3
$\begingroup$

Will wave function collapse without measurement?

Since all matters are described by wave functions, then in principle, I should be able to describe wave function collapse by Schrodinger's equation. (I don't know how exactly though)

But then here comes a counter question I would like to ask here:

Given all matters are described by wave functions, and now we take the observer who measures into a new system without changing anything; however, considering:

1.) there is a wavefunction collapse. Such things should not disappear simply by picking a new imaginary "lines of system".

2.) But there is no one measuring outside our new system. (in other words, assuming the former action of measurement is simply some interaction, time evolution of the "over all" wavefunction)

So my question is Will wave function collapse without measurement?


I suppose this question somehow runs down to asking what exactly measurement is. but I am giving it a try to see if this question can be answered with or without exact meaning of measurement.

$\endgroup$
  • 1
    $\begingroup$ Possible duplicate of How does a Wavefunction collapse? $\endgroup$ – John Rennie Nov 22 '15 at 7:39
  • 1
    $\begingroup$ You might be interested to search for lots more on wavefunction collapse on this site. $\endgroup$ – John Rennie Nov 22 '15 at 7:40
  • $\begingroup$ Probably not, I have just made my question clearer, please feel free to see if it makes sense or a duplicate. $\endgroup$ – Shing Nov 22 '15 at 8:25
  • 1
    $\begingroup$ Possibly closer duplicate to what is an observer in QM than what John suggested. $\endgroup$ – Kyle Kanos Nov 22 '15 at 12:11
  • 1
    $\begingroup$ actually not quite the duplicate, because that question was asking what exactly a measurement is. I was asking if a wave function can collapse with NO measurement. Knowing what a wavefunction collapse is one way to answer this question, however, I am open minded to another way to answer this question. $\endgroup$ – Shing Nov 23 '15 at 6:34
5
$\begingroup$

Following von Neumann, the measurement process is just a special type of interaction between two systems, that follows special rules when it comes to averaging a specific observable $X$.

Let $H$ be a Hilbert space, $(\Omega,\mathscr{B})$ a Borel space, with $\Omega\subseteq \mathbb{R}$ and $\mathscr{B}$ a Borel $\sigma$-algebra on $\Omega$. By means of the spectral theorem, a self-adjoint observable is uniquely determined by a projection valued measure $X:\mathscr{B}\to \mathcal{L}(H)$ such that $X(\Omega)=1$. Here $\mathcal{L}(H)$ denotes the bounded operators on $H$. A state of the system $\rho$ is a positive trace one operator (density matrix).

In the state $\rho$, the probability that the outcome of a measurement of $X$ is in the subset $B\in \mathscr{B}$ of possible values is given by $$\mathrm{Prob}(X\in B; \rho)=\mathrm{Tr}[\rho X(B)]\; .$$

Now let $K$ be another Hilbert space, $\sigma$ a state on $K$. Then the formula $$\mathrm{Tr}[\rho E_\sigma(x)]=\mathrm{Tr}[(\rho\otimes\sigma) x]$$ where $x\in \mathcal{L}(H\otimes K)$, and $\rho$ is a state on $H$, defines a map from the operators on $\mathcal{L}(H\otimes K)$ to the operators on $\mathcal{L}(H)$ such that for any operator $x=a\otimes 1$, where $1$ is the identity operator, $E_\sigma(a\otimes 1)=a$.

