How gravity affects wave function collapse

[EvilDonut]

Apologies if these questions have been answered before - I didn't have any luck with google. Something has bugged me for a while about quantum experiments like Schrödinger's cat where a 'black box' system is theorized to be in a superposition of states until observation causes wave function collapse - how does this explain the fact that a system can be observed by looking at its 'gravitational signature'? i.e. A dead cat lying down will have a weaker gravitational field than a live cat standing up to an observer above the box.

I know the actual cat experiment isn't the best example as the superposition conclusion was rejected (right?) but this extends to single particles too...hypothetically, isn't information on the position carried by the particle's 'gravitational signature'.

Secondly...is there a limit on the strength (or rather, weakness) of the gravitational field? i.e. is field strength discrete and thus is there a point when the gravitational field strength due to an object can be said to be actually zero, or is it 'continuous' and decays over an infinite distance but never reaches zero, as the inverse square formula implies?

If it is continuous rather than discrete, then would I be right to say that positional changes in even a single atom would technically be 'visible' to all observers anywhere in the universe (allowing for propagation speed of gravity). And, if so, how could anything ever be in a superposition of states?

 

[Simon Bridge]

Welcome to PF;
Generally gravitation acts as a perturbation on the wavefunction and has no affect, beyond this, on "collapse".
One could consider a gravitatonal interaction to be a measurement on the system though.

hypothetically, isn't information on the position carried by the particle's 'gravitational signature'.

Short answer: no. Why would it? (See "correspondence principle".)

However: your terms are not clear - what do you mean by "gravitational signature"?

is there a limit on the strength (or rather, weakness) of the gravitational field? i.e. is field strength discrete and thus is there a point when the gravitational field strength due to an object can be said to be actually zero, or is it 'continuous' and decays over an infinite distance but never reaches zero, as the inverse square formula implies?

Whether gravity comes in finite increments like charge does is the question of "quantization of gravity". It gets asked a lot - I understand that there are several models for quantum gravity with mixed successes but nothing conclusive has come up so far. Gravity is treated using General Relativity - in QM systems the effect is usually much smaller than the the other interactions the system undergoes - hence it becomes as continuous purturbation on the potential.

If it is continuous rather than discrete, then would I be right to say that positional changes in even a single atom would technically be 'visible' to all observers anywhere in the universe (allowing for propagation speed of gravity).

No. You cannot tell if the changed gravity you feel is due to a change in relative motion or a change in position of some mass. You need more information.

And, if so, how could anything ever be in a superposition of states?

Like: a particle may be confined to a strong gravitational field and prepared in a superposition of gravitational energy states.

It is presumably possible to prepare a state which is a superposition wrt some operator, but not to the gravitational interaction.

 

[EvilDonut]

However: your terms are not clear - what do you mean by "gravitational signature"

Probably not the best phrase for it, I guess ultimately that just means gravitational field strength, which could be detected by a (very sensitive!) bit of equipment, and the differences in field strength used to infer information about changes in the state of a system.

My specific thinking was - imagine something like an unstable particle in isolation. Would it be valid to say that the particle is "in a superposition of the states 'decayed' and 'not decayed'"? However, the gravitational field strength at the point of an observer will be different for each of these two states.

I take your point about not being able to tell the difference between change in gravitational field strength and acceleration in practice - I suppose I was assuming that the observer remains stationary, but I'm not sure that is a valid assumption for this little thought experiment.

One could consider a gravitational interaction to be a measurement on the system though.

I guess that would answer the first question. In this case, wouldn't such a gravitational interaction affect every single particle in the universe, propagating out from the source at the speed of light...and thus every particle is kind of an observer.

 

[Bill_K]

Apologies if these questions have been answered before - I didn't have any luck with google. Something has bugged me for a while about quantum experiments like Schrödinger's cat where a 'black box' system is theorized to be in a superposition of states until observation causes wave function collapse - how does this explain the fact that a system can be observed by looking at its 'gravitational signature'? i.e. A dead cat lying down will have a weaker gravitational field than a live cat standing up to an observer above the box.

I'd say you're right, EvilDonut. The usual discussion of the Schrodinger's cat experiment relies on the box contents being completely isolated from the outside world, and this is obviously impossible for the gravitational field. Good point.

