Tensor product of Hilbert spaces

Since Hilbert spaces have inner products, one would like to introduce an inner product, and thereby a topology, on the tensor product that arises naturally from the inner products on the factors. Let ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}}$  be two Hilbert spaces with inner products ${\displaystyle \langle \cdot ,\cdot \rangle _{1}}$  and ${\displaystyle \langle \cdot ,\cdot \rangle _{2},}$  respectively. Construct the tensor product of ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}}$  as vector spaces as explained in the article on tensor products. We can turn this vector space tensor product into an inner product space by defining $${\displaystyle \left\langle \phi _{1}\otimes \phi _{2},\psi _{1}\otimes \psi _{2}\right\rangle =\left\langle \phi _{1},\psi _{1}\right\rangle _{1}\,\left\langle \phi _{2},\psi _{2}\right\rangle _{2}\quad {\mbox{for all }}\phi _{1},\psi _{1}\in H_{1}{\mbox{ and }}\phi _{2},\psi _{2}\in H_{2}}$$  and extending by linearity. That this inner product is the natural one is justified by the identification of scalar-valued bilinear maps on ${\displaystyle H_{1}\times H_{2}}$  and linear functionals on their vector space tensor product. Finally, take the completion under this inner product. The resulting Hilbert space is the tensor product of ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}.}$ 

The tensor product can also be defined without appealing to the metric space completion. If ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}}$  are two Hilbert spaces, one associates to every simple tensor product ${\displaystyle x_{1}\otimes x_{2}}$  the rank one operator from ${\displaystyle H_{1}^{*}}$  to ${\displaystyle H_{2}}$  that maps a given ${\displaystyle x^{*}\in H_{1}^{*}}$  as $${\displaystyle x^{*}\mapsto x^{*}(x_{1})\,x_{2}.}$$ 

This extends to a linear identification between ${\displaystyle H_{1}\otimes H_{2}}$  and the space of finite rank operators from ${\displaystyle H_{1}^{*}}$  to ${\displaystyle H_{2}.}$  The finite rank operators are embedded in the Hilbert space ${\displaystyle HS(H_{1}^{*},H_{2})}$  of Hilbert–Schmidt operators from ${\displaystyle H_{1}^{*}}$  to ${\displaystyle H_{2}.}$  The scalar product in ${\displaystyle HS(H_{1}^{*},H_{2})}$  is given by $${\displaystyle \langle T_{1},T_{2}\rangle =\sum _{n}\left\langle T_{1}e_{n}^{*},T_{2}e_{n}^{*}\right\rangle ,}$$  where ${\displaystyle \left(e_{n}^{*}\right)}$  is an arbitrary orthonormal basis of ${\displaystyle H_{1}^{*}.}$ 

Under the preceding identification, one can define the Hilbertian tensor product of ${\displaystyle H_{1}}$  and ${\displaystyle H_{2},}$  that is isometrically and linearly isomorphic to ${\displaystyle HS(H_{1}^{*},H_{2}).}$ 

The Hilbert tensor product ${\displaystyle H_{1}\otimes H_{2}}$  is characterized by the following universal property (Kadison & Ringrose 1997, Theorem 2.6.4):

A weakly Hilbert-Schmidt mapping ${\displaystyle L:H_{1}\times H_{2}\to K}$  is defined as a bilinear map for which a real number ${\displaystyle d}$  exists, such that $${\displaystyle \sum _{i,j=1}^{\infty }{\bigl |}\left\langle L(e_{i},f_{j}),u\right\rangle {\bigr |}^{2}\leq d^{2}\,\|u\|^{2}}$$  for all ${\displaystyle u\in K}$  and one (hence all) orthonormal bases ${\displaystyle e_{1},e_{2},\ldots }$  of ${\displaystyle H_{1}}$  and ${\displaystyle f_{1},f_{2},\ldots }$  of ${\displaystyle H_{2}.}$ 

As with any universal property, this characterizes the tensor product H uniquely, up to isomorphism. The same universal property, with obvious modifications, also applies for the tensor product of any finite number of Hilbert spaces. It is essentially the same universal property shared by all definitions of tensor products, irrespective of the spaces being tensored: this implies that any space with a tensor product is a symmetric monoidal category, and Hilbert spaces are a particular example thereof.

Two different definitions have historically been proposed for the tensor product of an arbitrary-sized collection ${\textstyle \{H_{n}\}_{n\in N}}$  of Hilbert spaces. Von Neumann's traditional definition simply takes the "obvious" tensor product: to compute ${\textstyle \bigotimes _{n}{H_{n}}}$ , first collect all simple tensors of the form ${\textstyle \bigotimes _{n\in N}{e_{n}}}$  such that ${\textstyle \prod _{n\in N}{\|e_{n}\|}<\infty }$ . The latter describes a pre-inner product through the polarization identity, so take the closed span of such simple tensors modulo that inner product's isotropy subspaces. This definition is almost never separable, in part because, in physical applications, "most" of the space describes impossible states. Modern authors typically use instead a definition due to Guichardet: to compute ${\textstyle \bigotimes _{n}{H_{n}}}$ , first select a unit vector ${\textstyle v_{n}\in H_{n}}$  in each Hilbert space, and then collect all simple tensors of the form ${\textstyle \bigotimes _{n\in N}{e_{n}}}$ , in which only finitely-many ${\textstyle e_{n}}$  are not ${\textstyle v_{n}}$ . Then take the ${\displaystyle L^{2}}$  completion of these simple tensors.^[2][3]