Now, given an observable $X$ on a Hilbert space $H$ with value space $(\Omega,\mathscr{B})$, a measuring process for $X$ is a quadruple $(K,\tilde{X},\sigma,U)$ consisting of an Hilbert space $K$, an observable $\tilde{X}$ on $K$ with value space $(\Omega,\mathscr{B})$, a state $\sigma$ on $K$, and a unitary evolution $U$ on $H\otimes K$ satisfying the relation: \begin{equation} \tag{1} X(B)=E_\sigma\bigl(U^*(1\otimes \tilde{X}(B))U\bigr) \end{equation} for any $B\in \mathscr{B}$. The Hilbert space $K$ represents the apparatus system, the obervable $\tilde{X}$ is meant to be the position of the pointer on the scale of the measuring apparatus. The measurement is carried out by the interaction between $H$ and $K$ described by the unitary evolution $U$ that takes the form $$U=e^{-it (H_{1}\otimes 1+1\otimes H_{2}+ H_{int})}\; ,$$ where $t\in\mathbb{R}$ is the time necessary for the measurement.

The requirement (1) is equivalent to saying that the probability distribution $\mathrm{Prob}(X\in B;\rho)$ of an outcome being in $B\in\mathscr{B}$ must coincide with the probability $\mathrm{Prob}(\tilde{X}\in B;U(\rho\otimes\sigma)U^*)$ of the evolved system at time $t$.

Now this is the mathematical formulation of a measurement process. We have to observe that the process modifies the state $\rho$, "collapsing it" to the state $\rho^B$ after measurement, given by $$\rho^B= \frac{1}{\mathrm{Tr}[\rho X(B)]}\mathrm{Tr}_K \bigl[U(\rho\otimes\sigma)U^* (1\otimes \tilde{X}(B))\bigr]\; ,$$ where $\mathrm{Tr}_K$ stands for the partial trace on $K$. However, this is not a physical operation, but a convenience that we use because we would like to consider, after measurement, the system $H$ alone, and not the whole system $H\otimes K$. However, for the whole system nothing has happened different from the usual Schrödinger evolution.

Of course you may think of evolving the system $H\otimes K$ by $U$ independently of the measurement process. And starting with a state $\rho\otimes \sigma$, you will get after time $t$ the state $U(\rho\otimes\sigma)U^*$. However, it is only after partially averaging on $K$ the observable $(1\otimes \tilde{X}(B))$ that we obtained the collapsed state $\rho^B$. And the latter operation is indeed the act of measuring. So I do not see how the state would collapse without being involved in the act of measuring.

$\endgroup$
  • 1
    $\begingroup$ Bloody hell, that is one hell of an answer! :) $\endgroup$ – gented Nov 22 '15 at 21:16
  • $\begingroup$ Thanks so much for the detailed answer, but I am just starting taking my quantum physics, so the abstract math part I don't quite understand (but please keep it, so that I may revisit it someday) So basely you are saying measurement is some special interaction between two systems, and therefore, picking up a new system doesn't change anything at all, hence wave functions still collapse as long as the interaction exists right there? $\endgroup$ – Shing Nov 23 '15 at 7:04
  • $\begingroup$ Any interaction modifies the state of a system (in this case the global system observed+apparatus is modified by the measuring interaction), yielding a new state. On that state we perform the measurement, "looking up at the pointer on the scale" (this modifies the state further, since we have to take into account that the outcome happens with a certain probability; that is given by the partial trace of the pointer observable $\tilde{X}$). The state obtained after evolution, and taking into account the probability of the obtained measurement (projecting on the subspace of states corresponding.. $\endgroup$ – yuggib Nov 23 '15 at 7:13
  • $\begingroup$ @Shing to that measurement) is the collapsed state. Now, to obtain exactly the same type of collapse, we need to take the exact same process. However, the same procedure may be done for other observables, or in general be thought as a way to determine the state of a part of your system, given a global composite system in interaction. Nevertheless, collapse is not something that "wavefunctions do" (wavefunctions are an instrument to describe reality, not reality itself), but another useful instrument to describe the physics of some systems (like the ones involved in measurements). $\endgroup$ – yuggib Nov 23 '15 at 7:17
0
$\begingroup$

Will wave function collapse without measurement?

Yes and no. Collapse is a sometimes useful fiction. Sometimes you can pretend a subsystem has evolved into a particular state, a collapsed state.

