Question on Completeness of a set of Eigenstates

[JK423]

I know that the set of eigenstates of an operator forms a complete set, a basis. Which means that any state can be written as a linear sum of the eigenstates.
But is the word 'any' correct? I haven't seen a proof of the above completeness theorem, so i don't know what the mathematical requirements are.
For example, consider the eigenstates of the Hamiltonian H₀ that correspond to an infinite square well [0,L]. These eigenstates vanish for x<0 and x>L. But that's a little restrictive! Because i cannot expand ANY state in this basis, for example those states that are also defined in the x<0, x>L region. Am i wrong?
One last question. If your Hamiltonian of your system is H=H₀+V, where H₀ is a random Hamiltonian and V the potential. Since the eigenstates of H₀ form a complete set, then i can expand any state of the system in this basis, even though the system is described by H and not H₀. Is this correct?

 

[unusualname]

'any' = 'any possible'

you're solving a linear differential equation, and the eigenstates form a basis for the solution space, not the whole hilbert space.

 

[tom.stoer]

You have to be careful on which Hilbert space the Hamiltonian is defined.

Regarding the infinite square well with [0,L]: the appropriate Hilbert space is L²[0,L]; the eigenstates of H span this Hilbert space, that means any (square integrable) function with support on [0,L] can be expanded w.r.t. the eigenstates of H; but it does not makes sense to ask for x outside this interval!

Regarding you last question (H°, H and V): yes this is correct.

 

[JK423]

Thanks both of you!

tom.stoer, can you clarify what you mean by 'support' ?
<< .. function with support on [0,L]>>

Also, how can i define the Hillbert space that the eigenstates of a Hamiltonian span? Just from the definition domain of the operator? For example, if the operator is defined in (-00,+00) then every state can be exanded in its eigenstates?
I don't have a good knowledge on the purely mathematical side of the theory..

 

[sweet springs]

Hi.

For example, consider the eigenstates of the Hamiltonian H₀ that correspond to an infinite square well [0,L]. These eigenstates vanish for x<0 and x>L. But that's a little restrictive! Because i cannot expand ANY state in this basis, for example those states that are also defined in the x<0, x>L region. Am i wrong?

Let potential energy of the system V=-D inside the well , 0 outside the well. We have unbounded continuous energy eigenstates E>0 and bounded discrete eigenstates E<0. Even if D is infinite, we have unbounded energy states. How's that?

Regards.

 

[sweet springs]

Hi.

One last question. If your Hamiltonian of your system is H=H₀+V, where H₀ is a random Hamiltonian and V the potential. Since the eigenstates of H₀ form a complete set, then i can expand any state of the system in this basis, even though the system is described by H and not H₀. Is this correct?

Though I do not understand your 'random Hamiltonian', if we regard H0 as unperturbed Hamiltonian, V as perturbation, then usual perturbation treatment is applicable. Both H and H0 create complete sets of basis.

Regards.

 

[xepma]

Thanks both of you!

tom.stoer, can you clarify what you mean by 'support' ?
<< .. function with support on [0,L]>>

Also, how can i define the Hillbert space that the eigenstates of a Hamiltonian span? Just from the definition domain of the operator? For example, if the operator is defined in (-00,+00) then every state can be exanded in its eigenstates?
I don't have a good knowledge on the purely mathematical side of the theory..

You pretty much give the answer to your own question here. The Hilbert space is defined such that the Hamiltonian you are working with is self-adjoint. In other words, we want the Hilbert space to be spanned by the eigenstates of the Hamiltonian we are working with. If we wish to work with a larger Hilbert space, then the original Hamiltonian is no longer self-adjoint -- we would need to redefine it such that it becomes self adjoint again.

The same statement is: if we consider a state which is not an element of the Hilbert space we are working with (i.e. that Hilbert space on which the Hamiltonian is self-adjoint) then in general we will not be able to write this state as a sum over eigenstates of the Hamiltonian.

As for your question on perturbation theory: suppose you have a Hamiltonian of the form H = H0 + V. What's implicitly assumed here is that the operators H, H0 and V are all self-adjoint on the Hilbert space we are working with. So all these operators have a complete set of eigenstates in the Hilbert space -- just like sweet springs mentioned.

 

[tom.stoer]

Support on [0,L] means that the function is defined on [0,L] and vanishes outside this interval identically. Therefore one can savely restrict to this interval.

Having support in [0,L] does not imply that f must vanish at 0 and L! This is an additional physical requirement and means that one restricts to a subset of L²[0,L] only!
The problem is that in QM most people are not aware of the fact that there are these mathematical subtleties.

If you restrict the Hilbert space to L²[0,L] you can calculate

\int_0^L dx \bar{g}(x) f(x)

Now as f(x) vanishes at 0 and L one can allow for g(x) not to vanish there; nevertheless all matrix elements are perfectly well defined and do not pick up contributions from the boundary. But this indicates that g(x) lives in a larger space than f(x)! This is the reason why we have to distinguish between symmetric and self-adoint operators and why a space and its dual need not be identical (which is trivial for finite-dimensional vector spaces)

 

[JK423]

Thanks all of you for your answers..
tom.stoer, which book would you recommend in order to study the mathematical side of the theory?

 

[tom.stoer]

I do not know about any QM book were you can find that stuff.

I studied mathematics (functional analysis) as well and had a German book which discussed QM related topics. Would this be helpful?

 

[JK423]

So, i should study functional analysis? That seems to be hard.. !

 

[tom.stoer]

In many cases you wwill not need it; but sometimes it becomes relevant.

 

[unusualname]

It's more important to understand the physical implications of the problem you are solving. The mathematical theorems about completeness etc are nice to know but must be carefully applied. Eg in your problem with a square well you have boundary conditions at the end points requiring the wave function to be zero there, so your eigenstates are a basis with similar restrictions and don't extend to a basis ouside this region.

In other cases you may have a mixture of discrete and continuous eigenvalues and the basis is then the combined set of bounded eigenstates for the discrete set and unbounded (non square integrable) eigenstates for the continuous set. In the discrete case you can expand an arbitrary wave function in terms of a series, in the continuous case you must use an integral (so an arbitrary wave function is the sum of a discrete series of bounded eigenstates + an integral of unbounded eigenstates)

eg http://en.citizendium.org/wiki/Hydrogen-like_atom

People like Dirac didn't bother with measure theoretical functional analysis when introducing stuff like delta functions (Heisenberg barely knew general matrix theory) but they were able to make great progress and solve important physical problems. This crops up a lot in modern physics, the method of renormalisation can be made mathematically rigorous but it hardly helps with physical applications.