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In this answer dealing with details of decay theory (incl. references) it is shown that

[Given] a system initialized at $t = 0$ in the state [...] $| \varphi \rangle$ and left to evolve under a time-independent hamiltonian $H$ [... its] probability of decay is at small times only quadratic, and the survival probability is slightly rounded near $t = 0$ before of going down [exponentially].

Is it correct that therefore it is also possible to prepare (initialize) an entire ensemble of $N \gg 1$ such states $| \varphi \rangle$, such that their survival probability is at small times only quadratic ?

Is it instead possible at all to prepare an ensemble of $N$ states (which would likewise "evolve under the Hamiltonian $H$") such that their survival probability is (at least to a good approximation) not quadratic but rather drops linearly as a function of the duration since completion of the preparation ?

In particular, if an ensemble of $2~N$ states $| \varphi \rangle$ had been given and (in the process of an extended preparation procedure) half of those (i.e. $N$ systems) had decayed, do the remaining/surviving $N$ systems together then constitute such an ensemble? What exactly is the survival probability of these given, momentarily remaining/surviving $N$ systems; as a function e.g. of $t_{\text{(extended prep.)}} := t - \tau_{1/2}$, where $\tau_{1/2} = \tau~\text{Ln}[2]$ is the specific overall "half-life" duration?

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user12262
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1 Answers1

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This comes down to what you understand the initial state $|\varphi⟩$ to be, and most importantly to how you define the survival probability.

In my previous answer, to keep things general I took $|\varphi⟩$ to be any square-integrable state. If you want to be more specific, however, you will run into trouble. As the Fonda paper I referenced mentions (Rep. Prog. Phys. 41, 587 (1978)), it is in general quite hard to define the unstable states of a given system. Things like "an undecayed uranium nucleus" have a slightly fuzzy edge to their definition.

Moreover, the initial setup

Consider a system initialized at $t=0$ in the state $|\psi(0)⟩=|\varphi⟩$

is really loaded words. It implies that you know what $|\varphi⟩$ is, that you can prepare it, and that you can prepare it instantly.

The crucial point, however, is what I mean by survival probability, which I defined as $$ P(t)=|⟨\varphi|\psi(t)⟩|^2=⟨\psi(t)|\varphi⟩⟨\varphi|\psi(t)⟩. $$ Talking about this quantity implies the ability to measure it, or at least measure quantities that are similar enough. This quantity is the overlap of the state at time $t$ with the initial state, so you are assuming that you can not only prepare $|\varphi⟩$ but also measure it.

Sometimes this is true, as in the optical lattice experiment I linked to - you put an atom in a specific lattice site and then you check whether it's still there. If you engineer things correctly, only one pure state will really be able to sit in that site. In general, however, your measurement is a lot less detailed: for radioactive decay, for instance, you're measuring statements like "no alpha particles have left this lump of metal", and which have a mathematical expression of the form $$ P'(t)=⟨\psi(t)|\hat\Pi|\psi(t)⟩ $$ where $\hat\Pi$ is a spatial projector. This will obviously behave differently!

This means that to observe the rounded edge of the survival probability, you need (i) very detailed preparation procedures, (ii) very detailed measurement procedures, and (iii) the ability to perform both quickly. This is a far cry from a lump of radioactive metal slowly decaying, and you won't detect things like $⟨\varphi|\psi(t)⟩$ in the latter.

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Emilio Pisanty
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