Since all matters are described by wave functions, then in principle, I should be able to describe wave function collapse by Schrodinger's equation. (I don't know how exactly though)

Just write down the state of the device used to do the measurement and use the correct Hamiltonian describing the interaction. Just like any time evolution. Describe the system, use the correct Hamiltonian.

The only difference between collapse and no collapse is whether you look at the entire wave function or whether you look at pieces and pretend the pieces are the whole thing.

Collapse is a useful fiction when the pretending is both simpler than not pretending and when the pretending also won't lead to mistakes.

Collapse isn't a physical process that happens. It is a useful fiction to oversimplify a real thing that already happened. The real thing that already happened is that the system evolved into pieces that are now independent and will remain independent.

Here is a simple example. You have a superposition of a Gaussian free particle going right and one going left and they are overlapped. Since they are overlapped, interactions in the overlap will depend on both Gauusian waves.

But we could wait. When we wait one Gaussian goes left (and spreads out) and the other goes right (and spreads out) and we can start to appreciate the fiction that there is just one on the right going right and one on the left going left. But right now it would be premature, because they might hit a barrier and reflect back and start to overlap again.

However the wavefunction is defined in configuration space and there are other particles. So the wave going right could reach a region of configuration space where that particle is close to another particle and could deflect it. This means the whole eave in configuration space has turned, the wave heading in the $\hat x_1$ direction now is deflected to go a bit in the $\hat x_2$ direction as well. But that wave going left get close to another particle and so that waves gets deflected in the $\hat x_3$ direction.

And these different particles described by $x_2,y_2,z_2$ and $x_3,y_3,z_3$ can themselves bunch into particles described by $x_4,y_4,z_4$ and $x_5,y_5,z_5$ (with particle two interacting with four, and particle 3 interacting with particle 5). And it can snowball. Eventually the two waves have been deflected in so many different directions in the hugely dimensional configuration space that even if the original waves hit real barriers and reflect, it is unlikely to the extreme that the other wave hits a barrier in the exact place and time to direct it so the two overlap.

So effectively they can each act like they are the only wave. No one tried to measure any specific operator. But the system has evolved into pieces that can now be treated independently. And so it becomes acceptable and even potentially useful to pretend that it has evolved into one of those two waves.


Thanks for the detailed explanation, but I have heard of that such Hamiltonian would not be unitary, therefore, it is still questionable?

If you write down the wavefunction of the subject and the device, then the correct Hamiltonian describing their interaction will give a unitary time evolution. And it will make the state evolve into an entangled state where the subject and device are entangled with the different projections of the subject are entangled with different states of the device.

For instance you could evolve $\left(\alpha\psi_{s1}+\beta\psi_{s2}\right)\otimes\psi_{d0}$ into $\alpha\psi_{s1}\otimes\psi_{d1}+\beta\psi_{s2}\otimes\psi_{d2}$. This is a unitary evolution. And it is most definitely different than collapse which evolves $\left(\alpha\psi_{s1}+\beta\psi_{s2}\right)\otimes\psi_{d0}$ into $\psi_{s1}\otimes\psi_{d1}$ with probability $|\alpha|^2$ and into $\psi_{s2}\otimes\psi_{d2}$ with probability $|\beta|^2$.

However, when the device state is coupled to an environment then the two parts, $\alpha\psi_{s1}\otimes\psi_{d1}$ and $\beta\psi_{s2}\otimes\psi_{d2}$ can each act like their own world and act as if they were the only part of the wave function.

The issue is that when you treat the subject and the device and the environment all with quantum mechanics and all with the unitary time evolution given by the actual Hamiltonian you get something that acts the same way as collapse. So let's compare the two.

Under collapse you insist the world is described by $\alpha\psi_{s1}\otimes\psi_{d1}$ or is described by $\beta\psi_{s2}\otimes\psi_{d2}$ (overall scale doesn't actually affect any predictions, you can compute $\langle\psi|\hat O|\psi\rangle/\ langle\psi|\psi\rangle$ instead of $\langle\psi|\hat O|\psi\rangle$).

