-2
$\begingroup$

All wavefunction collapses are events which reveal some kind of classical information or thermodynamically visible 'event'. Is this true? In that case, what is the evidence of a 'not' collapse. A 'not' collapse is a projection of a position wavefunction to 0 at the position measurement screen at any given time that it has not arrived at the screen.

'revealing that the particle's location is not at the screen' is a bit of a silly argument because location isn't physically defined at all, only probabilities of collapses

What I have concluded is that collapses don't have to be visible to the experimenter (meaning to the classical scale universe), neither immediately (delayed choice) or at all ('not' collapse)

Does the solution lie in 'mixed states'?

$\endgroup$
17
  • 7
    $\begingroup$ "All wavefunction collapse are defined as 'events' which reveal some kind of classical information or thermodynamically visible 'event'." [citation needed] Collapse is interpretation-dependent and does not exist at all in some interpretations. Are you confusing measurement and collapse here (measurements induces wavefunction collapse in interpretations that believe in it, but the two are not synonymous)? $\endgroup$ – ACuriousMind Mar 21 at 21:12
  • $\begingroup$ what's the difference between measurement and collapse. how do you define measurement, if not synonymous with collapse or the physical evidence that it leaves behind? (which i have discovered as not always necessary) $\endgroup$ – ggmate Mar 21 at 21:19
  • 1
    $\begingroup$ @gmmate Referring to your comment: "how do you define measurement, if not synonymous with collapse...?" Measurement can be defined independently of quantum theory. Just think about what needs to happen physically in order for you to gain and retain information about the thing that is "being measured." To define collapse and appreciate its interpretation-dependence, think about how you tell quantum theory which measurement-outcome you actually experienced, given that quantum theory is unable to predict which one you'll experience. $\endgroup$ – Chiral Anomaly Mar 21 at 23:02
  • 1
    $\begingroup$ The best way to define measurement from a qm perspective is to say it is the event of a coupling of the wavefunction with a classical system. if this coupling is strong enough it leaves behind visible evidence. the coupling causes a collapse of the wavefunction which is projective if the appropriate hilbert space and hermitian operator is chosen (not a free choice), with probabilitiy given by <psi|P|psi> $\endgroup$ – ggmate Mar 21 at 23:20
  • 1
    $\begingroup$ The standard name for what you call a "not" collapse is interaction-free measurement. $\endgroup$ – benrg Apr 27 at 2:11
1
$\begingroup$

I am not completely clear what your ‘not’ collapse is, but it still seems to be a collapse of the wave function. Indeed, if there is a ‘not’ observable then the wave function must collapse into an eigenstate of the ‘not’ operator at each time when we measure the ‘not’ observable.

We can never observe an uncollapsed wave function i.e. a wave function that is not in an eigenstate of whatever observable we are measuring. On the other hand, wave function collapse occurs any time a quantum system interacts in a non-reversible way with its environment. So there are many, many wave function collapses that are never observed (unless you count the whole environment as an observer).

For example, one of the big challenges of quantum computing is preventing the wave function of a set of qubits from collapsing “accidentally” before we have finished running a program. Ideally, we only want the qubits’ wave function to collapse when we intentionally make a measurement to observe one or more classical bits when our program has completed. But at that point wave function collapse is unavoidable.

$\endgroup$
6
  • $\begingroup$ *non reversible way. So it is possible to hypothetically observe any collapse? I mean if a collapse leaves behind no visible evidence, it's still a collapse $\endgroup$ – ggmate Mar 22 at 10:26
  • 1
    $\begingroup$ @ggmate Observing the outcome of wave function collapse is a measurement. If you make a measurement of observable $X$ and you observe a value $k$ you know that when you took your measurement the wave function was in an eigenstate of the operator corresponding to $X$ with eigenvalue $k$. But if the wave function of a quantum system collapses (due to interacting with the environment) and then you only make a measurement of the system some time later, information about the state of the wave function before the measurement is lost (because the measurement process is non-reversible). $\endgroup$ – gandalf61 Mar 22 at 11:01
  • $\begingroup$ "We can never observe an uncollapsed wave function" A superposition state is a phenomenological fact realized by our inability to write down deterministic quantum laws. We observe the uncollapsed wave function when we repeat an identically prepared experiment and get different results despite our best efforts to get the same results. $\endgroup$ – Andrew Apr 27 at 1:43
  • $\begingroup$ The original question isn't asking if some wavefunction collapses go unnoticed. That's not very interesting. Many unnoticed "wavefunction collapses" leave evidence in some way (e.g. particle trails left in ice cores in Antarctica which go unnoticed until they are unearthed). You just have to be clever enough to recover that information. The original question seems to be asking about wavefunction collapse that leaves no evidence. My answer shows some wavefunction collapses hide their evidence in superposition states, which is interesting because they left undeterministic evidence. $\endgroup$ – Andrew Apr 27 at 2:28
  • $\begingroup$ @Andrew We can infer the existence of the uncollapsed wave function from repeated observations, but we never observe it directly. When we make a single measurement of the spin of an electron in a given direction, we always get a definite quantised value - say “up” or “down” - not something in between “up” and “down”. $\endgroup$ – gandalf61 Apr 27 at 3:28
1
$\begingroup$

