Product Bases

A product basis is the basis of a composite Hilbert space obtained by pairing basis vectors of each subsystem. If $\mathcal H_A$ has basis vectors labeled by $i$ and $\mathcal H_B$ has basis vectors labeled by $j$, then a product basis of $\mathcal H_A\otimes\mathcal H_B$ is

$$\lvert i\rangle_A\otimes\lvert j\rangle_B.$$

A product basis tells you exactly which local outcomes each joint coefficient refers to. It turns a state of two subsystems into a two-index array without assuming that the state itself factors.

Helpful background. Tensor Products of Hilbert Spaces explains bilinearity and the product inner product.

Basis of Subsystem A

Let subsystem $A$ have an orthonormal basis

$$\{\lvert i\rangle_A\}_{i\in I}.$$

In a finite-dimensional example, $I$ may be the set $\{0,1,\ldots,m-1\}$. For a harmonic oscillator, $I$ may be the nonnegative integers. For a continuous position basis, the label is a continuous variable and one must treat the basis in the generalized Dirac sense rather than as a countable Hilbert-space basis.

An arbitrary state of $A$ may be expanded as

$$\lvert\psi\rangle_A = \sum_i a_i\lvert i\rangle_A.$$

Basis of Subsystem B

Similarly, let subsystem $B$ have an orthonormal basis

$$\{\lvert j\rangle_B\}_{j\in J},$$

so that

$$\lvert\phi\rangle_B = \sum_j b_j\lvert j\rangle_B.$$

The labels $i$ and $j$ do different jobs: $i$ indexes the $A$ basis, while $j$ indexes the $B$ basis. Reusing one symbol for both factors is a common source of avoidable mistakes.

Product Basis of A and B

The product basis consists of all pairs

$$\lvert i\rangle_A\otimes\lvert j\rangle_B, \qquad i\in I,\ j\in J.$$

When the subsystem order has been declared, this may be abbreviated as

$$\lvert ij\rangle_{AB}.$$

Orthonormality follows from the tensor-product inner product:

$${}_{AB}\langle ij\vert kl\rangle_{AB} = {}_{A}\langle i\vert k\rangle_A\, {}_{B}\langle j\vert l\rangle_B = \delta_{ik}\delta_{jl}.$$

Completeness is expressed as

$$\sum_{i,j} \lvert ij\rangle\langle ij\rvert = I_{AB}$$

For countable bases, the partial sums converge strongly: they converge when applied to each Hilbert-space vector, not in operator norm. State expansions converge in the Hilbert norm.

Coordinate Representation of Composite States

Every vector in the composite Hilbert space can be expanded as

$$\lvert\Psi\rangle = \sum_{i,j} c_{ij}\, \lvert i\rangle_A\otimes\lvert j\rangle_B.$$

The coefficients $c_{ij}$ form a two-index array. Normalization is

$$\sum_{i,j}|c_{ij}|^2=1.$$

A product state has coefficients that factor:

$$c_{ij}=a_i b_j.$$

If no such factorization exists, the pure state is entangled across the $A|B$ split. For finite-dimensional bipartite pure states, this is equivalent to the coefficient matrix having rank greater than one.

A qubit and a three-level system

Take $A$ to have basis $|0\rangle,|1\rangle$ and $B$ to have basis $|0\rangle,|1\rangle,|2\rangle$. With the B label changing fastest, the six basis states are

$$|0,0\rangle,\ |0,1\rangle,\ |0,2\rangle,\ |1,0\rangle,\ |1,1\rangle,\ |1,2\rangle.$$

For example,

$$\begin{aligned} |\Psi\rangle &=\frac{|0,2\rangle+i|1,0\rangle}{\sqrt2},\\ [\Psi]_{AB} &=\frac1{\sqrt2}(0,0,1,i,0,0)^T. \end{aligned}$$

Measuring the product basis gives only the pairs $(0,2)$ and $(1,0)$, each with probability $1/2$. The six-dimensional column still remembers the relative phase $i$, which another measurement can detect. The coefficient matrix has two nonzero entries in different rows and columns, so it has rank two and the state is entangled.

Multi-Index Notation

For many subsystems,

$$\mathcal H = \mathcal H_1\otimes\mathcal H_2\otimes\cdots\otimes\mathcal H_N,$$

a product basis vector is written

$$\lvert i_1 i_2\cdots i_N\rangle.$$

This compact label hides an ordered tensor product:

$$\lvert i_1\rangle_1 \otimes \lvert i_2\rangle_2 \otimes\cdots\otimes \lvert i_N\rangle_N.$$

The order must be stated. Changing order changes the coordinate representation even when the abstract Hilbert spaces are isomorphic.

