Does the wave function spread more quickly after it is observed?

[Sciencemaster]

TL;DR Summary

If a particle is observed and its wave function collapsed, will its wave function spread out across space more quickly than before it was observed or than it would were it not observed?

For the sake of this question, I am primarily concerned with the position wave function. So, from my understanding, the wave function seems to 'collapse' to a few states apon measurement. We know this because, if the same particle is measured again shortly after this, it will generally remain in the same place (similar to what is described by Zeno's paradox). However, from what I have seen, it seems like the wave function generally spreads more quickly after measurement, which means that the likelihood of the particle being measured in the 'middle' decreases, and the probable locations of it being measured spread out. I think that this is due to the uncertainty in position becoming small enough that the uncertainty in momentum must increase to accommodate the Heisenberg Uncertainty Principle--which in turn causes the wave packet to spread more quickly, as there is some 'probability' that it moves more quickly in either direction, and there are more possible momenta--as the measurement--and thus collapse--of the wavefunction makes the wave packet 'tighter' and makes the uncertainty in position smaller. Is this correct that the wave function of a particle will spread more quickly after it is measured than it does in the previous state (assuming that the measurement decreases the uncertainty in position)? If so, can modern measurement devices measure a particle to the point that this occurs? Does this occur due to an increase in deviation of the momentum probability distribution? Will the wave spread faster if a wave packet is 'tighter' when the scenario starts, but is not measured?

A good way of visualizing this idea is through this PHET simulation I came across online: https://phet.colorado.edu/en/simulation/quantum-tunneling.

Notice that the wave packet spreads out more quickly after it is collapsed than it does before it is, or than it does after the same amount of time if you don't collapse it. Is this accurate to our understanding in real life (As in, experimental observation)?
Side question, is this simulation reasonably accurate to how a wave packet works in real life?

 

[atyy]

Although it may appear that the wave function spreads out more quickly after collapse than before collapse, the spread after and before collapse are governed by the same equation (the Schroedinger equation). The Schroedinger equation does not apply only at the moment of collapse. And yes, we do know that the Schroedinger equation is accurate.

https://www.nobelprize.org/uploads/2018/06/haroche-lecture.pdf

 

[PeroK]

Summary:: If a particle is observed and its wave function collapsed, will its wave function spread out across space more quickly than before it was observed or than it would were it not observed?

This is the basis of one explanation of diffraction. A particle moving in the x-direction must have some spread of position in the y-direction (but this may be relatively slowly increasing) and a small spread in y-momentum. If the particle passes through a narrow slit, then this acts as a tight measurement of y-position and the particle then has a larger uncertainly in y-momentum after the slit. Which leads to a increasing uncertainty in y-position.

That is single-slit diffraction.

 

[martinbn]

Although it may appear that the wave function spreads out more quickly after collapse than before collapse, the spread after and before collapse are governed by the same equation (the Schroedinger equation). The Schroedinger equation does not apply only at the moment of collapse. And yes, we do know that the Schroedinger equation is accurate.

https://www.nobelprize.org/uploads/2018/06/haroche-lecture.pdf

The equation is not all that matters for the evolution, the state at a given time is also important. Take for example the heat equation because it is more intuitive. If the temperature at a given time is constant it will remain constant, if it is not there will be heat flow, although in both cases the equation is the same. So if the state is different after collapse it may spread out differently even though the equation is the same.

 

[AndreasC]

Depends, if it is observed to be in an energy eigenstate then it shouldn't spread out at all because energy eigenstates are stationary states (for time invariant hamiltonians). If you measure the position then that's usually what happens especially when you measure the position with great accuracy, since then you get a very large momentum uncertainty. But I don't think it is a rule set in stone that applies to every case, I suppose there can be a "rough" position measurement that doesn't make it spread faster.

 

[AndreasC]

OK I see the skeptical emojis have come out so I'm guessing I probably said something dumb lol

 

[PeroK]

OK I see the skeptical emojis have come out so I'm guessing I probably said something dumb lol

It was more that I couldn't follow what you meant.

 

[AndreasC]

It was more that I couldn't follow what you meant.

Ah alright. Langauge barrier is an issue for me sometimes. I edited my post to make it slightly clearer because I figured it was a bit confusing.

 

[Nugatory]

OK I see the skeptical emojis have come out so I'm guessing I probably said something dumb lol

Not dumb as a general comment about observables in general, but not applicable in this case: the state after a position measurement is not an energy eigenstate .

