# Quantum leap

1. Jan 3, 2009

### sudhirking

How and why does quantum leap occur? And (please do not mind my 'confusion') how can an electron "absorb" a photon? If in this singularity, all photons were prematurly absorbed and ,in response, release, how are photons created. Before, i thought electrons jumped levels due to other quantized electromagnetic fields, and that they return due to atomic stability- which propagates a transverse wave if used continuosly. Also, how can quantum leap move 'behind the space' of other energy levels and why can't it move throught the levels. Finally, even if we moves through the fourth dimension, it always exactly transports throughtout a vector's time, which in part shows no credibility in the probablity created by this idea. Is it due to quantum design and entanglement for quantum leap to occur?

2. Jan 4, 2009

### olgranpappy

A lone electron can not absorb a photon. On the other hand, an electron in the field of a nucleus (e.g., hydrogen), can absorb a photon... but perhaps it is more correct to say that the atom absorbs the photon and just that the electron is promoted to a higher energy level.

3. Jan 4, 2009

### sudhirking

So where do all the photons come from? What cause there radiation and how are they electromanetically produced?

4. Jan 4, 2009

### sudhirking

why? Also, in electrodynamics what occurse when you have an electron next to a nuetron. Please assume an independent universe with only these two curvatures.

What causes the photons to be absorbed by the atoms as a whole?? There has to be some quantum effects occuring at those stages.. what are they?

5. Jan 4, 2009

### malawi_glenn

The last month this question has been discussed wildley in this forum, how photons are created and absorbed. I suggest you search for these threads and read them.

6. Jan 4, 2009

### sudhirking

thank you, but from what I've read from the forum, apart from the fierce argument, it seems to me that you are suggesting that phton be a virtual particle, and therefore hold a ceratin degreee of magnitude rather than spatial cordinates.

Last edited: Jan 4, 2009
7. Jan 4, 2009

### olgranpappy

purely kinematic reasons; it is impossible to conserve momentum and energy in the process if the electron is free. Sure you can draw the Feynman diagram, but the transition probability of the process is zero by conservation of energy and momentum.
Consider an atom in a 2s excited state. In a semi-classical theory the transition rate is zero because there simply is no electromagnetic field around. In QED the transition rate can be calculated in the usual way to lowest order in the perturbation
$$U\sim\bold{p} \cdot \bold{A}\;,$$
where A is an operator containing creation and annihilation operators of the photon field. The transition rate is still very "small" because the process is "forbidden" in the dipole approximation. the transition rate from 2p to 1s is bigger and can be calculated by well-known method via QED. But, again, in the semi-classical theory the rate is zero for the reason given above.

For absorption the rates can be calculated using a semi-classical thoery. See, for example, Bethe's book "Intermediate Quantum Mechanics".