Mathematical description of Hilbertspace

[moriheru]

I am looking for the mathematical description of Hilbertspace. I am talking about a mathematical description not something like Hilbert space is a infinite or finite dimensional vector space commonly used in QM, it was named after David Hilbert a german mathematician...I know this is a very fundamental question, but anyway...Thanks for any replies!

 

[aleazk]

Check these links, and in this order:

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

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

http://en.wikipedia.org/wiki/Metric_(mathematics)#Metrics_on_vector_spaces

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

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

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

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

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

 

[moriheru]

Thanks, helps lots!

 

[aleazk]

Anyway, better to check some book after that. A standard reference is:

Methods of Modern Mathematical Physics I: Functional Analysis - M.Reed, B.Simon.

 

[Fredrik]

You already have a good answer, but I'm bored, so I'm going to spell some of it out. I'll do it in the opposite order though. Aleazk's order is the one that makes the most sense when you study the terms. I'll just answer your question, and then answer the expected follow-up questions.

A Hilbert space is a complete inner product space. An inner product space is a vector space equipped with an inner product. An inner product can be used to define a norm through the formula $\|x\|=\sqrt{\langle x,x\rangle}$. A norm can be used to define a metric through the formula $d(x,y)=\|x-y\|$. Because of this, inner product spaces can be viewed as metric spaces. When we say that an inner product space is "complete" (a term that really only applies to metric spaces), we mean that the metric space with the metric defined as above is complete.

A metric space is a set X equipped with a function $d:X\times X\to\mathbb R$ that satisfies the following properties (which enable us to think of d(x,y) as the distance between x and y):

1. For all $x,y\in X$, we have $d(x,y)\geq 0$.
2. For all $x,y\in X$, we have $d(x,y)=0$ if and only if $x=y$.
3. For all $x,y\in X$, we have $d(x,y)=d(y,x)$.
4. For all $x,y,z\in X$, we have $d(x,y)=d(x,z)+d(z,y)$.

A metric space is said to be complete if every Cauchy sequence in it is convergent. A sequence $(x_n)_{n=1}^\infty$ in a metric space X is said to be Cauchy, or to be a Cauchy sequence, if for all $\varepsilon>0$, there's a positive integer $N$ such that for all integers $n,m$,
$$n,m\geq N\ \Rightarrow\ d(x_n,x_m)<\varepsilon.$$ A sequence $(x_n)_{n=1}^\infty$ in a metric space X is said to be convergent if there's an $x\in X$ such that for all $\varepsilon>0$, there's a positive integer $N$ such that for all integers $n$,
$$n\geq N\ \Rightarrow\ d(x_n,x)<\varepsilon.$$ The element $x$ in this statement is then said to be a limit of the sequence.

It's easy to show that every convergent sequence is Cauchy, but it's not true in general that every Cauchy sequence is convergent. That's why a term like "complete" is useful. The set of rational numbers $\mathbb Q$ is an example of a metric space that isn't complete. There are Cauchy sequences in $\mathbb Q$ that aren't convergent with respect to the metric of $\mathbb Q$. But those sequences are also Cauchy sequences in $\mathbb R$, and they are convergent with respect to the metric of $\mathbb R$. Their limits are irrational numbers.

Kreyszig is probably easier than Reed & Simon. (This is based only on the reputations of these books).

 

[dextercioby]

There's no useful way to learn about Hilbert spaces (for the purpose of using them in physics), if you neglect measure theory. A book which combines them nicely for the purpose of QM is Prugovecki's "Quantum Mechanics in Hilbert Space" (1st, or 2nd edition, doesn't matter).

 

[moriheru]

Thanks good answers! I am guessing most of you are familiar with number theory (yes I know this is a Quantum physics forum), so just to thorugh this in...Where can I read up on sedenions (16-dimensional numbers) or quaternions (pauli matrices and so on) and hypercomplex numbers in general?

