Selection of domain and codomain for wave function

1. Jan 10, 2015

Stephen Tashi

A wave function is a function in a Hilbert space of functions. When representing a physical sytem, what are the conventions for defining the domain and codomain of the wave function?

For example, is the domain always the scalar variable representing time? Is the codomain a set of finite n-tuples of numbers? If, so are the entries in these n-tuples always chosen to be quantities that have familiar physical interpretations like momentum, charge, etc ?

2. Jan 10, 2015

atyy

The quantum state is a ray in Hilbert space (which is essentially a vector space with a dot product). Usually we choose a unit vector to represent the ray, but any constant multiple of the vector represents the same ray. In the case of spin 1/2, we can choose the Hilbert space to be 2 dimensional, and we can take basis vectors to be |spin up> and |spin down>. Then an arbitrary vector in the Hilbert space is a linear combination of those two basis vectors. The meaning of the basis vectors is not necessarily a quantity with a "classical counterpart" such as position, since there is no classical counterpart to spin 1/2.

3. Jan 10, 2015

micromass

Last edited by a moderator: May 7, 2017
4. Jan 10, 2015

Stephen Tashi

From that one example, I visualize a wave function as having domain "time" and codomain the set of rays in a finite dimensional space. Does a wave function ever have a codomain with infinite dimensions?

In classical physics representing the mass of a real object requires representing a mass density function rather than a point mass. The information in a mass density function is an uncountable set of odered tuples of number (x,y,z,m). So to describe the mass density at time t ,for each time t, you have an uncountable set of ordered pairs (x,y,z,m(t)) since (x,y,z) varies over all of 3-d space. Mass density as a function a function of time has domain time and codomain an uncountable set of 3-tuples of number.

Presumably QM doesn't treat classical mass density functions, but does a wave function handle situations analagous to time varying density functions?

5. Jan 10, 2015

dextercioby

Typically for wavefunctions (Schrödinger picture), they are mappings $\psi (\vec{r},t)$, that is $\psi : \mathbb{R}^3 \times \mathbb{R} \mapsto \mathbb{C}$.

6. Jan 10, 2015

atyy

Yes, quantum state can be infinite dimensional. For example, the wave function of a non-relativistic particle in 1 spatial dimension $\psi(x)$, where x is a continuous index indicating position, is the representation in the position basis of the particle's quantum state.

Continuous indices and infinite dimensional Hilbert spaces are mathematically tricky for me, so I'll leave most questions to others who can answer it. For heuristic purposes, I always think of the Hilbert space as being large but finite dimensional.

The quantum state, defined as ray in Hilbert space, can be thought of as "the complete physical state of a single quantum system". There are other ways of thinking about it, but this is fine, as long as one remembers that (1) a measurement occurs when a classical apparatus interacts with a quantum system (2) quantum mechanics only predicts the expectation values of measurement outcomes, which are readings of a classical measurement apparatus. In other words, only the classical part of the world has the status of absolute reality (same as classical physics), while the quantum state is not necessarily real, and just a tool to make predictions about measurement outcomes.

Yes, the quantum state can change with time. There are (at least) two equivalent pictures of quantum mechanics. In the Schroedinger picture, the quantum state changes with time, and the time evolution of the quantum state is governed by the Hamiltonian operator via the Schroedinger equation.

7. Jan 10, 2015

Staff: Mentor

The key thing to remember about the wavefunction is its simply a representation in the position basis of the key object - the state. It can be repesented in other basis as well such as energy or momentum.

As others have alluded to it's a map from position and time to the complex plane.

But if you really want to venture into deep waters you need to look into Rigged Hilbert Spaces:
http://physics.lamar.edu/rafa/webdis.pdf

If you haven't come across distribution theory before the following will greatly help:
https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

Makes Fourier transforms a snap as well.

Rigged Hilbert spaces is basically Hilbert spaces with distribution theory stitched on it - loosely speaking of course.

Thanks
Bill

Last edited by a moderator: May 7, 2017
8. Jan 10, 2015

Stephen Tashi

Then is it better to say "a wavefunction" than "the wavefunction" (similar to nicety of saying "an antiderivative" instead of "the antiderivative"). Given a physical situation where there are several things of interest (e.g. momentum, spin, charge etc. ) or several particles, do you need several complex numbers to represent the quantum state or are these somehow representable in a singel complex number?

9. Jan 10, 2015

Staff: Mentor

I don't think it makes any difference.

Its really a linear algebra thing.

A vector space is defined by a set of axioms and the elements of such a space are vectors. However they can be expanded in terms of many different basis - each equally valid. The choice is simply one of what solves the problem better.

The dimension is the number of terms in that expansion. So called pure quantum states are elements of a vector space. The tricky thing about QM, and that is where the deep waters of Rigged Hilbert Spaces comes in, is you can have a continuous basis.

For this sort of stuff get a hold of Ballentine - QM A Modern Development and read the first two chapters. Although not related to this issue it would be a good idea to read chapter 3 as well because it clarifies exactly what's going on with the dynamics of QM which if you don't understand is really a result of symmetry can seem a bit ad-hoc.

Thanks
Bill

10. Jan 11, 2015

Stephen Tashi

To mathematician, if someone says "I have a function", the question of "What are its doman and codomain?" is completely straightforward to answer (unless the person claiming to have the function doesn't actually have a well defined function).