No. The wavefunction collapse is independent of any 'irreversible event'. It can be verified through further interactions at a later time.

Measurement of classical variables does not collapse the wavefunction. Interaction of a quantum system with a classical system collapses it. The attempt to measure a classical variable is just a tool we can use to approximate the way a wavefunction collapses, and therefore find the probabilities of these collapses. It collapses roughly to an eigenstate of the corresponding (perhaps discretized) Hermitian operator. The evidence this collapse leaves behind, our ability to physically check what kind of collapse occurred therefore corresponds roughly to classical knowledge, but no actual classical information is retrieved upon collapse.

I say roughly because certain collapses corresponding to 'classical observables' are impossible. The dirac delta function is not a valid wavefunction. Even if we discretize the operator to projections, this does not solve the problem. For example, the momentum of a particle in an infinite well cannot collapse to a projection of the momentum space without breaking the boundary conditions. But if you attempt to measure the momentum some kind of collapse does happen. However we can still approximate the collapse as a projection onto the momentum space if it is sharp enough.

$\endgroup$
1
  • $\begingroup$ "Measurement of classical variables does not collapse the wavefunction. Interaction of a quantum system with a classical system collapses it." I see no difference between the words "Measurement" and "interaction". How are they not equivalent in every way? Also, particle interactions cause collapses of the wavefunction. This is literally how we make measurements of stuff classical systems can't interact with directly in particle accelerators. Rather than confuse yourself with dirac delta functions try to understand simple measurements of particle spin up or down first. $\endgroup$ – Andrew Apr 26 at 22:36
0
$\begingroup$

No, not all wavefunction collapses have to be evident in some way. Your conclusion "that collapses don't have to be visible to the experimenter" is correct. To make the language clear, we say a "collapsed" wavefunction is one that is in a particular eigenstate, while a "not-collapsed" wavefunction is one that is in a superposition or linear combination of eigenstates.

To be clear, it is the transition called "wavefunction collapse" that is not phenomenological and open to interpretation. Conditions of "single state", "superposition" and "quantum entanglement" are phenomena. However, wavefunction collapse doesn't need to be understood to answer the original question. If an example can be provided where the wavefunction is already collapsed (i.e. in a single state) according to some observer, but not evident (i.e. in a superposition) according to a different observer, then the answer to the original question is "no, not all collapses of the wavefunction have to be evident in some way".

Quantum entanglement is the phenomenon where this occurs. When particles A and B are entangled, A and B may be in a superposition state relative to the laboratory observer, but not relative to each other.

Edit after considering comments:

Example: particle spin

Consider a system of two particles where the spin (up or down) of each is measured. We can prepare the system in two ways: 1) the spins are uncorrelated or 2) the spins are correlated, aka an entangled state.

In the first experiment, we can prepare the system so that there is a 25% chance of both spins to be measured up, 25% chance of the first up and the second down, 25% chance of the first down and the second up, and 25% chance of both down. We can represent this symbolically as: $$\left|\psi_1\right> ={1\over 2} \left|\uparrow \right>_A \left|\uparrow \right>_B +{1\over 2} \left|\uparrow \right>_A \left|\downarrow \right>_B -{1\over 2} \left|\downarrow \right>_A \left|\uparrow \right>_B +{1\over 2} \left|\downarrow \right>_A \left|\downarrow \right>_B$$

In the second experiment, we can prepare the system so that there is a 50% chance of finding one particle up and the other down. We can represent this symbolically as: $$\left|\psi_2\right> ={1\over \sqrt{2}} \left|\uparrow \right>_A \left|\downarrow \right>_B -{1\over \sqrt{2}} \left|\downarrow \right>_A \left|\uparrow \right>_B$$

