Outer Products, Projection Operators, and Completeness

An outer product $|u\rangle\langle v|$ is an operator that acts on a ket $|\psi\rangle$ by first taking the inner product with $|v\rangle$ and then producing the ket $|u\rangle$ scaled by that number: $(|u\rangle\langle v|)|\psi\rangle = |u\rangle(\langle v|\psi\rangle)$. It maps the whole space onto the one-dimensional subspace spanned by $|u\rangle$.

In a fixed orthonormal basis, the outer product $|j\rangle\langle k|$ is represented by a matrix with a single 1 in the (j,k) position and 0 elsewhere. The set of all such matrices forms a basis for the space of operators on an n-dimensional Hilbert space, which therefore has dimension $n^2$.

Operators can be expanded in this basis. If $A$ is any linear operator, its matrix elements are $A_{jk} = \langle j|A|k\rangle$, and we may write $A = \sum_{j,k} A_{jk} |j\rangle\langle k|$. This expansion shows that every operator is a linear combination of outer products.

The composition of outer products follows the rule $(|u\rangle\langle v|)(|w\rangle\langle x|) = |u\rangle \langle v|w\rangle \langle x| = \langle v|w\rangle |u\rangle\langle x|$. The inner product $\langle v|w\rangle$ is just a scalar, so the result is another outer product.

A projection operator onto a normalized state $|\phi\rangle$ is $P_\phi = |\phi\rangle\langle\phi|$. Applied to any state $|\psi\rangle$, it produces the component of $|\psi\rangle$ along $|\phi\rangle$: $P_\phi|\psi\rangle = |\phi\rangle\langle\phi|\psi\rangle$. The resulting ket is parallel to $|\phi\rangle.$

Geometrically, $P_\phi$ projects onto the line spanned by $|\phi\rangle$. It is the operator analog of casting a shadow. If $|\psi\rangle$ is already proportional to $|\phi\rangle$, the projection leaves it unchanged; if $|\psi\rangle$ is orthogonal to $|\phi\rangle$, the projection gives the zero ket.

The expectation value $\langle\psi|P_\phi|\psi\rangle$ equals $|\langle\phi|\psi\rangle|^2$, which is precisely the probability of finding the system in state $|\phi\rangle$ when it is prepared in $|\psi\rangle$. Thus projection operators encode both geometry and probability.

Projection operators onto higher-dimensional subspaces are sums of one-dimensional projectors. If $\{|\phi_1\rangle, \dots, |\phi_m\rangle\}$ is an orthonormal set spanning a subspace $W$, then $P_W = \sum_{i=1}^m |\phi_i\rangle\langle\phi_i|$ projects onto $W$.

A projection operator $P$ satisfies two defining properties. First, it is idempotent: $P^2 = P$. Applying the projection twice has the same effect as applying it once, because the output already lies in the target subspace. Second, it is Hermitian: $P^\dagger = P$.

Idempotence follows directly from the definition $P_\phi^2 = (|\phi\rangle\langle\phi|)(|\phi\rangle\langle\phi|) = |\phi\rangle\langle\phi|\phi\rangle\langle\phi| = |\phi\rangle\langle\phi| = P_\phi$, using $\langle\phi|\phi\rangle = 1$. Hermiticity follows because $(|\phi\rangle\langle\phi|)^\dagger = |\phi\rangle\langle\phi|$.

Conversely, any operator that is both Hermitian and idempotent is a projector onto some subspace. The subspace is its range, and the orthogonal complement is its kernel. This characterization is often the easiest way to verify that a given operator is a projector.

These two properties ensure that the eigenvalues of a projector are only 0 and 1. Eigenvectors in the target subspace have eigenvalue 1, while states orthogonal to the subspace have eigenvalue 0. This binary spectrum matches the yes-no nature of a projection measurement.

For a Hermitian operator $A$ with distinct eigenvalues $\lambda_i$ and orthonormal eigenvectors $|\lambda_i\rangle$, the spectral projector onto the eigenspace for $\lambda_i$ is $P_i = |\lambda_i\rangle\langle\lambda_i|$ when the eigenvalue is nondegenerate. If $\lambda_i$ is degenerate, $P_i$ is the sum of projectors over a basis of that eigenspace.

