Wavefunction in the context of quantum physics

[cianfa72]

TL;DR

On the definition of wavefunction as representation of the quantum state in an implied basis.

In the context of quantum physics the state of a quantum system is represented by a vector, say $\ket \psi$, in the appropriate/relevant Hilbert space.

When we talk of wavefunction $\psi()$ we are really talking of some representation of the vector $\ket \psi$. In other words, strictly speaking, a wavefunction is never a vector in Hilbert space. Just to simplify, the wavefunction is the set of coefficients (possibly uncountable many) of the expansion of an element of the Hilbert space in a given/implied basis.

What do you think about?

 

[PeterDonis]

strictly speaking, a wavefunction is never a vector in Hilbert space

This is not correct. Wave functions are elements of a Hilbert space--for example, the wave function of a single spinless particle moving in one spatial dimension is an element of the Hilbert space of square integrable functions of one variable.

the wavefunction is the set of coefficients (possibly uncountable many) of the expansion of an element of the Hilbert space in a given/implied basis.

You're just describing what a function is in different language. But functions can be elements of a Hilbert space all by themselves, without having to adopt any such interpretation of them. Any set of objects that satisfies the mathematical axioms of a Hilbert space, is a Hilbert space.

 

[cianfa72]

This is not correct. Wave functions are elements of a Hilbert space--for example, the wave function of a single spinless particle moving in one spatial dimension is an element of the Hilbert space of square integrable functions of one variable.

Yes, however to me this is a bit confusing. In your example, the particle (the quantum system) is in a quantum state in the relevant infinite-dimensional Hilbert space. The position representation returns a wavefunction of the "one spatial dimension" while the momentum representation returns a wavefunction as well this time as function of the "one dimensional linear momentum".

Take another example: a qubit. Its state is an element of $\mathbb CP^1$. Is the vector (or better the ray) $\ket \uparrow$ a wavefunction?

 

[PeterDonis]

In your example, the particle (the quantum system) is in a quantum state in the relevant infinite-dimensional Hilbert space.

Yes.

The position representation returns a wavefunction of the "one spatial dimension" while the momentum representation returns a wavefunction as well this time as function of the "one dimensional linear momentum".

Yes. Both of these functions will be square integrable functions of one variable, so they are both elements of Hilbert spaces.

You are correct that choosing the position or momentum representation can be viewed as a choice of basis, since you can Fourier transform between them. That is part of the physical interpretation. It does not change the fact that both wave functions (the position one and the momentum one) are elements of Hilbert spaces (they are actually the same Hilbert space, mathematically, the space of square integrable functions of one variable--you can think of them as two copies of that mathematical Hilbert space, related by a Fourier transform, that both describe the same physical system).

Take another example: a qubit. Its state is an element of $\mathbb CP^1$. Is the vector (or better the ray) $\ket \uparrow$ a wavefunction?

No. There is no space of functions that has the particular structure of the qubit Hilbert space.

 

[cianfa72]

I'm not sure whether what follows fits better in a math subforum. Take the Hilbert space of square integrable functions of one variable (the relevant notion of integrability is Lebesgue integral and to get an Hilbert space one needs to consider the equivalence classes of functions that may only differ on a set of zero Lebesgue measure on $\mathbb R$).

The set $\{e^{i\omega t} \}$ is supposed to be a maximal orthonormal set, i.e. an Hilbert basis for that space (note that in general, even for separable Hilbert spaces, this isn't the same as Hamel/algebraic basis). But the point is: none of those functions are actually square integrable !

Are we able to provide a "good" Hilbert basis for that Hilbert space ?

 

[renormalize]

Are we able to provide a "good" Hilbert basis for that Hilbert space ?

https://en.wikipedia.org/wiki/Rigged_Hilbert_space

 

[cianfa72]

https://en.wikipedia.org/wiki/Rigged_Hilbert_space

Apart from this, can we explicitly exhibit an Hilbert basis (i.e. a maximal orthonormal set) consisting of square integrable functions over $\mathbb R$ ?

 

[gentzen]

Apart from this, can we explicitly exhibit an Hilbert basis (i.e. a maximal orthonormal set) consisting of square integrable functions over $\mathbb R$ ?

