# 9. Tensor products#

In the mathematical formalism of quantum mechanics, the state of a
system is a (unit) vector in a Hilbert space, as mentioned in
Hilbert spaces and operators. Likewise, a *composite* of two
quantum systems can be described by the *tensor product* of the
corresponding Hilbert spaces.

## 9.1. Definition#

In this section, \(\HH_1\) and \(\HH_2\) are two Hilbert spaces.

(Tensor product)

The *tensor product* of \(\HH_1\) and \(\HH_2\) is a Hilbert space
\(\HH_1\otimes\HH_2\). This space contains vectors of the form

where \(\ket\alpha\) is in \(\HH_1\) and \(\ket\beta\) is in \(\HH_2\). These
vectors are called *pure tensors*. Importantly, \(\HH_1\otimes\HH_2\)
also contains linear combinations of such pure tensors (possibly with
infinitely many terms); these are called (non-pure) tensors.

The inner product between two pure tensors \(\ket\alpha\ket\beta\) and \(\ket{\alpha'}\ket{\beta'}\) is by definition \(\braket{\alpha|\alpha'}\braket{\beta|\beta'}\). (Note that the first inner product is taken in \(\HH_1\) and the second in \(\HH_2\)!)

Take \(\HH_1=\HH_2=\CC^2\) and denote the standard basis by \(\ket0,\ket1\). (This can be viewed as the state space of a single qubit.) There are four possibilities to pick one basis vector in \(\HH_1\) and one in \(\HH_2\), giving rise to four pure tensors

in \(\HH_1\otimes\HH_2\), where \(\ket{ij}\) is shorthand for \(\ket i\ket j\).

An example of a non-pure tensor in \(\HH_1\otimes\HH_2\) is

In Exercise 9.1, you will show that it is indeed impossible to write this as a pure tensor.

Warning

The tensor product behaves very differently from the ‘normal’ product
(or *direct sum*) of two vector spaces. For example, if \(\HH_1=\CC^m\)
and \(\HH_2=\CC^n\), then the direct sum of \(\HH_1\) and \(\HH_2\) has
dimension \(m+n\), while the tensor product has dimension \(mn\).

Note

In quantum mechanics, a non-pure tensor in \(\HH_1\otimes\HH_2\)
represents an *entangled state* of a composite quantum system.
In Example 9.1, the composite state consists of two qubits.
The non-pure state (9.2) is known as a *Bell state*.

## 9.2. Orthonormal bases#

We can easily write down a basis for a tensor product \(\HH_1\otimes\HH_2\) using bases of \(\HH_1\) and \(\HH_2\). It turns out that the ‘obvious’ definition works well with inner products.

(Orthonormal basis of a tensor product)

If \((\ket{e_i})_i\) is an orthonormal basis of \(\HH_1\) and \((\ket{f_j})_j\) is an orthonormal basis of \(\HH_2\), then \((\ket{e_i}\ket{f_j})_{i,j}\) is an orthonormal basis of \(\HH_1\otimes\HH_2\).

Take \(\HH=\CC^2\) with basis \((\ket0,\ket1)\) and equipped with the standard inner product. Then the basis (9.1) is an (orthonormal) basis for \(\HH\otimes\HH\).

More generally, an orthonormal basis for the \(n\)-fold tensor product \(\HH\otimes\cdots\otimes\HH\) consists of all \(n\)-bit strings.

## 9.3. Operators on a tensor product#

If \(A\) is an operator (i.e. a linear map) on \(\HH_1\) and \(B\) is a linear map on \(\HH_2\), then there is a linear map \(A\otimes B\) on \(\HH_1\otimes\HH_2\). This is defined on pure tensors by

However, just like not every tensor is a pure tensor, there are many operators on \(\HH_1\otimes\HH_2\) that cannot be written in the form \(T\otimes U\) (where \(T\) and \(U\) are operators on \(\HH_1\) and \(\HH_2\), respectively).

In quantum computing, the CNOT gate acts as follows on the standard basis of a two-qubit system:

## 9.4. Exercises#

Show that the tensor (9.2) in \(\CC^2\otimes\CC^2\) cannot be written as a pure tensor \(\ket\alpha\ket\beta\).

More generally, can you find a criterion for when a state

\[ a_{00}\ket{00}+a_{01}\ket{01}+a_{10}\ket{10}+a_{11}\ket{11} \]can be written as a pure tensor?

Show that the tensor (9.2) has norm 1.

Consider two operators \(A\) and \(B\) on \(\CC^2\) given in matrix form as

Write down the matrix of the operator \(A\otimes B\) on the standard basis (9.1) of \(\CC^2\otimes\CC^2\).

Write down the matrix of the CNOT operator (see Example 9.3) on the standard basis (9.1) of \(\CC^2\otimes\CC^2\).