# Wave function

1. Apr 30, 2007

### eoghan

Hi all,
why a wave function has to be a continuous function?

2. Apr 30, 2007

### StatMechGuy

What's the differential equation that wave functions satisfy? Could there be any point where it becomes discontinuous?

3. May 1, 2007

### Demystifier

Wave function does NOT need to be continuous. The derivatives may be well defined even if it is not continuous (for example, consider the step-function). Note also that, by the Fourier theorem, a superposition of plane waves (which is a solution to the free Schrodinger equation) may represent a function which is not continuous at certain points.

4. May 1, 2007

### Repetit

Wouldn't a discontinous wave function imply abrupt changes in probability density? How can that be physical? I mean, if there's a certain probability for a particle to be at location A there would be a discontinuity in the probability of being in a location A' immidiately adjacent to A.

5. May 1, 2007

### ZapperZ

Staff Emeritus
Under what condition would the wavefunction not be continuous? I can see how the derivative may not be continuous, such as for a delta function barrier resulting in a derivative jump, but this isn't the question posted by the OP.

Zz.

6. May 1, 2007

### jambaugh

A wave function needs to be "integrable" but it need not be continuous.
The Dirac delta function for example is neither continuous nor is it truely a function per se. It is rather a "distribution".

The physical predictions come from quantities obtained by integrating the wave function over all space. The value of the function at a given point has no meaning, only the integrals of values over regions. Thus two wave functions differing only by the value at one or a few points are considered to represent the same physical system.

You can imagine that the wave function for a given physical mode is a representative of an equivalence class of all wave functions (continuous or not) which yield the same physical predictions. By imposing continuity conditions you select (almost?) unique representatives of the given equivalence class.

Recall that in the definition of trans-finite cardinal numbers, the number of real valued functions on an interval or over a region is uncountably infinite, $$\aleph_1$$ while the number of continuous functions is countably infinite $$\aleph_0$$.

Ultimately the wave-functions are simply ways of representing elements of an infinite dimensional Hilbert space and that Hilbert space is a mathematical construct wherein which we can express the lattice of logical propositions about a physical system (the lattice of subspaces corresponds to the lattice of refined physical assertions about the system). It is this lattice logic along with the representation of the action of the group of relativity transformations (rotations, translations, changes of electro-magnetic ground potentials, etc) on the Hilbert space from which we get the physics.

Remember that there is no physical observable corresponding to the proposition "wave function is continuous". Thus there should be no physical meaning in distinguishing continuous and discontinuous wave functions.

Add to this that all actual experiments only occur up to some limit of precision (this independent of uncertanty principle issues) and so we can never resolve all the way down to the scale at which a discontinuity is replaced with an analytic smooth transition of values near a point. We can make such smooth transitions of values arbitrarily steep.

Regards,
J. Baugh

7. May 1, 2007

### ZapperZ

Staff Emeritus
But under what circumstances is the Dirac delta function as "solution" to either the Hamiltonian, or the time-dependent Schrodinger equation? I'm framing the OP's question in terms of that.

If we are invoking lattice space, etc., then all bets are off. Most of these systems don't have a solvable "wavefunction" to start with. In condensed matter, we certainly do not solve the Hamiltonian to get one, but rather formulate the most appropriate many-body ground state.

So if we plop the Hamiltonian in front of someone and ask him/her to produce a scenario in which the solution of it produces a discontinuous wavefunction, what would that be? What potential function and accompanying boundary conditions would produce such a thing? We already have an example of discontinuous first derivative using a Dirac delta function as the potential. Do you have one for a discontinuous wavefunction itself?

Zz.

8. May 1, 2007

### jostpuur

I didn't get into detail yet, but wouldn't this occur with some patological potentials? $$U(x)=\lambda \partial\delta(x)$$ for example. I don't think that would be completely unphysical. An intuitive interpretation of that potential would be, that there is a sharp spike of a magnitude $$-\infty\lambda$$ on the left side of the origo, and other spike of a magnitude $$\infty\lambda$$ on the right side. That could represent some real situation.

