13.6.2: Tensor Product—Composite Systems

The notation we have introduced so far deals with single qubit systems. It is nonetheless desirable to have a notation that allows us to describe composite systems, that is systems of multiple qubits. A similar situation arises in set theory when we have two sets \(A\) and \(B\) and we want to consider them as a whole. In set theory to form the ensemble set we use the cartesian product \(A×B\) to represent the ensemble of the two sets. The analogue in linear algebra is called tensor product and is represented by the symbol \(\otimes\). It applies equally to vectors and matrices (i.e. kets and operators).

From a practical standpoint, the tensor product concatenates physical systems. For example, a two particle system would be represented as \(|\; particle \;1\rangle \otimes |\; particle \;2\rangle\), and the charge and the spin of a particle would be represented also by the tensor product \(|\; charge\rangle \otimes |\; spin\rangle\). If we have two qubits \(|\; \psi \rangle\) and \(|\; \phi \rangle\), the system composed by these two qubits is represented by \(|\; \psi \rangle \otimes |\; \phi \rangle\).

Athough they share a similar goal, cartesian and tensor product differ in the way elements of the ensemble are built out of the parts. Cartesian product produces tuples. So if \(A\) and \(B\) are two sets of numbers, an element of their cartesian product \(A × B\) is a pair of numbers \((a, b)\) such that \(a\) belongs to \(A\) and \(b\) belongs to \(B\). It is a simple concatenation.

The elements of a tensor product are obtained from the constituent parts in a slightly different way. For example, consider two \(2×2 matrices\)

\[\mathbb{A}=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \quad \mathbb{B}=\left(\begin{array}{ll}
\alpha & \beta \\
\gamma & \delta
\end{array}\right) \tag{13.26} \]

the tensor product yields a \(4×4 matrix\)

\[\mathbb{A} \otimes \mathbb{B}=\left(\begin{array}{cccc}
a \alpha & a \beta & b \alpha & b \beta \\
a \gamma & a \delta & b \gamma & b \delta \\
c \alpha & c \beta & d \alpha & d \beta \\
c \gamma & c \delta & d \gamma & d \delta
\end{array}\right) \tag{13.27} \]

Although it may not be obvious at first sight, this way of constructing the tensor product is consistent with the way we do matrix multiplication. As an operation, it has a very interesting feature, it outputs 4×4 matrices out of 2×2 matrices, but not all 4×4 matrices can be generated in this way (for the mathematically inclined reader, the tensor product operation is not surjective, or “onto”), it is this feature of the tensor product that will motivate the discussion about entanglement, probably the most peculiar feature of quantum mechanics.

You should take some time to get used to the tensor product, and ensure you do not get confused with all the different products we have introduced in in the last two sections.

1. the dot product (\(\langle k | q\rangle\)) yields a complex number;
2. the exterior product (\(|\; k\rangle \langle q \;|\)) yields a square matrix of the same dimension that the Ket;
3. The tensor product (\(|\; k\rangle \otimes |\; q\rangle\) ) is used to examine composite systems. It yields a vector (or matrix) of a dimension equal to the sum of the dimensions of the two kets (or matrices) in the product.

Tensor Product in bracket notation

As we mentioned earlier, the tensor product of two qubits \(|\; q_1\rangle\) and \(|\; q_2\rangle\) is represented as \(|\; q_1\rangle \otimes |\; q_1\rangle\). Sometimes notation is abridged and the following four representations of the tensor product are made equivalent

\[\left|\;q_{1}\right\rangle \otimes\left|\;q_{2}\right\rangle \equiv\left|\;q_{1}\right\rangle\left|\;q_{2}\right\rangle \equiv\left|\;q_{1}, q_{2}\right\rangle \equiv\left|\;q_{1} q_{2}\right\rangle \tag{13.28} \]

For n qubits, it is frequent to abbreviate notation giving to each qubit \(|\; q\rangle\) an index:

\[\left|\;q_{1}\right\rangle \otimes\left|\;q_{2}\right\rangle \otimes \ldots \otimes\left|\;q_{n}\right\rangle=\bigotimes_{j=1}^{n}\left|\;q_{j}\right\rangle \tag{13.29} \]

The dual of a tensor product of kets is the tensor product of the corresponding bras. This implies that in the abridged notations, the complex conjugate operation turns kets into bras, but the labels retain their order

\begin{align*}
\left(\left|\;q_{1} q_{2}\right\rangle\right)^{\dagger} &=\left(\left|\;q_{1}\right\rangle \otimes\left|\;q_{2}\right\rangle\right)^{\dagger} \\
&\left.=\left\langle q_{1}\;\right| \otimes \left\langle q_{2}\;\right|\right) \\
&\left.=\left\langle q_{1} q_{2}\;\right|\right). \tag{13.30}
\end{align*}

As a consequence, the result of the dot product of two composite systems is the multiplication of the individual dot products taken in order

\[\begin{align*}
\left\langle q_{1} q_{2}\ \mid w_{1} w_{2}\right\rangle &=\left(\left\langle q_{1}\;\left|\;\otimes\left\langle q_{2}\;\right|\right)\left(\left|\;w_{1}\right\rangle \otimes\left|\;w_{2}\right\rangle\right)\right.\right.\\
&=\left\langle q_{1} \mid w_{1}\right\rangle \otimes\left\langle q_{2} \mid w_{1}\right\rangle \tag{13.31}
\end{align*} \nonumber \]

confusion often arises in the second term, where in the absence of the parenthesis it is easy to get confused by \(\langle q_2 \;||\; w_1\rangle\) and interpret it as a \(\langle|\rangle\) (note the two vertical separators in the correct form), and then try to to take the dot products inside out, instead of taking them in parallel as it should be done.