What's the source of EM radiation

[Naveen3456]

An electron and a positron annihilate and energy is released in the form of quantized EM (gamma etc.) rays:

Now some childish yet inquisitive questions;

1. Where were there quanta prior to annihilation, inside the electron or positron (or both)?

2. Were these quanta really present before or were created at the instant of annihilation?

3. If they were present already, what 'mechanism- I don't know any other word' led them out?

4. If they were created at the instant, did the matter wave associated with the electron and positron turn into the electromagnetic wave (quanta)?

 

[mfb]

1. Where were there quanta prior to annihilation, inside the electron or positron (or both)?

2. Were these quanta really present before or were created at the instant of annihilation?

The photons did not exist before, they are created in the annihilation process.

did the matter wave associated with the electron and positron turn into the electromagnetic wave (quanta)?

Sort of.

 

[Naveen3456]

The photons did not exist before, they are created in the annihilation process.
Sort of.

How come matter wave which has some intrinsic mass (however small) turn into something that has no intrinsic mass?

Did this matter wave gave its speed (or velocity) to the photon (which then continues to maintain that speed as if by magic, without any perpetual source of energy behind it for the maintenance of this speed.)

Plz be a little bit elaborate...

 

[mfb]

How come matter wave which has some intrinsic mass (however small) turn into something that has no intrinsic mass?

Quantum field theory. There is no (known) deeper reason "why" things happen, it is just an observation.

Did this matter wave gave its speed (or velocity) to the photon

No.

(which then continues to maintain that speed as if by magic, without any perpetual source of energy behind it for the maintenance of this speed.)

Without any force acting on them, all things keep their velocity, momentum and energy.

 

[ftr]

I don’t know your level of education, but at least you need to be at 2nd year college level (science or engineering to have any hope of understanding with good deal of effort.

Even in classical physics all we do is model. Like express properties of electron such as mass and charge by matching experimental results to formulas. When you get to QM similar activity takes place, however, the objects and their mathematical representation become more abstract. The functions which are used to model then are converted using Fourier transforms and manipulated to compute interactions. Spontaneous emission is modeled using what is called First quantization.

http://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

 

[Naveen3456]

Quantum field theory. There is no (known) deeper reason "why" things happen, it is just an observation.

Is there any time lag in the emission of a photon?

What if some super-super high photography (just suppose) is done and we see an emerging photon part by part.

If it’s instantaneous, how come time is not involved in it?

No.

Do you mean to say that at its 'birth' only the photon is traveling at the speed of light? It must have taken some time to reach this speed. How can this be explained?

Without any force acting on them, all things keep their velocity, momentum and energy.

What's the reason (mechanism) for this 'keeping' of velocity, momentum and energy?

Suppose a photon enters an area of space where there is absolute zero ( or near absolute zero) temperature, would it still keep its speed?

 

[bhobba]

Is there any time lag in the emission of a photon? If it’s instantaneous, how come time is not involved in it?

As one guy said you really need to be at least at second year college level to even begin to understand it - if not all you will get is hand-waving type stuff.

But in that vein here is what I think is the best way of looking at it - in Quantum Field Theory electric and magnetic fields involve the exchange of what are called virtual photons - changing fields create real photons by, in a rough way, shaking some of those virtual particles loose so it becomes real - in Quantum processes in general you have a probability of something happening rather than an actual elapsed time.

Thanks
Bill

 

[xAxis]

Is there any time lag in the emission of a photon?

No

What if some super-super high photography (just suppose) is done and we see an emerging photon part by part.

You made here two false assumptions.
1. We cannot see photon
2. Photons are point particles, they don't have parts.

If it’s instantaneous, how come time is not involved in it?

If it's instantaneous, no time between anihilation and radiation, time is not involved in it. It would be strange if time was involved

Do you mean to say that at its 'birth' only the photon is traveling at the speed of light? It must have taken some time to reach this speed. How can this be explained?.

How could you explain accelerating something that has no mass (photon in this case)?

What's the reason (mechanism) for this 'keeping' of velocity, momentum and energy?.

It's been known for about 350 years now that no force or mechanism is required to keep velocity of any object (first Newton's Law)

Suppose a photon enters an area of space where there is absolute zero ( or near absolute zero) temperature, would it still keep its speed?

Yes. The speed of light is constant in vacuum, regardless of the temperature, or anything else.