 

[AndreasC]

Not dumb as a general comment about observables in general, but not applicable in this case: the state after a position measurement is not an energy eigenstate .

Yes, but the original question wasn't uniquely about position measurements.

 

[Sciencemaster]

Yeah, my original question was primarily--but not uniquely--about position measurements. After reading through some of these replies, something caught my attention in the first one. As stated by the second comment, just because two different scenarios follow the same equation does not mean that they must move in the same way. Take kinematics for example, the position of a car starting at the origin will be different after some elapsed time depending on its velocity, acceleration, and so forth. Just because the particle follows the same equation before and after collapse does not mean that it moves in the same way (spreads at the same rate). I know that this is one explanation for the single-slit diffraction of a probability wave, but can the spreading of the wave function really be attributed to its collapse? What if we were to collapse said wave function by means of something other than a slit? How about this thought experiment: if we have a photon go through a clear film that records its position when it passes through, would it spread faster after passing through than it would otherwise? What if at some point we were to use some sort of field detector or something of the sort to determine where the photon is from a distance? Would it spread due to the collapse afterward?

 

[atyy]

Yeah, my original question was primarily--but not uniquely--about position measurements. After reading through some of these replies, something caught my attention in the first one. As stated by the second comment, just because two different scenarios follow the same equation does not mean that they must move in the same way. Take kinematics for example, the position of a car starting at the origin will be different after some elapsed time depending on its velocity, acceleration, and so forth. Just because the particle follows the same equation before and after collapse does not mean that it moves in the same way (spreads at the same rate). I know that this is one explanation for the single-slit diffraction of a probability wave, but can the spreading of the wave function really be attributed to its collapse? What if we were to collapse said wave function by means of something other than a slit? How about this thought experiment: if we have a photon go through a clear film that records its position when it passes through, would it spread faster after passing through than it would otherwise? What if at some point we were to use some sort of field detector or something of the sort to determine where the photon is from a distance? Would it spread due to the collapse afterward?

Depending on exactly what measurement is carried out, for the same wave function before the measurement process, one may get different wave functions immediately after the collapse (which occurs immediately after the measurement outcome is registered). The wave function immediately after collapse serves as the (new) initial state whose time evolution (spreading or whatever else) is governed by the Schroedinger equation.

So between measurements, quantum mechanics is like the car of classical physics. One simply has the initial conditions and the dynamics, and that governs everything that follows. The difference is that in quantum mechanics, measurement is a special process that produces a "classical" outcome that you read on your measuring apparatus, and a "quantum" outcome that is the state after the collapse. Quantum mechanics has two dynamical laws: (i) the Schroedinger equation, which applies between measurements, and (ii) collapse, which occurs when a measurement result becomes known.

 

[atyy]

I think that this is due to the uncertainty in position becoming small enough that the uncertainty in momentum must increase to accommodate the Heisenberg Uncertainty Principle--which in turn causes the wave packet to spread more quickly, as there is some 'probability' that it moves more quickly in either direction, and there are more possible momenta--as the measurement--and thus collapse--of the wavefunction makes the wave packet 'tighter' and makes the uncertainty in position smaller. Is this correct that the wave function of a particle will spread more quickly after it is measured than it does in the previous state (assuming that the measurement decreases the uncertainty in position)? If so, can modern measurement devices measure a particle to the point that this occurs? Does this occur due to an increase in deviation of the momentum probability distribution? Will the wave spread faster if a wave packet is 'tighter' when the scenario starts, but is not measured?

The spread of the wave function has nothing to do with the uncertainty principle. The spread of the wave function is governed by the Schroedinger equation.

 

[hutchphd]

The spread of the wave function has nothing to do with the uncertainty principle. The spread of the wave function is governed by the Schroedinger equation.

Of course it is governed by Shroedinger equation. It therefore has something to do with the uncertainty principle. Your categorical statement is incorrect on its face.

 

[atyy]

Of course it is governed by Shroedinger equation. It therefore has something to do with the uncertainty principle. Your categorical statement is incorrect on its face.

Only in a hair-splitting sense.

 

[hutchphd]

It is logically inconsistent. It needn't be so categorical, and it is either correct or not. Not hair-splitting in my book, sorry.

