What Happens When a Photon’s Superposition Encounters a Black Hole?

[Isambard]

TL;DR Summary

A photon that travels lightyears through the universe before the wave function collapse

When a photon travels through space it is spreading out like a fan while in its superposition (except that it is spreading also vertically in addition to horizontally). So, what happens if for instance the right outer edge of the photon's superposition is captured by the gravity of a black hole? Will the rest of it continue its path while the part that was captured by the gravity, and which will orbit the black hole some times before it escape (without its wave function collapsing), continue the journey a little behind and perhaps in a different direction?
(Yes, sometimes the wave function will collapse, but it clearly does not happen to all photons since we otherwise would not have been able to see the light that has escaped its orbit around the black hole)

 

[Delta2]

I don't understand your post very well but from what I know, according to mainstream QM/QFT photons don't have a well defined wave function.

 

[EPR]

Why would the photon be in superposition while the black hole is allegedly not? Where do you draw the line between objects in superposition and objects with definite properties?

 

[Isambard]

To clarify what I meant. A photon travels through the universe as a wave. If a part of this wave is captured by the gravity of a black hole without its wave function collapsing (and without passing the event horizon), and then continue its journey (perhaps after orbiting it a few times), it could perhaps continue in another direction than it was originally moving in.
Imagine the wave as a large rubber sheet being constantly stretched horizontally and vertically without ever tearing. A part of this sheet is being held back for a moment and is then stretched in a different direction, for instance a little to the left, than the rest of the sheet which continue straight forward.
It is perhaps not the best comparison, but the point is just to illustrate what I mean.

"Where do you draw the line between objects in superposition and objects with definite properties?"

In the famous double slit experiment, a photon is no longer in superposition when it hits the wall as particles instead of wave. If I have misunderstood the idea of superposition, just replace the word with "wave" instead.

 

[EPR]

A photon is a fully relativistic quantum system which doesn't have a wavefunction. You should use QED/QFT to study its behavior as it's the theory that describes how matter and light interact.

 

[Isambard]

A photon is a fully relativistic quantum system which doesn't have a wavefunction. You should use QED/QFT to study its behavior as it's the theory that describes how matter and light interact.

But this is not about how matter and light interact. It's about light being affected by gravity.

As for waves; if a bullet from a gun had the same properties as a photon, and you were firing against a target, and you had a camera that somehow could film it in slow motion before it was hitting the target (in the famous double slit experiment the detectors make the wave function collapse and the photon act like a particle, but in this scenario they don't affect the outcome) you would see a whole swarm of "ghost bullets" in front of the target until on of them suddenly becomes "real" and all the others disappear, leaving behind a single hole.
That's the explanation I have read countless times. And that's what I mean with a wave.

 

[vanhees71]

Fortunately you don't need photons in the general-relativistic context but just classical electromagnetic waves. Forget photons as long as you don't really need them, because they are only described using relativistic quantum field theory, which is not an easy subject.

In textbooks on cosmology they usually tell about "photons" in a very crude way. What is meant is just the solution of Maxwell's equations in a FLRW spacetime using the eikonal (aka geometric-optics) approximation. Then all boils down to the change of the four-vector $k$ parallel transported along a (radial) geodesic.

Generally in bad books about QT, usually popular-science books, they talk about "superpositions" in sentences like this in #1. They say "the system is in a superposition." This leaves any physicist who usually learns QM in the 4th semester pretty at a loss, because it is a completely meaningless sentence. You can write any vector in terms of a "superposition" of other vectors. If you have an arbitary basis of the vector space you can write the vector as a linear span of these basis vecctors, and this decomposition is unique. Thus, if you say "the system is in a superposition" you need to first define a basis to make this superposition unique.

Further what's in a superposition wrt. a basis is what's called a pure quantum state, which can be specified by a vector in Hilbert space (conveniently normalized to 1) and then written as a linear combination of the given basis vectors. Then and only then the sentence makes some mathematical sense.

