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 \(\CC^2\otimes\CC^2\).