Note that in the first experiment $${\left<\uparrow \right|_B\left|\psi_1\right>\over \left|\left<\uparrow\right|_B\left|\psi_1\right>\right|} ={1\over \sqrt{2}} \left|\uparrow \right>_A -{1\over \sqrt{2}}\left|\downarrow \right>_A$$ is the state of A in B's frame if B is measured to have spin up, while $${\left<\downarrow\right|_B\left|\psi_1\right>\over \left|\left<\downarrow\right|_B\left|\psi_1\right>\right|} ={1\over \sqrt{2}} \left|\uparrow \right>_A +{1\over \sqrt{2}} \left|\downarrow \right>_A$$ is the state of A in B's frame if B is measured to have spin down.

Likewise, in the second experiment $${\left<\uparrow \right|_B\left|\psi_2\right>\over \left|\left<\uparrow\right|_B\left|\psi_2\right>\right|} =-\left|\downarrow \right>_A$$ is the state of A in B's frame if B is measured to have spin up, while $${\left<\downarrow\right|_B\left|\psi_2\right>\over \left|\left<\downarrow\right|_B\left|\psi_2\right>\right|} = \left|\uparrow \right>_A$$ is the state of A in B's frame if B is measured to have spin down.

We can say that in preparing the second experiment (the process that reduced the system from $\left|\psi_1\right>$ to $\left|\psi_2\right>$) the particles became "entangled" (without getting into the details of the process of entanglement or wavefunction collapse). Yet, consider what this entanglement means to particle B. In the first experiment, when the particles were not entangled, regardless of the spin of B measured by the lab, particle A is in a superposition state - that's how B would describe A. Entanglement (taking $\left|\psi_1\right>$ to $\left|\psi_2\right>$) reduces the number of superposition states, but it does not completely collapse the wavefunction in the laboratory frame. However, from B's perspective, in the second experiment, regardless of the spin of B measured by the lab, particle A is in a single state - hence A's wavefunction has collapsed in B's frame.

Therefore, we can say that "entanglement" is a type of measurement of A in B's frame, aka interaction between A and B, aka "wavefunction collapse" of A in B's frame. An entangled B has a deterministic view of A. I.e. the wavefunction collapse of A is evident to an entangled B. However, in this entangled state, the lab observer still must describe the state of the combined system as probabilistic. I.e. the wavefunction collapse is not evident to the lab frame before it measures the spin of either A or B. The lab observer could prolong this uncertainty indefinitely by never measuring the spin of A or B. Therefore, we have discovered that entanglement is a type of wavefunction collapse, relative to an entangled particle, which may indefinitely be non-evident to the lab observer. Hence, the answer to the original question is "no". We have not said, nor do we need to say, anything about the process of "wavefunction collapse" - we are only referencing the states of the prepared systems before or after entanglement before the lab's measurements, and after the lab's measurements. The actual process of wavefunction collapse is not relevant to the question since we can answer the question using the well-defined states between the debatable process of wavefunction collapse.

$\endgroup$
7
  • $\begingroup$ the state of the wavefunction is not relative $\endgroup$ – ggmate Apr 26 at 19:58
  • $\begingroup$ @ggmate In an entangled state B no longer sees A as a superposition. In that sense it is relative. Schrodinger's cat doesn't see itself as a superposition of dead and alive, it is the laboratory that sees the cat in the superposition state. Consider more carefully what that tensor product means when you write the entangled wavefunction down. $\endgroup$ – Andrew Apr 26 at 20:05
  • $\begingroup$ i thought there was a very clear distinction between what is considered the hilbert space and what is considered the classical environment $\endgroup$ – ggmate Apr 26 at 20:26
  • $\begingroup$ in the case of entangled particles, the wavefunction collapse is quite evident (when one particle is measured) $\endgroup$ – ggmate Apr 26 at 20:30
  • $\begingroup$ @ggmate "in the case of entangled particles, the wavefunction collapse is quite evident (when one particle is measured)" There are two collapses: 1) when one particle is measured, and 2) (the one you're overlooking) is the entanglement. The entanglement does nothing to collapse the wavefunction for the lab observer, yet it does for the particles. There is nothing non-classical about entanglement - it is as fundamental to original formulations of QM as the concept of superposition. $\endgroup$ – Andrew Apr 26 at 20:55
0
$\begingroup$

Any dependency on time or position of the wave function is governed by the Schrödinger equation. I have never seen any mathematical account of wave function collapse, so the conclusion must be that it does not happen. It is something people dreamt up in an interpretation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.