In the 't Hooft Cellular Automation Interpretation it is declared, that for a system described by some template state $\lvert \psi \rangle$ and for any ontological measure outcome $\langle a \rvert$ the absolute square of inner product $\langle a | \psi \rangle$ is a probability to find the system in $\langle a \rvert$ state.

But if we remember, that classical states are not just ontological states, but sets of ontological states — because classical, macroscopic states do not describe state precisely, then we can face the following difficulty.

Consider macrostate $X$, which can be realized in one of many ($N$) sub-microscopic ontological states $\langle a_i \rvert$, $i = 1 \ldots N$. Suppose we know that current system state can be described using $\lvert \psi \rangle$ template state. Then if we want to calculate probablity to find the system in $X$ state, we should compute $\sum_{i=1}^N \lvert \langle a_i | \psi \rangle \rvert^2$. But this is not how we do real computations in quantum mechanics. In quantum mechanics we describe classical outcome with some pure state $\langle X \rvert$ and just compute one projection $\langle X | \psi \rangle$.

So my question is: how $\sum_{i=1}^N \lvert \langle a_i | \psi \rangle \rvert^2$ becomes $\lvert \langle X | \psi \rangle \rvert^2$ and what is $\langle X \rvert$ in 't Hooft interpretation?

  • $\begingroup$ Are you sure that this is a feature of the 't Hooft Cellular Automation Interpretation. Or could it be that by the macrostate you do not mean a pure state but an Ensemble of states, which should be described by a density matrix $\rho = \sum_i |a_i><a_i|$. Then, you obtain the expectation value of some pure $|\psi\rangle$ through $Tr( \rho |\psi\rangle\langle\psi|) = \sum_i |\langle a_i|\psi\rangle|^2$. $\endgroup$
    – Cream
    Dec 16, 2020 at 16:12
  • $\begingroup$ In CAI a measured stated = a classical state is, indeed, a macrostate, which is an ensemble of states. But in "ortodox" quantum mechanics a measured stated is a pure state in a basic case. This is not a big issue, because in CAI all template states, even pure, are ensembles. But details are not clear. $\endgroup$
    – warlock
    Dec 16, 2020 at 18:24

2 Answers 2


The previous answers are basically correct. In the CA interpretation you use the rules for computing something exactly as in “real” quantum mechanics; you may do exactly the same unitary transformations, go from one basis to another, solve the Schroedinger equation there, etc. The only difference is that the CA interpretation assumes the existence of one particular basis that represents the true world. If you stick to that basis, the Schroedinger equation never produces superpositions, it will always bring you back to elements of the basis states, if that’s where you started. This ensures that, in this very special basis, everything happens as in classical mechanics. In many simple quantum models, one can identify such a basis, often there are many possible different choices. And then the next step is to assume that one of these basis sets represents ‘reality’.

“Real" quantum mechanics reappears as soon as you make unitary transformations to other basis sets that seem to be more suitable for calculations, although these basis sets do not represent single real states anymore. For doing statistics, this is just fine, even if you don’t see the fine structure of reality anymore; just as in a postcard picture of a beach where you don’t see the sand grains anymore.

One twist added to the story later is that you should accommodate for the notion of energy in this picture, where energy just stands for the frequencies of some ultra-fast jittering variables. Their motion is so fast (corresponding to many TeV’s) that these energetic subsystems are always close to their lowest energy eigenstate. This is how they decouple, in practice, but they may still have effects on the slower, visible, variables. These then can develop observable properties that seem not to commute with one another (as can be checked by simple model calculations), and this may well be the way by which non-commuting observable operators enter into the picture.

  • $\begingroup$ I'm glad I didn't make a complete fool out of myself with my answer, because I have only a very basic understanding of your approach. Can I ask: I understand that we want a preferred basis such that time evolution only takes us from one basis element to another, but how is it possible to continuously evolve through orthogonal basis states, without passing through some intermediate states? $\endgroup$ Dec 18, 2020 at 20:45
  • $\begingroup$ ADDENDUM to my previous comment: I can see how that might work if the Hilbert space has a continuous basis, but would we run into a problem if part of the spectrum is discrete, such as for a particle in a finite well? Or, worse yet, for a 2D Hilbert space of a q-bit? $\endgroup$ Dec 18, 2020 at 22:32
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    $\begingroup$ I should add that a more instructive detailed model was described in my recent preprint: "Fast Vacuum Fluctuations and the Emergence of Quantum Mechanics", arxiv:2010.02019 . $\endgroup$ Dec 20, 2020 at 9:48

Short answer: the situation you describe corresponds, in regular quantum mechanics, to an "imprecise" measurement, i.e. not to a projection onto a single state, but to a projection onto a multidimensional subspace of the Hilbert space.

The ontological states correspond, in regular quantum mechanics, to some specific quantum states (which are assumed to be the "true" possible states of the system, as opposed to all other quantum states). So what would the probability to end up in one of several ontological states correspond to?

In quantum mechanics we describe classical outcome with some pure state $\langle X \rvert$ and just compute one projection $\langle X | \psi \rangle$.

We only do that if the measurement procedure is "precise", i.e. it's able to distinguish that specific pure state from all other states. But the situation with the "macrostate" you describe is different, it corresponds to a "coarse" measurement procedure, which regular quantum mechanics is also able to handle easily.

So how does it handle calculating the probability to end up in one of many possible states? Simple, we just project our state $| \psi \rangle$ onto the subspace spanned by all of those states. And the rest is as before - whether the measurement is precise or coarse, the probability is obtained by squaring the length of the projection.

The result is that the total probability is equal to the sum of the probabilities for all of the individual (orthonormal) states that our measurement procedure can't distinguish:

$\sum_{i=1}^N \lvert \langle a_i | \psi \rangle \rvert^2$,

which is of course the exact formula from CAI you wanted to recover.

  • $\begingroup$ But squaring the length of the projection does not equal to the sum of the probabilities for all of the individual states. We would suppose something like ''random phases assumption" to get equality. Does CAI really rely on the random phase assumption? $\endgroup$
    – warlock
    Dec 20, 2020 at 13:14
  • $\begingroup$ It does equal that, provided the individual states are orthogonal, which they are in your case. We don’t need anything like a random phase assumption. It's basically just the Pythagorean theorem. (For example, if we project a 3D vector onto the XY plane, the length of the projection squared will be equal to the square of the projection onto the X axis plus the square of the projection onto the Y axis) $\endgroup$ Dec 20, 2020 at 21:30
  • $\begingroup$ ADDENDUM: I added a guess what the source of the confusion might be, but I deleted it because I didn't want to seem presumptuous. Just to make sure we are on the same page, by "length" I of course mean the square root of the inner product of a vector with itself. $\endgroup$ Dec 20, 2020 at 21:53
  • $\begingroup$ I got it now. I read the word "subspace" several times, but somehow I imagined a completely different thing instead of a subspace. $\endgroup$
    – warlock
    Dec 20, 2020 at 23:13

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