# Boundary Conditions and Hilbert Space

1. Nov 5, 2006

### Manchot

Hey all,

Last year, I took my university's undergraduate QM sequence. We mainly used Griffiths' book, but we also used a little of Shankar's. Anyway, I decided to go through Shankar's book this year, in a more formal treatment of QM. After the first chapter, I already have some questions that I hope can be answered:

1. When extending the results from finite-dimensional vector spaces to the infinite-dimensional spaces, it was assumed that most of the same results would hold. I really don't have a big conceptual problem with this, except in the case of matrix elements. Shankar shows that Hermiticity of the matrix elements of an operator in an infinite-dimensional space is not a sufficient condition for that operator to be Hermitian; namely, the surface term must also vanish for all vectors in the space. I can see the mathematical reason for this, but conceptually, what is the reason? My guess is that because the basis vectors are not well-defined at the endpoints, the matrix elements are not well-defined there as well (e.g., in position space, the basis vector at the endpoint is only "half" a delta function). Is this line of reasoning correct?

2. All throughout QM last year, I was deeply troubled by the fact that the boundary conditions were applied during the calculation of eigenvectors rather than after. For example, in the calculation of the states of the infinite square well, I didn't like that the endpoint-vanishing conditions were applied when calculating the eigenvectors (and quantizing them) as opposed to summing over all the continuous range of eigenvectors and then applying the same results (getting delta functions at the quantized values). I never really understood where this comes from, so I'd like to now venture a guess. Does this just arise from the definition of physical Hilbert space? As the surface term must vanish for the Hamiltonian to be Hermitian, one might conclude that to get a physical system, one must force the surface terms to be zero for all vectors in the space (including the eigenvectors, of course). That would explain a lot to me, but not quite everything. For example, what about the continuity of the wavefunction? Again, that could be explained away by incorporating continuity into the definition of physical Hilbert space. However, I wouldn't see the motivation behind doing this. Finally, what about the near-continuity of the derivative of the wavefunction? Since it depends on the value of the Hamiltonian at certain points, I wouldn't think that this could be logically incorporated into the definition of Hilbert space very easily.

Can anyone resolve these issues for me? If so, I would greatly appreciate it.

2. Nov 7, 2006

### Manchot

Ok, I've done some thinking on the matter, and I've decided that a near-adequate definition of Hilbert space is the space of square-integrable continuous functions. This forces the eigenvectors of the Hamiltonian to be continuous, and therefore satisfy the continuity boundary conditions. Nevertheless, it excludes the possibility of plane wave solutions, as well as the position and momentum basis vectors. Can anyone provide a better, more adequate definition for me?

3. Nov 8, 2006

### dextercioby

The matter you're addressing is rather subtle. First of all, the idea of infinite dimensional topological vector spaces automatically brings in issues of convergence and continuity. The concept of "hermiticity" for matrix elements is still the same (the matrix should be equal to the complex conjugate of its transpose), however operator issues are not that easy to generalize from finite to infinite dimensional spaces. Hermiticity (or symmetry) is not enough and we'll need at least essential self-adjointness for an accurate description of observables into Hilbert space context.

There's someting more. Boundary conditions are crucial in this discussion, because our Hamiltonians, when quantized using various prescriptions, are only symmetric and, in principle, have several self-adjoint extensions (measure of which are its deficiency indices) and different boundary conditions imply different self-adjoint extensions and basically different physics.

The cases when Hamiltonians of quantum systems have eigenvectors in the Hilbert space one initially chose for a mathematical description of the problem are very fortunate, since that means that the operators have a pure point spectrum. In most cases, the Hamiltonians also have a continuous spectrum (or only continuous spectrum) and the rather difficult and vast mathematics of Hilbert spaces is not enough to a get that mathematical description of reality that every theory of physics provides.

Daniel.

4. Nov 29, 2006

### Manchot

Ok, here's a dumb question. When you do a measurement of a Hermitian operator, we say that the wavefunction collapses to one of that operator's eigenvectors. Obviously, if you do a position or momentum measurement, the resulting wavefunction cannot be in Hilbert space, since the eigenvectors are non-normalizable. Physically, then, what happens? Say you did a position measurement. In position space, does the wavefunction really collapse to a delta function, or does it just collapse to something close to one (e.g., a really tall Gaussian)?

5. Nov 29, 2006

### masudr

Physically, there is no such thing as a perfect/ideal measurement that would give you the position of some object; usually we get a distribution of positions.

6. Nov 30, 2006

### dextercioby

There's the Heisenberg UP that prevents us from getting exact measurements for the position of a particle at quantum level.

Daniel.

7. Nov 30, 2006

### Manchot

Yeah, but I thought that the UP didn't really matter for that, since a collapsed delta function would have a position uncertainty of zero and a momentum uncertainty of infinity, which could mathematically still cancel out to be higher than hbar/2.

8. Nov 30, 2006

### vanesch

Staff Emeritus
If you want to know about all that, I guess it is a good idea to pick up von Neumann's "mathematical foundations of quantum mechanics"...

Warning: the typesetting of that old book is horrible.

9. Nov 30, 2006

### dextercioby

Well, the problem is that $$\infty \cdot 0$$ is not really defined for a physicist and so it makes no sense to compare it with a real number.

Daniel.

10. Dec 8, 2006

### Gza

The issue here is the fact that once you perform a "perfect" position measurement, the wavefunction is now in an eigenstate of the position operator, so speaking about the measurement of the momentum from that point on makes little sense, since (obviously) position and momentum don't commute
([x,p] =i*hbar)