Tensor Products and the Composition Postulate

A tensor product combines the state spaces of two distinguishable subsystems. It contains independent preparations and their superpositions, including entangled states. Here we introduce the notation needed for the composition postulate; tensor products of Hilbert spaces develops the full construction and its infinite-dimensional extension.

Helpful background. State vectors explains superposition; composite systems identifies the physical subsystem split.

Combining Two State Spaces

For specified subsystems $A$ and $B$,

$$\mathcal H_{AB}=\mathcal H_A\otimes\mathcal H_B.$$

A product vector is written $|\psi\rangle_A\otimes|\phi\rangle_B$, or simply $|\psi\rangle_A|\phi\rangle_B$. The operation is linear in each factor:

$$(a|0\rangle+b|1\rangle)\otimes|\phi\rangle =a|0\rangle|\phi\rangle+b|1\rangle|\phi\rangle.$$

For finite dimensions $d_A,d_B$, the joint dimension is $d_A d_B$: each basis label of $A$ can be paired with every basis label of $B$. A qubit and a three-level system therefore have six joint basis vectors, $|00\rangle,|01\rangle,|02\rangle,|10\rangle,|11\rangle,|12\rangle$.

Reading Product Coordinates

With two qubits ordered as $|00\rangle,|01\rangle,|10\rangle,|11\rangle$,

$$\begin{pmatrix}a\\b\end{pmatrix} \otimes \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}ac\\ad\\bc\\bd\end{pmatrix}.$$

For example, $|+\rangle_A|1\rangle_B=(|01\rangle+|11\rangle)/\sqrt2$ has coordinates $(0,1,0,1)^T/\sqrt2$. Every vector in the joint space is a sum of product-basis vectors, but not every vector is one product. That distinction leads to entangled states.

Fix the factor and basis ordering before forming matrices. Reordering the factors is a coordinate identification using a swap map; physically applying a SWAP operation is a separate action on the state.

Quick Check

Write $|1\rangle_A|+\rangle_B$ in the same ordering. Is it the same state as $|+\rangle_A|1\rangle_B$?

Solution

Its coordinates are $(0,0,1,1)^T/\sqrt2$. The states differ: the first has $A$ certainly in $|1\rangle$, while the second has $B$ certainly in $|1\rangle$. The tensor-factor labels specify which subsystem has which preparation.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, ch. 2.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chs. 1–2. Author’s manuscript.