 

[Naveen3456]

1. We cannot see photon

We surely see light, don't we? Even single photons are detectable?

If it's instantaneous, no time between anihilation and radiation, time is not involved in it. It would be strange if time was involved

I think Einstein said that nothing can happen at a speed greater than the speed of light.

 

[mfb]

We surely see light, don't we? Even single photons are detectable?

Sure, but they do not have any parts to observe.

I think Einstein said that nothing can happen at a speed greater than the speed of light.

It also happens at a single place, there is no speed involved.
Note that this is just an attempt to describe a process in words, to avoid details of quantum field theory.

 

[Naveen3456]

It also happens at a single place, there is no speed involved.
Note that this is just an attempt to describe a process in words, to avoid details of quantum field theory.

Even if 'it' happens at a single place, it is for sure that one process (state) ends and another process (state) begins.

This ought to involve time as far as i think.

 

[mfb]

Not if both things happen at the same time.
But this discussion won't lead to anything. In quantum field theory, nothing propagates faster than light, it is possible to show that.

 

[LayMuon]

How come matter wave which has some intrinsic mass (however small) turn into something that has no intrinsic mass?

.

The important thing is the energy conservation. Mass is a form of energy: E = mc^2, where m is the dynamical mass. It is the rest mass of photon that is zero.

 

[Born2bwire]

We surely see light, don't we? Even single photons are detectable?

Sure, by absorbing the photon. We can't track a photon, we can only state that it was created here (most likely) and then detected (thereby annihilated) over here at this point in time.

 

[LayMuon]

Even if 'it' happens at a single place, it is for sure that one process (state) ends and another process (state) begins.

This ought to involve time as far as i think.

i think it does. think of em radiation by an excited atom. it takes some time for an electron cloud to change it's shape to the final state, and during that shape-shifting EM is being emitted.

 

[Bill_K]

i think it does. think of em radiation by an excited atom. it takes some time for an electron cloud to change it's shape to the final state, and during that shape-shifting EM is being emitted.

Nope, that's false too. Your classical intuition misleads you. Photon emission by an excited atom is also instantaneous. There is no elapsed time required for the electron to readjust. It is either in the initial state or the final state - there is no intermediate stage.

 

[LayMuon]

Nope, that's false too. Your classical intuition misleads you. Photon emission by an excited atom is also instantaneous. There is no elapsed time required for the electron to readjust. It is either in the initial state or the final state - there is no intermediate stage.

But if you right down and solve schroedinger equation the process is continuous, the the instantaneity question reduces to how to interpret wave function.

 

[Bill_K]

The process that's "continuous" is just the growing of the probability that the system is in the final state. Regardless of what your favorite interpretation is, the system will always be found to be in one or the other of the discrete states. There is no period during which "the electron cloud is changing shape."

 

[bhobba]

But if you right down and solve schroedinger equation the process is continuous, the the instantaneity question reduces to how to interpret wave function.

Not so. The solutions of the Schrödinger equation gives the possible energy states and in general they are discrete - most certainly so in an atom. If it emits or absorbs a photon it changes to another energy state and that change is not continuous - nor can it be since the solution to the Schrödinger equation are the only allowable states. This is part of the weirdness of QM - hard to wrap your mind around but is, as far as we can tell today, how the world works.

As an aside a massive amount of work has been done over the years on why QM is like that and I recently came across a paper that for me really got to the heart of the matter about what's going on - check it out - very interesting:
http://arxiv.org/pdf/0911.0695v1.pdf

Basically it would seem if you have entanglement then you must have QM - it more or less follows from some very reasonable assumptions that leads to only two choices - QM and bog standard probability theory - but what distinguishes QM is it allows entanglement.

Thanks
Bill

 

[Jano L.]

Laymuon:
i think it does. think of em radiation by an excited atom. it takes some time for an electron cloud to change it's shape to the final state, and during that shape-shifting EM is being emitted.

Bill_K:
Nope, that's false too. Your classical intuition misleads you. Photon emission by an excited atom is also instantaneous. There is no elapsed time required for the electron to readjust. It is either in the initial state or the final state - there is no intermediate stage.

Bill, what is the reason for you assertions? I concur with Laymuon, if we consider the basic equations of the theory, they are differential equations with no indication that changes in the atom are instantaneous jumps. Such instantaneous jump would violate wave equation for the EM field.

