I haven't seen this done before as tempting as it seems and was wondering if there was some way to disprove the below? In short can one think of the measurement as the sudden approximation (Eq $84$ onwards or see wiki) applied twice? I use the notation:

$$ P(| A \rangle \to | B \rangle ) = \text{Probability of State $|A \rangle $ going to State $|B \rangle$}$$

And also:

$$ P(| A \rangle \to \sum | B \rangle ) = \text{Sum of probabilities of State A going to one of the eigenstates of the observable of state $|B \rangle$}$$


I define the measurement as a short lived interaction $H_{int}$ produced due to the measurement device.

Let the system I am interested in have the Hamiltonian $H$ obeying:

$$ H | E_n \rangle = E_n |E_n \rangle$$

The Hamiltonian of the short lived system (during the measurement) be $H'= H + H_{int}$ obeying:

$$ H' |E'_n \rangle = E'_n |E'_n \rangle $$

Thus we have the following sequence of Hamiltonians:

$$ H \to H' \to H $$

Let the initial state be $| \psi \rangle $ then after $H \to H'$ to find it in say state $| E' \rangle$ the probability is given by the sudden approximation.

$$ P(|\psi \rangle \to | E_n' \rangle ) = |\langle E_n' | \psi \rangle|^2 $$

Now, again let us assume the final state is $| \psi' \rangle$ after $H' \to H$. Then the probability of us arriving in that state is:

$$ P(|E_n' \rangle \to |\psi') = |\langle E_n' | \psi '\rangle|^2$$

Hence, the probability of going from $|\psi \rangle \to |E'_n \rangle \to |\psi' \rangle $ is:

$$P(|\psi \rangle \to |E'_n \rangle \to |\psi' \rangle) =|\langle E_n' | \psi \rangle|^2 |\langle E_n' | \psi '\rangle|^2 $$

Let's say I repeat the experiment many times and go through all $E_n'$ and somehow know the outcome $|\psi' \rangle$ but not the intermediate $E'_n$. Then:

$$ P(|\psi \rangle \to \sum |E'_n \rangle \to |\psi' \rangle) = \sum_{n} | \langle \psi' |E'_n \rangle \langle E'_n | \psi \rangle|^2 $$

Using triangle inequality and $\sum_n |E'_n \rangle \langle E'_n | = \hat 1 $:

$$ P(|\psi \rangle \to \sum | E'_n \rangle \to |\psi' \rangle) = |\langle \psi | \psi' \rangle |^2 + c(\psi \to \psi') \geq | \sum_n \langle \psi | E'_n \rangle \langle E'_n | \psi ' \rangle |^2 $$

where $c$ is a constant:

$$ c(\psi \to \psi') = \sum_{i \neq j} \langle \psi | E_i \rangle \langle E_j | \psi ' \rangle \langle E_j | \psi \rangle \langle \psi ' | E_i \rangle $$

Now let's say we do the experiment many times and include all possible final outcomes of $| \psi ' \rangle $. Then what happens to $c$?

$$ \sum_{\psi'} c(\psi \to \psi') = \sum_{i \neq j} \sum_{\psi'} \langle \psi | E_i \rangle \langle E_j | \psi ' \rangle \langle E_j | \psi \rangle \langle \psi ' | E_i \rangle $$

But since $\sum_{'} |\psi' \rangle \langle \psi' | = \hat 1$:

$$ \sum_{\psi'} c(\psi \to \psi') = \sum_{i \neq j} \langle \psi | E_i \rangle \langle E_j | \psi \rangle \delta_{ij} = 0$$


$$ P(|\psi \rangle \to \sum | E'_n \rangle \to \sum |\psi' \rangle) = | \langle \psi | \psi' \rangle |^2 $$


After going through the derivation of Wikipedia. It seems indeed the Born Rule itself is used to derive the didactic/sudden approximation. Precisely when the wikipedia page where it defines the parameter $\zeta$:

To determine the validity of the adiabatic approximation for a given process, one can calculate the probability of finding the system in a state other than that in which it started.

$$ \zeta = \underbrace{\langle 0 | U^\dagger (t_1 ,t_0) U (t_1 ,t_0) | 0\rangle}_{1} - \underbrace{\langle 0 | U^\dagger (t_1, t_0) | 0 \rangle \langle 0 | U (t_1, t_0) | 0 \rangle}_{\text{Probability of finding state in $|0 \rangle$ after time evolution $t_1-t_0$}} $$

Unless, there is some other way to get this result. In light of this I would like to ask if this is a valid way to model the measurement (as a consequence of double sudden approximations).

  • $\begingroup$ is this to be understood as trying to derive the Born rule from some other principle? If so, it looks like you are assuming it during the calculation, e.g. when you write the probability of transitions as $|\langle E_n'|\psi'\rangle|^2$ $\endgroup$ – glS Apr 2 at 15:52
  • $\begingroup$ and how is that probability derived in the context of the sudden approximation?... $\endgroup$ – glS Apr 2 at 15:54
  • 1
    $\begingroup$ The sudden approximation says you should just ignore the intermediate Hamiltonian and pretend the whole evolution follows the normal Hamiltonian. How does double-ignoring the intermediate Hamiltonian change anything? Also, where are you getting the idea that there are probabilistic transitions implied by the sudden approximation; the whole thing should still be unitary? $\endgroup$ – Craig Gidney Apr 2 at 15:55
  • $\begingroup$ @MoreAnonymous I don't understand why that brings in probabilities. The reference you linked describes nothing like that. $\endgroup$ – Craig Gidney Apr 2 at 15:58
  • $\begingroup$ @glS You were right but I think the question is still valid. $\endgroup$ – More Anonymous Apr 2 at 17:22

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