Doing the actual math tells us the world is described by $\alpha\psi_{s1}\otimes\psi_{d1}+\beta\psi_{s2}\otimes\psi_{d2}$ but rather than getting excited about how different these look on paper we can investigate how they look in the lab.

And the point is that the second, correct, one will look like the first one. A world described by $\alpha\psi_{s1}\otimes\psi_{d1}+\beta\psi_{s2}\otimes\psi_{d2}$ will have two parts, a $\alpha\psi_{s1}\otimes\psi_{d1}$ part and a $\beta\psi_{s2}\otimes\psi_{d2}$ part. And each will act like it is the only part in the world. So the $\alpha\psi_{s1}\otimes\psi_{d1}$ part will evolve as if the whole world is described by $\alpha\psi_{s1}\otimes\psi_{d1}$. And the $\beta\psi_{s2}\otimes\psi_{d2}$ part will evolve as if the whole world is described by $\beta\psi_{s2}\otimes\psi_{d2}$.

So the actual state $\alpha\psi_{s1}\otimes\psi_{d1}+\beta\psi_{s2}\otimes\psi_{d2}$ will experimentally be indistinguishable from a prediction that you evolve into $\alpha\psi_{s1}\otimes\psi_{d1}$ or into $\beta\psi_{s2}\otimes\psi_{d2}$.

The biggest barriers to people that don't sit down and just do the math for the actual interaction (if you sat down and did it there is zero way you could think it wasn't unitary) is that people are used to being told that there are two different evolutions, a unitary evolution from the Hamiltonian, and a non unitary collapse.

And the reality is that all science requires a way to relate aspects of the mathematical models to experimental and observational outcomes. And the point is that the people that want to have a collapse only relate the collapse to observational and experimental outcomes and they simply don't discuss what the experimental and observational outcomes of a regular wave function would be.

Once you discuss what the observational and experimental outcomes of a regular wavefunction look like, then you see the regular unitary time evolution makes something that gives the observational and experimental outcomes we actually see.

So each side then invokes Occam's razor.

Collapse people say that it is silly to have other worlds (the $\alpha\psi_{s1}\otimes\psi_{d1}$ part should pretend like there is no $\beta\psi_{s2}\otimes\psi_{d2}$ part and the $\beta\psi_{s2}\otimes\psi_{d2}$ part should pretend like there is no $\alpha\psi_{s1}\otimes\psi_{d1}$ part) and since they object to any part having ignorance they postulate that there should be only one part and so make up a whole new evolution just so each part can be the only world in the universe rather than just being an independent part of the universe.

No collapse people on the other hand say it is silly to have two totally different types of time evolution when the regular unitary time evolution makes predictions that exactly match what we actually see. They instead predict that the universe evolves to have parts that act independently and thus each becomes ignorant to what the other one is and is doing. So the $\alpha\psi_{s1}\otimes\psi_{d1}$ can pretend like there is no $\beta\psi_{s2}\otimes\psi_{d2}$ part (and vice versa) for the simple reason that they act independently.

The collapse people are really just modern day solipsists. And they can claim they just use Occam's razor, even though the no collapse people use it too.

If solipsism is appealing to you, then you will like collapse and the price to pay is to have two evolutions. And you'll never see any evidence of a collapse other than solipsism because there is no trigger or time when it happens because you postulated a whole new time evolution that had zero experimentally different consequences and all it did was allow you to assert that there can't be a larger universe than the part you are coupled to.

If solipsism isn't appealing to you then you will find it silly to make up a physical collapse process that makes zero new predictions. And you will be forced to conclude that the universe has many parts/worlds and experimentally each part can only get data about itself and that's an experimental limitation that can never be overcome.