The changes in atoms can possibly be very quick due to interaction with the other atoms and radiation, but they are unlikely to be instantaneous if described by differential equations. The emission of radiation is often quite slow - the spontaneous emission from hydrogen atom takes times of order of a nanosecond, a very long time in which the electronic density oscillates back and forth cca millions of times (period in order of femtoseconds). Or take the Rabi oscillations - if the laser is tuned to resonance with some pair of levels and the strength of the electric field is weak, we can change the state of the atoms and make the $\psi$ function to oscillate between the ground state and the excited state very slowly (with frequency $\boldsymbol \mu_{12}\cdot \mathbf E_0 / \hbar$).

 

[bhobba]

Bill, what is the reason for you assertions? I concur with Laymuon, if we consider the basic equations of the theory, they are differential equations with no indication that changes in the atom are instantaneous jumps. Such instantaneous jump would violate wave equation for the EM field.

Sorry mate - can't follow that one. The energy solutions are discreet for a central potential like an atom. Jumps are modeled as a perturbation and perturbation theory is pretty clear - before it's in one of a discrete set of states - the same with after. Exactly why do you think its continuous? Do you know of some kind of exact solution rather than using perturbation theory that shows it is continuous?

Thanks
Bill

 

[Jano L.]

The energy solutions are discreet for a central potential like an atom. Jumps are modeled as a perturbation and perturbation theory is pretty clear - before it's in one of a discrete set of states - the same with after.

The eigenfunctions of the hydrogen Hamiltonian are indeed discrete. That does not imply that they are the only $\psi$ functions. We can sum up two such eigenfunctions to obtain another function which is perfectly valid solution of the time-dependent Schroedinger equation. Such functions are necessary if we want to explain dispersion/absorption starting from Schroedinger's equation, since only for such functions the electric current density oscillates. Current density does not oscillate if the function is equal to an eigenfunction.

Mathematically it is similar to oscillations of a bell. A bell also has discrete modes and eigen-frequencies, but that does not mean that it jumps from one mode to another instantaneously. If you kick the bell, it will ring in a complicated way, which can be described by sum of eigen-oscillations, and due to dissipation it can make transition in a continuous way to some preferred mode.

 

[bhobba]

The eigenfunctions of the hydrogen Hamiltonian are indeed discrete. That does not imply that they are the only $\psi$ functions. We can sum up two such eigenfunctions to obtain another function which is perfectly valid solution of the time-dependent Schroedinger equation.

Yea - but that linearity only applies to the same eigenvalue. Think back to your basic linear algebra - the sum of eigenvectors belonging to different eigenvalues are in general not eigenvectors. Different eigenvalues mean different observational energy. The observed energies are discreet and never observed to be a different value.

Thanks
Bill

 

[LayMuon]

Yea - but that linearity only applies to the same eigenvalue. Think back to your basic linear algebra - the sum of eigenvectors belonging to different eigenvalues are in general not eigenvectors. Different eigenvalues mean different observational energy. The observed energies are discreet and never observed to be a different value.

Thanks
Bill

We are considering time-dependent schroedinger equation, not just an eingenvalue problem. it is just a mathematics that this time dependent wave function can be presented as linear sum of time independent eigenfunctions with time dependent coefficients. Discreteness of observation can be thought of as due to external interference.

 

[Jano L.]

Different eigenvalues mean different observational energy. The observed energies are discreet and never observed to be a different value.

But we do not discuss process of measurement of energy. That would be formidably difficult both experimentally and theoretically. We were discussing the question whether the interaction of atom with light involves instantaneous jumps.

 

[LayMuon]

We can do the following: discard all eigenvalue stuff, storm-liouville math phys, let's just think we have partial differential equation with time dependence, we are solving numerically with boundary conditions, and that's it. The remaining question would be how to interpret constructed wave function. And it is certainly highly unlikely to encounter any sudden jumps.

 

[Bill_K]

The continuous wavefunction is obtained as an integral over discontinuous solutions. We often do this when solving PDEs, e.g. the use of Green's Functions. The system evolves in state A until t = t₀, decays instantaneously to state B, and from then on evolves in B. As Feynman told us, we integrate these partial amplitudes to get the total amplitude.

 

[Naty1]

Naveen:

We surely see light, don't we? Even single photons are detectable?

maybe not in the way you think...You 'see' a baseball about the time if leaves a pitcher's hand
and can track it's progress across home plate...each of the snapshots are recorded by you as a photon/many photons hits you eye.

