If the initial state of the system is:

$|\psi(0)\rangle$ = $|0_{cat}0_{gas}\rangle$

Where $|00\rangle$ is the state where the gas has yet to decay and the cat is still alive. Then say the state evolves to:

$|\psi\rangle$ = $|00\rangle + |11\rangle$

What would be the unitary operation to get:

$U|\psi(0)\rangle$ = $|\psi\rangle$?

I am trying to get this via matrices (4x4 matrices) but I have yet to find the solution. Thanks in advance!

  • $\begingroup$ Linked. $\endgroup$ Jan 10, 2021 at 19:39
  • $\begingroup$ Also linked. $\endgroup$ Jan 11, 2021 at 22:50
  • $\begingroup$ Yes but none of the links nor below were the answer I was looking for. I plan on adding my answer in the coming weeks. @CosmasZachos $\endgroup$
    – mikanim
    Feb 3, 2021 at 20:50

4 Answers 4


There are two separate issues you're running into:

(1) Irreversible decay can't occur for a two-state system. Quantum mechanics has unitary evolution, and with unitary evolution, the only behavior you can get from a two-state system is a time evolution in which it oscillates back and forth. A process like the radioactive decay of a nucleus has to be modeled with a continuum of states for the states after the decay.

(2) Again, quantum mechanics has unitary evolution, so information is conserved. Decoherence can't occur unless you have some way for information to flow out of the cat-in-the-box system and into the environment. So for this reason as well, you can't have anything as simple as a two-state system.


Call $|00\rangle=|i\rangle$ and $|11\rangle=|f\rangle$. The w.f. collapse, the "observation", happens at the trigger nuclei level, and their decays are irreversible (and, a fortiori, so is the cat's sadistic death). The live cat will never interfere with the dead cat, so it is a red herring in the problem--the very reason this paradigm is notoriously uninstructive. So you don't have a 4-state system (can't have undecayed triggers with dead cats, etc...), but could you have a 2-state system?

But even the trigger particles' decays are not a "small" unitary evolution, like an archetypal quantum flip-flop, even though they are quantum mechanical. Clearly, the decayed state will not interfere with the undecayed one, or reverse to the initial undecayed state. So you'd have a hell of a time modeling the decay unitarily.

That is, modeling the persistence amplitude $$\langle i|U(t)|i\rangle ~~~ ?\sim ? ~~~e^{i\phi} e^{-\Gamma t/2} $$ so as to have persistence probability
$$ e^{-\Gamma t} $$ could not work with any unitary 2×2 operator. The reason is the effective 2×2 Hamiltonian is non-hermitean.

Moreover, transition from $|f\rangle$ to $|i\rangle$ is zero, as well. In fact, simple models of unstable quantum systems are notoriously tricky.

NB (Geeky) To see how the nohermitean Hamiltonians describing such decays arise out of larger open systems, see, e.g., Bertlmann et al 2006 or Phys Rev A73 (2006) 054101. In density matrix language, $\rho(t)=\operatorname{diag}(e^{-\Gamma t}, 1-e^{-\Gamma t})$, the purity varies with time, $$ \operatorname{Tr} \rho(t)^2= 1- 2e^{-\Gamma t} +2e^{-2\Gamma t}, $$ going from 1 at $t=0$, to <1 for intermediate t, and back to 1 at $t=\infty$.

This cannot be unitary evolution.


From studying the evolution operator and the conditions that are expected of it, we find that it has the form $U(t) = e^{-i\hat{H}t/\hbar}$ with $\hat{H}$ the hamiltonian of the system. This Hamiltonian is itself the result of the potentials applied to the system, so say you want the cat to be in the box, you can model it as a box shaped well, with infinit potential at its boundaries. I suggest reading on the Heisenberg picture to undertand the evolution operator and why it has this form. https://en.wikipedia.org/wiki/Heisenberg_picture


Use this matrix as an unitary matrix:

$$U = \left( \begin{matrix} 1/\sqrt2 & 0 & 0 & 1/\sqrt2 \\ 0 & 1/\sqrt2 & 1/\sqrt2 & 0 \\ 0 & 1/\sqrt2 & -1/\sqrt2 & 0 \\ 1/\sqrt2 & 0 & 0 & -1/\sqrt2 \end{matrix} \right) $$

  • 1
    $\begingroup$ That's not a unitary matrix. Its determinant is 4. $\endgroup$ Aug 16 at 16:45
  • $\begingroup$ I factorized 1/sqrt(2). If you affect that, the determinant would be 1. Moreover, UU*=1. $\endgroup$ Aug 16 at 17:44

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