Yes
The Hermite functions are such a Hilbert basis. They are eigenfunctions of the Fourier transform. They are well localized in both position and frequency space, and this is not untypical for a Hilbert basis. If you scale them, you get another Hilbert basis, which no longer consists of eigenfunctions of the Fourier transfrom. You can also translate them in position or frequency space (or both), and get yet another Hilbert basis.

This might feel arbitrary, non-unique and somehow ugly, but in a certain sense not really surprising:

Perhaps the easiest way to see this is in Schwartz space, as a rigged Hilbert space of square-integrable functions on R^3. The "distributions" in the dual space (the space of tempered distributions) are the easiest to interpret and write down, for example, the constant 1, or the distributions x, y, or z, or the four-dimensional subspace of linear distributions, etc.

But actually, one would like to attribute meaning to the functions themselves, especially these "low-frequency" ones. Unfortunately, they must be low-frequency in both position and frequency space, which makes it much more difficult to simply write them down or characterize them nicely. But they are still important (the functions themselves), even if they are mathematically somewhat more cumbersome than the distributions of the dual space.

 

[PeterDonis]

The set $\{e^{i\omega t} \}$ is supposed to be a maximal orthonormal set

Not in the set of square integrable functions, since those functions are not square integrable, as you note:

none of those functions are actually square integrable !

Yes, which means they don't form a basis of the Hilbert space of square integrable functions--they can't. Yes, many QM textbooks gloss over this inconvenient fact.

If you want to actually do what you appear to be thinking of with some degree of rigor, you need to adopt something like the rigged Hilbert space formalism that @renormalize referred to.

 

[cianfa72]

But actually, one would like to attribute meaning to the functions themselves, especially these "low-frequency" ones. Unfortunately, they must be low-frequency in both position and frequency space, which makes it much more difficult to simply write them down or characterize them nicely. But they are still important (the functions themselves), even if they are mathematically somewhat more cumbersome than the distributions of the dual space.

Sorry, I'm not sure what that link refers to. Does the part in bold above refer to Hermite functions ?

 

[gentzen]

Sorry, I'm not sure what that link refers to. Does the part in bold above refer to Hermite functions ?

In the link, the part in bold above refers to functions from the Schwartz space. The Schwartz space is the function space of all functions whose derivatives are rapidly decreasing.

The Hermite functions are functions from the Schwartz space, so in this sense the part in bold also refers to them. But it is not just the part in bold that explains why they feel more complicated than "the simplest" distributions of the dual space. They are among the simplest low-frequency functions in the Schwartz space, and one would like to attribute physical meaning to them. They are the eigenfunctions of the quantum harmonic oscillator, so there is good hope that they should have some physical meaning. But the only meaning I can find is that they are well localized in both position and frequency space.

 

[cianfa72]

In the link, the part in bold above refers to functions from the Schwartz space. The Schwartz space is the function space of all functions whose derivatives are rapidly decreasing.

Ok, furthermore such Hermite functions (that are defined on the entire R, are measurable w.r.t. the Lebesgue measure on R and the square of any of them has Lebesgue integral equals to 1) are members of the Hilbert space of square integrable functions and constitute a maximal orthonormal set (i.e. an Hilbert basis for the separable Hilbert space of square integrable functions of one variable).

 

[gentzen]

yes

 

[cianfa72]

Ok, so why aren't these Hermite functions employed in quantum physics ? (Or rather, I've never seen them).

 

[gentzen]

Ok, so why aren't these Hermite functions employed in quantum physics ?

They are the eigenfunctions of the quantum harmonic oscillator, so there is good hope that they should have some physical meaning.

(Or rather, I've never seen them).

I guess there are not too many things to do with them. You could analyse their position-momentum uncertainty, you could analyse the 2D or 3D harmonic oscillator and its radial symmetry. And maybe two or three other things, but then it soon gets boring.

Plane waves are simply more interesting. There is a reason why Dirac reinvented distributions, and why mathematicians picked up the math and developped a theory of distributions.

 

[PeterDonis]

I've never seen them

Sure you have, just not under that name. The Hermite functions are Gaussians with polynomial prefactors--indeed, the zero-order Hermite function (which is the ground state of a harmonic oscillator) just is a Gaussian. Gaussians are used in QM all the time.