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Arbitrarily late finite phase recurrences

Statement

Let $H = H^{*}$ with energy unit J act on a finite-dimensional complex Hilbert space, let $ℏ > 0$ have unit J s, and let $U (t) = e^{- i t H / ℏ}$ for time $t$ measured in s. For every time unit $t_{0} > 0$ measured in s, there is a sequence $t_{n} \to + \infty$ and phases $λ_{n} \in C$ , $∣ λ_{n} ∣ = 1$ , such that $∥ U (t_{n}) - λ_{n} I ∥ \to 0$ , where $∥ T ∥ : = \sup {∥ T v ∥ : ∥ v ∥ \leq 1}$ is the operator norm induced by the Hilbert-space norm. Consequently, for every density matrix $ρ$ and bounded observable $A$ , the expectation $Tr (U (t_{n}) ρ U (t_{n})^{*} A)$ tends to $Tr (ρ A)$ . If $H$ has only one distinct energy, exact recurrences occur at every time.

Facts & Assumptions

Given: A finite-dimensional complex Hilbert space, $H = H^{*}$ , $ℏ > 0$ , and a fixed $t_{0} > 0$ .

[F1]

Finite unitary dynamics and stationary Gibbs states: the distinct real eigenvalues of $H$ define the spectral unitary group $U (t) = e^{- i t H / ℏ}$ .

Proof

Given: The finite-dimensional spectrum and unitary group above.

1.1givenF1construct

If the Hilbert space is zero-dimensional, its unique endomorphism is the identity and $U (t) = I$ ; take $t_{n} = n t_{0}$ and $λ_{n} = 1$ , so every operator-norm error is zero. No density operator exists on that space, making the density consequence vacuous there. On a nonzero space, if $H$ has one distinct energy $E_{1}$ , then $U (t) = e^{- i t E_{1} / ℏ} I$ , which proves the exact-recurrence clause. Otherwise list its $r \geq 2$ distinct energies as $E_{1}, \dots, E_{r}$ and put $θ_{j} = (E_{j} - E_{1}) t_{0} / (2 π ℏ)$ for $2 \leq j \leq r$ . For each integer $M \geq 1$ , partition the torus $[0, 1)^{r - 1}$ into $M^{r - 1}$ half-open cubes of side $1 / M$ and consider the $M^{r - 1} + 1$ points $k θ m o d Z^{r - 1}$ for $0 \leq k \leq M^{r - 1}$ . Two lie in the same cube, so their nonzero integer difference $q_{M}$ satisfies $1 \leq q_{M} \leq M^{r - 1}$ and each $q_{M} θ_{j}$ is within $1 / M$ of an integer.

2.1step 1.1algebra

The eigenvalue phase ratios of $U (q_{M} t_{0})$ to its $E_{1}$ phase therefore differ from $1$ by at most $2 π / M$ , since $∣ e^{i x} - 1 ∣ \leq ∣ x ∣$ . The difference $U (q_{M} t_{0}) - λ_{M} I$ is diagonal in the energy basis, so its induced operator norm is the maximum modulus of its diagonal entries and is at most $2 π / M$ , for $λ_{M} = e^{- i q_{M} E_{1} t_{0} / ℏ}$ . If the positive integers $q_{M}$ are unbounded, choose a subsequence tending to infinity; its times and phase errors give the required sequence. If they are bounded, some fixed positive integer $q$ occurs for arbitrarily large $M$ ; then every $q θ_{j}$ is an integer, so $U (q t_{0}) = λ I$ exactly and its positive integer multiples give exact arbitrarily late recurrences.

3.1step 2.1algebra∎

For any density $ρ$ , the expectation difference is bounded by $∥ U (t_{n})^{*} A U (t_{n}) - A ∥ \leq 2 ∥ A ∥ ∥ U (t_{n}) - λ_{n} I ∥$ , because scalar phases cancel in conjugation. The bound tends to zero by step 2.1.

Depends on

Finite unitary dynamics and stationary Gibbs states

Arbitrarily late finite phase recurrences

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources