Supposedly all physically realisable transformations are either linear or non-deterministic (measurements are not linear transformations, but they are non-deterministic, from the perspective of the observer that entangles with the observed system)

There is however at least one example where the application of the quantum Zeno effect seems to avoid this rule: consider an electron spin in some initial state

$$ | \Psi \rangle = | + \rangle_s + | - \rangle_s $$

where $ | + \rangle_s$ is some starting axis, and $| - \rangle_s$ is the opposing direction along that same axis

Now choose a final axis $| + \rangle_f$. On the 2D sphere there is at least one shortest path from the tip of the arrow along the $|+ \rangle_s$ direction, to the tip of $| + \rangle_f$. This path is labelled $P_{ s^{+} \rightarrow f^{+} }$. Likewise there is at least one shortest path from $|-\rangle_s$ toward $|+ \rangle_f$. This path is labelled $P_{ s^{-} \rightarrow f^{+} }$

Now consider a spin measurement apparatus with adjustable axis that can measure the spin direction at a finite rate, but fast enough that we can be certain that after an initial measurement, the spin evolution tracks the apparatus adjustable axis.

Now I prepare the apparatus such that the adjustable axis begins along the $|\pm\rangle_s$ axis, and according to the result from the first measurement (either $|+ \rangle_s$ or $|- \rangle_s$) the apparatus chooses either path $P_{ s^{+} \rightarrow f^{+} }$ or path $P_{ s^{-} \rightarrow f^{+} }$. As this process is repeated by increasing the measurement frequency as the apparatus axis moves along the chosen path, it seems that regardless of the initial uncertainty in the original state, the final state is in a well-defined axis and direction arranged beforehand

The above process does not seem to be representable by a linear unitary matrix, as no matter what the original values in the state vector are, the final state will be of the form $(1,0)$ in the $|\pm \rangle_f$ basis

Isn't this a problem? Am I overestimating the power of the quantum Zeno effect to keep a state from spreading?

  • 1
    $\begingroup$ It appears to me that you have merely stated the well-known fact that a time-evolution with repeated collapse (which is what you're doing in this model where you don't model the apparatus quantumly but just assume it projects onto the measurement states) is not linear/unitary. I'm not sure what exactly the question about that is. $\endgroup$
    – ACuriousMind
    Commented Jun 14, 2016 at 13:34
  • $\begingroup$ we have two initial different states that after the combined transformation end up in the same final state. An unitary transformation should always be biyective maps between vector spaces $\endgroup$
    – lurscher
    Commented Jun 14, 2016 at 23:54

1 Answer 1


Among other things, you wrote:

...and according to the result from the first measurement (either $|+ \rangle_s$ or $|- \rangle_s$) the apparatus chooses either path $P_{ s^{+} \rightarrow f^{+} }$ or path $P_{ s^{-} \rightarrow f^{+} }$.

Because the chosen path depends on the result of a measurement, your "transformation" isn't one transformation but actually two different transformations. It is the case because the result of the first measurement isn't determined (with certainty) without a measurement of the initial state, and the measurement produces a random bit.

So you haven't found any (single) transformation given by a nonlinear operator. So far, I assumed that the apparatus is considered to be a part of the observer.

You might also choose to describe the electron including the apparatus as a "big external system with many particles". In that case, the spin-up and spin-down states would evolve to entangled states of the electron and the apparatus and the operator describing this evolution would be as linear as you can get.

  • $\begingroup$ regardless, the combined total transformation turns both possible initial states in a single final state known with high certainty. Why does it matter that the chosen path depends on the result of the first measurement? $\endgroup$
    – lurscher
    Commented Jun 14, 2016 at 23:50
  • $\begingroup$ It matters because if you want to describe the transformation as "one transformation", you have to include the first measurement into this "one transformation", and when you do so, it implies that you must treat the apparatus as a part of the observed system. When you do so, quantum mechanics unambiguously implies that the "combined total transformation" is given by a linear operator on the big Hilbert space (including both the spin and the apparatus). You are just inconsistently mixing the assumption "the measurement is a part of the transformation" with "it's not". $\endgroup$ Commented Jun 15, 2016 at 6:50

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