Given the Hilbert space $H$ of a single particle, we know we can write \begin{equation} H = H_{spatial}\otimes H_{other} \end{equation} where $H_{spatial}$ is spanned by the possible position states of the particle (or by a Fourier transform, momentum states), and $H_{other}$ is the Hilbert space corresponding to other observables (e.g. spin). My questions are as follows:

(1) How 'fine' a tensor decomposition can we obtain of $H$?

My guess would be into Hilbert spaces $\{H_{k}\}$ where $H_{k}$ corresponds to some set $K$ of observables on the particle: for any two observables $A_{k}$ and $B_{k}$ in $K$, the sum of dimensions over each of their eigenspaces will be $dim(H_{k})$. Moreover, there may be some nontrivial commutation relations between them (but simultaneously diagionalisable observables should lie in the same $H_{k}$?). However, for observables corresponding to distinct $H_{k}$, there will be no nontrivial commutation relations between them.

(2) What are the implications measurement-wise?

If we make a measurement corresponding to one of the constituent spaces, does the particle still exist in a superposition corresponding to the other spaces (e.g. could we measure position yet still have the particle in a superposition of spin states etc.)?

(3) Can there ever be entanglement between the constituent tensored Hilbert spaces?

Like it says on the tin: are there any known instances where this can happen?


Second thought: Actually, maybe simultaneously diagonalisable observables should generically be assigned separate tensor components of the Hilbert space (e.g. spin and angular momentum)? Makes sense for that example at least. When might this prescription (if correct) fail?

  • $\begingroup$ In (1) you say "the sum of dimensions over each of their eigenspaces will be 𝑑𝑖𝑚(𝐻𝑘)". Shouldn't it be the product and not the sum ? $\endgroup$
    – Carla
    Apr 9, 2020 at 8:48

1 Answer 1


(1) How `fine' a tensor decomposition can we obtain of $H$?

Without any additional constraints, the answer is: as fine as you want.

For example, it can be factorized as $$ H = H_x\otimes H_y\otimes H_z\otimes H_\text{spin} \tag{1} $$ where ${\cal H}_x$ corresponds to the particle's freedom of motion in the $x$ direction, and similarly for the $y$ and $z$ factors. In this decomposition, the spin factor $H_\text{spin}$ is finite-dimensional (dimension $2j+1$ for a particle of spin $j$), and the other factors are infinite-dimensional. Beware that the word "dimension" will be used here for two different things: the number of dimensions of space, and the number of dimensions of the Hilbert space. The latter number is infinite.

To see that the decomposition (1) is legitimate, use the fact that any single-particle wavefunction may be written $\psi(x,y,z,m)$, where $x,y,z$ are real variables and $m\in\{1,2,...,2j+1\}$ is a discrete variable, and then use the fact that any such function may be expressed as a (continuous) linear combination of factorized functions $\alpha(x)\beta(y)\gamma(z)\mu(m)$. The factorization (1) is "obvious" in the sense that we implicitly already have that factorization in mind any time we write the wavefunction as a function of $x,y,z,m$.

But the same Hilbert space $H$ can also be written as a product of arbitrarily many factors. The key to understanding this is to remember two things:

  • $H$ is an infinite-dimensional separable Hilbert space, which means that it contains a countable set of basis vectors whose linear combinations can approximate any other state-vector arbitrarily well. An example of a countable basis for $H_x$ is the set of wavefunctions of the form $x^n\exp(-x^2)$ for $n\in\{0,1,2,3,...\}$.

  • All infinite-dimensional separable Hilbert spaces are isomorphic to each other, meaning that they are identical as Hilbert spaces even though we may write them in very different-looking ways. For example, $H$ is isomorphic to $H_x$.

If this unique infinite-dimensional separable Hilbert space is denoted $H_\infty$, then these two facts imply the following isomorphisms: \begin{align} H &= H_\infty \\ H &= H_\infty\otimes H_\infty \\ H &= H_\infty\otimes H_\infty\otimes H_\infty \\ H &= H_\infty\otimes H_\infty\otimes H_\infty\otimes H_\infty \\ H &= H_\infty\otimes H_\infty\otimes H_\infty\otimes H_\infty\otimes H_\infty \end{align} and so on. For example, the Hilbert space of a single particle in $3$-dimensional space is isomorphic to the Hilbert space of a single particle in $N$-dimensional space, for arbitrarily large $N$. Even better: the Hilbert space of a single particle in $3$-dimensional space is isomorphic to the Hilbert space of quantum chromodynamics, with its rich spectrum of mesons and baryons.

The important message here is that physical systems are not distinguished from each other by their Hilbert spaces. Instead:

  • The distinction between different physical systems in quantum theory is encoded in the association between specific physical observables and specific linear operators on the Hilbert space.