Spectral projectors are mutually orthogonal: $P_i P_j = 0$ for $i \neq j$, because eigenvectors belonging to distinct eigenvalues of a Hermitian operator are orthogonal. They are also complete: $\sum_i P_i = I$, where $I$ is the identity operator.

Using spectral projectors, any normal operator can be written $A = \sum_i \lambda_i P_i$. This is the finite-dimensional version of the spectral theorem. It expresses $A$ as a weighted sum of orthogonal projections, with the eigenvalues as weights.

In quantum measurement, $P_i$ is the projector associated with outcome $\lambda_i$. After measuring $\lambda_i$, the state collapses to $P_i|\psi\rangle / \sqrt{\langle\psi|P_i|\psi\rangle}$. Spectral projectors therefore connect the abstract spectral theorem to the measurement postulate.

If $\{|e_i\rangle\}$ is an orthonormal basis for a finite-dimensional Hilbert space, the completeness relation states that $\sum_i |e_i\rangle\langle e_i| = I$. Acting on any ket $|\psi\rangle$, this gives $\sum_i |e_i\rangle\langle e_i|\psi\rangle = \sum_i \psi_i |e_i\rangle = |\psi\rangle$, recovering the basis expansion.

The identity operator $I$ leaves every ket unchanged. The completeness relation rewrites $I$ as a sum of rank-one projectors. This is powerful because it lets us insert the identity anywhere in a Dirac expression without changing the value, opening the door to basis changes.

In infinite-dimensional spaces, completeness becomes $\int |x\rangle\langle x|\,dx = I$ or a sum over a countable basis. The spectral theorem generalizes this idea to integrals over projection-valued measures. All forms share the same idea: the whole space is pieced together from orthogonal parts.

Completeness is the reason that knowing all amplitudes $\langle e_i|\psi\rangle$ is equivalent to knowing the state $|\psi\rangle$. It also underlies Parseval's identity, which states that the squared norm of a ket equals the sum of the squared moduli of its expansion coefficients.

To compute the matrix elements of an operator $A$ in a new basis $\{|f_j\rangle\}$, insert two copies of the identity resolved in the old basis $\{|e_i\rangle\}$: $\langle f_j|A|f_k\rangle = \sum_{m,n} \langle f_j|e_m\rangle \langle e_m|A|e_n\rangle \langle e_n|f_k\rangle$. The numbers $U_{jm} = \langle f_j|e_m\rangle$ form a unitary change-of-basis matrix.

The unitarity of $U$ follows from orthonormality of both bases: $\sum_m U_{jm}^* U_{km} = \delta_{jk}$. Thus the new matrix $A^\prime = U A U^\dagger$ is related to the old matrix by a unitary similarity transformation. This preserves eigenvalues, trace, and determinant.

Inserting the identity is also the key to evaluating overlap integrals and transition amplitudes. For example, $\langle\phi|A|\psi\rangle$ can be expanded in any convenient basis, often turning an abstract expression into a sum or integral that can be computed explicitly.

A common quantum example is changing from the computational basis to the Hadamard basis. Writing $|0\rangle = (|+\rangle + |-\rangle)/\sqrt{2}$ and $|1\rangle = (|+\rangle - |-\rangle)/\sqrt{2}$ is equivalent to inserting the identity $\sum_{s=\pm} |s\rangle\langle s|$ into each ket.

Let $\{|0\rangle, |1\rangle\}$ be an orthonormal basis and let $|\psi\rangle = \frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle$. Compute $P_0|\psi\rangle$, $P_1|\psi\rangle$, and verify that $P_0 + P_1 = I$.

Step 1 -- Project onto $|0\rangle$. $P_0 = |0\rangle\langle 0|$. Applying it gives $P_0|\psi\rangle = |0\rangle\langle 0|\psi\rangle = \frac{1}{\sqrt{3}}|0\rangle$.

Step 2 -- Project onto $|1\rangle$. $P_1 = |1\rangle\langle 1|$. Applying it gives $P_1|\psi\rangle = \sqrt{\frac{2}{3}}|1\rangle$.

Step 3 -- Sum. Adding the two projected components gives $\frac{1}{\sqrt{3}}|0\rangle + \sqrt{\frac{2}{3}}|1\rangle = |\psi\rangle$. Therefore $(P_0+P_1)|\psi\rangle = |\psi\rangle$, confirming that $P_0+P_1$ acts as the identity on this state and hence on the whole space.