# How is energy transferred from a wave to a photon?

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• curiouschris
In summary: The energy it has is represented by its frequency. Thus if only 10% of the photons wave passes through the slit then the photon should lose 90% of its energy and as energy...This is how the photon behaves in the double slit experiment.
curiouschris
TL;DR Summary
Can someone better explain the transfer of energy during the collapse of a wave function in quantum mechanics, in particular I am thinking of the double slit experiment?
This is just a curiosity to me. My interest is from the position of a layman (as you will see from my description below).

In the double slit experiment it shows a wave passing through both slits and interfering with itself to create an interference pattern.

This is how I understand it...

From the visualisation of the experiment much of the wave is simply absorbed (or reflected) by the front walls of the experiment with the remaining portions of the wave passing through the slits and spreading out and then interfering with itself, finally the wave collapses into a photon and registers on the back wall of the experiment as a photon. This happens many times over until the pattern is created as defined by the probabilities of where the photon is.

However it is never explained what happens to the energy the wave contains.

Does the photon contain all the energy of the original wave? How is that the case if much of the wave is 'blocked' as described above?

curiouschris said:
Does the photon contain all the energy of the original wave?
For a single photon source, yes, although this wording sounds like you are making a distinction between the photon and the original wave. They are the same thing for a single photon source.

curiouschris said:
How is that the case if much of the wave is 'blocked' as described above?
Many of the single photons are absorbed by the wall. The ones that get through the slit are just a fraction of the ones emitted by the source.

vanhees71
Dale said:
Many of the single photons are absorbed by the wall. The ones that get through the slit are just a fraction of the ones emitted by the source.
This explanation assumes a light source like a torch, with trillions of photons blasting out in all directions. However I am thinking of where individual photons are shot at the experiment.

curiouschris said:
This explanation assumes a light source like a torch, with trillions of photons blasting out in all directions. However I am thinking of where individual photons are shot at the experiment.
Please don’t quote me out of context. That is dishonest and rather silly since anyone can read my actual statement. I said “single photon source” explicitly, multiple times.

curiouschris said:
This explanation assumes a light source like a torch, with trillions of photons blasting out in all directions.
It does no such thing. The explanation given applies to a single photon source, i.e., a source that emits single photons one at a time, just fine.

vanhees71
Dale said:
Please don’t quote me out of context. That is dishonest and rather silly since anyone can read my actual statement. I said “single photon source” explicitly, multiple times.
Sorry if I misunderstood. When I read "single photon" I understood you as meaning individual photons, like would happen in the case of a torch, rather than a single photon per unit time.

The only thing I can think of given the illustrations used to show the double slit experiment is that where a photon's probability is that it will pass through the slit the waveform collapses to a photon and then emerges from the other side of the slit and reforms as a wave to ultimately interfere with photons from the other slit.

So I am most interested with what happens to the wave front at the point when the photon enters the slit.

curiouschris said:
the waveform collapses to a photon and then emerges from the other side of the slit and reforms as a wave
I think this may be your problem. The photon is the wave. There is no collapsing to a photon and reforming to a wave because they are the same thing.

curiouschris said:
So I am most interested with what happens to the wave front at the point when the photon enters the slit.
The photon wavefunction propagates through the slit as waves normally do.

vanhees71
Dale said:
The photon wavefunction propagates through the slit as waves normally do.
This is the crux of my problem, Where I am struggling to understand the implications in what I see and what is said or implied.

Imagine a sea wall with a break in it. When a wave hits the sea wall a small portion of the wave will pass through the break and continue on forming what is a new wave front.

The rest of the wave hits the sea wall and dissipates its energy against the sea wall.

The portion that goes through the break in the sea wall only has a small percentage of the energy of the original wave, so if the break is 10% of the sea wall that portion of the wave only has 10% of the energy of the original wave.

So now consider the double slit experiment.

For a photon. The energy it has is represented by its frequency. Thus if only 10% of the photons wave passes through the slit then the photon should lose 90% of its energy and as energy (frequency) in a photon is related to its colour then the photon should change colour as it passes through the slit. However it doesn't (as far as I understand) therefore it has the same energy as the the original photon. It has somehow reclaimed the energy that was "lost" when the wave front hit the apparatus.

curiouschris said:
Imagine a sea wall with a break in it.
This is a classical process. You cannot use intuitions from a classical process to understand a quantum process.

curiouschris said:
if only 10% of the photons wave passes through the slit then the photon should lose 90% of its energy
No. The "wave" for a photon, a quantum object, is a wave of probability: each individual photon has a 10% probability of passing through the slits instead of hitting the wall. But the photons that do pass through the slits pass through with all of their energy.

mattt
PeterDonis said:
No. The "wave" for a photon, a quantum object, is a wave of probability: each individual photon has a 10% probability of passing through the slits instead of hitting the wall. But the photons that do pass through the slits pass through with all of their energy.
So when illustrating the double slit experiment the wave shown is not the photons energy but in fact its probability.