Qubit Basis Ordering

For two qubits ordered as $A,B$, this volume uses

$$\lvert 00\rangle,\quad \lvert 01\rangle,\quad \lvert 10\rangle,\quad \lvert 11\rangle.$$

Thus

$$\lvert\Psi\rangle = c_{00}\lvert00\rangle +c_{01}\lvert01\rangle +c_{10}\lvert10\rangle +c_{11}\lvert11\rangle$$

corresponds to the coordinate column

$$\begin{pmatrix} c_{00}\\ c_{01}\\ c_{10}\\ c_{11} \end{pmatrix}.$$

For three qubits ordered as $A,B,C$, the analogous order is

$$\lvert000\rangle,\ \lvert001\rangle,\ \lvert010\rangle,\ \lvert011\rangle, \ \lvert100\rangle,\ \lvert101\rangle,\ \lvert110\rangle,\ \lvert111\rangle.$$

Different software libraries may use different endian conventions. A calculation that moves between mathematical notation and code must state the convention explicitly.

Matrix Representation of Composite Operators

For bounded operators, or basis vectors in the domains of the unbounded operators involved, a local operator on $A$ has matrix elements

$$(O_A)_{ik} = {}_{A}\langle i\vert O_A\vert k\rangle_A.$$

On the composite product basis,

$${}_{AB}\langle ij\vert (O_A\otimes I_B) \vert kl\rangle_{AB} = (O_A)_{ik}\delta_{jl}.$$

For a product operator,

$${}_{AB}\langle ij\vert (A\otimes B) \vert kl\rangle_{AB} = A_{ik}B_{jl}.$$

These formulas are the index form of the tensor-product action. They are also the safest way to check whether a matrix implementation has the correct basis order.

Example: Two Harmonic Oscillators

If two distinguishable harmonic oscillators have number bases

$$\{\lvert n\rangle_1\}_{n=0}^{\infty}, \qquad \{\lvert m\rangle_2\}_{m=0}^{\infty},$$

then the product basis is

$$\lvert n,m\rangle = \lvert n\rangle_1\otimes\lvert m\rangle_2.$$

For uncoupled oscillators,

$$H_0 = \hbar\omega_1 \left(a_1^\dagger a_1+\frac12\right) +\hbar\omega_2 \left(a_2^\dagger a_2+\frac12\right),$$

the product number states are energy eigenstates. Here $a_1=a\otimes I$ and $a_2=I\otimes a$ already denote operators on the joint space. A coupling term can mix occupation pairs, although not every coupling changes this basis.

Example: Position and Spin

For a spin-$1/2$ particle,

$$\mathcal H = L^2(\mathbb R^3)\otimes\mathbb C^2.$$

A useful product basis is the generalized position-spin basis

$$\lvert\mathbf x\rangle\otimes\lvert\uparrow\rangle, \qquad \lvert\mathbf x\rangle\otimes\lvert\downarrow\rangle.$$

A state is then represented by two wavefunction components:

$$\Psi(\mathbf x) = \begin{pmatrix} \psi_\uparrow(\mathbf x)\\ \psi_\downarrow(\mathbf x) \end{pmatrix}.$$

This is a product-basis expansion, not a statement that the spin and position state must factor.

Continuous Bases

For two distinguishable spinless particles on a line, the position basis is formally

$$\lvert x_1\rangle\otimes\lvert x_2\rangle,$$

and the wavefunction is

$$\Psi(x_1,x_2) = \langle x_1,x_2\vert\Psi\rangle.$$

The resolution of identity is formal:

$$I = \int dx_1\,dx_2\, \lvert x_1,x_2\rangle \langle x_1,x_2\rvert.$$

As usual with position kets, this belongs to Dirac notation or a rigged-Hilbert-space treatment. For ordinary calculations, the resulting square-integrable wavefunctions live in $L^2(\mathbb R^2)$.

Common pitfalls

An expansion is not a factorization. Product-basis vectors span the whole joint space, including entangled states. Check whether $c_{ij}=a_i b_j$ for the whole coefficient array.

Continuous position labels are generalized. Position kets are not normalizable Hilbert vectors. Their expansions use integrals and distributional normalization rather than ordinary sums.

One ordering must serve states and operators. The ordering examples show how to permute both together.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, section 2.1.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapter 1. Author manuscript.

Exercises

1. In the default two-qubit basis order, what coordinate column represents the state

$$\frac{1}{\sqrt2} \bigl( \lvert01\rangle-\lvert10\rangle \bigr)?$$

Solution

The default order is

$$\lvert00\rangle,\quad \lvert01\rangle,\quad \lvert10\rangle,\quad \lvert11\rangle.$$

Therefore the coordinate column is

$$\begin{pmatrix} 0\\ 1/\sqrt2\\ -1/\sqrt2\\ 0 \end{pmatrix}.$$

2. For nonzero vectors $a,b$, show that the coefficients $c_{ij}=a_i b_j$ form a rank-one matrix.

Solution

The coefficient matrix is the outer product of the column vector $a$ and the row vector $b^T$:

$$C=ab^T.$$

Every column of $C$ is proportional to $a$, with proportionality coefficient $b_j$. Hence the column space is at most one-dimensional. If both factors are nonzero, the rank is one.

3. For a two-oscillator basis vector $\lvert n,m\rangle$, which oscillator is affected by $N\otimes I_2$, where $N=a^\dagger a$ is the single-oscillator number operator?

Solution

It acts on the first oscillator. On the basis vector,

$$(N\otimes I_2) \lvert n,m\rangle = n\lvert n,m\rangle.$$

The second occupation number $m$ is unchanged.