What Hilbert space for a spinless particle?

[David Olivier]

I'm looking for a rigorous mathematical description of the quantum mechanical space state of, for instance, a particle with no internal states.

At university we were told that it the Hilbert state of wave functions. They gave us no particular restrictions on these functions, such as continuity, apart from the fact that they should be quadratically integrable, so that we can define inner products. But then we are given a basis of kets of the form $| \vec r>$, representing "Dirac functions", that precisely are not quadratically integrable. We were told to shut up and calculate.

I'd like to clear this up a bit. I know the Dirac "functions" can be defined rigorously as distributions. But how does that fit into a Hilbert space?

Does someone know of a mathematically rigorous, but elementary, introduction to this issue?

 

[DrDu]

Dirac functions are indeed not elements of the Hilbert space. However, the Hilbert space may be enlarged to encompass these functions. The new space is called "rigged Hilbert space".
https://en.wikipedia.org/wiki/Rigged_Hilbert_space
Whether you want to do QM with rigged or ordinary Hilbert spaces is rather a matter of taste, i.e. whether you like the formal simplicity of Diracs formalism or whether you rather consider distributions as unphysical idealizations.

 

[David Olivier]

Thanks. I'll look it up.

If you want to stick to plain Hilbert spaces, what bases do you have? You don't have the "particle in a precise position" kets, nor the "particle with a precise momentum" ones; so what do you use?

 

[DrDu]

For example harmonic oscillator eigenstates.

 

[David Olivier]

OK, but that's not for a free particle.

Though I suppose that you could use them for a free particle (no potential), even though they would no longer be eigenstates of the Hamiltonian. They would still form a basis. But that looks like an awkward choice.

 

[David Olivier]

I've found this introduction to rigged Hilbert spaces: Rafael de la Madrid, "The role of the rigged Hilbert space in Quantum Mechanics" (free download). Seems easier to go through than the Wikipedia article.

 

[PeterDonis]

For example harmonic oscillator eigenstates.

As the OP says, those won't work for a free particle because they would just be plane waves, which are not normalizable and so are not in the (non-rigged) Hilbert space.

I suppose that you could use them for a free particle (no potential), even though they would no longer be eigenstates of the Hamiltonian. They would still form a basis.

Not if you want your basis states to be in the (non-rigged) Hilbert space.

 

[David Olivier]

As the OP says, those won't work for a free particle because they would just be plane waves, which are not normalizable and so are not in the (non-rigged) Hilbert space.

If I remember well, the wave functions of a harmonic oscillator are something like $\psi_0(x) = a_0 e^{-b x^2}$, $\psi_1(x) = a_1 x e^{-b x^2}$...

These are not plane waves, and are normalizable.

Do they not form a basis of the Hilbert space? It seems to me that the Hilbert space is independent from the Hamiltonian. Of course, the above wave functions will evolve differently in the absence of the quadratic potential, not being eigenstates anymore of the Hamiltonian, but as elements of the Hilbert space, if they form a basis in the presence of that potential, they form one in the absence of it too.

 

[DrDu]

It seems to me that the Hilbert space is independent from the Hamiltonian. Of course, the above wave functions will evolve differently in the absence of the quadratic potential, not being eigenstates anymore of the Hamiltonian, but as elements of the Hilbert space, if they form a basis in the presence of that potential, they form one in the absence of it too.

Yes, that's completely correct.
For operators like p or r, it is not necessary to have a set of eigenvectors in Hilbert space. It is sufficient that they can be expressed as an integral over a projection valued measure. Von Neumanns book is still a classic on this theory, and relatively easy to read.

 

[PeterDonis]

If I remember well, the wave functions of a harmonic oscillator are something like $\psi_0(x) = a_0 e^{-b x^2}$, $\psi_1(x) = a_1 x e^{-b x^2}$...

These aren't eigenstates of the harmonic oscillator Hamiltonian (or the free particle Hamiltonian, for that matter; they're not the same). They are coherent states, which are eigenstates of the harmonic oscillator annihilation operator $\hat{a}$. They are normalizable, yes, but I don't know if they can be used as a basis.

Eigenstates of the free particle Hamiltonian, which are the ones I was referring to, are plane waves, i.e., wave functions of the form $\psi = a e^{i k x}$. These are not normalizable.

Eigenstates of the harmonic oscillator Hamiltonian are eigenstates of the number operator, which is $\hat{a}^\dagger \hat{a}$. These are normalizable, and can be used as a basis. (I think I misspoke in my previous post, since these states might have been the basis @DrDu had in mind.)