 

[Sciencemaster]

I just used the Heisenberg Uncertainty Principle to show that decreasing uncertainty in position should increase the uncertainty in momentum (and thus velocity for particles with mass, and thus the rate of the spread of the wave function). Also, I understand what the Schrödinger equation does (describes what a particle does in terms of dynamics), but just because it describes all particles does not mean that the particles do not 'spread out' at different rates. So, say, if we put in the 'tighter' post-collapse wave function to the Schrödinger equation, will it now spread faster due to said 'tightness' (Not great terminology, but you get what I mean).

 

[hutchphd]

My point was that the two analyses will give consistent answers. (they are certainly not mutually exclusive) Obviously Schrödinger applied correctly will give a more detailed description.

 

[atyy]

I just used the Heisenberg Uncertainty Principle to show that decreasing uncertainty in position should increase the uncertainty in momentum (and thus velocity for particles with mass, and thus the rate of the spread of the wave function). Also, I understand what the Schrödinger equation does (describes what a particle does in terms of dynamics), but just because it describes all particles does not mean that the particles do not 'spread out' at different rates. So, say, if we put in the 'tighter' post-collapse wave function to the Schrödinger equation, will it now spread faster due to said 'tightness' (Not great terminology, but you get what I mean).

Would you still think that for the time evolution examples shown in C - H of the first figure of https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator?

 

[vanhees71]

The spread of the wave function has nothing to do with the uncertainty principle. The spread of the wave function is governed by the Schroedinger equation.

The spread of the wave function has everything to do with the uncertainty principle. The "spread" is mathematically nothing else than the standard deviation ("uncertainty") of the position of the particle $\Delta \vec{x}$. The corresponding momentum-space representation of the same state is given by the Fourier transform of the position-space wave function and also implies a standard deviation of momentum.

From this mathematical relation between position and momentum representation of the (pure) state QT predicts that measuring (on an ensemble) either position or (on another equatlly prepared ensemble!) the standard deviation of the momentum, results in confirming the uncertainty relation
$$\Delta x_i \Delta p_i \geq \hbar/2.$$
Of course, how the wave function and thus the probability distributions for the outcome of measurements of observables depends on time is governed by the Schrödinger equation, which implies of course the development of the corresponding "uncertainties" ("standard deviations"). Whether the position uncertainty increases of decreases of course depends on the initial state as well as on the Hamiltonian.

 

[atyy]

The spread of the wave function has everything to do with the uncertainty principle. The "spread" is mathematically nothing else than the standard deviation ("uncertainty") of the position of the particle $\Delta \vec{x}$. The corresponding momentum-space representation of the same state is given by the Fourier transform of the position-space wave function and also implies a standard deviation of momentum.

From this mathematical relation between position and momentum representation of the (pure) state QT predicts that measuring (on an ensemble) either position or (on another equatlly prepared ensemble!) the standard deviation of the momentum, results in confirming the uncertainty relation
$$\Delta x_i \Delta p_i \geq \hbar/2.$$
Of course, how the wave function and thus the probability distributions for the outcome of measurements of observables depends on time is governed by the Schrödinger equation, which implies of course the development of the corresponding "uncertainties" ("standard deviations"). Whether the position uncertainty increases of decreases of course depends on the initial state as well as on the Hamiltonian.

No, you have misunderstood our terms. Here "spread" refers to time evolution not to standard deviation. The time evolution is governed by the Schroedinger equation and the Hamiltonian.

 

[vanhees71]

Of course, but usually by "spread" I understand what's meant is how "broad" the position (or momentum) distribution of $|\psi(t,\vec{x})|^2$ or $|\tilde{\psi}(t,\vec{p})|^2$ is.

 

[atyy]

Of course, but usually by "spread" I understand what's meant is how "broad" the position (or momentum) distribution of $|\psi(t,\vec{x})|^2$ or $|\tilde{\psi}(t,\vec{p})|^2$ is.

Yes, that's the usual meaning of spread. However, the OP was arguing that the increase in spread (ie. time evolution) is due to the uncertainty principle, which is not correct.

 

[vanhees71]

You can as well have a decrease in spread. That's clear from time-reversal invariance of, e.g., the free-particle Hamiltonian $\hat{H}=\hat{p}^2/(2m)$. If you just start with a Gaussian wave packet with minimal uncertainty, i.e., $\Delta x \Delta p$ you get an increase in $\Delta x$ (while $\Delta p$ of course stays constant) with time. If you stop the wave packet at a certain time $t_0$ and then time-reverse the state, you get a wave packet whose $\Delta x$ decreases first (after a time $t_0$ you then reach the minimal-uncertatinty wave packet, and then from that time on $\Delta x$ increases again).

