# Continuity of the wave function

1. Sep 24, 2007

### luisgml_2000

I've heard some people say that the wave function and its first derivative must be continuous because the probability to find the particle in the neighborhood of a point must be well defined; other people say that it's because it's the only way for the wave function to be physically significant. There is even another hypothesis, which states it's a consequence of the eigenvalue equation

H$$\Psi$$=E$$\Psi$$

Which idea is the right one?

2. Sep 24, 2007

### meopemuk

I don't think that the idea of continuity of wave functions plays any significant role in quantum mechanics. After all, Dirac delta functions (which are not continuous) have a prominent place there. Without delta functions one cannot define eigenvectors of many important operators of observables, such as position and momentum. So, I would prefer to keep delta functions and abandon the idea of continuity.

Eugene.

3. Sep 24, 2007

### luisgml_2000

I think that the continuity of wave functions does have a great deal of relevance. For instance, in the potential well problem the continuity leads to the boundary conditions on the wave function, which are essential for solving the problem.

Every book I've seen until now just states that the wave function has to be square integrable (and I understand the reason for this), continuous and its first derivative a continuous function (except for discontinuous potentials), but the authors never give further explanations about it.

4. Oct 1, 2007

### Manchot

Well, I prefer to think of it as an axiom of the vector space comprising the set of states. Obviously, while this is perfectly acceptable within a logical framework, it's not very instructive. You can also infer continuity from the time-dependent Schroedinger equation. If a wavefunction was not continuous, it would have no first derivative at the points of discontinuity. Thus, it would have no second derivative at those points, the square of the momentum operator wouldn't exist, and the Hamiltonian would be undefined. Of course, assuming that the time derivative of the wavefunction is always well-defined, then this wavefunction would not satisfy the Schroedinger equation at the discontinuous points. Therefore, the discontinuous "wavefunction" cannot be a quantum state, and you can safely exclude all discontinuous functions from the vector space without a loss of generality.

5. Jan 6, 2012

### jewbinson

This may seem like a random post but I want to know a few things about the continuity of the wavefunction...I searched on google and up came this thread.

I get the jist of the post above. So when we consider the Schrodinger Equation, do we assume that a wavefunction must satisfy the Schrodinger Equation for ALL x (If so, why)? In this case, the above post is very convincing. If not, then surely we can make a wavefunction with lots of discontinuities so that the SE is satisfied by the wavefunction where the wavefunction is twice differentiable.

Edit: Do we usually assume that the wavefunction is twice differentiable for all x?

6. Jan 6, 2012

### ChmDudeCB

Continuity defines your boundary conditions as the wave function must exist in all of space.

7. Jan 6, 2012

### ChmDudeCB

If d/dx^2 doesn't exist then E is undefined and the particle doesn't exist.

It's nice because we choose wave functions that are continuously derivable. Sin/e^ix etc... A property of a wave is that it is ever changing. If your function stops changing at some point, it's not a wave function.

8. Jan 6, 2012

### Staff: Mentor

Yes.

It seems to me it would be rather difficult to use the SE to come up with useful predictions otherwise, unless we also had a good idea of exactly where it works and where it doesn't.

The analog of this question for classical mechanics is, "do we assume that Newton's Laws hold for all x (if so, why)?"

Last edited: Jan 6, 2012
9. Jan 6, 2012

### jewbinson

Ok thanks for your responses

10. Jan 6, 2012

### ChmDudeCB

No problem. Just remember, the probability of finding the particle in all of space is equal to 1. If you're not integrating over all of space with a continuous function then the the probability of finding the particle is no longer 1, however close to 1 it may be. Remember, quantum barriers only act to temporarily confine a particle. It can neither be stopped completely nor confined indefinitely.

11. Jan 6, 2012

### Ken G

Actually, I don't see any fundamental reason why either the wave function, or its first derivative, needs to be continuous (so I agree with meopemuk). For one thing, the wavefunction can still be square integrable without either of those constraints, as long as the discontinuities are not pathologically severe. This all relates to what I might call the "problem of initial conditions" in physics-- there is no such thing as a theory of initial conditions, one is allowed to use physics on any initial condition one wants. The physicist simply tailors the initial conditions to answer the question of interest, there is no such thing as the "real initial conditions" any more than there is a "real potential function." We just pick them-- indeed, the wave function's first derivative will not be continuous if we want to treat infinite potential wells (which we often do, as in the particle in the box idealization). One might say "but an infinite potential is only an idealized limit, there's no such thing in reality", but then I would just answer that all potential functions treated in physics are only idealized limits, this is not saying anything restrictive.

Now, one might then ask, if we have a wavefunction with a discontinuity in either the function or its derivative, how would it evolve, according to the Schroedinger equation? But that's no problem either-- the infinitely fast rate of change of the wavefunction is not a problem if it only lasts for an infinitesmal amount of time, and this kind of thing can be solved by using integrated forms of the Schroedinger equation, instead of its differential form (analogous to how shocks are treated in hydrodynamics).