 

[David Olivier]

(In reply to DrDu) In other words, the need for a basis is overdone! :)

 

[dextercioby]

I'm looking for a rigorous mathematical description of the quantum mechanical space state of, for instance, a particle with no internal states.

At university we were told that it the Hilbert state of wave functions. They gave us no particular restrictions on these functions, such as continuity, apart from the fact that they should be quadratically integrable, so that we can define inner products. But then we are given a basis of kets of the form $| \vec r>$, representing "Dirac functions", that precisely are not quadratically integrable. We were told to shut up and calculate.

I'd like to clear this up a bit. I know the Dirac "functions" can be defined rigorously as distributions. But how does that fit into a Hilbert space?

Does someone know of a mathematically rigorous, but elementary, introduction to this issue?

Some more comments at "A" level.

Regardless of the used interpretation, theoretically, a quantum system will always be specified by the mathematically conceivable states and observables. And here the word "mathematically" is crucial. If I had written "physically", you would have been in the position to ask me: how does one experimentally prepare and measure these states and observables? But since I've chosen to tell you about the mathematically meaningful states and observables, then let me do so.

A quantum system is mathematically defined by the Hamiltonian and the set of irreducible observables and the algebraic (addition, composition = multiplication) relations which help us construct from them the whole set of all possible observables. Depending on the system, the Hamiltonian may or may not be in the irreducible set. This set of all observables can be given the structure of an associative algebra with unity with respect to observables' addition. In this scenario, the set of irreducible observables of a quantum system is the center of this algebra, if we further endow it with the Lie product structure (commutator). For a free spinless particle in 1D, the irreducible observables are the coordinate x and momentum p, because the system is defined by H=p^2/2m. These 2 irreducible observables obey the so-called Born-Jordan commutator relation [x,p]=i.

If one then asks which are the possible PHYSICAL states for the free spinless particle in 1D, the answer is simple: it is the complex, separable, infinite dimensional Hilbert space (or, more pedantic, the set of points in the projective Hilbert space built from it is the set of pure states, as opposed to mixed states which can't be given the structure of a projective Hilbert space) in which the commutation relation can be mapped/realized. By the known Stone-von Neumann uniqueness theorem, this space is only L^2(R), if the particle is conceived to be unrestricted in motion. The possible MATHEMATICAL states are the ones which require the mapping of the B-J commutation relation into a rigged Hilbert space with L^(R) as the Hilbert space from it.

Anything else is derived from here. The set of states of "well-defined position" (and here we venture into the realms of interpretations of QM) for a free spinless particles is not made up of physical states, but of mathematical ones. The set of states of "well-defined energy" for a free spinless particles is not made up of physical states, but of mathematical ones etc.

 

[Physics Footnotes]

(In reply to DrDu) In other words, the need for a basis is overdone! :)

In principle, yes. Here's a rough outline of how it works von Neumann style. If it sounds like the approach appeals to you, I agree with @DrDu that von Neumann's original work "Mathematical Foundations of Quantum Mechanics" remains a wonderful introduction to the subject for the mathematically paranoid ;-)

Consider a finite dimensional Hilbert Space $\mathcal H$ spanned by a set of orthonormal vectors $|\lambda_i\rangle$ labelled by the eigenvalues of the corresponding Hermitian operator$$A=\sum \lambda_i|\lambda_i \rangle\langle \lambda_i|$$Now, for any element $|\psi\rangle \in \mathcal H$ we have:$$|\psi\rangle=\sum \langle \lambda_i|\psi\rangle|\lambda_i\rangle=\sum|\lambda_i \rangle\langle \lambda_i|\psi\rangle$$which, in terms of projection operators, gives $$I=\sum |\lambda_i \rangle\langle \lambda_i|$$All the work we need to do, it turns out, (as far as the statistical and spectral analysis of $A$ are concerned) can now be done with the family of projection operators $|\lambda_i \rangle\langle \lambda_i|$ (called, by virtue of that last relation, the Resolution of the Identity for $A$), which, together with the spectrum, is used to define the Projection Valued Measure (PVM) of $A$.

What's the point of all this? Well, it turns out that the notion of a PVM can be rigorously extended to infinite dimensional Hilbert spaces, and in particular copes perfectly with continuous spectra without any need of eigenvectors. This approach, invented by von Neumann in the late 1920s, allows us to do all quantum mechanical calculations rigorously within the Hilbert Space framework, by eliminating Dirac's delta function and all the Rigged Hilbert Space machinery that it requires for a rigorous treatment.