The difference between a single-particle system in $3$-dimensional space and a single-particle system in $42$-dimensional space has nothing to do with their Hilbert spaces; it has everything to do with which operators we designate as "position operators" (or "momentum operators") along specified spatial dimensions. When we use a function of the form $\psi(x,y,z,m)$ to represent an element of $H$, we are implicitly saying something about which operators we have designated to represent position operators along specified spatial dimensions — and also which operators represent the particle's intrinsic angular momentum.

(2) What are the implications measurement-wise?

This may or may not be answered by the preceding remarks about observables, depending on what kind of "implications" are of interest. When we represent an element of $H$ as a function $\psi(x,y,z,m)$, we are typically implicitly declaring that $-i\hbar\partial/\partial x$ represents the $x$-component of the particle's momentum, and similarly for the $y$ and $z$ components. The fact that these operators commute with each other is a necessary condition for associating them with different factors in some factorization of the Hilbert space. However, just because two operators commute with each other doesn't mean that we can or should associate them with separate factors. For example, in a two-dimensional Hilbert space, we can easily construct two operators that commute with each other; but we can't factorize a 2-d Hilbert space into smaller factors.

(3) Can there ever be entanglement between the constituent tensored Hilbert spaces?

Yes, absolutely. For example, with respect to the factorization (1), almost all wavefunctions $\psi(x,y,z,m)$ are entangled with respect to $x,y,z$. The only ones that aren't entangled are wavefunctions of the form $\psi(x,y,z,m)=\alpha(x)\beta(y)\gamma(z)\mu(m)$. The way the word is most often used, "entangled" just means that the given wavefunction (or state-vector) can't be factorized with respect to the given factorization of the Hilbert space. A wavefunction that is entangled with respect to one factorization may be non-entangled with respect to a different factorization. When we use a particular factorization of the Hilbert space, it's usually because we are implicitly saying something about the observables associated with those factors, and that's what makes entanglement interesting.

The moral of this story is that Hilbert space is a featureless place, devoid of any physical content on its own, and we can factorize an infinite-dimensional Hilbert space in countless different ways, none of which have any physical significance on their own. In quantum theory, the association between physical observables and linear operators is what distinguishes different models from each other.

Of course, a factorization of the Hilbert space can facilitate describing such an association. Statements like "subsystems correspond to Hilbert-space factors" are common in the literature, and that statement is really an abbreviation for this one: "observables represented by linear operators on the $n$th factor are observables associated with the $n$th subsystem." The association bewteen physical observables and linear operators is what matters.

Long answer, simple message.

  • $\begingroup$ I didn't mean to restrict the breadth of the question to a completely generic Hilbert space. I get that the take-home message is that Hilbert space is putty without some constraints on the observables. For (1) then, the answer is as many observables as we care to keep track of. However, the important thing to note then is probably that we can only factorise in a certain manner if the Hamiltonian can be decoupled accordingly. This directly relates to the notion of entanglement w.r.t two observables as discussed in part (3) of your answer. $\endgroup$ Feb 26, 2019 at 22:13
  • $\begingroup$ The $H_{\infty}$ argument is nice. Mathematically speaking, part of question (1) was asking: if two operators $A$ and $B$ have nontrivial commutation relations, they must lie in the same factorisation space - but if they have trivial commutation relations with total sum of eigenspace dimensions as described in the question, are we free to choose between: (a) assigning them distinct factor spaces, or (b) assigning them the same factor space ? $\endgroup$ Feb 26, 2019 at 22:27
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    $\begingroup$ @SValera If you're looking for some way to choose an appropriate factorization based on a complete set of commuting observables, I'm not aware of any such connection. For example, a two-dimensional Hilbert space admits a complete set of (two) mutually commuting observables, but it can't be factorized. Conversely, if two operators $A$ and $B$ don't commute, that doens't imply that their effects are both restricted to the same factor; most operators simply don't respect the given factorization at all, whether or not they commute with each other. $\endgroup$ Feb 27, 2019 at 0:39
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    $\begingroup$ @SValera Mathematically, wavefunction collapse just the application of a projection operator, regardless of any factorization. Even if a projection operator acts nontrivially on only one factor, that doesn't mean that it leaves superpositions intact in other factors -- unless the state-vector itself is already factorized (that is, non-entangled). Yes, there are special states in which a measurement of position has no effect on the spin state, but for most states, position and spin are entangled with each other, and you can't measure one without affecting the other. $\endgroup$ Feb 27, 2019 at 0:43
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    $\begingroup$ @SValera Factorizations of the Hilbert space are often specified (not inferred) as a way of describing the physical significance of various operators. (The Hamiltonian usually acts non-trivially on all of those factors.) But in quantum field theory, a different approach is used to describe the physical significance of operators, one that doesn't use any factorization at all. A factorization is scaffolding used to construct simple models, not a feature to be inferred in models that are already given, and I don't know how any unique factorization can be inferred without additional information. $\endgroup$ Feb 27, 2019 at 0:51

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