# I Free Particle: Time dependence of expectation values Paradox

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1. Jun 14, 2016

### Joker93

It would be really appreciated if somebody could clarify something for me:
I know that stationary states are states of definite energy. But are all states of definite energy also stationary state?
This question occurred to me when I considered the free particle(plane wave, not a Gaussian packet). The time derivative of the mean value of the position is not constant in time but equal to hbar*k/m. But, plane waves are energy(and momentum) eigenstates for the free particle, that means that they are states of definite energy. So, what's the deal here?
Thanks!

Last edited: Jun 14, 2016
2. Jun 14, 2016

### avikarto

Saying that something is a stationary state is equivalent to saying that it is an eigenstate of the system's Hamiltonian. Then, it is clear to see why we say this is a state of definite energy:
H |a> = E |a>
The Hamiltonian for the free particle is just P^2/m. Its position expectation value isn't really important in considering its momentum.

3. Jun 14, 2016

### Joker93

Thanks for the reply, but saying that the position expectation value is not important does not mean that the "paradox" that there seems to be taking place here is not important as well. Since the plane wave is an energy eigenstate, i would expect every expectation value to not have a time dependence. But, clearly, this is not the case for the position's mean value.

4. Jun 14, 2016

### vanhees71

Note, however, that the energy eigenfunctions of free particles, which are in fact the plane wave solutions, do not represent states, because they are not square integrable. Only square integrable functions represent pure states in quantum mechanics not such generalized functions.

5. Jun 14, 2016

### Joker93

So, we can't even speak about expectation values in these types of states, right?

6. Jun 14, 2016

### kith

Let's suppose that the space where your free particle can be found is finite, so that we can normalize the wavefunction of interest. $\psi(x) = \frac{1}{\sqrt{N}}e^{ikx}$ is stationary under the action of the Hamiltonian $H = \frac{p^2}{2m}$. So the probability density for the position is time-independent and given by $|\psi(x)|^2 = \frac{1}{N} = const.$ So also the expectation value of the position is constant in time.

For plane waves which extend to infinity, the modulus squared is still a constant, so some of the intuition of the paragraph above is retained. However, as others have mentioned, plane waves are not normalizable and thus not valid wavefunctions in the first place.

You can also look at the situation from the point of view of what A. Neumaier explained in your other thread. If you increase $\sigma$, the Gaussian shape of the probability density gets more and more similar to that of a constant function.

7. Jun 14, 2016

### Joker93

Thanks for the reply, although i do not understand two things. First, i don't understand what you are trying to explain to me with your point about the σ getting larger and larger in a Gaussian. I didn't quite get what you are trying to say.
Secondly, as you and others imply, we can't even talk about expectation values when dealing with such "bad-behaved" functions, but as mr. Neumaier explained in my other post, we can indeed find the expectation value of the momentum(and also, via Ehrenfest's theorem, we can find the time derivative of the expectation value of position, and thus an expression of that expectation value too through integration). So, if we can't talk about expectation values, then why can we find them in some cases?

8. Jun 14, 2016

### Paul Colby

Isn't a stationary state one with a time harmonic phase as its only time dependence? Since the over all phase of the state vector is not significant this means the state is unchanged in time.

9. Jun 14, 2016

### Joker93

The time derivative of the expectation value of position is a constant, thus the expectation value doesn't depend on time via a harmonic phase. Also, what you are saying does not hold for mean values. For mean values, it does matter. It's when we have a harmonic(complex) phase in front of the function of the wavefunction that the phase does not matter.

10. Jun 14, 2016

### Paul Colby

Okay, I would say that is what I said except in terms of Hilbert type state vectors?

$\vert \psi(t)\rangle = e^{i\omega t}\vert\psi(0)\rangle$

11. Jun 14, 2016

### kith

Well, your initial problem seems to be that if you start with a plane wave as wavefunction and use Ehrenfest's theorem, you get a time-dependent $\langle x(t) \rangle$ which implies that the wavefunction can't be stationary. In the course of the thread, you realized that the root of the problem is that you used an invalid wavefunction. So basically, you can stop here and your question is answered.

But in order to better understand what's going on, you can look at your unphysical situation from two different, physically sensible starting points.
1) Restrict the spatial extension so that the plane wave becomes normalizable.
2) Start with a different wavefunction which reduces to the plane wave in a certain limit.

I would say that this is mainly semantics. As long as you are aware of the fact that plane waves are unphysical, you can use them as a tool to simplify the calculations for actual physical situations. If they don't work somewhere or lead to paradoxes, you need to keep in mind that the actual wavefunction looks something like A. Neumaier's function with a finite $\sigma$.

12. Jun 14, 2016

### Joker93

Excellent! Something else, if you don't mind:
By restricting the spatial extension, you mean integrating from -b to b (b is a real finite number)? But how can i know that my results are valid? And what if i integrate for something and get two different results for two different values of b?

13. Jun 14, 2016

### kith

For example, if you try to prepare particles in a state with a sharp momentum to do experiments in your lab, the particles are confined to the spatial extent of the preparation apparatus.

But note that the main motivation was a gedankenexperiment: I have a problem with invalid states, how can I easily get states which are in principle valid and have similar properties. To find an actual experimental arrangement to realize a certain state may be very difficult.

14. Jun 14, 2016

### Joker93

But, mathematically, how do i work with confining the spacial extension?

15. Jun 14, 2016

### kith

You could use a function which is equal to the plane wave in a certain region of space and zero everywhere else. In effect, this leads to integrals which are similar to what you considered in post #12.

16. Jun 14, 2016

### Swamp Thing

The trouble with infinite plane waves...! There is an interesting discussion on page 3 of this doc : http://arxiv.org/abs/1210.7265 A Bohmian perspective that uses "plane wave packets" as opposed to Gaussian ones.

17. Jun 15, 2016

### vanhees71

Yes, that's correct. However a generalized eigenstate of the Hamiltonian for a "eigenvalue" in the continuous part of the spectrum is not representing a state, because it's not a Hilbert-space vector but lives in another larger space, i.e., the space of distributions that are defined on a smaller dense subspace of the Hilbert space, where the self-adjoint operators are defined. The generalized eigenfunctions live in the dual space of this smaller subspace of the Hilbert space, which thus is larger than the Hilbert space itself. The mathematical theory behind this is called "rigged Hilbert space", which is a formalism that makes the physicists handwaving treatment of the spectral properties of unbound operators rigorous. A good source to study it is L. Ballentine, Quantum Mechanics - A modern development, Addison-Wesley.

18. Jun 15, 2016

### Joker93

Are you talking about an actual function that approximates a plane wave in some region(something like exp[-x^2]*exp[ikx] which is approximately equal to exp[ikx] for small x--or something like mr.Neumaier said in the other post--or maybe both are correct ways?) or you are talking about confining the plane wave in a box(by defining the function as being a plane wave for |x|<b and zero for |x|>b)?

19. Jun 15, 2016

### Joker93

Wow, really pedagogical article, and might also help someone that is not interested in the pilot-wave theory. Thanks!