[strogonoff]

> "when interacted with by another force"

But that is not the same, right? I mean, if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Honest question (to avoid any defensiveness, I should disclose that I don’t subscribe to panpsychism).

> or anything along those lines

Any suggestions?

[pa7x1]

For "collapse" to occur, the interaction of the quantum system must be with a classical system. A classical system being something big, noisy that can be well approximated by classical mechanics.

Simple interaction between two systems doesn't cause "collapse" it makes the two systems become entangled. Classical systems are a bit contagious in this sense, anything that gets entangled with them becomes classical.

To be a bit more precise, this distinction between classical and quantum is a bit our fault. Everything is quantum at a fundamental level, classical system is one for which we do have not have a precise knowledge of the state of the system, instead we have a coarse representation. This should make more obvious in which way "classicalness" is contagious. Since the knowledge of a part was coarse, the knowledge of the newly entangled system is also necessarily coarse.

[a1369209993]

> if it interacted with a force and no observation was made

Consider the classic two-slit interference experiment. Whether the electron goes through the left or right slit can be treated a single qubit. Use a controlled-NOT gate to copy that qubit onto a second storage location, without observing either. Optionally drop the second qubit into a black hole to head off any claims about supposed future observations. Allow the electron to continue. Do you still observe interference pattens as in the non-copying version of the experiment? Why or why not?

[aeternum]

Interesting throught experiment. I believe you would still see the interference since there are still two possible paths.

Using the word copy in conjunction with C-NOT is slightly misleading as the copies do not behave independently.

Tongue-in-cheek explanation: Maybe whoever wrote our simulation used shallow copy when they should have done a deep copy.

[a1369209993]

> Using the word copy in conjunction with C-NOT is slightly misleading as the copies do not behave independently.

That's what the word "copy" means. If you flip a coin and copy that bit, you will observe that those copies do not behave independently either. If you want independent bits, flip two coins.

Similarly, "erasing" a (qu)bit technically consists of performing a exchange operation between it and a known-zero bit. In typical electronic computers, this would generally involve diffusion-like exchanges between the voltage level in a memory capacitor (such as a FET gate) and that on the GND rail, which has a much greater effective number of bits and therefore will stay mostly zero, but eventually requires a thermodynamic expenditure of known-valued bits (aka negentropy) from some external source to maintain its voltage level / bit zeroness. (This is rather simplified; there's lots of other sources of known-zero and known-one bits getting depleted and replenished, and the exact accounting depends on how you interpret various physical states information-theoretically.)

[a1369209993]

> In typical electronic computers,

That should have had a "for example" in front.

[aeternum]

Calling it a copy seems to me to violate QM's no cloning theorem, but maybe it's just that we have slightly different definitions of 'copy'.

[a1369209993]

The no cloning theorem is just the rigorous quantum mechanical version of the fact that if you flip a coin and write down the result, then copy what you just wrote down, you don't get two independent coin flips.

[tim333]

Modern explanations tend to have interactions causing decoherence rather than observers and collapses.

[goatlover]

Doesn’t decoherence just spread entanglement out to the rest of the world such that you end up in a situation analogous to Many Worlds?

[pault]

> if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it?

Any two systems interacting will cause the collapse. It doesn't matter if the system is attached to a scientist or not.

> Any suggestions?

No, I'm a software developer, not a quantum physicist. :)

[strogonoff]

> Any two systems interacting will cause the collapse.

I suppose that means if a photon, say, is reflected by a mirror, that should collapse its wave function and any measurements after that should not have any effect on it?

Maybe it should be qualified what kind of interaction collapses wave function?

> I'm a software developer, not a quantum physicist.

Great, I’m not a quantum physicist either—yet here we are, talking about quantum physics!

[pault]

> I suppose that means if a photon, say, is reflected by a mirror, that should collapse its waveform and any measurements after that should not have any effect on it?

I'm not sure if that example is the right one to use, but yes, that's roughly my understanding.

[strogonoff]

Okay. I actually thought the paradox was that wave collapses only if the photon is influenced by measurement activity but I’s been a while since I was reading up on this topic so I may need a refresher.

I suppose if we can calculate exactly how a given force would influence a photon, that would be essentially the same as “measuring” it.

[sdenton4]

Turns out you need to run your quantum computer in near-zero-kelvin conditions to avoid accidental 'observations' from collapsing the precious waves you're using to factorize the number 6.

[simonh]

That’s not what Schroedinger’s equation says though, it can combine the wave functions of two quantum systems, say a particle and a force, and show how they evolve with no problem, in fact that literally what it does. Quantum mechanics simply doesn’t have a complete account of collapse to a specific state.

[aeternum]

It depends upon the QM interpretation you subscribe to. For example with many-worlds, there is no collapse (at all).