# What's your opinions on the Axiom of Choice?

#### Crosson

I did not mean to imply that there was a simple empirical connection between quantum mechanics and the axiom of choice. Suppose AOC is false, then all we can conclude is that there is at least one (infinite-dimensional) vector space V without a basis (good luck constructing it). We do not know, however, that V represents the configuration space of any quantum system. Even though we often assume a quantum system has a basis without displaying it explicitly, we have not contradicted anything unless we know that the "defective" space V in fact represents a real quantum system (which would force the dichotomy I mentioned above).

To put it another way: if you can test it experimentally, it's a natural science, not (pure) math.
That all depends on what you mean by "test it." You alluded to validity above, and I agree that the validity of mathematical theorems are independent of any empirical fact. The theorems are all tautologies after all; although we often leave out the antecedent of theorems like "the square root of 2 is irrational" we could choose to make the hypothesis explicit if so desired.

But as you mentioned, another "test" of mathematical results is their aesthetic value. Why do we place so much emphasis on the ring of integers? Of course it is because the integers are suitable and applicable, as you say. But there is not only one model of arithmetic, and so there remains the question of whether we can rule out various models based on empirical testing. Similarly, there is not only one model of set theory, so this leaves open the possibility that some of these models will be shown to be incompatible with empirical facts.

Last edited:

#### Ka Yan

Why is that the Banach-Tarski Paradox (Theorem, say) ture for R^n of n$$\geq$$3, but nor for R and R^2, please?

Last edited:

#### Hurkyl

Staff Emeritus
Gold Member
I just wanted to point out a factual error -- the notion of a vector space basis is slightly different than the notion of a Hilbert space basis. In particular, for the latter, every vector can be written uniquely as a possibly infinite sum of multiples of basis vectors.

#### Crosson

I just wanted to point out a factual error -- the notion of a vector space basis is slightly different than the notion of a Hilbert space basis. In particular, for the latter, every vector can be written uniquely as a possibly infinite sum of multiples of basis vectors.
Thanks for clarifying that in this context Hurkyl! I had not realized what an unintuitive theorem "every vector space has a basis" is, since I had thought that in an arbitrary Hilbert space H there is not necessarily a set of vectors B, every subset of which satisfies linear independence, such that each element of H can be written as a linear combination of the vectors in some finite subset of B. But this is what it means to be a basis in a purely algebraic sense, and since an arbitrary infinite-dimensional vector space does not necessarily have the structure to define infinite-sums this must be what is meant. Very surprising, makes AOC seem even less aesthetically satisfying for me!

#### Hurkyl

Staff Emeritus
Gold Member
Well, I would expect (but do not know) that the AoC is also equivalent to the claim that every Hilbert space has a Schauder basis. AoC is clearly equivalent to the claim that every topological vector space has a Schauder basis. (Consider the discrete topology)

(the term Schauder basis denotes one where every element is an infinite linear combination. The finitary version is called a Hamel basis)

Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space). I feel like I should already know the answer, but I'm not recalling it.

Last edited:

#### mathboy

So how much does quantum mechanics rely on Axiom of Choice? And are there any observable phenomenon in QM that depend on the accept of AoC?

#### Crosson

Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space).
The position and momentum representations do not yield bases, but the position and momentum operators do (with respect to a particular basis). The necessary conditions for the operators $\overhat{x},\overhat{p}$ are that they be self-adjoint and obey the commutator:

$$[\hat{x},\hat{p}] = i$$

And so in the position representation we have:

$$\hat{x} = x$$

$$\hat{p} = i \frac{\partial}{\partial x}$$

And in the momentum representation we have:

$$\hat{x} = i \frac{\partial}{\partial k}$$

$$\hat{p} = k$$

Now, to try an answer to your question I would say that the eigenvectors of the position operator in the position representation are:

$$\hat{x} = x \rightarrow eigenVectors(\hat{x}) = \{\phi_{x'}:\mathbb{R} \rightarrow \mathbb{C} | \phi_{x'}(x) \mapsto \delta(x - x') , x'\in \mathbb{R}\}$$

$$\hat{p} = i \frac{\partial}{\partial x} \rightarrow eigenVectors(\hat{p}) = \{\phi_{k}:\mathbb{C} \rightarrow \mathbb{R} | \phi_{k}(x) \mapsto e^{i k x} , k\in \mathbb{R}\}$$