To my mind this would then imply that the probability wave did in fact collapse as it entered the slit, as the probability the photon was going to pass through the slit was "realised" and the photon passed through the slit and then the probability started to spread out again (as a wave) till it encounters the next obstacle which is the rear wall or detector.

curiouschris said:
So when illustrating the double slit experiment the wave shown is not the photons energy but in fact its probability.
Yes.

curiouschris said:
To my mind this would then imply that the probability wave did in fact collapse as it entered the slit, as the probability the photon was going to pass through the slit was "realised" and the photon passed through the slit and then the probability started to spread out again (as a wave) till it encounters the next obstacle which is the rear wall or detector.
This works fine as far as predicting what will be observed at the detector for photons that do pass through the slits. However, it is important to resist the temptation to view the "collapse of the probability wave" as necessarily being an actual physical thing that happens. Different QM interpretations have very different views on this (and discussion of QM interpretations belongs in the interpretations subforum). As far as just making predictions is concerned, it is not necessary to adopt any particular interpretation.

Dale, vanhees71 and curiouschris
Thanks.

As an aside the probability is usually displayed as a curved wave. Thus as it propagates the wave becomes wider and wider. This then implies given enough distance the photon may be realized light years apart.

Does this mean the photon could "decide to appear where it wants" or is it just that we don't know which direction a particular photon went until it does make its appearance? In other words its not so much a probability wave but an ignorance wave. In that we are ignorant to the true direction of the photon until we actually measure it?

Or am I wandering into interpretation grounds?

curiouschris said:
am I wandering into interpretation grounds?
Yes.

curiouschris
I think that this discussion is again an example for the problems with the "naive photon picture", which originates from introducing photons as localized massless point particles, which originates from Einstein's very early ideas of 1905. The problem with that is that this model provides an entirely wrong picture, even on a qualitative level of understanding, and Einstein was well aware of this.

The only way to understand really, what a single-photon state is, is modern relativistic QFT (modern means it's the theory introduced as early as 1926 by Born and Jordan in there paper following Heisenberg's famous "Helgoland paper"). A single photon is a specific state of the quantized electromagnetic field and as such the right qualitative picture is that of an electromagnetic wave rather than any kind of localized point particle. In fact, it is impossible to localize a photon since there's no proper position operator for massless quanta with spin 1. All you can say about the photon, given its precise state, are probabilities for detect it at a given place (e.g., by use of a photoplate or more modern a CCD cam), and the probability distribution is given by the (properly normalized) energy density of the electromagnetic field. The difference to classical electromagnetic waves (which are quantum-fieldtheoretically described by socalled coherent states, for which the photon number is not determined but itself random and Poisson distributed) is that the photon can only be either detected as a whole or not detected, i.e., a single photon leaves precisely one spot on the observational screen (or a localized detector either clicks or doesn't click). There's not way to detect a fraction of a single photon.

mattt and Dale
curiouschris said:
Does this mean the photon could "decide to appear where it wants" or is it just that we don't know which direction a particular photon went until it does make its appearance? In other words its not so much a probability wave but an ignorance wave. In that we are ignorant to the true direction of the photon until we actually measure it?
Experimentally all we can do is detect it here or there and determine the probability. We cannot determine any motives or hidden truth.

vanhees71
curiouschris said:
Sorry if I misunderstood. When I read "single photon" I understood you as meaning individual photons, like would happen in the case of a torch, rather than a single photon per unit time.

The only thing I can think of given the illustrations used to show the double slit experiment is that where a photon's probability is that it will pass through the slit the waveform collapses to a photon and then emerges from the other side of the slit and reforms as a wave to ultimately interfere with photons from the other slit.

So I am most interested with what happens to the wave front at the point when the photon enters the slit.

This reference is not about a double slit setup; however it discusses the differences between the classical picture of individual light waves versus the quantum nature of photons in some detail. Specifically, the experiment deals with single photon sources, which is what you are inquiring about.

http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf

vanhees71 and Dale
curiouschris said:
In other words its not so much a probability wave but an ignorance wave. In that we are ignorant to the true direction of the photon until we actually measure it?
It's fundamentally a probability wave and not an ignorance wave. A basic principle of QM is that systems do not in general have well-defined quantities until they are measured. Talking about the "true direction" of a photon goes fundamentally against the principles of quantum mechanics.