[Basically], are you saying the whole wavefunction collapse is just something for "fit in data"?

I am saying no collapse and collapse make the same predictions, they just extract predictions from the math in different ways. Some people find it silly to make two different time evolutions when one of them can already be used to make the predictions that perfectly match what we see. Other people object to the fact that the math clearly predicts the universe evolving into parts that can't be aware of what the other parts are doing and would rather postulate a magical collapse process that makes it so the different parts are alone rather than ignorant.

No one makes different predictions. One group of people predicts parts that each act as if the part is alone. The other group of people predicts parts that are truly alone.

There is a slight possibility to distinguish them, in that the group that days they act " as if" they were alone, there is a super super tiny ability to interact with the other one, not a perfectly zero. So if the collapse people actually insisted that collapse happened at a particular time or place as a physical process and were willing to pay the no collapse people money to be proven wrong then it is in principle possible (though too hard to actually do with a reasonable level of resources and effort) to prove the collapse people wrong.

The problem is that the no collapse people say you can act "as if" collapse happened, exactly when it becomes too hard to detect the difference. So feel free to act as if collapse happened when its useful and appropriate to do so. Just don't get all excited about it. All it means is that the difference between collapse and what the math actually predicts is too hard to notice and therefore is safe to ignore.

$\endgroup$
  • $\begingroup$ Thanks for the detailed explanation, but I have heard of that such Hamiltonian would not be unitary, therefore, it is still questionable? besides, I don't quite understand a few parts of your answer. Basely, are you saying the whole wavefunction collapse is just something for "fit in data"? $\endgroup$ – Shing Nov 25 '15 at 12:34
  • $\begingroup$ @Shing I edited $\endgroup$ – Timaeus Nov 25 '15 at 15:19
-1
$\begingroup$

According to the standard presentation of QM, yes, there is a collapse.

The wave function evolves in space not in a local manner, in the way, say a particle travels along line, but in the way a balloon evolves when air is blown in.

A measurement, or interaction is local; the wave-function - the balloon collapses, say where I pricked it with a needle.

But this is where the puzzle resides: for the balloon collapses in a mechanical and causal fashion ie through tension.

But in the wave picture of QM, once the measurement or interaction occurs how does the wave collapse away from there? How is it informed - not through a more basic picture, because QM is the basic picture; it's this problematic of locality and realism that's explored in thought experiments like EPR and so on.

$\endgroup$
  • $\begingroup$ Absolutely everything is wrong about this answer. The wavefunction is not a function of space, it is a function of configuration space. It does evolve locally in configuration space (it is a partial differential equation). And collapse is a just an idealization of an interaction when you want to ignore some details about the interaction, such as how long it took and which things got correlated which other things, such as ignoring what specifically you did to interact with it. $\endgroup$ – Timaeus Nov 22 '15 at 17:56
  • $\begingroup$ @timeaus: isn't configuration space a function of space? Or are you saying that there is no relation between configuration space and space itself? Perhaps it worth reminding yourself just how a configuration space is built? $\endgroup$ – Mozibur Ullah Nov 22 '15 at 18:12
  • $\begingroup$ I'd also say your notion of cause is wrong; a wave evolves not because it is a PDE; but because we describe evolution through PDEs. $\endgroup$ – Mozibur Ullah Nov 22 '15 at 18:18
  • $\begingroup$ The wave is defined on configuration space, a vastly vastly larger space. It doesn't evolve in space, how can you possibly discuss, for instance, the antisymmetry of inter particle exchange without the larger configuration space? (Sure, in QFT you can use Fock space, but that's also a larger space.) $\endgroup$ – Timaeus Nov 22 '15 at 18:21
  • $\begingroup$ The value of the wave in a region of configuration space is determined solely by the values just a bit before and in a nearby (i.e. slightly larger and enveloping) region of configuration space. $\endgroup$ – Timaeus Nov 22 '15 at 18:23

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.