# I Why are Hilbert spaces used in quantum mechanics?

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1. May 27, 2016

### Frank Castle

In classical mechanics we use a 6n-dimensional phase space, itself a vector space, to describe the state of a given system at any one point in time, with the evolution of the state of a system being described in terms of a trajectory through the corresponding phase space. However, in quantum mechanics we instead use Hilbert spaces. What is the intuitive reasoning for why we use Hilbert spaces? Is it simply due to the non-commutative nature of observables and the fact that we now have operators acting on states to extract observable quantities and also time evolve them, and these operators naturally act on a Hilbert space?

2. May 27, 2016

### thephystudent

Note that 'Hilbert space' is a pretty general concept, every vector space with an inner product (and no 'missing points' e.g. like real numbers but not only the rational ones) is a hilbert space. Classical phase space is also a Hilbert space in this sense where positions and momenta constitute the most useful basis vectors.

In Quantum mechanics, the whole system cannot be described in terms of seperate particles each having different positions and momenta. Still, you can define some hilbert space describing the possible states of the particles. Then, one usually choses a basis in this quantum hilbert space as the eigenvectors of an operator of interest.

Last edited: May 27, 2016
3. May 27, 2016

### Frank Castle

Why do we distinguish between classical phase space and Hilbert spaces then? In all introductions that I've read on quantum mechanics , Hilbert spaces are introduced with little motivation and its also implied that they are a new space (that wasn't present in classical mechanics) in which quantum states exist.

4. May 27, 2016

### bhobba

See:

It's so matrix methods can be linked to the functions in Schrodinger equation. The wave-functions of Schrodinger's equation need to form a vector space which naturally leads to a Hilbert space.

If you want something deeper this explains it from a deep analysis of the logical foundations of QM:

Be warned - its what mathematicians call non trivial - meaning its hard.

One of the key results used is Pirons theorem:
https://www.quora.com/What-is-the-significance-of-Pirons-theorem

It shows quantum logic nearly, but not quite, implies a Hilbert Space is required.

Recently a new theorem, called Solers Theorem, gets us even closer:
https://golem.ph.utexas.edu/category/2010/12/solers_theorem.html

Its tantalizingly close to proving Hilbert spaces are required, but its still not there.

Thanks
Bill

Last edited by a moderator: May 7, 2017
5. May 27, 2016

### Frank Castle

Thanks for all the links, I shall have a read through and get back if I have any further queries.

Last edited by a moderator: May 7, 2017
6. May 27, 2016

### George Jones

Staff Emeritus
Classical phase space is the union (as sets) of a bunch of vector spaces (cotangent spaces), but is not itself a vector space, it is a vector bundle.

7. May 27, 2016

### thephystudent

You are absolutely right, I simplified too much. I'll delete my post
EDIT: too late to remove it

8. May 27, 2016

### Truecrimson

Phase spaces emphasize the symplectic geometry while Hilbert spaces emphasize the orthogonal geometry. The former leads to Hamilton's equations and the latter gives a statistical distance between two quantum states. IMHO, in the end it is a matter of convenience because quantum mechanics also has a symplectic structure and we can do statistical mechanics on a phase space.

Quantum mechanics can be formulated in a phase space by allowing negative distributions
https://en.wikipedia.org/wiki/Phase_space_formulation

Classical mechanics can be formulated in a complex Hilbert space with no non-commuting observables, I think. I'm not familiar with this approach.
https://en.wikipedia.org/wiki/Koopman–von_Neumann_classical_mechanics

Last edited: May 27, 2016
9. May 27, 2016

### sandy stone

This discussion is a bit over my head, but I was curious about one point. Do "ordinary" vector spaces allow complex values?

10. May 27, 2016

### Truecrimson

11. May 27, 2016

### Frank Castle

So are Hilbert spaces simply introduced such that Schrödinger's wave mechanics and Heisenberg's matrix mechanics are isomorphic within this space? The wave functions need to form a vector space due to the linearity of the Schrödinger equation, hence they should satisfy the vector space axioms?

