What to think of the convergence of the logarithmic integral and its repercussions in QM?

Question

Suppose we prepare a certain quantum system with Hilbert space ${\cal H}$, a self-adjoint Hamiltonian operator $H:{\cal H} \to {\cal H}$ whose spectrum is bounded below by $E_0\in \mathbb{R}$ (i.e. the energy is bounded below), and with initial unit-norm state vector $\psi_0\in {\cal H}$. The time-evolved state is $\psi(t):=e^{-iHt}\psi_0$.

A consequence of the spectral theorem for unbounded operators is that the probability amplitude $t \mapsto C(t):=\langle\psi(t),\psi_0\rangle$ to find the system in the original state is equal to a (Lebesgue-Stieltjes) integral of the following form $$C(t)=\int_{E_0}^\infty dE\,\omega(E)\,e^{-iEt}$$ where the (generalized, non-negative) function $\omega$ obeys $\int_{E_0}^\infty dE\,\omega(E)=\|\psi_0\|^2=1$. The fact that $C(t)$ is equal to an integral of this form implies via "a certain" Paley-Wiener theorem that $$\int_{-\infty}^\infty dt \frac{|\log(|C(t)|)|}{1+t^2}<+\infty$$ and this in turn rules out that $C(t)=O(\exp(-\Gamma t))$ as $t\to \pm\infty$: QM apparently can never accomodate the radioactive decay law, at least not 'all the way'.

That being said, "I don't know in which chapter of theoretical physics this result is at home". I lack intuition why exactly this result comes about. It doesn't immediately look like a quantum counterpart of the Poincaré recurrence theorem in classical mechanics, since compactness or finite volume of the accessible phase space is not required for this quantum result (1)(2). My question asks to provide that context and clarification.

(1) In quantum mechanics, I use "compact phase space" as slang to say that the Hamiltonian has a compact resolvent.

(2) Consider e.g. a free non-relativistic particle in 3D-space whose initial state is a Gaussian, $\psi_0(\mathbf{x})=\left(\frac{1}{(\pi)^{1/2}\sigma}\right)^{3/2}\exp\left(-\frac{x^2}{2\sigma^2}\right)$. Its time-evolved state is $\psi(\mathbf{x},t)=\left(\frac{\sigma}{(\pi)^{1/2}(\sigma^2+it)}\right)^{3/2}\exp\left(-\frac{x^2}{2(\sigma^2+it)}\right)$ and a calculation yields (something like) $C(t)=\left(\frac{2\sigma^2}{2\sigma^2+it}\right)^{3/2}$, which involves a much-slower-than-exponential decay in time.

N.B. I learned of this result through Mohsen Razavy's book titled "quantum theory of tunneling".

Answer 1

The exponential decay law is based on the assumptions that unstable particles decay randomly and that there are no processes which re-populate the unstable state. However, the latter assumption is not consistent with quantum mechanics.

Ultimately this comes down to the fact that if you want your Hamiltonian to be self-adjoint, then every allowed process needs to be accompanied by an "inverse process" of some kind. If you couple a hydrogen atom to a radiation field, then there has to be both emission and absorption. In this case, whatever mechanism you use to model the decay of an unstable parent into stable daughters must be accompanied by a mechanism for the stable daughters to recombine into the unstable parent. At long times, this inverse process will become non-neglible, and the decay will proceed at a slower-than-exponential rate.

There is a good discussion of non-exponential decay in this review article from 1978 (warning: paywall). If you can access that article, it discusses this precise issue in detail from both physical and mathematical standpoints.

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From the above-referenced article:

Let the system be in an initial (unstable) state $\psi_0$. We then have $$\psi(t) = e^{-iHt}\psi_0 = A(t)\psi_0 + \phi(t)$$ where $\langle \psi_0,\phi(t)\rangle = 0$. Applying $e^{-iH t'}$ to both sides gives $$e^{-iH(t+t')}\psi_0= A(t)e^{-iHt'}\psi_0 + e^{-iHt'}\phi(t)= A(t)A(t')\psi_0+ A(t)\phi(t')+e^{-iHt'}\phi(t)$$ Taking the inner product with $\psi_0$ kills the second term and yields

$$A(t+t') = A(t)A(t') + \langle \psi_0,e^{-iHt'}\phi(t)\rangle$$

Combined with the fact that $e^{-iHt}$ is taken to be unitary, this means that $A(t) \propto e^{-\alpha t}, \alpha\geq 0$ if and only that last term vanishes. But that last term describes the "re-scattering amplitude" - the amplitude to evolve back to $\psi_0$ from $\phi(t)$.

