Complete basis for quantum oscillator Hilbert space

1. Jun 13, 2007

smallphi

In the creation/annihilation operator picture the Hilbert state of a quantum harmonic oscillator is spanned by the eigenstates |n> of the number operator.

I've never seen a proof that:

1. the ground state |0> is unique
2. the states |n> form a complete basis i.e. any state in that Hilbert space can be decomposed into them

Any references or explanations?

2. Jun 14, 2007

vanesch

Staff Emeritus
It doesn't have to be ! You postulate that it is unique. However, what you can easily show is that if there are different ground states, then they give rise to independent "series" of excited states (in other words, you have in fact several harmonic oscillators in parallel).

Nothing stops one from having |0_1> and |0_2>, and having:
a+ |01> = |1_1> etc...
and
a+ |02> = |1_2> etc...

As such, (a+)^n (u |0_1> + v |0_2>) = Constant (u |n_1> + v |n_2> ), and you have a superposition (coefs u and v) of two parallel harmonic oscillators.

Again, this is (the same) hypothesis that you accept, namely that the unique ground state and its family of excited states reached with a+ is spanning the entire hilbert space of states.

You could just as well make another hypothesis, and, as above, introduce several independent ground states, with a (new) set of operators which gives transitions between them.

3. Jun 14, 2007

smallphi

The original Hilbert space is spanned by eigenverctors of the X (or P) operators (although those are not square integrable so don't belong to the space themselves). So the Hilbert space is fixed/defined from the very begining, you have no freedom re-defining it by postulating unique |0> or completeness of |n>'s.

Your explanation maybe works if you start with a HO Hamiltonian directly written in terms of creating and annihilation operators and the Hilbert space is yet not defined.

My question is related to the question how, given a fixed/pre-defined Hibert space we determine a 'complete set of commuting observables' i. e. a set of commuting operators whose common eigenvectors span the space. In the case of harmonic oscillator the complete set of commuting observables consists of only the number operator:

$$N = a^+ a$$

Last edited: Jun 14, 2007
4. Jun 14, 2007

Hans de Vries

The ground state can also be obtained by the requirement that the
annihilation operator acting on the vacuum gives zero, that is:

$${\hat a}|0 \rangle\ \equiv\ 0$$

thus

$$\sqrt{\frac{1}{2}}\left( \frac{\partial}{\partial \psi} + \psi \right) |0 \rangle\ =\ 0$$

which is satisfied by:

$$|0 \rangle \ =\ e^{-\psi^2/2}$$

Which is the usual (unnormalized) ground state.

A proof of the orthogonality is given in Arfken and Weber:
"Mathematical Methods for Physicists", chapter 13 which handles the
Hermite polynomials including the canonical quantization with creation,
annihilation and number operators.

Regards, Hans

Last edited: Jun 14, 2007
5. Jun 15, 2007

vanesch

Staff Emeritus
Ah, then that's easy (see Hans' proof for example). I was indeed considering an abstract harmonic oscillator setup, defined by the creation and annihilation operators, and not based upon the mechanical 1-particle model.
That the set of eigenstates of the hamiltonian is complete should be obvious given that it is a hermitean operator, so it gives automatically rise to a complete set of eigenvectors.
The |0> is unique then follows from an explicit calculation as a function of the defining states of the hilbert space, which is here the position operator eigenvectors - that's what Hans did.

Indeed, that's what I was considering, given that the mechanical example is rather straight-forward.

In fact, although in most intro QM courses, things are not presented that way, the Hilbert space is never "pre-given", but is always the result of a *postulated* complete set of commuting observables, for which one *postulates* a common set of eigenvectors and their associated eigenvalues.

You START by enumerating (as a postulate for your specific quantum-mechanical model) a basis, and you characterise that basis by postulating a complete set of operators, which have that basis as eigenvectors. THIS then defines your hilbert space, and all the rest you're going to do will have to be defined as a function of your postulated complete set of operators and your postulated basis.

For instance, you can define/postulate that you have a quantum-mechanical system with, say, 6 basis states, and 2 operators A and B.
Let us call the 6 basis states for the moment, |a>, |b>, |c>, |d>, |e> and |f>.
We next have to postulate what are their eigenvalues under A and B:

Say:
A |a> = 0, A |b> = |b>, A|c> = 0, A |d> = |d>, A|e> = 0, A|f> = |f>

B |a> = 0, B |b> = 0, B|c> = |c>, B |d> = |d>, B|e> = 2|e>, B|f> = 2|f>

In other words, A has 2 eigenvalues, 0 and 1, and the eigenspace of 0 is spanned by a,c and e.
B has 3 eigenvalues, 0, 1 and 2, and the eigenspace of 0, for instance, consists of a and b.

