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Unitarity and differentiation of the finite Hermitian matrix exponential

Statement

Let $d \geq 1$ , let $A = A^{*}$ be an endomorphism of $C^{d}$ with its standard complex inner product, and let $ℏ > 0$ . For real $t$ , define $E_{A} (t) : = e^{- i t A / ℏ}$ by the finite-dimensional matrix exponential defined on this page. Then $E_{A} (t)$ is unitary, $E_{A} (t + s) = E_{A} (t) E_{A} (s)$ for all real $s, t$ , and its operator-norm derivative is $\frac{d}{d t} E_{A} (t) = - \frac{i}{ℏ} A E_{A} (t) = - \frac{i}{ℏ} E_{A} (t) A .$

Statement references

Finite-dimensional matrix exponential

Facts & Assumptions

Given: $d \geq 1$ , $A = A^{*}$ on $C^{d}$ , $ℏ > 0$ , and real $s, t$ .

[F1]

Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space makes $A$ normal; Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely supplies an orthonormal eigenbasis. If $A v = λ v$ with $v \neq 0$ , self-adjointness gives $λ ∥ v ∥^{2} = ⟨ A v, v ⟩ = ⟨ v, A v ⟩ = \overline{λ} ∥ v ∥^{2}$ , so each eigenvalue $λ$ is real.

[F2]

Finite-dimensional matrix exponential defines $E_{A} (t)$ by an absolutely convergent matrix series. The scalar exponential satisfies $\exp (z + w) = \exp (z) \exp (w)$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential), $\exp^{'} (z) = \exp (z)$ (The complex exponential is entire and its complex derivative is itself), the real-variable chain rule applies (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ ), and $∣ \exp (i y) ∣ = 1$ for real $y$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

Proof

Given: [F1], [F2], and the data in the Statement.

1.1F1F2given

Choose an orthonormal eigenbasis $v_{1}, \dots, v_{d}$ of $A$ from [F1], with real eigenvalues $λ_{1}, \dots, λ_{d}$ . Applying the defining power series term-by-term to $v_{j}$ gives $E_{A} (t) v_{j} = \exp (- i t λ_{j} / ℏ) v_{j}$ . Since each scalar has modulus one by [F2], $E_{A} (t)$ maps an orthonormal basis to an orthonormal basis and is therefore unitary.

2.1F1F2step 1.1

On each $v_{j}$ , [F2] gives $E_{A} (t) E_{A} (s) v_{j} = \exp (- i (t + s) λ_{j} / ℏ) v_{j} = E_{A} (t + s) v_{j}$ . Equality on the basis proves $E_{A} (t) E_{A} (s) = E_{A} (t + s)$ .

3.1F1F2algebra∎

The same finite eigenbasis expression is a finite sum of scalar exponential functions. Differentiating each summand by [F2] and the chain rule gives $E_{A}^{'} (t) v_{j} = (- i λ_{j} / ℏ) E_{A} (t) v_{j}$ . Because the basis is finite, these scalar derivatives give the operator-norm derivative, and the equality on every $v_{j}$ yields $E_{A}^{'} (t) = (- i / ℏ) A E_{A} (t) = (- i / ℏ) E_{A} (t) A$ .

Unitarity and differentiation of the finite Hermitian matrix exponential

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources