L² Hilbert space, bound states, asymptotics of wave functions

1. May 27, 2010

tom.stoer

Hi,

I discussed this with some friends but we could not figure out a proof.

Usually when considering bound states of the Schrödinger equation of a given potential V(x) one assumes that the wave function converges to zero for large x.

One could argue that this is due to the requirement that the wave function is square integrable, i.e. an Element of L². But mathematically this is not necessary. One can construct wave functions with peaks, where the width of the peaks shrinks to zero, the distance between two peaks increases and the height of the peaks increases for growing x. That means that the wave function does not converge to zero, but nevertheless it remains square integrable (provided that the width decreases fast enough). This may seem artificial and the potential need not make sense physically, but it is perfectly valid mathematically.

What I have not used is the fact that we are talking about a bound state, i.e. it has an eigenvalue E<0 in the discrete spectrum of the Hamiltonian. Is there an argument using the discrete spectrum that forces the wave function to decay faster than 1/x or something like that? Or is it possible that something like these pathologically defined wave functions could be bound states of some strange Hamiltonian?

2. May 29, 2010

strangerep

IMHO, it doesn't remain rigorously square integrable in the limit of zero widths of the
peaks (since it's then defined only on a set of measure zero), so we must pass to
use of distributions, etc, for such cases.

3. May 29, 2010

Fredrik

Staff Emeritus
This is often claimed to be "completely obvious" even for arbitrary states, but it certainly isn't. This was discussed here.

Unfortunately I don't know the answer to that.

See #6 in the thread I linked to. Hurkyl gave an explicit example of a discontinuous function $\psi$ that's square integrable and doesn't satisfy $|\psi(x)|\rightarrow 0$ as $x\rightarrow\pm\infty$. Then he described an obviously correct method that would give us a smooth square integrable function with the same property. So there seems to be no need to involve distributions.

4. May 29, 2010

tom.stoer

I tend to agree that the construction mentioned above requires distributions (as the width shrinks to zero), but even then it's not a problem.

There a coupe of different shapes using basically the same methods. I was thinking about a sum of gaussians with increasing distance and decreasing width such that the sum remains square integrable, (strictly) smooth everywhere and non-zero.

Then it should be possible to select any E as energy eigenvalue, devide the Schrödinger equation by this eigenfunction and solve for V(x). One can therefore reconstruct the potential. But of course this does not show if E is in the continuous or in the discrete spectrum.

My feeling is that there must be some argument based on the Hamiltonian and the fact that we are talking about a bound state i.e. about the discrete spectrum which forces the wave functons to converge to zero, but I have absolutely no idea how to show this. My feeling is simply due to the fact that I have never seen a bound state which does not have this behaviour.

5. May 29, 2010

unusualname

Wave functions are equivalence classes in a hilbert space of measurable functions, any pathological function you might construct which doesn't decay to 0 at +-infinity is equivalent (modulo a set of measure 0) to a well behaved continuous one that does.

You can't attribute any physical meaning to the value of the wavefunction on a set of points of measure zero, so such patholgical wave function constructions have no physical meaning, they are mathematically represented by an equivalence class containing a well-behaved function.

6. May 29, 2010

tom.stoer

I don't agree. If you use such constructions you may arrive at distributions which are not equivalent to continuous functions. The delta function is one example.

But nevertheless your idea might goe into the right direction.

7. May 29, 2010

Fredrik

Staff Emeritus
We only need a distribution when we want the width of a specific peak to be zero. What Hurkyl constructed is definitely a function, not a distribution.

It's not. See the example I linked to. There are smooth square integrable functions that don't go to zero as $x\rightarrow\pm\infty$. Each peak has a finite width, so you can't change it into a function that goes to zero as $x\rightarrow\pm\infty$ just by changing its values on a set of measure zero.

8. May 29, 2010

tom.stoer

Convincing; I agree!

9. May 29, 2010

jostpuur

If you want to prove that $\psi(x)\to 0$ when $x\to\pm\infty$, you must use some assumtion about $V(x)$. An easy example:

$$V(x) = \left\{\begin{array}{ll} 0, & x < -L\\ \textrm{something complicated}, & -L\leq x \leq L\\ 0, & L < x\\ \end{array}\right.$$

and assume that $\psi_E(x)$ is an eigenfunction with energy $E<0$. Now

$$\psi_E(x) = \left\{\begin{array}{ll} C_1\exp\big(\frac{\sqrt{2m|E|}}{\hbar}x\big),& x<-L\\ \textrm{something complicated}, & -L\leq x\leq L\\ C_2\exp\big(-\frac{\sqrt{2m|E|}}{\hbar}x\big), & L < x\\ \end{array}\right.$$

so $\psi(x)\to 0$ when $x\to\pm\infty$ is obvious, even though the entire wave function was not solved

If the $V(x)$ does not have this nice form where it becomes a constant outside some domain, then I'm not sure what kind of results exist. If the potential is approximately constant, perhaps the wave functions are approximately exponential functions?

Actually, precisely at this moment, I came up with one conjecture. If there exists constants $V_M$ and $L$ such that $V(x) > V_M$ for all $|x|>L$, and if $E < V_M$, then the wave functions would go to zero. I'm not yet sure how to prove it, but it looks plausible.

10. May 29, 2010

unusualname

Ok, correct, but that's in the space of all square integrable funtions, we're restricted to the subspace of functions which are solutions of the schrodinger equation.

I wonder if an easy physics solution would do here rather than a full analytical derivation, suppose we say that anything defined on a set whose total size is less than the planck scale can be physically disregarded.

That's like nature's definition of measure 0

Right, try constructing a square integrable function which doesn't decay to zero now

11. May 30, 2010

tom.stoer

@jostpuur: of course your simple potentials solve this problem; but there are other potentials with exponential or gaussion decay, e.g. 1/r and r²