12. May 27, 2016

### Truecrimson

Let's restrict ourselves to finite dimensional spaces first because of complications in the infinite dimensional case. A finite dimensional Hilbert space is just a vector space with an inner product, which we use to compute the transition probability between two states.

13. May 28, 2016

### vanhees71

Sure, you can have vector spaces with any field (in the sense of the mathematical structure, in German called a "Körper") as scalars.

Any $n$-dimensional complex vector space is equivalent (via a basis) with the vector space $\mathbb{C}^n$ of $n$-tupels of complex numbers.

14. Jun 4, 2016

### lavinia

If the square of the modulus of the amplitude of the wave function is going to represent a probability distribution then the wave function must be square integrable. That is: it lies in the Hilbert space of square integrable complex valued functions.

Last edited: Jun 4, 2016
15. Jun 6, 2016

### Frank Castle

Is this the main motivation for the usage of Hilbert spaces in quantum mechanics, or are there other motivating qualities?

16. Jun 7, 2016

### vanhees71

Well, the most convincing argument for using a mathematical structure in physics is that it works, i.e., it is successful in describing what's objectively and reproducibly observed in Nature, and the Hilbert space is precisely the mathematical structure that is needed to describe matter from quarks and leptons up to neutron stars. That's why it's used.

17. Jun 7, 2016

### Frank Castle

Yes, I understand the reasoning at that level, but I'm struggling a bit with the mathematical reasoning as to why one uses Hilbert spaces in quantum mechanics? Is it simply because the normed vector space structure complies with the linearity of the Schrödinger equation and allows one to construct probability densities and a notion of unit story. Additionally, completeness guarantees that one can use calculus and thus the Schrödinger equation and that wave functions are bounded and their decomposition onto sets of basis vectors are convergent?!

18. Jun 7, 2016

### bhobba

If you want the physical reason and not from QM foundations its got to do with the requirement to have continuous transformations between pure states:
https://arxiv.org/pdf/quant-ph/0101012v4.pdf

Guys - I know I link to that paper a lot. But I really do beieve its very important as far as quantum foundations go. QM is almost pulled out of thin air.

Thanks
Bill

19. Jun 7, 2016

### ShayanJ

I don't think mentioning a paper a lot is a bad thing. Its just that when you mention a paper, you should be careful not to present it as a mainstream opinion in the scientific community if its not.

Its correct that its an interesting view but its far from satisfactory and also I think it introduces more problems than it solves. Assuming that QM is only another kind of probability theory, a mathematical theory applied to physics, just makes the philosphical issues of QM worse than before! Its definitely not a price I'm eager to pay only to derive QM from some axioms! I really prefer the out-of-thin-air QM to this proposal.

P.S.
Also it doesn't treat the continuous observbles.

20. Jun 7, 2016

### lavinia

I am new to this stuff but Leonard Susskind says that unlike classical state space, quantum state space is a vector space - which means that linear combinations of states are also states. So the square integrablility of states is not the only use of Hilbert space. Quantum Mechanics also uses its linear structure.

Hilbert space is the only normed linear space that has an inner product. In Quantum Mechanics, inner products of two states represent transition amplitudes from one quantum state to the next. So these transition amplitudes are orthogonal projections of one state onto another. So quantum Mechanics uses the inner product feature of Hilbert space as well.

Measurements in Quantum Mechanics are linear operators on the vector space of quantum states. For each operator there is an orthonormal basis of " eigen states "(eigen vectors of the operator). The idea of orthonormality only makes sense in a Hilbert space.

These eigen vectors are the possible states that a ensue after a measurement. When measuring a property of a quantum state, the outcome of the measurement is uncertain. It can be any of the eigen states of the linear operator. The square of the probability that the measurement will land in a particular eigen state is the Hilbert space inner product of the quantum state with the eigen state.

So the full suite of features of Hilbert space are used to describe quantum states.

- I wonder if every vector in the Hilbert space is a possible quantum state.

Last edited: Jun 7, 2016