# 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.

21. Jun 18, 2007

### olgranpappy

Let's put your system in a large box. Now, if you want the amplitude to find each of the N photons at a given point in the box, how do you find it?

22. Jun 18, 2007

### Anonym

Your problem is that you are listening to yourself only. These are not a questions Smallphi asking (see above and below). Smallphi is math-ph., was born so (like Quantum River). This is another Galaxy for you. You apparently don’t able to understand that before “postulating extra degrees of freedom which are oscillator-independent” one should define the notion of internal degree of freedom (and it is already done in math-ph: C.N.Yang 50 years ago). However, I don’t want misinterpretation that you are master in that: We can’t work without physicists like you.

Smallphi, I already answered your questions. Read carefully J. von Neumann book and D.Hilbert paper around 1912 I guess if you need the origin. To understand better the situation, read E. Schrödinger, Die Naturwissenschaften, 48,807,(1935); 48,823,(1935); 48,844,(1935) (English tr. in W&Z). E. Schrödinger explain that even non-relativistic QM is intrinsically relativistic like CED because the problem of time operator. See also F.J. Dyson “Feynman’s proof of the Maxwell equations”, Am. J. Phys., 58, 209 (1990).

The Hilbert space is the mathematical language that uses the complex state vectors and the complex scalar products only. We have four numerical systems available: 1-dim real, 2-dim complex, 4-dim real quaternion and 8-dim real octonions, which allow 1, 3 and 7 relative phases correspondingly. That in turn allows defining the mathematical languages similar to the Hilbert space (Hilbert module). Since in general in order to construct Fock spaces you need to use the Kronecker products, the requirement that the scalar product belong to the same numerical system is unjustified.

Thus you end up with: four structures with the real, complex, quaternion and octonion state vectors with the real scalar product (dispersion free field theories; J. von Neumann: classical physics); complex, quaternion and octonion state vectors with the complex scalar product (quantum physics); quaternion and octonion state vectors with the quaternion scalar product (I don’t know what it is); octonion state vectors with the octonion scalar product (I have no idea what it is: Be Resheet).

Regards, Dany.

P.S. The only case known to me when you have 2-fold ground state in absence of local gauge interaction is in octonion QM with the complex scalar product.

P.P.S. I intentionally do not discuss with you the question: 'complete set of commuting observables' since I consider it the key question for formulation of consistent relativistic QM and I want to do that with my own hands.

Last edited: Jun 18, 2007
23. Jun 18, 2007

### George Jones

Staff Emeritus
Last edited by a moderator: May 2, 2017
24. Jun 18, 2007

### Anonym

Thank you, I’ll do it. Did he explain the results of A. Einstein analysis of black body radiation curve? ("A man ought to read just as inclination leads him; for what he reads as a task will do him little good.").

Regards, Dany.

25. Jun 18, 2007

### Hans de Vries

This threat is way to abstract and mixing up all kinds of things.
So, let’s try to handle this as simple as possible:

1) Canonical ('second') quantization is not a quantization of the wave-
function in space but that of phi(x,t) as an internal coordinate.

2) The form of the wave-function in space is always assumed to be a
sinusoidal plane wave.

3) Second quantization is intended to be an explanation why the vacuum
can only have 1, 2, 3, ... n of such sinusoidal plane waves, for each
frequency, representing 1, 2, 3, ...n particles of that frequency.

CANONICAL QUANTIZATION OF THE KLEIN GORDON FIELD
==========================================

Now we get to our harmonic oscillators. You can literally see them
immediately by looking at this picture:

http://chip-architect.com/physics/Klein_Gordon.jpg

This is a mechanical representation (with springs and masses) of the
one dimensional real Klein Gordon equation:

$$\frac{\partial^2 \psi}{\partial t^2}\ =\ c^2\frac{\partial^2 \psi}{\partial x^2} - m^2 \psi$$

Where the horizontal axis is x and the vertical axis is psi. The wave-
functions of this mechanical system behave exactly as deBroglie waves,
including the phase speed of c^2/v.

The vertical springs c^2 determine the propagation speed and the
horizontal springs m^2 determine the mass of the particle. The vertical
springs make up an harmonic oscillator at each point in space.

Canonical quantization is now done by treating these harmonic oscillators
as quantum mechanical harmonic oscillators instead. The solutions
are the Gaussian Hermite functions which you can see here:

http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

The creation operator repeatedly applied to the ground state H0 gives
H1, H2, H3, ... Hn. The annihilation operator does the opposite and goes
the other way H3, H2, H1, H0. Operating on H0 gives 0. The number
operator gives back the value of n. The ground state H0 is denoted by |0>

The energy level of n particles is given by:

$$E_n \ =\ \left( n+\frac{1}{2} \right)\ \hbar \omega$$

Which quantizes the number of particles that can exist at a certain
frequency.

CANONICAL QUANTIZATION OF THE DIRAC FIELD
====================================

This is not that much different. Not in the last place because all the
individual spinor components obey the Klein Gordon equation.

CANONICAL QUANTIZATION OF THE EM FIELD
==================================

This is more tricker. Peshkin & Schroeder doesn't even discuss it. We
see the problem if we consider that the field is mass-less. The vertical
springs m^2 in our image above disappear and there is no more
harmonic oscillator. So, we need something else to take over this

The EM field can be described, considering gauge invariance, with four
equations:

$$\partial^2_t A^\nu\ =\ \partial_x^2 A^\nu +\partial_y^2 A^\nu +\partial_z^2 A^\nu \ +\ \partial^\nu \left( \partial_\mu A^\mu \right)$$

Where each value of $\nu = 0,1,2,3$ gives us one of the four equations.
It is clear that the last term should take over the role that the mass had
in the Klein Gordon equation in one way or another. However, the conserved
current rule for the EM field gives us:

$$\partial_\mu A^\mu \ =\ \partial_t V + \partial_x A_x + \partial_y A_y + \partial_z A_z \ =\ 0$$

Under this condition it is not possible to do a second quantization of the
EM field. Now, Gupta and Bleuler came up with a less restrictive rule:

$$\partial_\mu A^\mu \ \neq\ 0, \qquad \mbox{but} \quad \partial_\mu A^{(+)\mu} |\psi \rangle \ =\ 0$$

The latter means that if we consider only positive frequency parts then
these operating on a field should have a zero effect. This is what made
it possible to do a second quantization of the EM field.

that's it.

Now personally I'm not fully satisfied with this solution, because the
source of the EM field, The charge/current density is conserved.
That is:

$$\partial_\mu J^\mu \ =\ 0$$

And it is hard to see how this conserved current can give rise to a non-
conserved EM field, and worse: If the EM field is expressed as a plane
wave, going in the z-direction, for instance: (0,Ax,Ay,0), then all derivatives
in x and y are zero, and this also sets the last term to zero, the term we
need for second quantization, but OK. A good treatment of the EM field
can be found in:

Lewis H Ryder, Quantum Field Theory, chapter 4

Regards, Hans

Last edited: Jun 18, 2007