8
$\begingroup$

The first postulate of Quantum Mechanics as I've learned can be stated as:

The states of a quantum system can be described by vectors in a Hilbert space.

I've seem also some people also requiring the Hilbert space be separable. Anyway, the point here is: we are associating a Hilbert space to quantum systems in order to describe their states. Usually, in basic treatments of Quantum Mechanics, when dealing with a spinless particle, the Hilbert space is usually taken to be $L^2(\mathbb{R}^n, d^nx)$, and states are taken to be equivalent to wavefunctions.

On the other hand, there is the more general situation on which the wavefunction is just the position representation of the state. And in truth there are situation on which there is no wavefunction. This led me to wonder how does one really choose the Hilbert space.

There is, however, the following theorem:

Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, then they are isometrically isomorphic if and only if they have the same dimension.

In that case, the question of "which Hilbert space to choose" seems to be nonsensical. After all, choosing one with a certain dimension, any other would work as well. This makes it quite strange also to associate a Hilbert space to a system, if this association is not unique.

Still, there is the question of the dimension. I believe that one could argue that this could be determined by the observables. But still, one would need to know the spectrum of the operator before even defining the space upon which it is defined!

Anyway, I'm quite confused with these questions. How does one choose a Hilbert space to associate with a system? How does one deal with this ambiguity given by that theorem? If the observables determine the space, how does this happen if one can only define the operators after knowing the space?

Improve this question
$\endgroup$
5
  • 3
    $\begingroup$ Hint: Infinite-dimensional cases are difficult ;) $\endgroup$
    – Danu
    Sep 29, 2015 at 15:07
  • $\begingroup$ I see @Danu. The first case the book I'm studying deals with is the Hilbert space for a particle with spin, which is two-dimensional. But he constructs it based on the possible outcomes of the measurement of spin, i.e. based on the spin operator. Is that the usual procedure? $\endgroup$
    – Gold
    Sep 29, 2015 at 15:18
  • $\begingroup$ What does the $d^nx$ in $L^2(\mathbb{R}^n, d^nx)$ mean? $\endgroup$
    – enigmaticPhysicist
    Sep 29, 2015 at 19:48
  • $\begingroup$ "On the other hand, there is the more general situation on which the wavefunction is just the position representation of the state."—How is that any different from the example you describe one paragraph before? $\endgroup$
    – enigmaticPhysicist
    Sep 29, 2015 at 19:51
  • $\begingroup$ This $d^nx$ stands for the usual measure. It is different in the following sense: when one considers $L^2(\mathbb{R}^n, d^nx)$, we are assuming the states are the wavefunctions. In that case, if we pick $\psi \in L^2(\mathbb{R}^n, d^nx)$ it is really a function $\psi : \mathbb{R}^n \to \mathbb{C}$. On the second case $\mathcal{H}$ is not necessarily a space of functions. If, one does have a position operator $X : \Omega\subset \mathcal{H}\to \mathcal{H}$ with eigenvectors $|x\rangle$, then one can associate a wavefunction $\psi\in L^2(\mathbb{R}^n,d^nx)$ by $\psi(x) = \langle x|\psi\rangle$. $\endgroup$
    – Gold
    Sep 29, 2015 at 19:56

3 Answers 3

Reset to default
7
$\begingroup$

There are many approaches. But first I want to make sure you know that when you have $n$ spin zero particles in a 3d space and have a wavefunction that the function is from $\mathbb R^{3n},$ i.e. from configuration space. But also I want you to know when someone says infinite dimensional Hilbert Space, they mean the size of a set of mutually orthonormal vectors.

Now when you have some particles and know their spin you could start with a specific concrete set of wavefunctions from configuration space into a tensor product of the spin states. You can then use the inner product on the spin states to get an inner product of the wavefunctions. And this is basically like picking a basis when do vector analysis.

Given the wavefunctions you could then talk about the operators. The operators form a vector space, then have a norm defined on them, they have an adjoint operation. And some times $A-\lambda I$ is not invertible as an operator (for instance when $\lambda$ is an eigenvalue of $A$). And it is possible to reverse the process.