 

[bhobba]

I'd like to clear this up a bit. I know the Dirac "functions" can be defined rigorously as distributions. But how does that fit into a Hilbert space? Does someone know of a mathematically rigorous, but elementary, introduction to this issue?

Its called a Rigged Hilbert Space.

Unfortunately mathematically rigorous and elementary are not found in this area. Attached is a rigorous treatment - but far from elementary. Its Dr Madrid's PhD Thesis on it and goes into much more detail than the articles by him you have probably read eg areas like Nuclrar spaces etc.

However the following, while not fully explaining what going on is approachable:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

It should really be known by all physicists, and indeed applied mathematicians. Its worth it for its treatment of Fourier Transforms alone which otherwise becomes bogged down in issues of convergence,

Thanks
Bill

 

[DrDu]

These aren't eigenstates of the harmonic oscillator Hamiltonian (or the free particle Hamiltonian, for that matter; they're not the same). They are coherent states, which are eigenstates of the harmonic oscillator annihilation operator $\hat{a}$. They are normalizable, yes, but I don't know if they can be used as a basis.
@DrDu had in mind.)

No, these are really eigenstates of the harmonic oscillator and not coherent states. You can easily show that e.g. $\psi_1(x)$ isn't an eigenstate of $\hat{a}$.

 

[atyy]

(In reply to DrDu) In other words, the need for a basis is overdone! :)

If you measure position, you still need the position states which are Dirac Deltas and not in the Hilbert space. You need the position Delta functions, simply because they come along as part of the definition of an exact measurement of position.

The modification of rule-of-thumb basic QM that is needed is that after an exact position measurement, the wave function does not collapse into a position eigenstate (since that doesn't belong to the Hilbert space). The collapse rule is given in Eq 3 and 4 of https://arxiv.org/abs/0706.3526, and an example of the wave function after an exact position measurement is discussed in section 2.3.2 on p7.

 

[DrDu]

If you measure position, you still need the position states which are Dirac Deltas and not in the Hilbert space. You need the position Delta functions, simply because they come along as part of the definition of an exact measurement of position.

The modification of rule-of-thumb basic QM that is needed is that after an exact position measurement, the wave function does not collapse into a position eigenstate (since that doesn't belong to the Hilbert space). The collapse rule is given in Eq 3 and 4 of https://arxiv.org/abs/0706.3526, and an example of the wave function after an exact position measurement is discussed in section 2.3.2 on p7.

I don't think there is a modification of any QM rules needed. The point is rather that we can't measure a continuous variable arbitrarily precisely, so the limit of delta functions is unphysical.

 

[David Olivier]

Attached is a rigorous treatment - but far from elementary. Its Dr Madrid's PhD Thesis on it and goes into much more detail than the articles by him you have probably read eg areas like Nuclrar spaces etc.

Thanks. As noted above, I already had found a (much shorter) pdf by the same author, "The role of the rigged Hilbert space in Quantum Mechanics". What I now lack is time to digest all this.

 

[DrDu]

Consider a finite dimensional Hilbert Space $\mathcal H$ spanned by a set of orthonormal vectors $|\lambda_i\rangle$ labelled by the eigenvalues of the corresponding Hermitian operator$$A=\sum \lambda_i|\lambda_i \rangle\langle \lambda_i|$$Now, for any element $|\psi\rangle \in \mathcal H$ ...

I would like to dwell on this a little bit more.
Let's generalize this to infinite dimensions and take the position operator as an example. Then,

$X=\int dx x \delta(x'-x) \delta(x''-x)=\int dx x \delta(x'-x) \delta(x''-x')= \delta(x''-x') \int x d \theta(x-x').$
The last step should be familiar from classical statistics, where we express the integral over a probability density function by an integral with the cumulative distribution function as integrand.

 

[atyy]

I don't think there is a modification of any QM rules needed. The point is rather that we can't measure a continuous variable arbitrarily precisely, so the limit of delta functions is unphysical.

At least in the formalism, a sharp measurement of position is possible. After a sharp measurement of position, the wave function collapses to something that is not a position delta function.

 

[DrDu]

At least in the formalism, a sharp measurement of position is possible. After a sharp measurement of position, the wave function collapses to something that is not a position delta function.

I fear, to convice me, you would have to provide a sound reference.

 

[DrDu]

In this scenario, the set of irreducible observables of a quantum system is the center of this algebra, if we further endow it with the Lie product structure (commutator). For a free spinless particle in 1D, the irreducible observables are the coordinate x and momentum p, because the

Isn't the center of the algebra formed by all those operators which commute with all elements of the algebra? Then how can p and x be in the center if they don't even commute among themselves?