 

[vanhees71]

Well, there's a textbook forum, but anyway. The classical source for quaternions is

P. G. Tait, An elementary treatise of quarternions, Clarendon Press, Oxford (1867)

 

[moriheru]

Well, there's a textbook forum

I know, I just didn't want to start a new thread, as the topic seemed related.

The classical source for quaternions is
P. G. Tait, An elementary treatise of quarternions, Clarendon Press, Oxford (1867)

Thanks,
Do you know about octernions and quaternions?

 

[vanhees71]

Not really, i.e., I know that they exist and that they were common in the mid of the 19th century (Maxwell wrote his famous papers on electromagnetism using the language of quaternions) but never used them. Nowadays we usually use vector and tensor calculus as well as matrix representations of the Clifford algebras when needed (e.g., Dirac matrices).

 

[dextercioby]

Thanks good answers! I am guessing most of you are familiar with number theory (yes I know this is a Quantum physics forum), so just to thorugh this in...Where can I read up on sedenions (16-dimensional numbers) or quaternions (pauli matrices and so on) and hypercomplex numbers in general?

This is a little off-topic for the subject at hand. You can (hopefully) find answers in the Algebra section of the Mathematics forums.

 

[bahamagreen]

There's no useful way to learn about Hilbert spaces (for the purpose of using them in physics), if you neglect measure theory.

I know little about this, but it this because the metric space third and fouth properties Fredrik listed don't regard relativistic accelerations (or time at all) or something else?

 

[moriheru]

Why hilbert space why not any other metric space?I mean why does one use Hilbert space in QM, would the math change if one uses other metric spaces, can one still use Dirac notation?

 

[Fredrik]

Why hilbert space why not any other metric space?I mean why does one use Hilbert space in QM, would the math change if one uses other metric spaces, can one still use Dirac notation?

The definition of "metric space" doesn't even require that there's an addition operation on the set.

 

[Fredrik]

I know little about this, but it this because the metric space third and fouth properties Fredrik listed don't regard relativistic accelerations (or time at all) or something else?

The main thing that you need measure theory for is to find examples of infinite-dimensional Hilbert spaces. I don't think measure theory is as important as dextercioby is suggesting, because you can get pretty far without even specifying the Hilbert space or thinking about what an integral really is.

 

[homeomorphic]

The main thing that you need measure theory for is to find examples of infinite-dimensional Hilbert spaces.

Not even that. Any inner product space gives an example of a Hilbert space. Just take the completion. I think measure theory mainly comes in if you want to be more rigorous and explore the more mathematical aspects of things.

Why hilbert space why not any other metric space?

The inner product is how you calculate amplitudes, so you need an inner product. Requiring completeness allows stuff like taking projections. A good example is the space of square integrable functions, which can be interpreted as wave functions. They form a vector space because you can add them and multiply them by scalars, and there's an inner product that you get by multiplying functions together and then integrating. If you take the inner product of a function with itself, you get the normalization factor you need to multiply, so that you get probably 1 for the particle being found anywhere, so you need that integral to be finite.

 

[dextercioby]

The main thing that you need measure theory for is to find examples of infinite-dimensional Hilbert spaces. I don't think measure theory is as important as dextercioby is suggesting, because you can get pretty far without even specifying the Hilbert space or thinking about what an integral really is.

Oh, but it is. There's a nice theorem shat states loud and clear that any 2 complex separable Hilbert space are isomorphic, hence isomorphic to L^2(set, measure on set). Surely, you can go 'far' using abstract notions, but how far do you think the spectral theorems are, especially since they are exploited in any presentation of Quantum Mechanics.

 

[aleazk]

I mean why does one use Hilbert space in QM?

An answer to that question was provided by von Neumann's approach to the mathematical formalization of QM.