 

[PeroK]

Summary:: A photon that travels lightyears through the universe before the wave function collapse

When a photon travels through space it is spreading out like a fan while in its superposition (except that it is spreading also vertically in addition to horizontally). So, what happens if for instance the right outer edge of the photon's superposition is captured by the gravity of a black hole? Will the rest of it continue its path while the part that was captured by the gravity, and which will orbit the black hole some times before it escape (without its wave function collapsing), continue the journey a little behind and perhaps in a different direction?
(Yes, sometimes the wave function will collapse, but it clearly does not happen to all photons since we otherwise would not have been able to see the light that has escaped its orbit around the black hole)

This and your subsequent posts are filled with so many misconceptions and confused ideas that they are impossible to answer.

What you are posting is almost a parody of modern physics!

 

[Isambard]

Fortunately you don't need photons in the general-relativistic context but just classical electromagnetic waves. Forget photons as long as you don't really need them, because they are only described using relativistic quantum field theory, which is not an easy subject.

In textbooks on cosmology they usually tell about "photons" in a very crude way. What is meant is just the solution of Maxwell's equations in a FLRW spacetime using the eikonal (aka geometric-optics) approximation. Then all boils down to the change of the four-vector $k$ parallel transported along a (radial) geodesic.

Generally in bad books about QT, usually popular-science books, they talk about "superpositions" in sentences like this in #1. They say "the system is in a superposition." This leaves any physicist who usually learns QM in the 4th semester pretty at a loss, because it is a completely meaningless sentence. You can write any vector in terms of a "superposition" of other vectors. If you have an arbitary basis of the vector space you can write the vector as a linear span of these basis vecctors, and this decomposition is unique. Thus, if you say "the system is in a superposition" you need to first define a basis to make this superposition unique.

Further what's in a superposition wrt. a basis is what's called a pure quantum state, which can be specified by a vector in Hilbert space (conveniently normalized to 1) and then written as a linear combination of the given basis vectors. Then and only then the sentence makes some mathematical sense.

Thanks for replying and the explanation about superpositions, but since we are talking about lightyears here, the question is still about how light travels through space.

Maybe it is simpler if we take it one step at the time: A photon travels through space as a wave, which is a simplified version used by textbooks. Is there a limit for how much this wave can spread?
If you look at the attached image, you see that the wave in the experiment is spreading and stops when it hits the wall. What if there was no wall there? And no other obstacles behind the wall either? The wave could continue to spread into darkness forever. Would it continue to be wider and wider the further it travelled?
Of course, in the illustration it has passed through two slits, but I assume a single photon that travel through space will spread out even without passing any slits (again, according to available articles).

 

[PeterDonis]

A photon travels through space as a wave

No, it doesn't. As you have already been told, a photon does not have a well-defined wave function. That means you cannot properly think of it as a wave traveling through space. You can think of classical light that way, but that requires ignoring all quantum aspects of light. (Which is actually fine for actual astronomical applications.)

In particular, the mental picture you have of a photon wave in a gravitational field only works for classical light waves.

If you really want to try to model the quantum aspects of light in a gravitational field, you need to use quantum field theory in curved spacetime, which is a very different model.

 

[vanhees71]

What travels is the electromagnetic field. The naive photon picture invoked in GR textbooks (concerning the famous deflection of light on the Sun) and cosmology textbooks (concerning the important redshift-distance relation ship in FLRW spacetimes) is in fact a classical solution of the Maxwell equations in a given spacetime background (Schwarzschild and FLRW solution of the Einstein field equations of GR) in the eikonal approximation. All that's done is to solve for the corresponding light-like geodesics of these spacetimes. The physical meaning is that of a "light ray" as in geometrical optics. It describes the change of the four-wave-vector of an electromagnetic wave from the emission of the em. wave (a far distant star or galaxy) to the position of observation. From this the deflection angle and the frequency of the wave as measured by an observer at the observation point can be deduced in the usual way making use of an appropriate reference frame (vierbein) at the place of the observer.

 

[Isambard]

This and your subsequent posts are filled with so many misconceptions and confused ideas that they are impossible to answer.

What you are posting is almost a parody of modern physics!

It's called metaphors to help you visualize the scenario.

 

[PeterDonis]

It's called metaphors to help you visualize the scenario.

Physics is not done in metaphors. It's done in math. You need to go learn the math.

Thread closed.