So you cannot observe an individual photon as it is in transit...only when it impinges on your eye and is converted into an electrical signal you brain registers. If you 'see rays of the sun', for example, what you are REALLY observing are reflections off particles suspended in air...not the photons in transit...

Another 'funny' [perhaps] aspect of photons id that they don't change speed in a vacuum...they are at 'c' or they don't exist...photons don't accelerate like matter particles. That's why we model them as appearing and disappearing [emission, absorption] 'instantaneously'...there is no elapsed time interval when matter particles annihilate, photons accelerate up to light speed and then appear...its instantaneous.

Nobody knows exactly WHY things are that way any more than we know why we are carbon based life, why photons carry the electromagnetic force, unit charge has the value we observe, ...nor why we even observe four different forces [EM, strong,weak, gravity] or why there even IS an EM force...that's just the way THIS universe is built...

 

[bhobba]

But we do not discuss process of measurement of energy. That would be formidably difficult both experimentally and theoretically. We were discussing the question whether the interaction of atom with light involves instantaneous jumps.

We can do the following: discard all eigenvalue stuff, storm-liouville math phys, let's just think we have partial differential equation with time dependence, we are solving numerically with boundary conditions, and that's it. The remaining question would be how to interpret constructed wave function. And it is certainly highly unlikely to encounter any sudden jumps.

Both of you are missing the point. Let's back track a bit here. To do the analysis properly you need to treat the EM fields quantum mechanically. When that is done photon number eigenvectors form a complete orthonormal basis of the state space. As the state of the EM field changes so do the 'weightings' of photon numbers of its representation - hence the probability of getting a particular number of photons in an observation. Of course the state changes continuously but in doing so the probability of finding a photon, or photons, changes as well and this gives rise to phenomena like spontaneous emission etc where the system makes a sudden change and emits a photon. We can calculate the probability of this but its inherent in the proper QM formalism that systems will make sudden jumps corresponding to changes in state.

Thanks
Bill

 

[strangerep]

A simplified concrete example may help illustrate the difference between measurements, ensembles, and blind math...

Suppose we consider a transition between two possible energy eigenstates (ignoring the radiation field): an initial state $\psi_i$ and a final state $\psi_f$ (both assumed normalized). The general wave function is
$$\psi ~=~ a(t) \psi_i + b(t) \psi_f ~,~~~~
\mbox{where}~~ a(0)=1 ,~ b(0)=0 ,~ a(\infty)=0 ,~ b(\infty)=1 ~,
~~|a(t)|^2 + |b(t)|^2 = 1 ~.
$$
The expectation value for (say) the radial position of the state is
$$
\def\<{\langle}
\def\>{\rangle}
\<r\> ~=~ \int_0^\infty\!\! dr\; (a\psi_i + b\psi_f)^* r (a\psi_i + b\psi_f)
$$
Writing out the energy eigenstates $\psi_i$ and $\psi_f$ more explicitly, they are
$$ \psi_i ~\equiv~ \psi_i(r) e^{it E_i/\hbar} ~,~~~~
\psi_f ~\equiv~ \psi_f(r) e^{it E_f/\hbar} ~,
$$
Hence
$$
\<r\> ~=~ |a|^2\<r_i\> ~+~ |b|^2\<r_f\>
~+~ 2 Re \Big( \bar a b \, e^{it \Delta E/\hbar} \Big) ~,~~~~
(\Delta E := E_i - E_f) ~.
$$
Assuming $a,b$ to be slowly varying with time, we find an approximate time-dependence during the transition of
$$
\<r\> ~\sim~ \cos(t \Delta E/\hbar) ~.
$$
Hence the expectation of radial position of the electron has a time-dependence in general (and moreover it oscillates at the same frequency as the emitted radiation). (Of course, this assumes we're considering only a timescale which is small compared to the time variation in $a,b$.)

But in any individual (ideal) measurement only $\psi_i$ or $\psi_f$ is observed (assuming the apparatus is for measuring energy states). That's basic QM when you've got discrete eigenstates.

Since energy and position do not necessarily commute in general, measurements by an ideal position-measuring apparatus might in principle suggest different conclusions but I have no idea how such an apparatus could be constructed in practice that would not itself interfere with the transition and associated radiation.