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

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

Product bases are the bookkeeping layer behind coordinate arrays, matrix representations, qubit registers, two-particle wavefunctions, oscillator products, and angular-momentum addition.

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 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

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

in the finite or countable orthonormal case.

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.

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

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 states are energy eigenstates. A coupling term can mix states with different pairs of occupation numbers.

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 Mistakes

• Reversing subsystem order while keeping the same coordinate column.
• Confusing compact labels such as $\lvert01\rangle$ and $\lvert10\rangle$.
• Treating the coefficient array $c_{ij}$ as basis-independent.
• Forgetting that continuous basis labels require integrals and distributional normalization.
• Assuming a product-basis expansion is itself evidence of a product state.
• Using a software tensor-order convention without matching it to the written basis order.

Cross-Links

• Tensor Products of Hilbert Spaces
• Operators on Composite Systems
• Local and Global Observables
• Tensor Product Ordering
• Direct Sums versus Tensor Products
• Tensor Product Exercises
• Notation and Subsystem Labels
• Mathematical Toolkit: Tensor Products
• Two Spin-1/2 Particles

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.

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. 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 $a_1^\dagger a_1\otimes I_2$?

Solution

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

$$(a_1^\dagger a_1\otimes I_2) \lvert n,m\rangle = n\lvert n,m\rangle.$$

The second occupation number $m$ is unchanged.