All this is part of the *postulates of your specific quantum-mechanical model*.

The next thing to do is to define a dynamics over that space, by giving a Hamiltonian as a function of A and B, and next you can introduce all kinds of observables, which should also be expressed as a function of A and B.

In fact, the above simple model is just a 2-spin system, which is the composition of a spin-1/2 and a spin-1 system, and A and B are shifts of the respective sigma_z: for instance: A = sigma1_z + 1/2, and B = sigma2_z + 1.

Now, there is a specific class of QM models, which is inspired by classical mechanics, and which says the following:

take the configuration space of the classical system, and its generalized variables q1,...qn. Associate with each point of configuration space (defined by the point {q1,q2,...qn}) one single basis vector, which we call |q1,q2,...qn>, and consider n operators Q1, ... Qn as the complete set of commuting observables. The eigenvalue of Qk on |q1,...,qk,...,qn> equals qk by postulate.

So the quantum-mechanical model "inspired" by a classical model, on the above example, is simply given by a hilbert space which associates a basis vector to each point in configuration space, and a complete set of commuting observables corresponding to the set of generalized coordinates, which have this basis as common eigenbasis, and which have as eigenvalue the value of the generalized coordinate of the corresponding point in configuration space.

The momentum operators {P1,...,Pn} are defined as being those operators that satisfy the canonical commutation relations with the set of operators {Q1,...Qn}, and are hence defined in function of {Q1,...Qn}.

But we can think of other quantum-mechanical models. Spin systems are such (they are fundamentally not inspired by a classical system). A priori, the harmonic oscillator is also different, because (and it was in that frame that my previous comments were made) you can define it abstractly by introducing a single operator N, and a complete (by postulate) set of eigenvectors {|n>}, such that N |n> = n |n>, n >= 0.
It then turns out that you can write N = (a+) a and that these are ladder operators.
Now, it turns out also, that this system is 1-1 with a classical model, which is the 1-dimensional mechanical harmonic oscillator, but that's "a result after the fact".

6. Jun 15, 2007

Anonym

May you be accurate just for the difference? Most close to the classical harmonic oscillator are the states which are eigenstates of the annihilation operator and not the eigenstates of the number operator. “Unfortunately”, they do not form the orthonormal basis for the Hilbert space. It is still open problem.

Regards, Dany.

7. Jun 15, 2007

smallphi

I want confirmation of two facts I gathered from this discussion:

1. Any hermitian operator in given Hilbert space has eigenvectors that form a complete basis of the space. Can someone refer me to a proof of that?

2. In QFT we quantize say a scalar field by saying that the ground state |0> is the state annihilated to zero by the annihilation operators of all modes. We don't have to prove that this state exists at all or is unique because we POSTULATE this is so effectively building the Hilbert space.

I once asked a professor 'How do we know the ground state (the vacuum) of the field exists?'. And he replied 'How do you know the ground state of harmonic oscillator exist?'.

8. Jun 15, 2007

smallphi

I want confirmation of two facts I gathered from this discussion:

1. Any hermitian operator in given Hilbert space has eigenvectors that form a complete basis of the space. Can someone refer me to a proof of that?

2. In QFT we quantize say a scalar field by saying that the ground state |0> is the state annihilated to zero by the annihilation operators of all modes. We don't have to prove that this state exists at all or is unique because we POSTULATE this is so effectively building the Hilbert space.

I once asked a professor 'How do we know the ground state (the vacuum) of the field exists?'. And he replied 'How do you know the ground state of harmonic oscillator exist?'.

9. Jun 15, 2007

smallphi

I want confirmation of two facts I gathered from this discussion:

1. Any hermitian operator in given Hilbert space has eigenvectors that form a complete basis of the space. Can someone refer me to a proof of that?

2. In QFT we quantize say a scalar field by saying that the ground state |0> is the state annihilated to zero by the annihilation operators of all modes. We don't have to prove that this state exists at all or is unique because we POSTULATE this is so effectively defining/building/constructing the Hilbert space.

I once asked a professor 'How do we know the ground state (the vacuum) of the scalar field exists?'. And he replied 'How do you know the ground state of harmonic oscillator exist?'.

Last edited: Jun 15, 2007
10. Jun 15, 2007

Fra

This is not the case, in general.

Like Vanesch mentioned you'd have to ask - what hilbert space?

OTOH, given a hermitian operator, it induces a minimal hilbert space where it's eigenfunctions are a complete basis. For proof, search for kolmogorov's dilation theorem in some functional analysis book.