 

[bhobba]

$X=\int dx x \delta(x'-x) \delta(x''-x)=\int dx x \delta(x'-x) \delta(x''-x')= \delta(x''-x') \int x d \theta(x-x').$
The last step should be familiar from classical statistics, where we express the integral over a probability density function by an integral with the cumulative distribution function as integrand.

Hmmm. The multiplication of two Dirac Delta functions. That's rather interesting. Its legit mind you - but is mathematicaly a bit hairy:
https://arxiv.org/abs/1308.0257

Do physicists really want this sort of mathematical sophistication?

Thanks
Bill

 

[DrDu]

Do physicists really want this sort of mathematical sophistication?

Thanks
Bill

Here, the point is rather the result than the way to derive it. At the end we can express the operator X in terms of projectors, which here are simple Heaviside theta functions in position space. The defining property of a projector P is that it is idempotent :$P^2=P$ from which it follows that it has only eigenvalues 0 or 1.
These projectors appear as measure in the integral, hence projector valued measure.
In contrast to the products of delta functions from which we started, the projectors are the most well behaved operators we can think of in Hilbert space. As we see, they are sufficient to characterize the whole position operator.

 

[PeterDonis]

these are really eigenstates of the harmonic oscillator

Eigenstates of the harmonic oscillator what?

Eigenstates of the harmonic oscillator Hamiltonian are eigenstates of the number operator, $\hat{a}^\dagger \hat{a}$. In the position representation, these are fairly complicated expressions involving Hermite polynomials. They aren't just functions of the form $a e^{-bx^2}$ (although factors of that form appear in them).

 

[PeterDonis]

You can easily show that e.g. $\psi_1(x)$ isn't an eigenstate of $\hat{a}$.

Yes, you're right, I misspoke. Coherent states in the position representation have a factor in them that looks like $a e^{-bx^2}$, but that's not all there is to them.

 

[DrDu]

Eigenstates of the harmonic oscillator what?

Eigenstates of the harmonic oscillator Hamiltonian are eigenstates of the number operator, $\hat{a}^\dagger \hat{a}$. In the position representation, these are fairly complicated expressions involving Hermite polynomials. They aren't just functions of the form $a e^{-bx^2}$ (although factors of that form appear in them).

The first Hermite polynomial is a constant, the second one proportional to x, so this are HO eigenfunctions

 

[PeterDonis]

The first Hermite polynomial is a constant, the second one proportional to x, so this are HO eigenfunctions

Ah, ok, I didn't understand that that was what the earlier post was referring to.

 

[David Olivier]

Thanks for the discussion, a lot of which is above my current level. For the moment, I have at least learned that my trivial question wasn't so trivial.

To get back to basics: I still don't know what the one-dimensional state space of a spinless particle can look like. Let's say we don't try to squeeze any $\delta$'s into it, nor any $e^{ikx}$'s. What do we have? The space of continuous $\mathbb R \to \mathbb C$ functions? That's certainly too large. We want the dot product, so we need the functions to be quadratically integrable. We also want $P = -i \frac \partial {\partial x}$ to be an endomorphism, hence they must be derivable. Actually, for this they must be $C^\infty$ (because the resulting function must satisfy the same conditions). For $X$ to be an endomorphism, we need not only our $\psi(x)$ to be quadratically integrable, but also $x \psi(x)$, $x^2 \psi(x)$ and so on. How do we impose this on our functions?

I've given it some thought, and come up with the following condition, which I think works and isn't too restrictive: we want our functions $\psi$ to be $C^\infty$ and such that, for each of them, there exist some positive reals $a$ and $b$ such that for all $x$, we have $|\psi(x)|^~ < a e^{- b x^2}$. This does ensure, it seems to me, that for any $n \in \mathbb N$, $x^n \psi(x)$ will be not only derivable and quadratically integrable, but also satisfy all those same conditions. Does this make sense?

 

[DrDu]

I've given it some thought, and come up with the following condition, which I think works and isn't too restrictive: we want our functions $\psi$ to be $C^\infty$ and such that, for each of them, there exist some positive reals $a$ and $b$ such that for all $x$, we have $|\psi(x)|^~ < a e^{- b x^2}$. This does ensure, it seems to me, that for any $n \in \mathbb N$, $x^n \psi(x)$ will be not only derivable and quadratically integrable, but also satisfy all those same conditions. Does this make sense?

What, if the particle you want to describe is located on the moon?

 

[David Olivier]

What, if the particle you want to describe is located on the moon?

No problem. Coefficient $b$ can be arbitrarily small; it must just not be zero.

