Hilbert space in quantum mechanics

Question

I think in quantum mechanics we assign to each system a specific Hilbert space i.e. if systems are different then their Hilbert spaces are different. Is this true? If not why? For differernt system I mean their hamiltonians are different.

Answer 1

Short Answer. Different quantum systems often share the same Hilbert space. In particular, as we'll see in a moment, whether or not their Hilbert spaces are the same (isomorphic) depends only on the dimensions of these spaces.

Details.

There are two key theorems that together, give a decisive answer to this question. The first is from finite-dimensional linear algebra, and the second is from functional analysis.

Theorem 1. Any two finite-dimensional vector spaces of the same dimension over the same field are isomorphic.

Theorem 2. Any two separable, complex Hilbert spaces are isomorphic.

The term separable here means that the Hilbert space admits a countable, orthonormal basis.

In the standard treatment of quantum mechanics, we demand that

QM Fact. The state space of a quantum system is a complex Hilbert space. If it is infinite-dimensional, then it is separable.

If we combine Theorems 1 and 2 with QM Fact, then we get the following result as a corollary:

Corollary. If two quantum systems have states spaces of the same dimension, then these state spaces are isomorphic.

In other words, the only thing that mathematically distinguishes the Hilbert space structures of distinct quantum systems is the dimensions of these Hilbert spaces.

Example. The state space of a free particle moving in one dimension is the Hilbert space of complex-valued, square-integrable functions on the real line: $L^2(\mathbb R)$. The state space of a free particle moving in three dimensions is the Hilbert space of complex-valued, square-integrable functions on three-space: $L^2(\mathbb R^3)$. Both of these spaces are separable, complex Hilbert spaces, so they are isomorphic to one another. This does not mean that they are equal (they consist of different kinds of functions), and it does not mean that they have the same physical interpretation, but it does mean that as Hilbert spaces, their structure is the same; they are two different "versions" of the same beast.

The point about physical interpretations being distinct is the key one that illuminates by distinct physical systems can have isomorphic Hilbert spaces. To illustrate this further, we consider another example.

Example. Consider the Hilbert space $\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}$ for a composite system of two spin-$1/2$ particles on one hand, and consider the Hilbert space $\mathcal H_{3/2}$ for a system of a single spin $3/2$ particle on the other. Now recall that the dimension of the spin $s$ Hilbert space $\mathcal H_s$ obeys \begin{align} \dim(\mathcal H_s) = 2s+1. \end{align} It follows that \begin{align} \dim(\mathcal H_\frac{1}{2}\otimes\mathcal H_\frac{1}{2}) = \dim(\mathcal H_\frac{1}{2})^2 = 2^2 = 4 \end{align} and \begin{align} \dim(\mathcal H_\frac{3}{2}) = 2(\tfrac{3}{2})+1 = 4 \end{align} In other words, the dimension of the state space of two spin-$1/2$ particles matches that of the state space of a single spin-$3/2$ particle, so by the corollary above, they are isomorphic. But notice that their physical interpretations are completely different! Isomorphic Hilbert spaces are being used to model two physically distinct systems.

Answer 2

I think in quantum mechanics we assign to each system a specific Hilbert space i.e. if systems are different then their Hilbert spaces are different. Is this true? If not why? For differernt system I mean their hamiltonians are different.

I found really interesting the question and @joshphysics, however I desagree with his point of view.

It is true that spaces are often isomorphic, but I'd not interpret that as been the same space.

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Consider a QM system of two (free) spin one-half particles. The Hilbert space of a particle is $\mathcal{H}_{\tfrac{1}{2}}$, and the one of the system would be $\mathcal{H}_{\tfrac{1}{2}}\otimes \mathcal{H}_{\tfrac{1}{2}}$. I usually think that as two copies of (equivalent but different) spaces... Indeed that's the form they are treated mathematically, operators of the first Hillbert space do not act on the second and viceversa.

Disclaimer: Of course all this changes if one introduces an interaction, because one can always find irreducible representations of $G\times G$ under $G$ (for a given group of transformation $G$)

I believe it is clear that inequivalent systems have different different Hilbert spaces, that's why I explain only systems of equivalent "particles"