1. Nov 5, 2005

### thesm

I had a search for an answer but I turned up nothing, if this has been covered before could someone point me in the right direction?

To the question.

I'm studying QM at the moment but I'm having trouble with two of the postulates. Is the constraint that the wavefunction must be continuous and smooth (continuous first derivatives) based on anything more, or is it simply an axiom of the theory? To me it seems like the condition was placed because it means you then have enough equations to determine the coefficients of the wavefunction.

If it is one of the axioms of the theory I'd be interested to know of any experimental ways of demonstrating that this is the case.

Thanks
Ryan

(long time lurker, first time poster)

2. Nov 5, 2005

### Kruger

"If it is one of the axioms of the theory I'd be interested to know of any experimental ways of demonstrating that this is the case."

tunnel effect

3. Nov 5, 2005

### thesm

Ok, so it's an axiom, thanks. I got the impression from my lecturer that the continutity conditions should make sense physically, like that it should be obvious that a wavefunction is smooth and continuous, but it wasn't making any sense like that to me.

And of course, quantum tunneling, thanks again.

4. Nov 5, 2005

### Dr Transport

I am not a mathemetician, but if I remember correctly a second order differential equation has to has continuous solutions along with a continuous 1st derivative.

5. Nov 5, 2005

### thesm

Ok, I feel stupid now.
I was going to ask why should this apply on the boundary, since if the wave function was undefined outside this then the continuous/smooth condition couldn't be added, but one of the postulates is that the wave function is defined everywhere.
Thanks for the responses.

6. Nov 5, 2005

### Hurkyl

Staff Emeritus
A more abstract way of looking at it is to say that the operators are what's really important, and continuously differentiable functions are a good class of things upon which they can act.

7. Nov 7, 2005

### dextercioby

Well, first thing's first:

the wavefunction has nothing to do with the postulates of (nonrelativistic) QM.

Then, once u pick the irreducible representation of the Born-Jordan CCR-s to be the "wave mechanics" formalism, then the Hilbert space for one particle becomes $L^{2}\left(\mathbb{R}^{3} \right)$. That's where the physics takes place. Since the free particle hamiltonian is an unbounded, densly defined, self-adjoint linear operator which, up to a constant is

$$-\nabla^{2} [/itex] , then the normalization condition and the differential form of the common operators (linear momentum, angular momentum, hamiltonian) impose certain restrictions upon the element of the Hilbert space. $C^{2}$ class on a certain domain of $\mathbb{R}^{3}$ is a good requirement on the wavefunction. And of course, being a Schwartz functional is another one. A good survey on this matter is found in Reed & Simon. Daniel. 8. Nov 7, 2005 ### lalbatros Probability current will be continuous, if the gradient of the wave function has a smooth derivative. Would that not be a good reason? (see http://electron6.phys.utk.edu/qm1/modules/m4/probability.htm [Broken] for the maths) Last edited by a moderator: May 2, 2017 9. Nov 7, 2005 ### ZapperZ Staff Emeritus Just for completeness sake, it may be interesting to know that if the potential term in the time-independent Schrodinger equation is a Dirac delta function, the solution will have a "derivative jump", i.e. the first derivative will have a finite discontinuity, even when the wavefunction itself remains continuous. Zz. 10. Nov 8, 2005 ### Galileo It seems to me obvious that the wavefunction should be continuous and differentiable. To say it is discontinuous at a certain point in space would be physically meaningless, since we have no measurable access to phenomena in an infinitely small region as a 'point', which has no dimensions. We can't measure such sudden change in such a function over an interval of, say, $10^{-50}$m. Any function which varies significantly over a very small interval will give the same results as a discontinuous one. For example, a square well potential is physically impossible, but it will do to describe a potential which varies (continuously and smoothly) over a distance which is short compared to the wavelength of the particle. If you want 'mathematical proof' of this, you can do as Griffiths does and integrate the (time-independent) SchrÃ¶dinger equation over a small interval: [tex]-\frac{\hbar^2}{2m}\int_{-\epsilon}^{+\epsilon}\frac{d^2\psi}{dx^2}dx+\int_{-\epsilon}^{+\epsilon}V(x)\psi(x)dx=E\int_{-\epsilon}^{+\epsilon}\psi(x)dx$$
The first integral is $d\psi / dx}$ evaluated at the two endpoints. The last integral is zero in the limit $\epsilon \to 0$, since $\psi$ is bounded and the width of the 'window' goes to zero. So:
$$\lim_{\epsilon \to 0} \Delta \left(\frac{d\psi}{dx}\right)=\frac{2m}{\hbar^2}\lim_{\epsilon \to 0}\int_{-\epsilon}^{+\epsilon}V(x)\psi(x)dx$$

V(x) is normally (and in any physical case) finite and thus $d\psi / dx$ is continuous. If V(x) is infinite however (as in the case of a delta function potential), the derivative of $\psi$ can undergo a discontinuity.

Last edited: Nov 8, 2005