The Exercise

An operator $A$ on a finite-dimensional complex Hilbert space is normal when it commutes with its adjoint, $AA^\dagger = A^\dagger A$ . The exercise asks for the polar decomposition of a normal operator in the outer product representation. The polar decomposition writes $A$ as a unitary $U$ times a positive operator $J$ , and the exercise wants both factors as sums of outer products $|i\rangle\langle i|$ over an orthonormal basis $\{|i\rangle\}$ in which $A$ itself reads $\sum_i \lambda_i |i\rangle\langle i|$ .

Diagonal Operators

Let $E$ be a Hilbert space of finite dimension $d$ , over the complex or the real numbers. The diagonal operator $D$ with entries $w_i$ in an orthonormal basis $\{|i\rangle\}$ of $E$ is the operator on $E$ that sums the projectors onto the basis vectors, each weighted by its entry: $D = \sum_i w_i\, |i\rangle\langle i| .$ (1)^✦

The phase $\operatorname{ph}(z)$ of a number $z$ is $z$ divided by its modulus, and $1$ when $z = 0$ : $\operatorname{ph}(z) = \begin{cases} z/|z|, & z \neq 0, \\ 1, & z = 0 . \end{cases}$ (2)^✦

The right singular vectors $|v_0\rangle, \dots, |v_{d-1}\rangle$ of an operator $T$ on $E$ are an orthonormal basis of $E$ made of eigenvectors of the Hermitian operator $T^\dagger T$ .^✦ The absolute value $|T|$ of $T$ is the operator on $E$ diagonal in that basis, with the singular values $s_0, \dots, s_{d-1}$ of $T$ as its entries: $|T| = \sum_{i=0}^{d-1} s_i\, |v_i\rangle\langle v_i| .$ (3)^✦

The Claim

Theorem 1. Let $A$ be a normal operator on a complex Hilbert space $E$ of finite dimension $d$ . Then there are an orthonormal basis $|0\rangle, \dots, |d-1\rangle$ of $E$ and complex numbers $\lambda_0, \dots, \lambda_{d-1}$ with $A = \sum_{i=0}^{d-1} \lambda_i\, |i\rangle\langle i| .$ (4)^✦ Let now $A = \sum_i \lambda_i |i\rangle\langle i|$ be diagonal in an orthonormal basis $\{|i\rangle\}$ of a finite-dimensional real or complex Hilbert space, and let $J = \sum_i |\lambda_i|\, |i\rangle\langle i|$ and $U = \sum_i \operatorname{ph}(\lambda_i)\, |i\rangle\langle i|$ . Then $A$ factors in both orders, $A = JU ,$ (5)^✦ $A = UJ ,$ (6)^✦ and $J$ is the absolute value of $A$ : $J = |A| .$ (7)^✦

For a normal $A$ , the basis and the numbers $\lambda_i$ of (4) turn (5) and (6) into the decomposition the exercise asks for.

What Is Checked

The claim is stated over the diagonal operator (1) in an orthonormal basis, the phase (2) of a number, and the absolute value (3) of an operator.

For any orthonormal basis and any numbers $\lambda_i$ , the factor of moduli is positive and the factor of phases is unitary: $J \ge 0 ,$ (8)^✦ $U^\dagger U = U U^\dagger = I .$ (9)^✦