It might seem that the condition $\exists a>0, b>0, \forall x \in \mathbb R, |\psi(x)|^~ < a e^{- b x^2}$ gives a special role to the origin $x = 0$. But this is not so. It is equivalent to $\exists a>0, b>0, \forall x \in \mathbb R, |\psi(x)|^~ < a e^{- b (x-A)^2}$ for any real $A$. It is really only a condition about how the function behaves at infinity (+ or - infinity), which is what we need to ensure that it is quadratically integrable (and remains so after all iterations of $X$ and $P$).

 

[Physics Footnotes]

To get back to basics: I still don't know what the one-dimensional state space of a spinless particle can look like. Let's say we don't try to squeeze any $\delta$'s into it, nor any $e^{ikx}$'s. What do we have? The space of continuous $\mathbb R \to \mathbb C$ functions? That's certainly too large. We want the dot product, so we need the functions to be quadratically integrable. We also want $P = -i \frac \partial {\partial x}$ to be an endomorphism, hence they must be derivable. Actually, for this they must be $C^\infty$ (because the resulting function must satisfy the same conditions). For $X$ to be an endomorphism, we need not only our $\psi(x)$ to be quadratically integrable, but also $x \psi(x)$, $x^2 \psi(x)$ and so on. How do we impose this on our functions?

Short Answer: The space of admissible (pure) states of a spinless particle in one dimension (in the position representation) is taken to be the Hilbert Space $L^2[\mathbb R]$ of (Lebesgue) square-integrable functions from $\mathbb R$ into $\mathbb C$. (Strictly speaking there are some nuances about equivalence classes, but that would be a distraction from your real question.)

Longer Answer: The square-integrability requirement secures, as you might guess, the statistical interpretation of the wave function. The use of a fancy Lebesgue integral, as opposed to a more pedestrian Riemann one, ensures you get a well-behaved Hilbert Space (which is partly justified physically and partly mathematically).

The sorts of constraints you are playing around with are not at all necessary for a wave function to represent a plausible physical state, however they do become the sorts of things you have to think about when specifying the domains of individual operators.

The momentum operator $P=-i\frac{d}{dx}$ you mention, for example, is indeed not defined on the entire Hilbert Space (which not only contains non-smooth functions but also wildly discontinuous ones, by the way!). However, even the plain old position operator $Q=x$ is not defined everywhere on $L^2[\mathbb R]$, since multiplication by the independent variable will send some functions out of the Hilbert Space.

The main point I would make here though, is that the restriction of the domain of an operator corresponding to an observable is not due to certain states being unphysical (which seems to be the way you're thinking), but rather it is a necessary technical requirement to make the operator self-adjoint, which ensures it has the right measure-theoretic properties to yield consistent expectation values. And that, by the way, is a very subtle and difficult task; PhDs have been granted for working out the domain of self-adjointness of a single operator!

 

[David Olivier]

Thanks. I was not really trying to impose restrictions to ensure physicality; rather just trying to get things to work mathematically; and in particular ensure that the operators map the set to itself.

I'm not sure my solution works, though; the "Hilbert space" I described may not be complete.

 

[Physics Footnotes]

Thanks. I was not really trying to impose restrictions to ensure physicality; rather just trying to get things to work mathematically; and in particular ensure that the operators map the set to itself. I'm not sure my solution works, though; the "Hilbert space" I described may not be complete.

That's right. Sticking to well-behaved functions will not, in general, give you a Hilbert Space.

By the way, following on from what I explained above, the 'endomorphism' requirement you mention is correct, but only insofar as an operator must map functions in the Hilbert Space (linearly) to other functions in the Hilbert Space. When an operator is only defined on a proper subset $\mathcal D \subset \mathcal H$, it is perfectly ok if the operator projects wave functions out of $\mathcal D$, so long as they stay in $\mathcal H$. This happens especially when you start needing to impose boundary conditions on wave functions to make an operator self-adjoint.

Also, keep in mind that domains always matter when an operator is unbounded, since no such operator can be defined on the entire Hilbert Space, while remaining self-adjoint.

Again, if you like sorting out these kinds of subtleties, I think you'd enjoy von Neumann's book.

 

[DrDu]

No problem. Coefficient $b$ can be arbitrarily small; it must just not be zero.

When you consider a free particle in an initial wavefunction $\sim \exp(-bx^2)$, this wavefunction will remain Gaussian, but b will decrease with time t like 1/t . So for any fixed lower bound on b, an initially arbitrary localized wavefunction will violate it, for sufficiently long times. This is one of the fundamental observations to proove the validity of scattering theory.