I'm used to the way physicists mangle mathematics, so perhaps it takes several physics books to answer the question. :)

Before tackling Ballentine, I'll try the Wikipedia

Wave function:
One guess about the meaning of that statement is that a wave function essentially "is" a quantum state, in the sense that it contains all the information that defines a quantum state. It is merely a "representative" in the same sense that one point in Euclidean space has a representative in one coordinate system and a different representative in a different coordinate system. If one of the variables in the wave function is time then "a quantum state" is more than just the properties of the system at a particular time. The state would include the present and the future time evolution of those properties, until the state is disturbed by something,. Presumably the wave function can't be relied upon to tell the past of the system beyond the point where the wave function was created by disturbance of some other wave function.

So we know that the codomain of a wave function "may be" chosen to be a vector space over the field of complex numbers. Whether this vector space is finite dimensional or not isn't specified.

That answers my question about what to do if a system has many physical aspects or many particles. To handle that situation you represent these things as variables in the domain of the wave function. How you establish the dimension of the vectors in the codomain isn't revealed. I assume it depends on physical laws.

The two uncertainties in this post are the reliability of the Wikipedia and the reliability of my interpretation .

11. Jan 11, 2015

atyy

Yes, generally we have an initial preparation procedure, and the wave function only exists after that. Between measurements, the wave function evolves according to the Schroedinger equation. When a measurement is made, the wave function collapses to another wave function with some probability given by the Born rule.

The simplest form of the collapse rule is the von Neumann projection postulate. For continuous variables, it is trickier. The correct rule is found in:
http://arxiv.org/abs/0706.3526
http://arxiv.org/abs/0810.3536
A different conception of the quantum state is used in those papers called the density operator, which allows one to deal with statistical mixtures of the pure quantum states that are rays. Roughly, if a pure state is a vector, the density operator corresponding to the pure state is the outer product of the vector and its dual vector.

It depends on what properties we are aware of or include in the model. For example, in the Bell experiments we often only write two dimensions (the two basis vectors can be taken to be horizontal and vertical polarization) of the quantum state of each photon.

Last edited: Jan 11, 2015
12. Jan 11, 2015

Staff: Mentor

Yes and no.

I am tempted to say to find out what sense that is meant read Ballentine - but will say a bit more.

Hopefully you have studied linear algebra so this makes sense. Suppose we have a reasonable linear operator (obeys the premises of the spectral theorem) in some vector space and we find its eigenvalues and eigenvectors. In QM the observables are linear operators (they are Hermitian so obey the spectral theroem) and one of the axioms of QM is the possible outcomes of the observation associated with the observable are its eigenvalues. The eigenvectors form an orthonormal basis of the vector space. We can represent any vector in that basis. This is the representation associated with the linear operator - or in QM the observable. All the wave-function is, is the representation of the state associated with the position observable - the twist here is the eigenvectors and eigenvalues form a continuum. You can represent it in terms of any observable you want - energy, momentum, spin - anything.

Thanks
Bill

13. Jan 11, 2015

dextercioby

If you have spin (let's say s), then the (Schrödinger position basis) wavefunction will still have $\mathbb{R}^3 \times \mathbb{R}$ as domain, but $\mathbb{C}^{2s+1}$ as codomain.

14. Jan 11, 2015

atyy

To understand both of these as consistent, can we also (unconventionally, but correctly) treat spin $s=1/2$ as having $\mathbb{R}^3 \times \mathbb{R}^3 \times \mathbb{R}$ domain and $\mathbb{C}$ as codomain?

15. Jan 11, 2015

kith

I think this is a bit misleading. In almost all cases, the codomain of the wavefunction is simply $\mathbb{C}$. Only if you combine spin and position degrees of freedom you get kind of an exception (a so-called spinor wavefunction). The reason for this is that if you consider only spin degrees of freedom, you wouldn't talk about a wavefunction at all. The term "wavefunction" is only used with reference to position and momentum.

More general is the concept of the quantum state as a general vector |Ψ>. The probability for obtaining a state |φ> after a measurement is the absolute square of the inner product |<φ|Ψ>|². If we have a continuum |x> of possible measurement outcomes, we define the wavefunction as the inner product Ψ(x) := <x|Ψ>. This is the way most phycicsts think about it. For a mathematician, this is probably hard to swallow because the vectors |x> aren't normalizable. However, there exists a less naive version of what I wrote which involves rigged Hilbert spaces. You can read about this in Ballentine's book.

Last edited: Jan 11, 2015
16. Jan 11, 2015

dextercioby

No, because the wavefunction is necessarily 2x2 matrix-valued.

EDIT: 2x1, of course.

Last edited: Jan 11, 2015
17. Jan 11, 2015

atyy

Yes, you are right. Let me try something different. I would like to think of the wave function at a particular time as $\psi(x,i)$ where x is the continuous position index, and i is a discrete index from {0,1}. In that case can I think of $\mathbb{R}^3 \times \{0,1\}$ as the domain and $\mathbb{C}$ as codomain?

18. Jan 11, 2015

dextercioby

Yes, but a 'spin variable' is really a stretching of a term. Surely, you can put the numbers 1/2 or -1/2 as a 'variable' along with x or p, but in the end, you're single out a particular component of the spinor (sorry, I meant about 2x1 matrix).