 

[Sciencemaster]

Would you still think that for the time evolution examples shown in C - H of the first figure of https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator?
[/QUOTE]
No, because both of these examples have the same standard deviation (the square of both has the same uncertainty in position).

 

[Sciencemaster]

Yes, that's the usual meaning of spread. However, the OP was arguing that the increase in spread (ie. time evolution) is due to the uncertainty principle, which is not correct.

Yeah, what I meant by spread was the change in, say, the standard deviation as time passes. What I am trying to find out (I am the OP) is whether or not the rate at which the standard deviation of the wave function of a particle increases will itself inherently increase if the particle is itself measured to a small enough uncertainty in position. As a secondary motivation, I would also like to find out--if so--why, but first I have to find out if it occurs.

 

[Sciencemaster]

You can as well have a decrease in spread. That's clear from time-reversal invariance of, e.g., the free-particle Hamiltonian $\hat{H}=\hat{p}^2/(2m)$. If you just start with a Gaussian wave packet with minimal uncertainty, i.e., $\Delta x \Delta p$ you get an increase in $\Delta x$ (while $\Delta p$ of course stays constant) with time. If you stop the wave packet at a certain time $t_0$ and then time-reverse the state, you get a wave packet whose $\Delta x$ decreases first (after a time $t_0$ you then reach the minimal-uncertatinty wave packet, and then from that time on $\Delta x$ increases again).

The problem with this is that this question is, at its core, at least partially about wave function collapse, which is both non-time reversible and discontinuous (if I am not mistaken, feel free to correct me if I am wrong). Other than that, I suppose that you are right, but "normally" the wave packet spreads out over time rather than shrinking. My way of thinking is that this is because, with a small enough uncertainty in position due to measurement, the momentum values (i.e. the Fourier transform in momentum space) will have to itself spread out during the measurement, in turn making the wave function inherently 'spread out' faster after collapse to a small/"tight" enough wave packet. The core of my question is about whether or not this is accurate--or at least partially correct--and if not, what does happen in such a scenario?

 

[Sciencemaster]

Perhaps we should plug in a general-case wave packet to the Schrodinger equation with a fairly wide spread, then collapse it by making it have a "tighter" wave packet, and then seeing how it evolves. I wish that there were a nice, visual simulation to simply plug this into, as that would make it easier to analyze (In my opinion), and to really, intuitively comprehend (Again, in my opinion). Unfortunately, the only remotely good simulation of the wave function that I have found online is the one linked at the beginning, and that one is basic.

 

[Sciencemaster]

Of course, I could be wrong, I am human after all. The skeptical emoji is making me nervous...

 

[atyy]

The problem with this is that this question is, at its core, at least partially about wave function collapse, which is both non-time reversible and discontinuous (if I am not mistaken, feel free to correct me if I am wrong). Other than that, I suppose that you are right, but "normally" the wave packet spreads out over time rather than shrinking. My way of thinking is that this is because, with a small enough uncertainty in position due to measurement, the momentum values (i.e. the Fourier transform in momentum space) will have to itself spread out during the measurement, in turn making the wave function inherently 'spread out' faster after collapse to a small/"tight" enough wave packet. The core of my question is about whether or not this is accurate--or at least partially correct--and if not, what does happen in such a scenario?

As both @vanhees71 and I have said, the uncertainty principle has nothing to do with the time evolution of the wave function. So the intuition you propose in the OP is wrong.

 

[PeroK]

Of course, I could be wrong, I am human after all. The skeptical emoji is making me nervous...

The Internet is awash with simulations and lectures on the free particle wave-packet:

 

[Sciencemaster]

Honestly, I was hoping for an interactive simulation. I was hoping to create something like the first video, but then partway through collapse the wave function. Reflecting on this further, I can better say this as the collapse of the wave function creating a wider spread in the momentum space wave function through the Fourier Transform, which does indeed seem to happen, please correct me if I am wrong.

 

[atyy]

Honestly, I was hoping for an interactive simulation. I was hoping to create something like the first video, but then partway through collapse the wave function. Reflecting on this further, I can better say this as the collapse of the wave function creating a wider spread in the momentum space wave function through the Fourier Transform, which does indeed seem to happen, please correct me if I am wrong.

That's correct. A wave function with a narrow spread in position has a broad spread in momentum. That is the uncertainty principle. However, this does not say anything about how the spread changes with time.