# Non-observance and the Schrödinger equation

I was thinking today about configurations where one measures that a certain observable is not in a certain state.

I was getting confused about what this means for decoherence. If I observe a detector and I measure when a particle does not interact with it, then, I don’t understand how this can be entirely equivalent to allowing the particle to interact with further macroscopic objects (f.i. detectors, my brain) in such a way that the wave functions collapse. I’m detecting when it doesn’t interact so, I’m not interacting with it..

If the Schrödinger equation yields solutions that show the probability as the square of the amplitude, then the ‘negative’ Schrödinger equation’s solution is an operator $$\sqrt(1-x^2)$$ applied to the normal solution.

Under what conditions is that still a solution of Schrödinger equation? And is it possible to define hermitian operators that give the probability of “not observing” a property?

I don’t see how physically decoherence of non-observance can happen in the same way as regular observing, and at the same time it feels like it has to, although this may just be another aspect of QP that defies intuition.

If the Schrödinger equation yields solutions that show the probability as the square of the amplitude, then the ‘negative’ Schrödinger equation’s solution is an operator √(1−x2) applied to the normal solution.

This expression doesn't make sense on dimensional grounds. If you intend $$x$$ to be the wavefunction, then $$x^2$$ has units, so you can't subtract it from 1.

I was getting confused about what this means for decoherence. If I observe a detector and I measure when a particle does not interact with it, then, I don’t understand how this can be entirely equivalent to allowing the particle to interact with further macroscopic objects (f.i. detectors, my brain) in such a way that the wave functions collapse.

Decoherence isn't the collapse of a wavefunction.

And is it possible to define hermitian operators that give the probability of “not observing” a property?

Yes. You can define a the kind of does-it-have-this-property operator as a projection operator $$P$$, which is one that has eigenvalues all equal to 0 or 1. In Mackey, The Mathematical Foundations of Quantum Mechanics, 1963, these are referred to as "questions." The logical negation of the operator is defined as $$1-P$$.

The act of measuring (an eigenvalue of) $$1-P$$ requires an interaction, just as the act of measuring $$P$$ requires in interaction. For example, in the Stern-Gerlach experiment, $$P$$ would be an operator whose eigenvalue is 0 on the spin-down state and 1 on the spin-up state. The measurement to implement $$1-P$$ is implemented using exactly the same apparatus as the measurement to implement $$P$$.

• - I believe I mentioned that x is not the solution of the wave function equation but the amplitude of a solution, so you can in a way subtract 1 from its square Jul 3 '20 at 5:13
• - Thanks for pointing me to "questions". However, this seems to be a very specific case. I am really talking about "observing that an observable is not within x1,x2". I don't see how this interacts with the particle in the same way as other observings. Jul 3 '20 at 5:14