In this approach, the central objects are the yes-no propositions one can make about a system, i.e., propositions about the system that only admit an answer of yes or no (e.g., P="the position of the particle is x=5"). The porposition by itself is not enough, you have to ask: "when the system is in the state S, what's the truth value, or at least a probability, of elementary porposition P?". In classical mechanics, the set of all these propositions can be seen as equivalent to the Borel sigma-algebra of phase space, which can be seen as a distributive lattice (a 'Boolean algebra'; I'm simplifying things, there are more assumptions it its definition!) A key property in this structure is that all elementary propositions are compatible, i.e., given any two propositions, is meaningful to ask about their truth values simultaneously: the classical logical connectives 'or' and 'and' are well defined for these pairs.

A key property in QM is the existence of incompatible elementary propositions: for them, is meaningless to ask about their truth values simultaneously. Clearly, it becomes problematic if we want to use the classical Borel sigma-algebra for modeling the set of elementary propositions in QM. How do we realize explicity the set of elementary propositions in QM then? von Neumann's brilliant insight was that we can see them as orthogonal projectors on a Hilbert space: compatible propositions correspond to commuting projectors and incompatible propositions correspond to non-commuting projectors.

By assuming certain properties on the lattice (the quantum one), one can ask if the Hilbert space realization is the only possible one (up to lattice isomorphisms, of course). The answer is involved and I'm not an expert on it. I think you get some kind of 'generalized Hilbert space' over a division ring. Further assumptions presumably lead to R, C or H (the quaternions).

 

[moriheru]

An answer to that question was provided by von Neumann's approach to the mathematical formalization of QM.

In this approach, the central objects are the yes-no propositions one can make about a system, i.e., propositions about the system that only admit an answer of yes or no (e.g., P="the position of the particle is x=5"). The porposition by itself is not enough, you have to ask: "when the system is in the state S, what's the truth value, or at least a probability, of elementary porposition P?". In classical mechanics, the set of all these propositions can be seen as equivalent to the Borel sigma-algebra of phase space, which can be seen as a distributive lattice (a 'Boolean algebra'; I'm simplifying things, there are more assumptions it its definition!) A key property in this structure is that all elementary propositions are compatible, i.e., given any two propositions, is meaningful to ask about their truth values simultaneously: the classical logical connectives 'or' and 'and' are well defined for these pairs.

A key property in QM is the existence of incompatible elementary propositions: for them, is meaningless to ask about their truth values simultaneously. Clearly, it becomes problematic if we want to use the classical Borel sigma-algebra for modeling the set of elementary propositions in QM. How do we realize explicity the set of elementary propositions in QM then? von Neumann's brilliant insight was that we can see them as orthogonal projectors on a Hilbert space: compatible propositions correspond to commuting projectors and incompatible propositions correspond to non-commuting projectors.

By assuming certain properties on the lattice (the quantum one), one can ask if the Hilbert space realization is the only possible one (up to lattice isomorphisms, of course). The answer is involved and I'm not an expert on it. I think you get some kind of 'generalized Hilbert space' over a division ring. Further assumptions presumably lead to R, C or H (the quaternions).

Thanks aleazk

 

[atyy]

There is a slightly different way to understand the system that aleazk mentioned. Instead of thinking of QM as giving properties of systems, it gives the outcomes of measurements. Thus the results you see do not measure intrinsic properties of the system, but also contain the interaction of the measuring system on the apparatus. Then QM can be compatible both with a lattice that is the classical Boolean algebra ('properties'), as well as the quantum lattice ('measurement outcomes'). At least for non-relativistic physics, a Bohmian-type interpretation with hidden variables is an example of the former description, and standard Copenhagen QM is an example of the latter description. The latter description is usually more practical, because a hidden variable description is not unique as long as QM is not experimentally falsified, and in most cases quite cumbersome compared to standard QM (eg. http://arxiv.org/abs/0711.4770).

We had a thread on the quantum lattice recently started by microsanfil, and bhobba pointed to this write-up by John Baez: https://golem.ph.utexas.edu/category/2010/12/solers_theorem.html.