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 $H\_1$ and $H\_2$ be two Hilbert spaces with inner products $\backslash langle\backslash cdot,\; \backslash cdot\backslash rangle\_1$ and $\backslash langle\backslash cdot,\; \backslash cdot\backslash rangle\_2,$ respectively. Construct the tensor product of $H\_1$ and $H\_2$ 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

$$\backslash left\backslash langle\backslash phi\_1\; \backslash otimes\; \backslash phi\_2,\; \backslash psi\_1\; \backslash otimes\; \backslash psi\_2\backslash right\backslash rangle\; =\; \backslash left\backslash langle\backslash phi\_1,\; \backslash psi\_1\backslash right\backslash rangle\_1\; \backslash ,\; \backslash left\backslash langle\backslash phi\_2,\; \backslash psi\_2\backslash right\backslash rangle\_2\; \backslash quad\; \backslash mbox\{for\; all\; \}\; \backslash phi\_1,\backslash psi\_1\; \backslash in\; H\_1\; \backslash mbox\{\; and\; \}\; \backslash phi\_2,\backslash psi\_2\; \backslash in\; H\_2$$

and extending by linearity. That this inner product is the natural one is justified by the identification of scalar-valued bilinear maps on $H\_1\; \backslash times\; H\_2$ 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 $H\_1$ and $H\_2.$

The tensor product can also be defined without appealing to the metric space completion. If $H\_1$ and $H\_2$ are two Hilbert spaces, one associates to every simple tensor product $x\_1\; \backslash otimes\; x\_2$ the rank one operator from $H\_1^*$ to $H\_2$ that maps a given $x^*\backslash in\; H^*\_1$ as

$$x^*\; \backslash mapsto\; x^*(x\_1)\; \backslash ,\; x\_2.$$

This extends to a linear identification between $H\_1\; \backslash otimes\; H\_2$ and the space of finite rank operators from $H\_1^*$ to $H\_2.$ The finite rank operators are embedded in the Hilbert space $HS(H\_1^*,\; H\_2)$ of Hilbert–Schmidt operators from $H\_1^*$ to $H\_2.$ The scalar product in $HS(H\_1^*,\; H\_2)$ is given by

$$\backslash langle\; T\_1,\; T\_2\; \backslash rangle\; =\; \backslash sum\_n\; \backslash left\; \backslash langle\; T\_1\; e\_n^*,\; T\_2\; e\_n^*\; \backslash right\; \backslash rangle,$$

where $\backslash left(e\_n^*\backslash right)$ is an arbitrary orthonormal basis of $H\_1^*.$

Under the preceding identification, one can define the Hilbertian tensor product of $H\_1$ and $H\_2,$ that is isometrically and linearly isomorphic to $HS(H\_1^*,\; H\_2).$

The Hilbert tensor product $H\_1\; \backslash otimes\; H\_2$ is characterized by the following universal property {{harv|Kadison|Ringrose|1997|loc=Theorem 2.6.4}}:

{{math theorem | There is a weakly Hilbert–Schmidt mapping $p\; :\; H\_1\; \backslash times\; H\_2\; \backslash to\; H\_1\; \backslash otimes\; H\_2$ such that, given any weakly Hilbert–Schmidt mapping $L\; :\; H\_1\; \backslash times\; H\_2\; \backslash to\; K$ to a Hilbert space $K,$ there is a unique bounded operator $T\; :\; H\_1\; \backslash otimes\; H\_2\; \backslash to\; K$ such that $L\; =\; T\; p.$}}

A weakly Hilbert-Schmidt mapping $L\; :\; H\_1\; \backslash times\; H\_2\; \backslash to\; K$ is defined as a bilinear map for which a real number $d$ exists, such that

$$\backslash sum\_\{i,j=1\}^\backslash infty\; \backslash bigl|\; \backslash left\backslash langle\; L(e\_i,\; f\_j),\; u\; \backslash right\; \backslash rangle\backslash bigr|^2\; \backslash leq\; d^2\backslash ,\backslash |u\backslash |^2$$

for all $u\; \backslash in\; K$ and one (hence all) orthonormal bases $e\_1,\; e\_2,\; \backslash ldots$ of $H\_1$ and $f\_1,\; f\_2,\; \backslash ldots$ of $H\_2.$

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 $\backslash \{H\_n\backslash \}\_\{n\backslash in\; N\}$ of Hilbert spaces. Von Neumann's traditional definition simply takes the "obvious" tensor product: to compute $\backslash bigotimes\_n\{H\_n\}$, first collect all simple tensors of the form $\backslash bigotimes\_\{n\backslash in\; N\}\{e\_n\}$ such that . 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 $\backslash bigotimes\_n\{H\_n\}$, first select a unit vector $v\_n\backslash in\; H\_n$ in each Hilbert space, and then collect all simple tensors of the form $\backslash bigotimes\_\{n\backslash in\; N\}\{e\_n\}$, in which only finitely-many $e\_n$ are not $v\_n$. Then take the $L^2$ completion of these simple tensors.Nik Weaver (8 March 2020). [https://mathoverflow.net/revisions/354457/4 Answer] to [https://mathoverflow.net/posts/354418/revisions Result of continuum tensor product of Hilbert spaces]. MathOverflow. StackExchange.Bratteli, O. and Robinson, D: Operator Algebras and Quantum Statistical Mechanics v.1, 2nd ed., page 144. Springer-Verlag, 2002.