 

[Simon Bridge]

I suppose I was assuming that the observer remains stationary, but I'm not sure that is a valid assumption for this little thought experiment.

The observer is always stationary in their own reference frame. That's not really helpful here.

I guess that would answer the first question. In this case, wouldn't such a gravitational interaction affect every single particle in the universe, propagating out from the source at the speed of light...and thus every particle is kind of an observer.

Everything is a potential observer and each interaction is a kind of observation, yes.
It takes a while to get a handle on QM measurement.

I think the part that is giving you trouble is the idea that a measurement "collapses the wavefunction" so that it is no longer (if it was before) in a superposition of eigenstates.

However, consider a particle in an infinite square well, we'll do this in 1D for simplicity of math.
We make a measurement of energy and find it is in the ground state ... whatever is was before it is in the ground state now and it stays that way provided there are no further interactions.

Now make a measurement of position - and find the particle a bit off to one side inside some range of x values. What is the wavefunction now? What happens if you make a new measurement of energy?

Basically the situations you deal with when you are first learning are over-simplified in order to illustrate some features you need to learn. The infinite square well is a case in point - I just talked about making measurements on the system but how can I do this if the well is infinite - how can there be an interaction with the outside for the measurement to take place? How does the information about the particle get to me?

But it makes sense because there are many possible interactions and the infinite well is an approximation only. Nothing is ever truly isolated - (cosmological considerations notwithstanding.) But that doesn't mean we cannot treat some systems as isolated to a high degree of accuracy. The gravity of Alpha-centauri affects pendulum experiments on Earth, but we don't need to take the effect into account to find the period of a pendulum.

 

[tom.stoer]

How gravity affects wave function collapse

I think the idea goes back to Penrose, so I would recommend to check his work. Caveat: as far as I can remember he has nothing else to say but just this - an idea.

 

[The_Duck]

Here is a similar Physics Stack Exchange question.

The conclusion is that you can get wave function collapse (decoherence) if the gravitational field of the "live" and "dead" states are different enough. How different do they have to be? You get wave function collapse when there has been a "measurement," meaning that some information about the two possible states of the system has been stored in the surrounding environment. To count as a "measurement," you have to be able to tell by looking at the *environment* which of the two states the measured system is in. For example, the environment might contain a scientific instrument with a needle that can point to ALIVE or DEAD. If you can tell by looking at that needle whether the cat is alive or dead, then a measurement has taken place.

So can we look at the motion of particles in the surrounding environment, deduce the gravitational field of the box, and thereby measure the health of the cat? Yes, but there is an obstacle. The state of the environment also has some quantum uncertainty. If the gravitational field of the box does not change much between the "live" and "dead" states, the difference between the effects of their gravitational fields will be completely drowned out by this quantum mechanical noise. We won't be able to deduce the state of the box from its gravity if its gravity does not change enough to overcome the quantum uncertainty in the environment.

So gravity-induced wave function collapse will only occur if the difference in gravitational fields is big enough. I've no idea how to do a practical calculation of when it will start to occur.

Note that there is not actually a hard boundary between "collapse" and "no collapse:" there is a smooth interpolation as the difference in gravitational fields increases. This is one reason people in the linked thread prefer to talk about "decoherence" rather than "wave function collapse."

Secondly...is there a limit on the strength (or rather, weakness) of the gravitational field? i.e. is field strength discrete and thus is there a point when the gravitational field strength due to an object can be said to be actually zero, or is it 'continuous' and decays over an infinite distance but never reaches zero, as the inverse square formula implies?

The field strength is not discrete. It's continuous and decays over an infinite distance but never reaches zero. We want to "quantize" gravity, but the main effect of that would be that *gravitational waves* would come in discrete packets called gravitons, just like electromagnetic waves come in discrete packets called photons.

If it is continuous rather than discrete, then would I be right to say that positional changes in even a single atom would technically be 'visible' to all observers anywhere in the universe (allowing for propagation speed of gravity). And, if so, how could anything ever be in a superposition of states?

Because of the caveat mentioned above: to count as a measurement the interaction between the system and the environment has to overcome the preexisting quantum uncertainty in the environment. Very weak effects of very distance particles can be ignored for this reason.