9. May 1, 2007

### ZapperZ

Staff Emeritus
But isn't this the delta function potential that I've mentioned at least twice in here? This results in a first derivative jump but not a discontinuous wavefunction, no?

Zz.

10. May 1, 2007

### jostpuur

Not really, it was supposed to be a derivative of a delta function. I though it could have solution like $$\psi(x)=((A-B)\theta(x)+B)e^{ipx/\hbar}$$, that would be discontinuous, but now when I checked that, I couldn't make it work. Perhaps my intuition was wrong.

11. May 1, 2007

### ZapperZ

Staff Emeritus
Perhaps I don't understand the profile of your potential function. To me, "sharp spikes" in the potential means delta functions. Your infinite delta functions, if they are at particular positions, then would be the very same ones that I have described. But the "derivative" of delta function is not something that I'm familiar with. Is it even well-defined? Such function do have finite integral, but not first derivatives, no?

Zz.

12. May 1, 2007

### jostpuur

Actually I have never seen derivatives of delta functions used in any reliable sources, but I did convince myself that they should be quite OK objects. If you approximate delta function with some finite peak $$g(x)$$, then its' derivative has a positive peak on the left side of the origo and a negative peak on the right side (I said this the wrong way in the previous post), but you cannot write $$\partial\delta(x)$$ as sum of two other delta functions, because the peaks are infinitesimally close to the origo. The integral $$\int g(x)f(x) dx$$ then is something like $$\approx A(-f(\Delta x) + f(-\Delta x))$$ for some small $$\Delta x$$, and multiplied with some constant A whose value is not obvious, but it seems reasonable to assume it could be $$-\partial f(0)$$. That it is what it becomes when you use integration by parts like this $$\int (\partial\delta(x))f(x)dx = -\int \delta(x)\partial f(x)dx$$.

Having made these remarks myself, I've come to believe, that $$\partial\delta(x)$$ is okey. I don't know about rigorous distribution theory.

13. May 1, 2007

### ZapperZ

Staff Emeritus
I think you need to be very careful here. You can't simply do this for "small" $\Delta(x)$. It would not be a delta function. You have to do this for infinitesimal $\Delta(x)$. If not, the whole idea of the delta function isn't valid since it has a "coordinate" extent.

It is rather dubious to do "mathematical derivation" of this because, as mathematician can tell you, the Dirac delta function is more of a "definition" rather than a rigorous mathematical function. So I am not sure if a derivative exists based on what it has been defined as.

Zz.

14. May 1, 2007

### StatMechGuy

The Dirac delta function is is technically a "functional" on a Hilbert space, since it isn't really very well defined outside of being in an integral.

I've seen "derivatives of the delta function" used before, but it was in an hand-waving argument for what a pulsed dipole that existed for an infinitesimal time would look like. I've never taken the derivative of the delta function unless it was going to vanish in the wash anyway (i.e. integration by parts trick).

15. May 1, 2007

### akhmeteli

The delta-function is one of so-called distributions or generalized functions (sorry, I don't have time to look for exact references). There is a rigorous mathematical theory of such functions (indeed, they are defined as functionals). Their derivatives are well-defined (and are also distributions). Moreover, distributions are typically (or always - don't remember now) infinitely differentiable. If I need to differentiate a distribution, I usually differentiate its Fourier expansion. The problem with distributions is that their product cannot be reasonably defined. Actually, this is sometimes regarded as the source of problems requiring renormalization in quantum field theory.

16. May 3, 2007

### jambaugh

WRT the delta-function, I don't think that "functional" is the right term.
Rather the integral which utilizes the delta function is a functional. Functional = a "function" with domain within a space of functions. Example the action of classical Lagrangian mechanics is a "function" of the particle path and so is a functional.

The proper term for the delta function is that it is a "test function" or a "distribution". It is a distribution in the sense that it acts as a measure on the real interval weighing one point (the origin) over all others.