But this doesn't mean there aren't bigger hilbert spaces, which your hermitian operator can't span.

Given a set of assumed commuting operators you can define a minimal hilbert space where the set of eigenfunctions is a complete base, in line with vanesch post.

This is math.

I personally think the interesting points (physical content) and the hardest part is to what accuracy these postulates/assumptions/axioms make real life easier or not. Clearly they are useful, but are they perfect? The problems of applying formalisms technical, and often complex, but usually readily solvable and provable. The harder parts is that you can't prove or disprove as easily. This is often I think the interesting parts.

/Fredrik

11. Jun 15, 2007

Anonym

Hans de Vries in post #4 presented the complete answer to your original question. The rest is irrelevant to your original question.

Regards, Dany.

12. Jun 15, 2007

Anonym

D. Hilbert, Spectral Decomposition Theorem, R.Courant and D.Hilbert, Methoden der Mathematischen Physik, Berlin, (1931); J. von Neumann, Mathematische Grundlagen der Quantenmechanik, Berlin, (1931).

The accurate description of V.A. Fock second quantization procedure you may find in any textbook on QM and QFT.

Find a professor that know QM and QFT, that is able to answer your questions and do not ask you questions.

Regards, Dany.

13. Jun 15, 2007

olgranpappy

P.S. you've never seen a proof that the ground state is unique in general (i.e., for *any* system, even fake pedegogical systems) because there is no proof. For example: Two perfectly isolated infinite square wells. Another more "realistic" example: the N-particle (N>1) groundstate of a *non-interacting* fermi gas.

14. Jun 15, 2007

olgranpappy

...on the otherhand, for an N-particle *non-interacting* bose gas the ground state is unique... because the ground state of one oscillator is unique (because it is *the* solution to a first order differential equation), and the ground state of N-particle bose gas occurs when all are in that single-particle ground state.

15. Jun 15, 2007

smallphi

A system of oscillators like bunch of particles in quadratic potential would have unique ground state because you can solve for the wave function in an already defined Hilbert space spanned by eigenstates of the position or momentum operators.

On the other hand, the EM field is also a 'bose gas' but the Hilbert space is not spanned in advance before you postulate the existence and uniqueness of the ground state. The Hamiltonian may look like bunch of 'oscillators' but you are missing the formal expressions of the creation and annihilation operators in terms of differential operators acting in some functional space like in the case of harmonic oscillator. Hence you can't solve for the ground state of each 'oscillator' and prove its unique. The only way would be to postulate it and see if that agrees with experiment.

Last edited: Jun 15, 2007
16. Jun 16, 2007

Anonym

No. It is an intermediate step usually used before the adequate mathematical framework is found.

Regards, Dany.

17. Jun 16, 2007

smallphi

Anonym, I don't understand your post.

18. Jun 16, 2007

Anonym

You can’t introduce the postulates arbitrarily in order to solve the particular problem. It is similar as in the history of CED. M.Faraday postulated closed lines of force to explain the experimentally observed behavior of the magnetic field. It led to the development of the vector algebra and the vector analysis which is the adequate mathematical framework for CED. Newton’s classical analysis was not adequate to that purpose.

By the way M. Faraday discovered the electrical charge quantization. I did not see the convincing demonstration of it even in U(2) electroweak phenomenological model but it is obviously the QED natural problem.

Regards, Dany.

P.S. The alternative way may be through the connection between the CED and QED as I mentioned in post #6.

Last edited: Jun 16, 2007
19. Jun 17, 2007

vanesch

Staff Emeritus

That's correct.

You want an example of a harmonic oscillator that has no unique ground state ?
Take a spin-1/2 particle in a harmonic potential with no spin term in the hamiltonian.

We now have TWO independent ground states, namely |0+> and |0->.

The whole creation and annihilation business works, and it is like the example in my earlier post:

If you start with ground state a |0+> + b |0->, you will end up in a |n+> + b |n-> after n creation operations.

So the CHOICE of having 1 ground state is a physical hypothesis for the model at hand.
In the case of a mechanical oscillator, it would come down to the hypothesis that the oscillator has
no extra degrees of freedom (internal degrees of freedom). Making the choice of several ground
states comes simply down to postulating extra degrees of freedom which are oscillator-independent.

20. Jun 17, 2007

smallphi

How exactly they hypothesis that the oscillator has no extra degree of freedom translates into math language?

I bumped on several occations on statements like 'the X and P operators are irreducible'.

I would like to see a proof of existence and uniqueness of ground state following from the mathematical statement that the system has no extra degree of freedom.