 

[tom.stoer]

So if the cat is jumping around (when the bottle with poisson is destroyed) her wave functions collapses faster compared to the situation when she is sleeping (b/c the sleeping cat is gravitationally rather similar to a dead cat ;-)

 

[cosmik debris]

I think the idea goes back to Penrose, so I would recommend to check his work. Caveat: as far as I can remember he has nothing else to say but just this - an idea.

Yes, Penrose wrote some papers with Stuart Hameroff in which they discuss this sort of thing, it is highly speculative I believe.

 

[kos]

Welcome to PF;
Generally gravitation acts as a perturbation on the wavefunction and has no affect, beyond this, on "collapse".
One could consider a gravitatonal interaction to be a measurement on the system though.

Short answer: no. Why would it? (See "correspondence principle".)
However: your terms are not clear - what do you mean by "gravitational signature"?

Whether gravity comes in finite increments like charge does is the question of "quantization of gravity". It gets asked a lot - I understand that there are several models for quantum gravity with mixed successes but nothing conclusive has come up so far. Gravity is treated using General Relativity - in QM systems the effect is usually much smaller than the the other interactions the system undergoes - hence it becomes as continuous purturbation on the potential.

No. You cannot tell if the changed gravity you feel is due to a change in relative motion or a change in position of some mass. You need more information. Like: a particle may be confined to a strong gravitational field and prepared in a superposition of gravitational energy states.

It is presumably possible to prepare a state which is a superposition wrt some operator, but not to the gravitational interaction.

What do you mean by

"No. You cannot tell if the changed gravity you feel is due to a change in relative motion or a change in position of some mass. You need more information. Like: a particle may be confined to a strong gravitational field and prepared in a superposition of gravitational energy states."

Please explain this idea in details and as for dummies

 

[Simon Bridge]

Where did I lose you?

 

[kos]

"You cannot tell if the changed gravity you feel is due to a change in relative motion or a change in position of some mass."

Why?
I know the idea and equivelence principle but I cannot relate it with your words in my imagination !

 

[kos]

Give me just an example

 

[stevendaryl]

Yes, Penrose wrote some papers with Stuart Hameroff in which they discuss this sort of thing, it is highly speculative I believe.

Yes, I have a lot of respect for Penrose, but in my opinion, his speculations about the nature of mind, and gravity-induced wave function collapse are pretty flaky.

 

[stevendaryl]

Give me just an example

Well, do you understand the basic idea of the equivalence principle? Imagine a rocket with its engines firing just strong enough to hover in place above the surface of a planet. The people on board the rocket feel gravity pulling them to the floor at a certain strength.

Now, imagine a different scenario: the same rocket, firing its engines in the same way, but now the rocket is in empty space, far from any planets. In this scenario, the power of the engines doesn't keep the rocket in place, but makes the rocket accelerate faster and faster. In this scenario, the people on board the rocket will again be pressed against the floor, but this time due to "g-forces".

The equivalence principle just says that if you don't look out the window (and if the variation in gravity or g-forces from the bottom of the rocket to the top is too slight to notice) then the two situations are indistinguishable: There is no experiment that can distinguish between the two.

A slight variation on this thought experiment is that a rocket hovering above a planet whose mass is increasing (maybe because meteors are hitting it) will be indistinguishable from a rocket in outer space whose engines are gradually increasing their power.

 

[Simon Bridge]

Do you not understand how a change in gravitational force could be due to movement?
If you are in an elevator, you experience your weight change ... if the elevator accelerates upwards, you get heavier. However, someone could have just moved a big mass under your feet. How will you be able to tell the difference without looking outside the elevator car?

 

[kos]

Ok , let's assume that someone with some technology,situated somewhere in the universe does surveillance via gravitational detector . It takes fingerprints of changes of gravitational field the same way we tap camera video recording but in this case the source is not electromagnetic radiation but gravity . So let's suppose his stuff is under some extraordinary conditions and it is settled in absoulute zero temperature without any movement of its system.So than he can spot on any point and tell how for instance the atom with its nucleus particles and electrons have changed from point A to point B without any other information but the difference in traces due to the changes in gravitational fields left by different positions occupied by particles respectively in points A and B . So comment this scenario please :)

 

[Simon Bridge]

It takes fingerprints of changes of gravitational field the same way we tap camera video recording...

