Spectral decomposition · Quantum Mechanics

Intuition

A Hermitian operator does nothing but multiply each of its eigen-directions by its eigenvalue. So it can be written as a sum: each eigenvalue times the projector onto its eigenvectors. Written this way, any function of the operator is the same sum with the function applied to the eigenvalues — squares, square roots and exponentials of operators become as easy as those of numbers.

A mixing desk has one slider per channel. Once the sound is split into channels, turning a slider acts on one channel and leaves the others alone. The spectral decomposition splits a state into eigen-channels and turns each by its eigenvalue.

Writing an operator through its eigenvalues

Every Hermitian operator on a finite-dimensional space has an orthonormal basis of eigenvectors $\ket{a_{n}}$ with real eigenvalues $a_{n}$ , and is then the weighted sum of their projectors. In that basis its matrix is diagonal. For a function $f$ defined on the eigenvalues, $f(\hat{A})$ is defined by applying $f$ to each eigenvalue.

\begin{aligned}\hat{A}&=\sum_{n}a_{n}\ket{a_{n}}\bra{a_{n}},\\ f(\hat{A})&=\sum_{n}f(a_{n})\ket{a_{n}}\bra{a_{n}}\end{aligned}

Using it

A degenerate eigenvalue collects its eigenvectors into one projector: $\hat{A}=\sum_{a}a\,\hat{P}_{a}$ , where projects onto all eigenvectors with eigenvalue .

Powers of an operator act on its eigenvalues

\hat{A}=\sum_{n}a_{n}\hat{P}_{n}\ \Longrightarrow\ \hat{A}^{k}=\sum_{n}a_{n}^{k}\hat{P}_{n}

Multiply the decomposition by itself. Products of different projectors vanish and each projector squares to itself, so the square keeps one term per eigenvalue with the eigenvalue squared. Induction gives every power, and sums of powers give every polynomial and power series.

Proof steps

\hat{A}^{2}=\Bigl(\sum_{m}a_{m}\hat{P}_{m}\Bigr)\Bigl(\sum_{n}a_{n}\hat{P}_{n}\Bigr)=\sum_{m,n}a_{m}a_{n}\hat{P}_{m}\hat{P}_{n}

Multiply out the decomposition.

\hat{P}_{m}\hat{P}_{n}=\delta_{mn}\hat{P}_{n}

Projectors onto orthogonal eigenvectors multiply to zero, and each squares to itself.

\hat{A}^{2}=\sum_{n}a_{n}^{2}\hat{P}_{n}

Only the terms with $m=n$ survive.

\hat{A}^{k+1}=\hat{A}^{k}\hat{A}=\sum_{n}a_{n}^{k}a_{n}\hat{P}_{n}

The same collapse turns each extra factor into one more power of the eigenvalue.

Applications

Practice

An Operator From Its Eigenvalues

Knowing the eigenvalues and the eigenvectors of a Hermitian operator is knowing the operator: it is the sum of each eigenvalue times the projector onto its eigenvector.

\hat{A}=\sum_{n}a_{n}\ket{a_{n}}\bra{a_{n}}

Try it

An operator has eigenvalue $+1$ on $\ket{+}$ and $-1$ on $\ket{-}$ , where $\ket{\pm}=\tfrac{1}{\sqrt2}(\ket{0}\pm\ket{1})$ . Which operator is it?