In physics we say that either of these sets of eigenvectors are an uncountable basis! The linear combinations become definite integrals covering the range of the eigenvalues:

$$f(x) = \int_{-\infty} ^{\infty} c_{x'} \phi_{x'}(x) dx' [/itex] where the coefficients are given by Fourier's trick: [tex] c_{x'} = \int_{-\infty} ^{\infty} f(x) \phi_{x'}(x) dx [/itex] I'm interested to learn more about what's really going on here. So how much does quantum mechanics rely on Axiom of Choice? And are there any observable phenomenon in QM that depend on the accept of AoC? The standard formulation of quantum mechanics is built on the fact that the set of eigenvectors of a self-adjoint operator in a Hilbert space will be an orthonormal basis. We are trying to determine whether this theorem depends on the axiom of choice. Even then I don't know any real experiments that verify the quantum state of a system to be an infinite-dimensional vector as opposed to high-finite dimensional one, but there is hopefully there is some major qualitative difference between these cases that could one day be observed. There was a time when it was suitable to treat matter as an uncountable continuum, an actual infinity. Even in the 1920s we knew that was far from the case so there was no fear (or joy) that the Banach-Tarski construction could be realized. But we still treat time and space as uncountable and complete in quantum field theory, and that is likely to be with us for sometime. Also as mentioned the quantum mechanical state of a single particle alone in the universe is an infinite-dimensional vector. It is only through these few remaining vestiges of believable actual infinities in physical reality that we ever have any hope of "testing" the axiom of choice. Last edited: #### morphism Science Advisor Homework Helper Incidentally... does the Hilbert space of square-integrable functions on the real line have an "obvious" basis? The position and momentum representations do not yield bases (the eigenstates are not even elements of the Hilbert space). I feel like I should already know the answer, but I'm not recalling it. There's the (normalized) Hermite functions (Gram-Schmidt {x^n exp(-x2/2)}). #### mathboy The standard formulation of quantum mechanics is built on the fact that the set of eigenvectors of a self-adjoint operator in a Hilbert space will be an orthonormal basis. We are trying to determine whether this theorem depends on the axiom of choice. Even then I don't know any real experiments that verify the quantum state of a system to be an infinite-dimensional vector as opposed to high-finite dimensional one, but there is hopefully there is some major qualitative difference between these cases that could one day be observed. There was a time when it was suitable to treat matter as an uncountable continuum, an actual infinity. Even in the 1920s we knew that was far from the case so there was no fear (or joy) that the Banach-Tarski construction could be realized. But we still treat time and space as uncountable and complete in quantum field theory, and that is likely to be with us for sometime. Also as mentioned the quantum mechanical state of a single particle alone in the universe is an infinite-dimensional vector. It is only through these few remaining vestiges of believable actual infinities in physical reality that we ever have any hope of "testing" the axiom of choice. Aha! George Jones, are you reading this? There is some hope after all. Could some physicist please take the plunge !? Last edited: #### ice109 now i haven't read the whole thread but why again are we discussing the truth value of an axiom? #### Hurkyl Staff Emeritus Science Advisor Gold Member Now, to try an answer to your question I would say that the eigenvectors of the position operator in the position representation are: [tex] \hat{x} = x \rightarrow eigenVectors(\hat{x}) = \{\phi_{x'}:\mathbb{R} \rightarrow \mathbb{C} | \phi_{x'}(x) \mapsto \delta(x - x') , x'\in \mathbb{R}\}$$

$$\hat{p} = i \frac{\partial}{\partial x} \rightarrow eigenVectors(\hat{p}) = \{\phi_{k}:\mathbb{C} \rightarrow \mathbb{R} | \phi_{k}(x) \mapsto e^{i k x} , k\in \mathbb{R}\}$$

In physics we say that either of these sets of eigenvectors are an uncountable basis!

The linear combinations become definite integrals covering the range of the eigenvalues:

[tex] f(x) = \int_{-\infty} ^{\infty} c_{x'} \phi_{x'}(x) dx' [/itex]

where the coefficients are given by Fourier's trick:

[tex] c_{x'} = \int_{-\infty} ^{\infty} f(x) \phi_{x'}(x) dx [/itex]

Well, the thing here is that neither of these "bases" contain a single element of your Hilbert space! The position and momentum operators, acting on our Hilbert space, actually do not have any nonzero eigenvectors.

There are at least two ways to make sense of this. One is via a "rigged Hilbert space", where in addition to our Hilbert space, we consider a smaller subspace of "test functions", its dual space of "generalized functions". The position and momentum operators have eigenvectors in this larger space of generalized functions. (distribution is another buzzword to look for)

Another is through a "direct integral" of Hilbert spaces, which generalizes the (finite) direct sum of Hilbert spaces in a way similar to how the ordinary integral can be viewed as generalizing finite sums of numbers. We can write our Hilbert space as a direct integral of a copy of C at each point in the real line. Now, your "basis vectors" are not elements of our Hilbert space of interest, but instead a choice of basis vectors for these individual Hilbert spaces.

#### Hurkyl

Staff Emeritus
Gold Member
now i haven't read the whole thread but why again are we discussing the truth value of an axiom?
Because people aren't used to separating the notion of a mathematical theory and their favorite interpretation of that theory.