Try to avoid metaphor when you try to make things clearer ... "fingerprint" and "tap" used here obscure meaning. There is no need to tap a camera feed in the usual sense of making a secondary connection to the wires for recording at a secondary site because (a) no camera, and (b) it's our detector so we can just look at the primary data. No idea how the concept of taking a fingerprint could be applied to this context. Fortunately, I don't think we need this sentence at all.

Measuring the local gravitational field is nothing like taking a photograph or movie. It's more like detecting wind-speed or the height of the tide at a spot. You can deploy several to get a sense of direction that a passing wave may come from.

So let's suppose his stuff is under some extraordinary conditions and it is settled in absolute zero temperature without any movement of its system.

I get we want to just assume an ideal detection system without going into details of how it works. That's fine - we are interested in gravity, not the equipment used to detect it. Just think of the detector as a box with a dial on it - the dial gives the local strength of gravity where the box is, and we consider this box to be as reliable and accurate as we need it to be.

However - there is no way to tell if the box is stationary: movement is relative.
The fact that it is detecting non-zero gravity tells us that it is not in an inertial reference frame.

So than he can spot on any point and tell how for instance the atom with its nucleus particles and electrons have changed from point A to point B without any other information but the difference in traces due to the changes in gravitational fields left by different positions occupied by particles respectively in points A and B .

... OK, so can the box, or a collection of boxes, be used to track the motion of a point gravitational source (lets not use "atom" ... we needn't be that specific) just by tracking changes in the gravitational field?

Just to clarify:
This would amount to "looking outside the elevator car" and making an interpretation on what you see.

This is just like you look out the window of a car and see the road, trees, etc flying past and you look at the fuel gauge and notice that petrol is getting used up, and so you reasonably conclude that it is the car that is moving wrt the ground and not the ground moving wrt the car. This does not invalidate the notion that absolute motion cannot be detected.

Similarly, if we want to know the origin of a gravitational change, we need to go look for it. We need external information.

 

[DrChinese]

Ok , let's assume that someone with some technology,situated somewhere in the universe does surveillance via gravitational detector . It takes fingerprints of changes of gravitational field the same way we tap camera video recording but in this case the source is not electromagnetic radiation but gravity . So let's suppose his stuff is under some extraordinary conditions and it is settled in absoulute zero temperature without any movement of its system.So than he can spot on any point and tell how for instance the atom with its nucleus particles and electrons have changed from point A to point B without any other information but the difference in traces due to the changes in gravitational fields left by different positions occupied by particles respectively in points A and B . So comment this scenario please :)

In my mind, no quantum collapse (due to gravity) can occur unless gravity is a quantum force. You could test that idea with a pair of entangled particles in a suitably strong gravitational field. If entanglement survives such a field, there must no quantum gravity. Either that, or the "observation by gravity" must have been erased.

http://arxiv.org/abs/0910.2322

"We propose a thought experiment to detect low-energy Quantum Gravity phenomena using Quantum Optical Information Technologies. Gravitational field perturbations, such as gravitational waves and quantum gravity fluctuations, decohere the entangled photon pairs, revealing the presence of gravitational field fluctuations including those more speculative sources such as compact extra dimensions and the sub-millimetric hypothetical low-energy quantum gravity phenomena and then set a limit for the decoherence of photon bunches and entangled pairs in space detectable with the current astronomical space technology."

 

[Igael]

http://arxiv.org/abs/0910.2322
"... decohere the entangled photon pairs, revealing the presence of gravitational field fluctuations ..."

The difference between quantum and random needs a statistic and is very small.
As Ligo works, the effect has a direction and affects differently the 2 beams. Here , how to compare the pairs and extract a result ?

 

[kos]

I cannot understand what if the box is or is not stationary ? Does it affect the results from measuring of gravitational field ?

 

[kos]

And what are the options for this complinaitary information ?

 

[kos]

May we remove completely the Heisenberg uncertainty having a measuring via gravity field ?
Second, if something is in absoulute zero temperature knowing relation between temperature and velocity of a single molecule for example knowing the exact values of velocities of all molecules and their number ( the same as the temperature of the system ) knowing the relation between accelaration ,velocity,energy stress tensor disturbence of space time ,( gravitatiinal field of the box ) we know can exactly determinate the value of the incoming changes of measured external fields ?
What about scenario ?