# Self-adjointness of H, particular example

1. Apr 30, 2013

### fluidistic

Hi guys,
I don't really know how to explain my question...
I've heard my professor saying something like "for $\hat H$ to be self-adjoint, Psi and its first derivative with respect to x must be absolutely continuous."

However when I consider the problem of the particle in a box, the fact that Psi(x) is worth 0 at the edges seems to be due to the fact that Psi(x) is continuous and since it's worth 0 outside the box, it's 0 on the edges. However in that case, the derivative of Psi(x) is not continuous: in a 1 dimension box the eigenfunctions are either sines or cosines depending upon their energy. The derivative evaluated at the edges of a cosine or a sine that is worth 0 at the edges is clearly not 0. However outside the box Psi(x)=0=constant so that Psi'(x)=0. Hence the non continuity of Psi'(x).
So if I understood well, this means that the Hamiltonian is not self-adjoint in this case. (I'm not even sure it's well defined since $V(\hat x )$ is infinite).

So I don't really understand what my prof. was saying. In which case(s) what he said does hold?

2. Apr 30, 2013

### Fightfish

When we speak of an operator being Hermitian, the domain matters; an operator that is Hermitian over one domain need not be Hermitian in another domain. This is fine and good for the infinite square well; H is Hermitian over the domain of the well. Outside the well, well, things become rather ill-defined.

The infinite square well is actually quite a problematic idealised scenario. It turns out that Ehrenfest's theorem actually fails for the infinite square well!

3. Apr 30, 2013

### DrDu

You are right that the infinite potential is not well defined. For an operator to be self-adjoint it has to be defined on a dense set of vectors in Hilbert space. However you assume that all the wavefunctions vanish outside the box, so you can't represent a wavefunction which has a finite value there. So this hamiltonian is only defined on the hilbert space of square integrable functions over the length L. There are various self-adjoint extensions of $H= -d^2/dx^2$ on that domain and the one with the wavefunction vanishing at the end points is only a particular special case.

4. Apr 30, 2013

### fluidistic

Ok thank you guys.
It's not really clear to me then why the eigenfunctions must vanish at the edges. I think it's mathematically explained in Messiah's book though.
Just to be clear DrDu, when you wrote
I don't really understand why you say "finite value". 0 is a finite value.

So if I understand well, I can't even say that the eigenfunctions vanish outside the box? They are not even defined, right?

5. Apr 30, 2013

### Jano L.

This is a requirement that we pose for simplicity to obtain some useful basis for some finite space interval of interest. In a more realistic case, the potential outside the interval is finite and the eigenfunctions are non-zero outside the well; however, as the potential on the outside is increased, the eigenfunctions there become smaller and in this process the eigenfunctions have limit 0 everywhere the potential increases to infinity.

As the eigenfunctions $\phi_n$ come out always continuous, if they become zero outside the interval, they have to be 0 on the boundary.

For some calculations, it is easier to assume $\phi = 0$ on the boundary than to consider some complicated finite potential and non-zero eigenfunctions. In short, we assume the boundary condition mainly in order to simplify the mathematics.

6. Apr 30, 2013

### fluidistic

I understand this reasoning but from it I don't understand why the continuity of the derivative of the psi_n's and thus psi' is ignored.
In the case of the finite potential well, the derivative of the psi_n's are always continuous on the edges. Suddenly when you take the limit when the potential becomes infinite, there must be a discontinuity in the derivative of the psi_n's. That is, if it makes sense to talk about the psi_n's in the region where the potential is worth infinity.
Apparently they are not worth 0 there, there aren't even defined if I understood well the first 2 replies. So in that sense the infinite potential well is quite different from the finite potential well when we take the potential to tend to infinity. Because in the latter case, the psi_n's and their first derivatives are always well defined everywhere in space.

7. May 1, 2013

### DrDu

With finite, we sometimes mean neither of infinitely large nor small magnitude (i.e. zero). The meaning is mostly clear from the context. Of course you can define eigenfunctions which vanish outside the box. However they won't form a complete set and therefore the hamiltonian won't be self-adjoint.

8. May 1, 2013

### tom.stoer

As said in the beginning the self-adjointness of an operator H depends on the domain. For the infinite box there are two choices:
[-∞,+∞] with L2[-∞,+∞]: then the eigenfunctions vanishing outside the box do not form a complete set
[a,b] with L2[a,b] where a and b represent the boundaries of the box; now the eigenfunctions vanishing outside the box DO form a complete set for square-integrable functions on [a,b], and it would be strange to consider x<a or x>b

9. May 1, 2013

### Jano L.

Yes, it is different. Infinite potential well cannot be properly handled in wave mechanics - the Hamiltonian operator would involve multiplication by infinity, which is invalid operation. We cannot circumvent this by forgetting about the infinite potential region: the sudden change of the wave function on the boundary of finite well is important.

