# Showing that wave functions are continuous

1. Apr 7, 2009

### AUMathTutor

Hello,

In my QM class last semester, I produced a proof that wave functions must be continuous (used for boundary conditions, etc.) It was an undergraduate level course, so I don't know how easy it would be to do if you had more in the way of theory...

But I've been wondering lately whether my proof was correct or not. I don't want to give away too many details, but I will say it was less than elegant (lots of cases). I was wondering if anybody knew of an easier way than I figured out, and if so, if you could take me through it step-by-step?

PM me if you want to make sure that I'm not trying to trick you into doing my HW for me. I'll tell you how I did it. Thanks!

2. Apr 7, 2009

### Tac-Tics

I'm pretty sure this isn't something you can prove. That physical quantities are continuous is taken as a fundamental assumption. So fundamental that it's not usually even specified.

The entire notion of continuity, in fact, roughly follows the intuition of (classical) measurements. Scientists have always understood their are limits to the accuracy at which you can measure something. A continuous function is defined as a function where the margin of error of the output can be made arbitrarily small by providing sufficiently accurate input.

On top of that, wave function are tied to probability distributions. The theory of probability is built on top of calculus, where functions have to more or less continuous.

3. Apr 7, 2009

### AUMathTutor

"The theory of probability is built on top of calculus, where functions have to more or less continuous."
This is where you can tell a mathematician apart from a physicist. In fact, general probability distributions - even ones for continuous random variables - needn't be continuous at all. Lots of functions are square-integrable yet discontinuous, the most notable of which is the unit jump function.

I did end up proving (or at least I hope I did) the continuity of the wave function, so I think it's possible. I mean, I proved it was the only mathematically consistent possibility there was (well, for a restricted case, the 1D TISE). I don't want to say much more for fear of influencing future posters.

Thanks for your input, though. When I asked my professor last semester about it, he said the same thing (wave functions have to be continuous because they're probability distributions) and then when I told him that probability distributions didn't have to be continuous, he responded that it was simply a postulate. So you're not alone, and in fact, I suppose that makes 2 against 1 now...

4. Apr 7, 2009

### dx

What were your assumptions? I don't see anything in the postulates of QM that can be used to prove this. Also, I assume you're talking about the boundary conditions for the infinite well. It's not true in general. Wave functions don't have to be continuous.

Last edited: Apr 7, 2009
5. Apr 7, 2009

### AUMathTutor

"Wave functions don't have to be continuous."
Really? I am assuming only those wave functions which obey the 1-D TISE. Are there examples of these which are not continuous? I'm not sure I was aware of any...

I thought that you imposed continuity on the infinite well because it was always the case. Is it not true for the SHO, the free particle, the finite square well, ...

I didn't make any other assumptions in my proof than the ones I've listed; namely, you have a wave function which obeys the 1-D TISE. Maybe the proof is messed up, in fact, it probably is. But then I would like to know why the heck you impose continuity on *any* wavefunction at all. I'm not sure I like it being a postulate... seems too convenient to me.

6. Apr 7, 2009

### dx

A free particle can be in a position eigenstate, which is a delta function. A delta function is not continuous in the usual sense, although the definition could probably be generalized.

It would be a lot easier to see what you're talking about if you post your work. PM it to me if you don't want to influence future posters.

7. Apr 7, 2009

### dx

Also, remember that the TISE makes no sense if you just substitue V = infinity in it. If you want to do it properly, you have to solve it for finite V, and then take the limit. If you do that, you will find the usual solution. With finite V, for the TISE to make sense, the wavefunction must be differentiable, so it obviously has to be continuous.

8. Apr 7, 2009

### Fredrik

Staff Emeritus
The Schrödinger equation only makes sense if the partial derivatives of $\psi$ exist. If they exist, the maps $t\mapsto\psi(\vec x,t)$ and $x_i\mapsto\psi(\vec x,t)$ with i=1,2,3, must all be continuous. Differentiability implies continuity.

Edit: I didn't see the post before this one before I wrote this.

Last edited: Apr 7, 2009
9. Apr 7, 2009

### AUMathTutor

"Also, remember that the TISE makes no sense if you just substitue V = infinity in it. If you want to do it properly, you have to solve it for finite V, and then take the limit. If you do that, you will find the usual solution. With finite V, for the TISE to make sense, the wavefunction must be differentiable, so it obviously has to be continuous."

I now remember thinking exactly this at the time that I was doing it, and that the problem only arose if you considered non-physical potentials in the infinite-square-well problem. In the back of my mind I knew that the equation only worked for physical, that is, finite cases, but I guess the high physics content confused my brain into thinking that it was alright to think of infinity like a real number. Whoops!

So I guess what I did was to show that it still had to be continuous at infinite potential, using a more "physical" method or something. God, that was a long time ago. I'll look for it and see what I find.

Thanks for all the help, guys! By the way, I remember that whatever I did involved integrating the Schrodinger equation and looking at boundary conditions to derive several proofs by contradictions for various cases of discontinuous wave functions and potentials.

Thanks again!

10. Apr 7, 2009

### peteratcam

Surely the point is that because the kinetic energy term in the hamiltonian is quadratic in derivatives, a discontinuous wave function has infinite kinetic energy? Think about evaluating the average kinetic energy for a state with discontinous wave function: an example which is analytically tractable is tanh(ax), taking limit a-> infinity to obtain a discontinuity.

Even with an infinite potential, the wavefunction is continuous! There seems to be some confusion on this point.

11. Apr 7, 2009

### Fredrik

Staff Emeritus
Saying e.g. that the potential is infinite outside the interval [0,L] is just a weird way to say that we're considering a physical system for which the appropriate Hilbert space is the set of functions $\psi:[0,L]\rightarrow\mathbb C$ that satisfy $\psi(0)=\psi(L)=0$, rather than $L^2(\mathbb R)$. The Schrödinger equation still holds, so the wave function is continous everywhere in its domain of definition.

Last edited: Apr 7, 2009
12. Apr 8, 2009

### alxm

Consider for instance a particle in a coulomb potential (hydrogenlike atom), the ground-state wave function is (ignoring constants):
$$\psi=e^{-r}$$ where $$r=\sqrt{x^2+y^2+z^2}$$

The first order derivative has a singularity at r=0:
$$\frac{\partial\psi}{\partial x} = -\frac{xe^{-\sqrt{x^2+y^2+z^2}}}{\sqrt{x^2+y^2+z^2}}$$

But: The overall wave function for any multi-particle wave function in configuration space is continuous and hence bounded, and the divergence at the potential energy singularities is canceled out by diverging kinetic energy terms.