Tensor product of Hilbert spaces explained

In mathematics, and in particular functional analysis, the tensor product of Hilbert spaces is a way to extend the tensor product construction so that the result of taking a tensor product of two Hilbert spaces is another Hilbert space. Roughly speaking, the tensor product is the metric space completion of the ordinary tensor product. This is an example of a topological tensor product. The tensor product allows Hilbert spaces to be collected into a symmetric monoidal category.[1]

Definition

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

H1

and

H2

be two Hilbert spaces with inner products

\langle,\rangle1

and

\langle,\rangle2,

respectively. Construct the tensor product of

H1

and

H2

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\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 \phi_1,\psi_1 \in H_1 \mbox \phi_2,\psi_2 \in H_2and extending by linearity. That this inner product is the natural one is justified by the identification of scalar-valued bilinear maps on

H1 x H2

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

H1

and

H2.

Explicit construction

The tensor product can also be defined without appealing to the metric space completion. If

H1

and

H2

are two Hilbert spaces, one associates to every simple tensor product

x1x2

the rank one operator from
*
H
1
to

H2

that maps a given

x*\in

*
H
1
asx^* \mapsto x^*(x_1) \, x_2.

This extends to a linear identification between

H1H2

and the space of finite rank operators from
*
H
1
to

H2.

The finite rank operators are embedded in the Hilbert space
*,
HS(H
1

H2)

of Hilbert–Schmidt operators from
*
H
1
to

H2.

The scalar product in
*,
HS(H
1

H2)

is given by\langle T_1, T_2 \rangle = \sum_n \left \langle T_1 e_n^*, T_2 e_n^* \right \rangle,where
*\right)
\left(e
n
is an arbitrary orthonormal basis of
*.
H
1

Under the preceding identification, one can define the Hilbertian tensor product of

H1

and

H2,

that is isometrically and linearly isomorphic to
*,
HS(H
1

H2).

Universal property

The Hilbert tensor product

H1H2

is characterized by the following universal property :

A weakly Hilbert-Schmidt mapping

L:H1 x H2\toK

is defined as a bilinear map for which a real number

d

exists, such that \sum_^\infty \bigl| \left\langle L(e_i, f_j), u \right \rangle\bigr|^2 \leq d^2\,\|u\|^2 for all

u\inK

and one (hence all) orthonormal bases

e1,e2,\ldots

of

H1

and

f1,f2,\ldots

of

H2.

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.

Infinite tensor products

Two different definitions have historically been proposed for the tensor product of an arbitrary-sized collection \_ of Hilbert spaces. Von Neumann's traditional definition simply takes the "obvious" tensor product: to compute \bigotimes_n, first collect all simple tensors of the form \bigotimes_ such that \prod_<\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 \bigotimes_n, first select a unit vector v_n\in H_n in each Hilbert space, and then collect all simple tensors of the form \bigotimes_, in which only finitely-many e_n are not v_n. Then take the

L2

completion of these simple tensors.[2] [3]

Operator algebras

Let

ak{A}i

be the von Neumann algebra of bounded operators on

Hi

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

A1 ⊗ A2

where

Ai\inak{A}i

for

i=1,2.

This is exactly equal to the von Neumann algebra of bounded operators of

H1 ⊗ H2.

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.

Properties

If

H1

and

H2

have orthonormal bases

\left\{\phik\right\}

and

\left\{\psil\right\},

respectively, then

\left\{\phik\psil\right\}

is an orthonormal basis for

H1 ⊗ H2.

In particular, the Hilbert dimension of the tensor product is the product (as cardinal numbers) of the Hilbert dimensions.

Examples and applications

The following examples show how tensor products arise naturally.

Given two measure spaces

X

and

Y

, with measures

\mu

and

\nu

respectively, one may look at

L2(X x Y),

the space of functions on

X x Y

that are square integrable with respect to the product measure

\mu x \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 x 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

L2(X) x L2(Y)\toL2(X x Y).

Linear combinations of functions of the form

f(x)g(y)

are also in

L2(X x Y).

It turns out that the set of linear combinations is in fact dense in

L2(X x Y),

if

L2(X)

and

L2(Y)

are separable.[4] This shows that

L2(X)L2(Y)

is isomorphic to

L2(X x 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

L2(X;H)

, denoting the space of square integrable functions

X\toH,

is isomorphic to

L2(X)H

if this space is separable. The isomorphism maps

f(x)\phi\inL2(X)H

to

f(x)\phi\inL2(X;H)

We can combine this with the previous example and conclude that

L2(X)L2(Y)

and

L2(X x Y)

are both isomorphic to

L2\left(X;L2(Y)\right).

Tensor products of Hilbert spaces arise often in quantum mechanics. If some particle is described by the Hilbert space

H1,

and another particle is described by

H2,

then the system consisting of both particles is described by the tensor product of

H1

and

H2.

For example, the state space of a quantum harmonic oscillator is

L2(\R),

so the state space of two oscillators is

L2(\R)L2(\R),

which is isomorphic to

L2\left(\R2\right).

Therefore, the two-particle system is described by wave functions of the form

\psi\left(x1,x2\right).

A more intricate example is provided by the Fock spaces, which describe a variable number of particles.

Bibliography

Notes and References

  1. [Bob Coecke|B. Coecke]
  2. Nik Weaver (8 March 2020). Answer to Result of continuum tensor product of Hilbert spaces. MathOverflow. StackExchange.
  3. Bratteli, O. and Robinson, D: Operator Algebras and Quantum Statistical Mechanics v.1, 2nd ed., page 144. Springer-Verlag, 2002.
  4. Book: p. 100, ex. 3.. Elements of the theory of functions and functional analysis. 2: Measure, the Lebesgue integral, and Hilbert space. A. N.. Kolmogorov. Andrei Kolmogorov. S. V.. Fomin. Sergei Fomin. 1960. Hyman. Kamel. Horace. Komm. Graylock. Albany, NY. 1961. 57-4134.