[Physics] How do tensor products and direct sums fit into quantum mechanics

hilbert-spacequantum mechanicsquantum-entanglement

I understand that at times tensor products or direct sums are taken between Hilbert spaces in quantum mechanics. I don't, however, know when this can be done or when it should be done. I would like this meticulously explained. In particular, where does this fit with respect to the axioms of quantum mechanics? I understand that there may be more than one set of axioms for quantum mechanics, so let's adhere to these (that may be faulty):

I. The state of a system is represented by a ray of a Hilbert space $H$, the vectors of which are called kets.

II. Corresponding to each observable $A$ is a Hermitian operator $\hat{A}$ on $H$.

III. When measuring an observable $A$ of a system in a state represented by a ray with normalized representative $\Psi$, the result of the measurement will be one of the eigenvalues $a$ of $A$ with probability $\left<\mathbf{a}, \Psi\right>$, where $\mathbf{a}$ is the normalized eigenvector corresponding to the eigenvalue $a$. Furthermore, the state of the system will be that represented by $\mathbf{a}$.

IV. A ket $\Psi$ representing a system will evolve according to the Schrodinger equation.

Best Answer

Tensor product

Writing the Hilbert space as a tensor product $$ \newcommand{\cH}{\mathcal{H}} \cH=\cH_A\otimes \cH_B $$ can be useful when we want to think of $\cH_A$ and $\cH_B$ as two complementary subsystems of the full system. Observables associated with subsystem $A$ act like the identity on the other factor $\cH_B$, and conversely. For example, an observable associated with subsystem $A$ has the form $O_A\otimes 1$. A general observable affects both $\cH_A$ and $\cH_B$. That is, a general observable is a sum of terms of the form $O_A\otimes O_B$ where the operators $O_{A/B}$ act only on $A/B$ respectively.

In particular, if $\cH=\cH_A\otimes \cH_B$, the Hamiltonian in the Schrödinger equation is a sum of terms of the form $O_A\otimes O_B$. In particular, terms of the form $O_A\otimes 1$ and $1\otimes O_B$ describe the dynamics of the $A$ and $B$ subsystems by themselves, and all other terms describe interactions between these subsystems.

Another example is a non-relativistic particle with spin: we can express the Hilbert space as a tensor product $\cH_X\otimes \cH_S$, where observables associated with the particle's location have the form $O_X\otimes 1$ and observables associated with its spin have the form $1\otimes O_S$. In this case, the different parts are usually called different "degrees of freedom" instead of different "subsystems." In the case of a non-relativistic particle, we can further factorize $\cH_X$ into three factors associated with the three dimensions of space. Again, we would normally call these "degrees of freedom" instead of "subsystems."

Very generally, we can define a "subsystem" or "degree of freedom" as a special collection of observables. The tensor product construction isn't needed for this, but it's often useful. If observables associated with different subsystems (or degrees of freedom) commute with each other, then the tensor product is often useful as a systematic way of mathematically delineating the different sets of observables: each set acts nontrivially on only one of the tensor factors.

The "subsystems" and "degrees of freedom" concepts are just vaguely-delineated special cases of a much more general idea: mutually commuting subsets of the set of observables. The same Hilbert space admits many different tensor-product factorizations. Which one is most useful (if any) depends on which operators we want to represent which physical observables -- the decisions we make when we're defining a model. A similar comment applies to the most common definitions/measures of "entanglement," because they refer to a given tensor product factorization.

Learning about the "split property" in quantum field theory reveals some limitations of the tensor product formulation. The split property is mentiond in this related post:

Should it be obvious that independent quantum states are composed by taking the tensor product?

The tensor product also has other uses. For example, for a single particle at rest in 3-d space, we can systematically express the spin-$j$ representation for any $j$ by taking the tensor product of $2j$ copies of the spin-1/2 representation and symmetrizing. We can think of this as a special application of the subsystem idea, because a symmetrized collection of $2j$ spin-1/2 particles has total spin $j$.

Axioms I-IV listed in the OP are the same whether or not $\cH$ is written as a tensor product, because those axioms are independent of what representation we use for the Hilbert space.

Direct sum

Writing the Hilbert space as a direct sum $$ \cH=\cH_1\oplus\cH_2 $$ is useful when we want to focus on a particular subspace of states. Using a non-relativistic single-particle model as an example, $\cH_1$ could consist of states in which a particle's wavefunction has support only within a given region $R$, and $\cH_2$ could consist of states with support in the complement of $R$.

More generally, given any discrete observable (such as the observable that asks "is the particle located in the region $R$ or not?"), we can write $\cH$ as the direct sum of that observable's eigenspaces. A direct-sum decomposition of the Hilbert space corresponds to a block-matrix representation of operators on the Hilbert space.

More esoterically, the direct sum is also useful for representing mixed states as vector states: every state, whether pure or mixed, can be expressed as a vector state in a sufficiently large Hilbert space, with the understanding the all observables have a block-diagonal form that doesn't mix the different direct-summands with each other. This fact is sometimes useful for proving theorems, and this fact can in turn be proven using the GNS Construction.

Again, axioms I-IV listed in the OP are the same whether or not $\cH$ is written as a direct sum, because those axioms are independent of what representation we use for the Hilbert space.

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