So if you start with a bunch of things that act like operators, i.e. that form a vector space, that have a multiplication, have a norm, have a star operator and they satisfy the rules you'd expect. Then you can call it a $C^*$ algebra. Then it turns out the standard assumptions in mathematics assert that there is a Hilbert space and a subset of the operators on the Hilbert space that as a vector space with a multiplication, a norm, and the adjoint (as the star operator) will act just like your $C^*$ algebra.

So you can start with a Hilbert Space and then define operators. Or you can start with a $C^*$ algebra and then define your Hilbert Space as one of the Hilbert Spaces that has a subalgebra of operators just like your $C^*$ algebra. The biggest deal is knowing how to connect to experimental results since either way you have an algebra, some operators and a Hilbert Space and any triple you have is going to look like the other triple you made starting from the other end.

As for a spectrum. Since the algebra has a multiplication you can discuss whether $A-\lambda I$ is invertible without needing to have vectors that are eigen to any operators, the object just have an inverse or it doesn't, from the set of things you can multiply.

You could imagine messing with matrices, adding, scaling, multiplying, taking adjoints. And discussing whether they have inverses, without every mentioning that there are vectors the matrices could act on. And then you can be more abstract and say the fact they are matrices wasn't important, just have things with a scaling operation, a norm operation, and adjoint operation, and a multiplication operation, that do the right things.

How does one choose a Hilbert space to associate with a system?

You could start with it, or start with your algebra. Or you can focus on connecting a Hilbert Space with the outcomes of certain experimental results. For instance the completion of the set of eigenvectors for a maximal set of compatible measurements.

How does one deal with this ambiguity given by that theorem?

Having different vector spaces that correspond to your physical setup is no worse than being able to choose your x y and z axis to point any direction you want in the lab. Truly no different. They are all isomorphic, and 3d vector space. And it isn't a problem.

If the observables determine the space, how does this happen if one can only define the operators after knowing the space?

The algebra doesn't need the space. You can have a bunch of square matrices without having column or row vectors. You can have the algebra with the scaling, adding, multiplying, morning, and adjointing without having the matrices. And you still can have a spectrum.

Improve this answer
$\endgroup$
4
$\begingroup$

Separable Hilbert spaces are all isomorphic to one other, since they are all isomorphic to $\ell_2$, despite the dimensions (this holds even in the infinite dimensional case). No matter how you realise it, you may just choose one Hilbert space and its properties will exhaust all you need to perform the calculations and derive observable quantities (they do not depend on the Hilbert space realisations). After all, the Schrödinger equation does not limit you to narrow down to this or that other space, rather, it just assumes the state to belong to any Hilbert space, plus additional constraints on the domain of definitions of the operators.

In case of non-separability usually the Hilbert space gets constructed ad hoc starting from example states which are required to span a Hilbert structure for physical reasons (scalar products and observable quantities must be calculated eventually). A good example can be found in Loop Quantum Gravity, where the Hilbert space is defined as the space of spin foams and is non-separable. For all the other special constructions case by case I have no experience (I guess someone else can come up with other particular examples).

Improve this answer
$\endgroup$
0
$\begingroup$

Choosing a Hilbert space is easy. It just needs to be mathematically convenient, while still completely characterising the system. (You could use $l_2$ to represent any quantum state, but you often don't because that is often mathematically very inconvenient.)

As for defining operators, it is true one cannot explicitly define an operator without defining its domain (i.e. the Hilbert space), but most operators have properties, like commutation relations, that must be true no matter what Hilbert space you choose. These properties greatly inform your choice of Hilbert space because, again, certain Hilbert spaces are mathematically far more convenient than others.

For example, if you require the use of two operators $A$ and $B$ with the property

$-i[A, B] = 1$

then the Hilbert space you choose cannot possibly be finite. This happens in quantum mechanics all the time, for example when $A$ is position and $B$ is momentum.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.