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intermediate · Programming · Quantum Phase Estimation

Eigenvalues and Phases

Before we can understand quantum phase estimation (QPE), we need a precise answer to a simpler question: what does a unitary operator do to its eigenvectors?

Eigenvalues of unitary operators

Recall that a vector u|u\rangle is an eigenvector of an operator UU with eigenvalue λ\lambda if

Uu=λu.U|u\rangle = \lambda\,|u\rangle.

For a general matrix the eigenvalue can be any complex number. But unitary operators obey UU=IU^\dagger U = I, which places a tight constraint on λ\lambda.

Claim: every eigenvalue of a unitary operator lies on the unit circle of the complex plane, i.e., λ=1|\lambda| = 1.

Proof: take the norm of both sides of Uu=λuU|u\rangle = \lambda|u\rangle:

Uu2=λ2u2.\|U|u\rangle\|^2 = |\lambda|^2\,\||u\rangle\|^2.

Because UU is unitary it preserves norms, so Uu=u\|U|u\rangle\| = \||u\rangle\|. Dividing both sides by u20\||u\rangle\|^2 \neq 0 gives λ2=1|\lambda|^2 = 1, hence λ=1|\lambda| = 1.

Any complex number with unit modulus can be written as eiϕe^{i\phi} for some real angle ϕ\phi. This angle is called the phase of the eigenvalue.

Writing the eigenvalue as a phase

Since every unitary eigenvalue satisfies λ=1|\lambda| = 1, we write

λ=e2πiφ,φ[0,1).\lambda = e^{2\pi i\,\varphi}, \qquad \varphi \in [0,\,1).

The factor of 2π2\pi is conventional: it makes φ\varphi a fraction of a full turn, so φ=0\varphi = 0 corresponds to λ=1\lambda = 1 and φ=12\varphi = \tfrac{1}{2} corresponds to λ=1\lambda = -1.

The eigenvector equation becomes

Uu=e2πiφu.U|u\rangle = e^{2\pi i\,\varphi}\,|u\rangle.

The number φ\varphi encodes everything about how UU acts on its eigenvector: it is a pure rotation of the global phase.

A concrete example: the TT gate

The TT gate (also called π/8\pi/8 gate) is defined by

T=(100eiπ/4).T = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\pi/4} \end{pmatrix}.

Its eigenvalues are the diagonal entries: λ0=1\lambda_0 = 1 (eigenvector 0|0\rangle) and λ1=eiπ/4\lambda_1 = e^{i\pi/4} (eigenvector 1|1\rangle).

For λ1=eiπ/4\lambda_1 = e^{i\pi/4} we identify

e2πiφ=eiπ/4    2πφ=π4    φ=18.e^{2\pi i\,\varphi} = e^{i\pi/4} \implies 2\pi\varphi = \frac{\pi}{4} \implies \varphi = \frac{1}{8}.

So the TT gate imparts a phase of φ=1/8\varphi = 1/8 (one-eighth of a full turn) to its 1|1\rangle eigenvector. QPE can recover this fraction 1/81/8 as an ordinary binary number.

Why this matters for algorithms

Many quantum algorithms reduce to estimating the phase φ\varphi of a known unitary applied to one of its eigenvectors. Examples include:

In each case the core question is the same: given UU and u|u\rangle, find φ\varphi such that Uu=e2πiφuU|u\rangle = e^{2\pi i\varphi}|u\rangle. The next lessons build the circuit that answers it.

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