free particle. Here, no energy-eigenstate is represented by a true Hilbert-space vector but by a generalized eigenket, which is only "normalizable" to a $\delta$ distribution,
$$\langle p|p' \rangle=2 \pi \delta(p-p').$$
The energy eigenstates are uniquely labeled by the momentum eigenstates in this case, because
$$\hat{H}=\frac{\hat{p}^2}{2m}.$$
Thus also any eigenstate of the Hamiltonian is not a representant of a state the particle can be in. A proper state is rather represented by a "wave packet", i.e., a true Hilbert-space vector
$$|\psi(t) \rangle=\int_{\mathbb{R}} \mathrm{d} p \frac{1}{\sqrt{2 \pi}} A(p) \exp(-\mathrm{i} E(p) t) |p \rangle, \quad E(p)=\frac{p^2}{2m}$$
which is normalizable,
$$\langle \psi(t)| \psi(t) \rangle=\frac{1}{2 \pi} \int_{\mathbb{R}} \mathrm{d} p \int_{\mathbb{R}} \mathrm{d} p' A^*(p) A(p') \exp[-\mathrm{i}(E(p')-E(p))] \langle p|p' \rangle =\int_{\mathbb{R}} \mathrm{d} p |A(p)|^2 \stackrel{!}{=}1.$$
Thus the function $A(p)$ must be square integrable.

A nice example is the Gaussian wave packet with
$$A(p)=N \exp[-\frac{(p-p_0)^2}{4 \sigma_p^2}].$$
You can show that the expectation values of the corresponding state $|\psi(t) \rangle$ of position and momentum are like the motion of classical particle with the average momentum $\langle p \rangle=p_0$,
$$\langle x(t) \rangle=\frac{p_0}{m} t$$.

6. Sep 29, 2015

### kith

To expand on what Nugatory said: if you know the exact state of your system before a measurement, you won't get additional information by performing the measurement. If you have a particle moving with a well determined momentum, you can of course learn something about its uncertain position by performing a position measurement. But in the process, you will also forget something about its momentum - this is expressed in the HUP.

Whether you view your initial state as an (approximate) momentum eigenstate or a superposition of position eigenstates depends on the measurement you are going to perform.