So why is it ever necessary to assume either the wavefunction or its derivative are continuous? It seems to me that the predictions of quantum mechanics involve doing integrals, so all that is really needed is that the necessary integrals exist, which is a less stringent requirement but perhaps more stringent ones are commonly adopted just out of pure convenience. Alternatively, perhaps what is really intended is that energy eigenstates have these continuity requirements, which is clearly true because they have to be stationary in time.

12. Oct 1, 2013

### marcelnv

The cool thing with Physicists is that they don't solve random problems blindly, just for the fun of mathematical exercise. All sort of mathematics could be done with discontinuous functions. However, Physicists define the wave function as a quantity that describes the probability to find a particle within a region in space. That said, the probability must be well define at a point, we can't afford to have ambiguities. So continuity of the wave function is fundamental in this respect. The continuity of the first derivative is mainly a consequence of the of the form of the wave equation. Again, Physicists always try to make all problems quantitatively meaningful!

13. Oct 1, 2013

### bhobba

Scratching my head about this one. Since it is assumed to obey Schrodinger's equation of course it must be differentiable and hence continuous.

Or am I missing something?

Thanks
Bill

14. Oct 1, 2013

### haael

First of all, wavefunctions are not ordinary numerical functions; they are distributions. Distributions are analytical (infinitely differentiable) by default, since they are defined as limits of series of analytical - hence continuous - functions. Dirac delta functions are also analytical/continuous in this sense.

The question: where it came from. I think it was a numerical tool in early physics. Equations describing point objects (like point forces, point charges or point particles) came as taking the limit of some equation in zero length. The physical equations are smooth in general, so the resulting object was the limit of a series of smooth functions. Later, the mathematical distribution theory was applied to it.
So, ultimately the continuity/smoothness of wavefunctions, comes from experiment.

Distributions don't have problems with differentials and being continuous, but they do have problems with multiplication. In some cases, multiplication is not well defined on distributions, namely it's not associative. I've heard the very quantization procedure from the mathematical point of view may be seen as assigning meaning to distribution multiplication.

15. Oct 1, 2013

### bhobba

That's true - but a lot of work has been done on the issue and evidently distributions can be further generalized so they can:
http://en.wikipedia.org/wiki/Colombeau_algebra

I like math but this is even getting over the top pure math wise for my tastes.

Thanks
Bill

16. Oct 1, 2013

### WannabeNewton

The momentum operator $P$ is a strictly unbounded operator whose domain makes essential reference to continuous differentiability. You can't just apply $P$ to any ray in $L^2$; a strictly unbounded operator has a domain which is by definition a proper dense subspace of $L^2$ that cannot be extended to all of $L^2$.

17. Oct 1, 2013

### PhilDSP

Even in the classical realm a wave function may be discontinuous at the boundary conditions. In fact, Maxwell adopted the prevalent definition of charge as a discontinuity of polarization, i.e. a place in space where the fields do not vary in time across the surface of the charge.

18. Oct 1, 2013

### Ken G

What I attempted to convey in my earlier post is that there is an important difference between a mathematical theory, and the testable predictions it makes. The former is mathematics, or mathematical physics if you prefer, whereas the latter is actual physics. Over and over in the history of science, we have failed to make this distinction, and ended up with egg on our faces. I suggest, unless we enjoy omelets, we get used to that difference! So what I mean is, the tested predictions of quantum mechanics could also be made by a different version of the formal theory, which used an integral formulation, and did not require that the wave function be continuous. Indeed, if it is claimed that the wave function must be infinitely differentiable, then we cannot handle particles in a box, one of the most common of all quantum mechanical idealizations that we can test with experiment.

Granted, one could simply exclude from the range of the function the domain where the potential is infinite, but one can imagine a limit where the potential height gets larger as its domain gets narrower, and in the limit becomes infinite over an infinitely narrow region. In an integral formulation, that would simply yield a kink in the first derivative of the wave function for an energy eigenstate, something we could easily model and address with experiment, but which we are hearing falls outside the realm of applicability of the current formulation of quantum mechanics. In practical applications, what we want is an integral constraint on a barrier like that, a constraint that simply sets the discontinuity in the slope of the eigenfunction.

19. Oct 1, 2013

### DrDu

Every continuous wavefunction solution $\psi$ pertaining to an hamiltonian H can be rendered into a discontinuous solution $U\psi$ of another Hamiltonian $\tilde{H}=U H U^+$ setting e.g. $U=\exp(i\pi \theta(x))=\theta(-x)-\theta(x)$. So it is quite a matter of taste.

20. Apr 17, 2014

### Demon117

I am scratching my head on all of this. I had heard, and I don't remember if it was from a textbook or from a professor that continuity of the wave function and it's derivative imply conservation of momentum across a boundary. Anyone care to elaborate or expel this idea?