More rigorously, if $e^{-iHt}$ is taken to be unitary and the spectrum of $H$ is not $\mathbb R$, then this is not possible unless for all $\psi_\perp$ orthogonal to $\psi_0$, $\langle \psi_\perp,e^{-iHt}\psi_0\rangle=0$ as well - that is, the decay process is also forbidden. Though I am not familiar with the proof, it can be found e.g. here.

Answer 2

Probably there is indeed not so much to be surprised about. My first observation is that hoping to derive the radioactive decay law rests on a hope of finding a coarse-grained description of the quantum system with the former obeying the Markov property (no memory): it is a lack of memory in the first place that underpins an exponential decay process.

Consider then a Markov process that is governed by a state-dependent energy function $x \mapsto V(x)$ (i.e. the process is a gradient flow relative to the free energy $F:=\langle V\rangle -\beta^{-1}S_{\text{Shannon}}$) with $V$ bounded below: recall that such a lower bound to the energy was one of the assumptions on the quantum side as well. If all kinetic and/or diffusion coefficients are suitably bounded above, such a Markov process will also have a hard time to reproduce an exponential decay law.

In stead of proving a general statement in this regard, let me give a simple example: consider the Markov jump process with state space $\mathbb{N}=\{0,1,2,...\}$ where the jump rate from site $n$ to site $n\pm 1$ equals $e^{-\beta (V(n\pm 1)-V(n))/2}$, $V(n)=V_0\delta_{n=0}$ where $V_0>0$ and all other jump rates (between non-adjacent sites) are zero. The picture is that $n=0$ is an unstable system state while the other sites correspond to decay states. Abbreviating $\alpha:=e^{\beta V_0/2}=:\gamma^{-1}$ the evolution equation for the probability $\rho_n(t)$ that the system is at site $n$ at time $t$ is given by $$\dot{\rho_0}(t)=-\alpha \rho_0(t) + \gamma \rho_1(t),\quad\dot{\rho_1}(t)=-(\gamma+1)\rho_0(t) + \rho_2(t)\\\dot{\rho_n}(t)=-2\rho_n(t) +\rho_{n-1}(t)+ \rho_{n+1}(t)\quad (n>1)$$ We will solve these equations w.r.t. the initial condition $\rho_n(0)= \delta_{n=0}$ and our goal is to show that $\forall \Gamma>0:\,\limsup_{t\to \infty}\exp(\Gamma t)\rho_0(t)=+\infty$.

Laplace-transforming the equation of motion, one gets the following coupled system of equations: $$\hat{\rho}_1(p)=\frac{(p+\alpha)\hat{\rho}_0(p)-1}{\gamma},\quad \hat{\rho}_2(p)=(p+1+\gamma)\hat{\rho}_1(p)-\alpha \hat{\rho}_0(p)\\ \hat{\rho}_{n+1}(p)=(2+p)\hat{\rho}_n(p)-\hat{\rho}_{n-1}(p)\quad (n>2)$$ The latter equation implies that there are constants $C_{\pm}(p)\in \mathbb{R}$ s.t. $$\forall n>2:\,\hat{\rho}_{n}(p)=C_+(p)\lambda_+^n(p) + C_-(p) \lambda_-^n(p),\quad \lambda_{\pm} = \frac{2+p \pm ((2+p)^2-4)^{1/2}}{2}$$ However, since $0<\lambda_-(p)<1<\lambda_+(p)$, the sum $\sum_{n=2}^{\infty}\hat{\rho}_{n}(p)$ is finite (for all $p>0$) iff $C_+(p)\equiv0$ and this summability is required since $$\sum_{n=2}^{\infty}\hat{\rho}_{n}(p)=\int_0^\infty dt\,e^{-pt}\underbrace{\text{Prob}[n(t)\geq 2]}_{\leq 1}\leq \frac{1}{p}.$$ Now, $C_+(p)\equiv 0$ iff $\frac{\hat{\rho}_2(p)}{\hat{\rho}_1(p)}\equiv \lambda_-(p)$. We could proceed to solve this equation for $\hat{\rho}_0$, but a shorter route is to assume (anticipating a contradiction) that there is a certain $\Gamma^*>0$ s.t. $\limsup_{t\to \infty}\exp(\Gamma^* t)\rho_0(t)<+\infty$. In that case, the Laplace-transform of $\rho_0$ exists for $p\in(-\Gamma^*,+\infty)$. Denote $x:= \hat{\rho}_0(p=0)$. Then one gets $$1=\lim_{p \to 0+}\lambda_-(p)=\lim_{p \to 0+}\frac{\hat{\rho}_2(p)}{\hat{\rho}_1(p)} = \frac{(1+\gamma)(\alpha x-1)-\alpha\gamma x}{\alpha x -1}$$ which, after rewritting, results in the contradiction $\gamma = 0$.