The proper way to understand infinite well is to understand finite well first and then analyze its limit. Which is quite an uncomfortable surprise, since the infinite well is much more easily solved :-)

10. May 1, 2013

### tom.stoer

I disagree; the infinite potential well on the intervall [a,b] with boundary condition ψ(a) = ψ(b) = 0 is perfectly well-behaved. The Hilbert space L2[a,b] of square-integrable functions allowes for self-adjoint operators (or self-adoint extensions). There is no infinity in the potential, the potential is zero everywhere, the region outside [a,b] simply does not exist. A slightly different but also well-defined problem are periodic boundary conditions where the [a,b] is compactified to a circle S1

The infinite square well is in that sense, when discussed on [a,b], not the limit of the finite square well.

11. May 1, 2013

### Jano L.

Yes, but the boundary condition seems arbitrary. Why not consider all $L_2(a,b)$ functions? That would be quite intuitive: since the outside "simply does not exist", the probability density should be allowed to be non-zero at $a$ or $b$.

One way to motivate the boundary condition $\psi(a) =0$ is to calculate first the finite wall and make the limit. Boundary condition $\psi \rightarrow \infty$ when $x\rightarrow \infty$ seems more natural there, as we expect that particle occurrence at large distances from the attracting well is quite unlikely.

12. May 1, 2013

### DrDu

It is easy to show that the operator $-d^2/dx^2$ is not self-adjoint when defined on all $L_2(a,b)$. Simply perform twice the integration by parts, and you will see that there are boundary terms left.

In functional analysis there are different kinds of convergence. While the low energy eigenvalues and eigenfunctions of the hamiltonian with a finite potential converge against the eigenvalues and functions of the infinite potential hamiltonian (for the eigenfunctions in norm, not point-wise), this convergence is not uniform. The eigenfunctions above the potential in the finite potential hamiltonian are always qualitatively different from the ones of the infinite well hamiltonian.

13. May 1, 2013

### Jano L.

I agree, self-adjointness is also a good reason to consider such boundary condition. For a mathematician. Considering how artificial the infinite well is physically and how subtle the mathematics of linear operators can get (each operator has its own garden of admissible functions!), I would rather keep with the limit argument in the first motivation of why consider such thing as infinite well in physics in the first place. For finite well, things are much more intuitive physically - the wave function is smooth, the boundary condition seems more natural, self-adjointness of H comes out naturally etc. It is perhaps an issue of taste/background, I admit.

As to the non-trivial convergence, I am surprised that there is no point-wise convergence. Can you elaborate? Outside the well there has to be point-wise convergence to 0, so the problem is inside the well?

14. May 1, 2013

### tom.stoer

The fundamental difference for the finite potential well (V=0 inside, V=V0 outside)
is that there are scattering eigenstates with energy eigenvalues E>V0

15. May 1, 2013

### DrDu

Seems you got me here.

16. May 1, 2013

### DrDu

Clearly. Thinking about the problem I would proceed as follows: We want to construct an operator defined on L2(R) which approximates the bound states of the finite box hamiltonian. So we could add delta functions at a and b to the finite box hamiltonian and increase their strength to infinity. Clearly the new operator is the infinite box operator on [a,b], but there are still scattering states, so the hamiltonian is self-adjoint on whole of L2(R). The two hamiltonians are just different self-adjoint extensions of the same Schroedinger operator and one can use Krein's formula to relate the resolvents of the two hamiltonians. So the spectrum of the two operators can be obtained from each other using perturbation theory.

Edit: Googling a little bit I found that this has obviously been done already:
http://www.sciencedirect.com/science/article/pii/002212368390085X

Last edited: May 1, 2013
17. May 2, 2013

### geoduck

So if I'm understanding what you're saying correctly, it is a sufficient condition for self-adjointness that the eigenfunctions over the domain form a complete L2 set over that domain?

Is it necessary and sufficient?

18. May 3, 2013

### tom.stoer

Do you mean that "an operator is self-adjoint if and only if its eigenstates form a complete set"?

Yes, any self-adjoint operator has eigenstates which form a complete set (math talks about the spectral theorem and the resolution of the identity).

No, I think there may be counterexamples where non-self-adjoint operators have a complete set of eigenstates; I am not sure whether the eigenstates of the annihilation operator a with a|z> = z|z> for arbitrary complex z serve as a valid example (b/c the states are not countable).

We do not necessarily need to talk about L2, spectral theory applies to other Hilbert spaces as well (even so the L2 case is most popular in QM).

In case of finite dimensional Hilbert spaces or bounded operators it's not so complicated, but in the general case one must be careful to distinguish symmetric, hermitean and self-adjoint operators. A nice example where the main difficulties can be studied is the momentum operator -i d/dx on the real half-line [0,∞[ (in math the relevant topics are self-adjoint extensions, Cayley transform and defect indices)

Last edited: May 3, 2013
19. May 3, 2013

### DrDu

I would not talk about eigenstates. A necessary (though not sufficient) condition for an operator being self-adjoint is that it's domain is dense in L2.
Besides that, it has to be symmetric and the domain has to coincide with the domain of the adjoint operator.
That's nicely explained on not too mathematical a level in Balentines book on quantum mechanics.