Let $\backslash mathfrak\{A\}\_i$ be the von Neumann algebra of bounded operators on $H\_i$ for $i=1,2.$ 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 $A\_1\backslash otimes\; A\_2$ where $A\_i\; \backslash in\; \backslash mathfrak\{A\}\_i$ for $i\; =\; 1,\; 2.$ This is exactly equal to the von Neumann algebra of bounded operators of $H\_1\backslash otimes\; H\_2.$ 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. This is one advantage of the "algebraic" method in quantum statistical mechanics.

If $H\_1$ and $H\_2$ have orthonormal bases $\backslash left\backslash \{\backslash phi\_k\backslash right\backslash \}$ and $\backslash left\backslash \{\backslash psi\_l\backslash right\backslash \},$ respectively, then $\backslash left\backslash \{\backslash phi\_k\; \backslash otimes\; \backslash psi\_l\backslash right\backslash \}$ is an orthonormal basis for $H\_1\backslash otimes\; H\_2.$ 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 $X$ and $Y$, with measures $\backslash mu$ and $\backslash nu$ respectively, one may look at $L^2(X\; \backslash times\; Y),$ the space of functions on $X\; \backslash times\; Y$ that are square integrable with respect to the product measure $\backslash mu\backslash times\backslash nu.$ If $f$ is a square integrable function on $X,$ and $g$ is a square integrable function on $Y,$ then we can define a function $h$ on $X\backslash times\; Y$ by $h(x,\; y)\; =\; f(x)\; g(y).$ The definition of the product measure ensures that all functions of this form are square integrable, so this defines a bilinear mapping $L^2(X)\; \backslash times\; L^2(Y)\; \backslash to\; L^2(X\backslash times\; Y).$ Linear combinations of functions of the form $f(x)\; g(y)$ are also in $L^2(X\; \backslash times\; Y).$ It turns out that the set of linear combinations is in fact dense in $L^2(X\; \backslash times\; Y),$ if $L^2(X)$ and $L^2(Y)$ are separable.{{cite book|at=p. 100, ex. 3.|title=Elements of the theory of functions and functional analysis|volume=2: Measure, the Lebesgue integral, and Hilbert space|first1=A. N.|last1=Kolmogorov|author-link1=Andrei Kolmogorov|first2=S. V.|last2=Fomin|author-link2=Sergei Fomin|orig-date=1960|translator-first1=Hyman|translator-last1=Kamel|translator-first2=Horace|translator-last2=Komm|publisher=Graylock|location=Albany, NY|year=1961|lccn=57-4134}} This shows that $L^2(X)\; \backslash otimes\; L^2(Y)$ is isomorphic to $L^2(X\; \backslash times\; Y),$ 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 $L^2(X;H)$, denoting the space of square integrable functions $X\; \backslash to\; H,$ is isomorphic to $L^2(X)\; \backslash otimes\; H$ if this space is separable. The isomorphism maps $f(x)\; \backslash otimes\; \backslash phi\; \backslash in\; L^2(X)\backslash otimes\; H$ to $f(x)\; \backslash phi\; \backslash in\; L^2(X;H).$ We can combine this with the previous example and conclude that $L^2(X)\; \backslash otimes\; L^2(Y)$ and $L^2(X\; \backslash times\; Y)$ are both isomorphic to $L^2\backslash left(X;\; L^2(Y)\backslash right).$

Tensor products of Hilbert spaces arise often in quantum mechanics. If some particle is described by the Hilbert space $H\_1,$ and another particle is described by $H\_2,$ then the system consisting of both particles is described by the tensor product of $H\_1$ and $H\_2.$ For example, the state space of a quantum harmonic oscillator is $L^2(\backslash R),$ so the state space of two oscillators is $L^2(\backslash R)\; \backslash otimes\; L^2(\backslash R),$ which is isomorphic to $L^2\backslash left(\backslash R^2\backslash right).$ Therefore, the two-particle system is described by wave functions of the form $\backslash psi\backslash left(x\_1,\; x\_2\backslash right).$ A more intricate example is provided by the Fock spaces, which describe a variable number of particles.

{{reflist}}

• {{Cite book | last1=Kadison | first1=Richard V. |author1link = Richard Kadison| last2=Ringrose | first2=John R. |author2link = John Robert Ringrose| title=Fundamentals of the theory of operator algebras. Vol. I | publisher=American Mathematical Society | location=Providence, R.I. | series=Graduate Studies in Mathematics | isbn=978-0-8218-0819-1 | mr= 1468229 | year=1997 | volume=15}}.
• {{Cite book | last1=Weidmann | first1=Joachim | title=Linear operators in Hilbert spaces | publisher=Springer-Verlag | location=Berlin, New York | series=Graduate Texts in Mathematics | isbn=978-0-387-90427-6 |mr=566954 | year=1980 | volume=68}}.

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