Let ${\displaystyle {\mathfrak {A}}_{i}}$  be the von Neumann algebra of bounded operators on ${\displaystyle H_{i}}$  for ${\displaystyle i=1,2.}$  Then the von Neumann tensor product of the von Neumann algebras is the strong completion of the set of all finite linear combinations of simple tensor products ${\displaystyle A_{1}\otimes A_{2}}$  where ${\displaystyle A_{i}\in {\mathfrak {A}}_{i}}$  for ${\displaystyle i=1,2.}$  This is exactly equal to the von Neumann algebra of bounded operators of ${\displaystyle H_{1}\otimes H_{2}.}$  Unlike for Hilbert spaces, one may take infinite tensor products of von Neumann algebras, and for that matter C*-algebras of operators, without defining reference states.^[3] This is one advantage of the "algebraic" method in quantum statistical mechanics.

If ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}}$  have orthonormal bases ${\displaystyle \left\{\phi _{k}\right\}}$  and ${\displaystyle \left\{\psi _{l}\right\},}$  respectively, then ${\displaystyle \left\{\phi _{k}\otimes \psi _{l}\right\}}$  is an orthonormal basis for ${\displaystyle H_{1}\otimes H_{2}.}$  In particular, the Hilbert dimension of the tensor product is the product (as cardinal numbers) of the Hilbert dimensions.

The following examples show how tensor products arise naturally.

Given two measure spaces ${\displaystyle X}$  and ${\displaystyle Y}$ , with measures ${\displaystyle \mu }$  and ${\displaystyle \nu }$  respectively, one may look at ${\displaystyle L^{2}(X\times Y),}$  the space of functions on ${\displaystyle X\times Y}$  that are square integrable with respect to the product measure ${\displaystyle \mu \times \nu .}$  If ${\displaystyle f}$  is a square integrable function on ${\displaystyle X,}$  and ${\displaystyle g}$  is a square integrable function on ${\displaystyle Y,}$  then we can define a function ${\displaystyle h}$  on ${\displaystyle X\times Y}$  by ${\displaystyle h(x,y)=f(x)g(y).}$  The definition of the product measure ensures that all functions of this form are square integrable, so this defines a bilinear mapping ${\displaystyle L^{2}(X)\times L^{2}(Y)\to L^{2}(X\times Y).}$  Linear combinations of functions of the form ${\displaystyle f(x)g(y)}$  are also in ${\displaystyle L^{2}(X\times Y).}$  It turns out that the set of linear combinations is in fact dense in ${\displaystyle L^{2}(X\times Y),}$  if ${\displaystyle L^{2}(X)}$  and ${\displaystyle L^{2}(Y)}$  are separable.^[4] This shows that ${\displaystyle L^{2}(X)\otimes L^{2}(Y)}$  is isomorphic to ${\displaystyle L^{2}(X\times Y),}$  and it also explains why we need to take the completion in the construction of the Hilbert space tensor product.

Similarly, we can show that ${\displaystyle L^{2}(X;H)}$ , denoting the space of square integrable functions ${\displaystyle X\to H,}$  is isomorphic to ${\displaystyle L^{2}(X)\otimes H}$  if this space is separable. The isomorphism maps ${\displaystyle f(x)\otimes \phi \in L^{2}(X)\otimes H}$  to ${\displaystyle f(x)\phi \in L^{2}(X;H)}$  We can combine this with the previous example and conclude that ${\displaystyle L^{2}(X)\otimes L^{2}(Y)}$  and ${\displaystyle L^{2}(X\times Y)}$  are both isomorphic to ${\displaystyle L^{2}\left(X;L^{2}(Y)\right).}$ 

Tensor products of Hilbert spaces arise often in quantum mechanics. If some particle is described by the Hilbert space ${\displaystyle H_{1},}$  and another particle is described by ${\displaystyle H_{2},}$  then the system consisting of both particles is described by the tensor product of ${\displaystyle H_{1}}$  and ${\displaystyle H_{2}.}$  For example, the state space of a quantum harmonic oscillator is ${\displaystyle L^{2}(\mathbb {R} ),}$  so the state space of two oscillators is ${\displaystyle L^{2}(\mathbb {R} )\otimes L^{2}(\mathbb {R} ),}$  which is isomorphic to ${\displaystyle L^{2}\left(\mathbb {R} ^{2}\right).}$  Therefore, the two-particle system is described by wave functions of the form ${\displaystyle \psi \left(x_{1},x_{2}\right).}$  A more intricate example is provided by the Fock spaces, which describe a variable number of particles.

• Kadison, Richard V.; Ringrose, John R. (1997). Fundamentals of the theory of operator algebras. Vol. I. Graduate Studies in Mathematics. Vol. 15. Providence, R.I.: American Mathematical Society. ISBN 978-0-8218-0819-1. MR 1468229..
• Weidmann, Joachim (1980). Linear operators in Hilbert spaces. Graduate Texts in Mathematics. Vol. 68. Berlin, New York: Springer-Verlag. ISBN 978-0-387-90427-6. MR 0566954..