Finally the derivatives of the delta function --which itself can be defined formally as the derivative of the Heavyside step function-- are used very often to express all linear operators on a function in terms of an integral:

$$(\frac{d}{dx})^n f(x)= \int_{-\infty}^\infty \delta^{(n)}(y-x)f(y)dy$$

These are perfectly well defined as distributions or test functions. In effect test functions are the closure of the space of continuous functions in the same way that the real numbers are the closure of the set of rationals. We write decimal expansions of, say pi, and this is just expressing this number as a limit of rationals.

Likewise think of the delta function (and its derivatives) as limits of smooth, continuous functions. Nothing weird about this and its just like the wave-functions themselves, they are mathematical objects we utilize to describe physical systems.

Don't get in a tizzy about the physical meaning of discontinuous functions. We use e.g. the discontinuous step function to represent densities of solid objects. The density is a large scale statistical quantity and gets meaningless when we resolve below the scale of the atoms of an object. Yet we would be silly to worry about this when we describe refraction of light through a prism.

Don't take the mathematical features too seriously. The mathematics is just the tool we use, not the physics itself.

Regards,
James Baugh

Last edited: May 3, 2007
17. May 3, 2007

### jostpuur

That notation looks strange. I've got used to $$\delta^{(n)}(x-x')$$ meaning delta-function in n dimension. Did that n now mean n:th derivative? If so, shouldn't there have been also a factor $$(-1)^{n}$$?

18. May 3, 2007

### akhmeteli

See http://mathworld.wolfram.com/GeneralizedFunction.html . By the way, there are references there to books by Vladimirov and Gel'fand/Shilov .
As I said, "generalized functions" is just another name of "distributions", such as the delta-function.
Let me emphasize the following phrase from Mathworld: "Generalized functions are defined as continuous linear functionals over a space of infinitely differentiable functions such that all continuous functions have derivatives which are themselves generalized functions." So they are typically DEFINED as functionals. Thus, distributions are functionals by definition. Of course, you may define distributions many other ways. But this is one of generally accepted ways to define distributions. You may say, following Wikipedia, that mathematicians identify distributions with functionals. Whether we, physicists, like it or not

19. May 8, 2007

### jambaugh

The parenthetic superscript indicates a multiple derivative while the superscript without parenthesis is a simple power. It is a power which is used for a product of delta functions for each coordinate in a plural dimension case. Hence:

$$\delta^3(r-r') = \delta(x-x')\delta(y-y')\delta(z-z')$$

while
$$\delta^{(3)}(x-x')= \frac{d^3}{dx^3}\delta(x-x') = \frac{d^4}{dx^4}\Theta(x-x')$$
which is common calculus notation for higher order derivatives.

As to the factor of -1, hmmm... maybe I forgot. Let's see...
$$\int_{-\infty}^{\infty} \delta^{(n)}(y-x)f(y)dy = f(y)\delta^{(n-1)}(y-x)|_{-\infty}^{\infty} - \int_{-\infty}^{\infty}\delta^{(n-1)}(y-x)f'(y)dy = \ldots$$
Yes, you are quite right! The formula should read:

$$\int_{-\infty}^{\infty} (-1)^n\delta^{(n)}(y-x)f(y)dy = f^{(n)}(x)$$

Thanks!
JB

Last edited: May 8, 2007
20. May 8, 2007

### jambaugh

Quite right, generalized functions aka test functions aka distributions are defined via functionals however to emphasize my point consider a perfectly well defined function e.g. the Gaussian:
$$f(x) = e^{-x^2}$$
It it by itself not a functional however it defines a functional via:
$$F[\psi] = \int_{\mathbb{R}}\psi(x)f(x)dx$$

Possibly I'm just being defensive but I think failure to make this distinction can lead to confusion. But to make your point, when we work with finite dimensional inner product spaces there is a one to one mapping between the "linear functionals" i.e. dual space and the "functions" i.e. original inner product space hence we identify functions with the respective dual elements.

In the infinite dimensional case the dual space is larger. Hence e.g. in L2 space, we must invent generalized functions to take the place of functions when used in this dual role for those extra dual elements.

Regards,
J. Baugh