# Particle interactions which are NOT considered observations?

When considering the term "observe" often used in discussions related to quantum entanglement and the collapse of a wave functions, I found text I quote at the bottom of this question (I've also linked to the source).

My question: What is an example of an interaction between particles (or waves) that DOES NOT constitute an "observation" of certain (or any) of the particle (or wave's) characteristics? Apologies in advance if this question is entirely misguided or invalid.

An "observation" does not require an actual mind to perceive it. Rather, a particle/wave is "observed" as having certain characteristics whenever it interacts with any other particle/wave in a way that would require it to have those characteristics. As in Eogan's example. if a particle/wave is involved in a collision, then its wavefunction must "collapse" into a subset that is compatible with it having been at that particular time and place and having the particular energy and vector required for the collision. In short, particles observe each other wherever and whenever they interact. It is actually quite difficult to get any large number of particles to interact without collapsing.

http://forums.xkcd.com/viewtopic.php?f=7&t=75527#p2789781

• From an experimental point of view an observation produces some sort of persistent record, either as a physical effect like on a piece of film or as a numerical output (on paper, in a computer memory etc.). An interaction all by itself will not produce such a record, i.e. we know less about the system than we could if we had made an equivalent observation. To me this is an information theoretical issue after the actual physical interaction. All else being equal to the interacting system it doesn't matter if we get an observation output, or not, it's a huge difference in our theory, though. – CuriousOne Sep 10 '15 at 16:21

All observations imply interactions, any measurement is the result of an interaction.

Do all interactions imply observation? It is a matter of how one defines "observe". In the above quote the verb is used as synonymous with "interacting", and the answer will be by construction "none".

If observe means a specific experimental setup to measure something, then in general,in parallel with the sought for interaction , for example Higgs production, a number of other interactions are taking place which are not "observed" in the setup, for example production of a Psi going undetected because the experiment is not set up to detect it/observe it.

• I think high energy measurements are a good example. The interaction at the interaction point is not an observation because we don't have a direct record. The signals in the detectors are interactions and observations. Would you agree with that? – CuriousOne Sep 10 '15 at 16:36
• @CuriousOne yes, all that is not set up to be measured can be called "not observed", but in principle it could be observed if one set up a particular experiment – anna v Sep 10 '15 at 17:09

You can do so by simply not "observing" your measuring device. Let's say you have an electron in the sate $\frac{1}{\sqrt{2}}\left(|\uparrow\rangle+|\downarrow\rangle\right)$ (which means you have a 50% chance of measuring it being spin up, and a 50% change of measuring it down. It's in a superposition of up and down), and that you have a measuring device which can measure zero [$|0\rangle$ for "hasn't made a measurement yet"], up ($|\uparrow\rangle$), or down ($|\downarrow\rangle$). The system starts in the state: $$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle+|\downarrow\rangle\right)\otimes |0\rangle$$ (the symbol "$\otimes$" called the tensor product is the way of concatenating states in quantum mechanics. $|\uparrow\rangle\otimes |0\rangle$ means 'the electron is in the quantum "up" state, and the measuring device is in the quantum "unmeasured" state.')

When you flip your measuring device on, the state evolves to:

$$\frac{1}{\sqrt{2}}\left(|\uparrow\rangle \otimes |\uparrow\rangle+|\downarrow\rangle\otimes |\downarrow\rangle\right)$$

where $|\uparrow\rangle\otimes |\uparrow\rangle$ means 'the electron is in the quantum "up" state, and the measuring device measured up"'.

This final state has the solution to all your problems! Imagine you are the measuring device inside a very well sealed box inside some mad scientist's laboratory. You can be up, down, or zero. To the person in the laboratory who has not interacted with you in any way, you are in state up or state down, each with 50% probability, and nothing more can be said.

The part of this formula that reads $|\uparrow\rangle\otimes|\uparrow\rangle$ should be seen as the version of you in the up state. To the version of "you" in the up state, the electron is in state $|\uparrow\rangle$ with 100% certainty, and you've destroyed its original superposition of $\frac{1}{\sqrt{2}}\left(|\uparrow\rangle+|\downarrow\rangle\right)$.

Likewise, to the version of you in the $|\downarrow\rangle\otimes|\downarrow\rangle$ world, the electron is in state $|\downarrow\rangle$ with 100% probability, and you've destroyed its original superposition.

The point is this:

1. The "collapse" of the wavefunction is nothing more than its interaction with the measuring device.
2. The interaction -- meaning the way the measuring device works -- chooses what to measure, and in doing so chooses what the state "collapses" into. In this case the interaction "chose" $\uparrow$ and $\downarrow$, but if it interacted with the system in a different way, it could have chosen to measure $\frac{1}{\sqrt{2}}(|\uparrow\rangle+|\downarrow\rangle)$ and $\frac{1}{\sqrt{2}}(|\uparrow\rangle-|\downarrow\rangle)$ (which in the electron experiment is equivalent to measuring "left" and "right" instead of "up" and "down").
3. The tricky thing is to conceptualize living inside/BEING a quantum state. The wavefunction of the person in the laboratory tells them that you have a 50% chance of being in either state, and in fact that is the most fundamental picture of the universe that the lab person can see, even though the person inside the experiment gets the result of the experiment and knows 100% the state of the electron, breaking its superposition.