To emphaise this. If we measure the position of an electron it is wrong to think that the electron was definitely where we found it before we measured its position. Instead, the electron has no well-defined "position" until we measure it.

In QM, all we can know given the state of a system is the probability of the various measurement outcomes. There is no deeper, hidden knowldege that QM does not tell us about.

Lord Jestocost and vanhees71
PeroK said:
It's fundamentally a probability wave and not an ignorance wave. A basic principle of QM is that systems do not in general have well-defined quantities until they are measured. Talking about the "true direction" of a photon goes fundamentally against the principles of quantum mechanics.
I think it's important to formulate this a bit more carefully. Of course, all this is a bit in danger to refer to "interpretation" and maybe we should open another thread in the interpretation section of this forum. Nevertheless, here I argue within the minimal-statistical interpretation, which is (as the name suggest) the minimum you need to use the quantum formalism to describe real-world experiments.

First one should have in mind that the basic notions of QT are "states" and "observables". First one has to define them operationally, i.e., how to relate them to real-world experiments. According to this a "state" refers to a preparation procedure and an "observable" to a measurement procedure.

This clarifies the fact that the observables are always well-defined in the sense that you can any measure observable of a quantum system, independent of the state this system has been prepared in.

In the formalism the state is described by a self-adjoint positive semidefinite operator ##\hat{\rho}## with ##\mathrm{Tr} \hat{\rho}=1##, the statistical operator, and the observables by self-adjoint operators ##\hat{A}##. The possible values an observable takes when (accurately) measured are the eigenvalues of these operators, and in general the value of the observable is indetermined, i.e., when you measure the observable on an ensemble of systems, all prepared in the state ##\hat{\rho}##, you'll get any possible value ##a## with some probability. Given a complete set of eigenvectors ##|a,\beta \rangle##, where ##\beta## denotes the eigenvalues of other observables, which are compatible with ##\hat{A}##, forming together a complete, independent set of observables, these probabilities are given by
$$P(a)=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle.$$
This implies that the preparation in the state ##\hat{\rho}## only implies a determined value ##a## of the observable ##\hat{A}## if and only if ##P(a)=1## for one eigenvalue ##a##, and then all other ##P(a')=0## for ##a' \neq a##.

The point is that it depends on the preparation (not the measurement!) whether an observable takes a determined value or not.
PeroK said:
To emphaise this. If we measure the position of an electron it is wrong to think that the electron was definitely where we found it before we measured its position. Instead, the electron has no well-defined "position" until we measure it.

In QM, all we can know given the state of a system is the probability of the various measurement outcomes. There is no deeper, hidden knowldege that QM does not tell us about.
This is of course correct (within the minimal statistical interpretation).

Dale and PeroK
After some discussion amongst the mentors several off topic posts have been deleted and the thread is reopened

## 1. How is energy transferred from a wave to a photon?

Energy is transferred from a wave to a photon through a process called absorption. When a wave, such as an electromagnetic wave, interacts with matter, the energy from the wave is absorbed by the particles in the matter. This absorption causes the particles to become excited and emit photons, which carry the energy of the original wave.

## 2. Can energy be transferred from a wave to a photon in other ways?

Yes, energy can also be transferred from a wave to a photon through a process called scattering. Scattering occurs when a wave interacts with particles in matter, causing the wave to change direction and transfer some of its energy to the particles. The particles then emit photons with the transferred energy.

## 3. What determines the amount of energy transferred from a wave to a photon?

The amount of energy transferred from a wave to a photon depends on the frequency and intensity of the wave. Higher frequency waves and more intense waves transfer more energy to the particles they interact with, resulting in higher energy photons being emitted.

## 4. Can energy be transferred from a photon back to a wave?

Yes, energy can be transferred from a photon back to a wave through a process called emission. When an excited particle emits a photon, the photon carries the energy away and can be absorbed by another particle, causing it to become excited and emit another photon. This process can continue, resulting in the transfer of energy from photons back to waves.

## 5. Is the transfer of energy from a wave to a photon a one-way process?

No, the transfer of energy from a wave to a photon is not a one-way process. Energy can be transferred back and forth between waves and photons through absorption, scattering, and emission processes. This exchange of energy is essential for many natural processes, such as photosynthesis and vision.

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