How long does it take an electric dipole to emit a photon?

  • Thread starter blenx
  • Start date
  • #1
30
0

Main Question or Discussion Point

Suppose there is an electic dipole that starts to oscillate with frequency ω at t=0, then how long does it take the electric dipole to emit a photon?

We know that the radiant power of such electric dipole calculated from quantum physics is [tex]P\left( \omega \right) = \frac{{{\omega ^4}}}{{3\pi {\varepsilon _0}{c^3}}}{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|^2}[/tex]. Does it mean that the time it take the electic dipole to emit a photon is [tex]\Delta t = \frac{{\hbar \omega }}{{P\left( \omega \right)}}[/tex]?([tex]{\boldsymbol{\hat d}}[/tex] is the dipole operator.)
 
Last edited:

Answers and Replies

  • #2
30
0
Does somebody know the answer?
 
  • #3
The probability of a photon emission from a single electron is related to the alpha constant, so I think you would have to have that constant somewhere in your calculation in order for it to be accurate. But honestly, that's just a hunch.
 
  • #4
30
0
The probability of a photon emission from a single electron is related to the alpha constant, so I think you would have to have that constant somewhere in your calculation in order for it to be accurate. But honestly, that's just a hunch.
Because [tex]{\boldsymbol{\hat d}} = e{\boldsymbol{\hat x}}[/tex], the expression of the radiant power actually has contained the alpha constant. What confuse me very much is that if you evaluate the Δt, you will find it quite large, which means that we have to wait for a very long time before the dipole emit a photon.
 
  • #5
What do you consider large? If you do the calculations correctly, it should be pretty small. Make sure you are plugging in the right constants.
 
  • #6
30
0
What do you consider large? If you do the calculations correctly, it should be pretty small. Make sure you are plugging in the right constants.
[tex]\Delta t = \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|}^2}}} \approx \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{e^2}{a^2}}} = \frac{{3{c^2}}}{{4\alpha {\omega ^3}{a^2}}}= \frac{{3{c^2}}}{{32\alpha {\pi ^3}{\nu ^3}{a^2}}}\approx \frac{{1.33 \times {{10}^{37}}}}{{{\nu ^3}}}[/tex]
where a is the Bohr radius, alpha is the fine-structure constant, ν is the frequency.

The frequency of the microwave is 300MHz~300GHz. Substitute it into the expression of the Δt, we immediately obtain Δt≈492s~4.92×(10^11)s, which means that if I call you with my cellphone, you will have to wait for Δt before you hear my voice.

Don't you think it a little long?
 
  • #7
893
26
What is the result of the integral of P(w) in w? What does it mean? I miss in this expression for P(w) some parameter that allows one to produce two dipole emiters with two different emission powers.

Maybe it contributes to the clearification of this issue.

Best regards

DaTario
 
  • #8
2,257
7
cellphones dont work by means of atomic emission.
they use antennas

I've always heard that an atom takes about 10^-8 sec to emit one photon of energy.

since you are using the bohr radius why dont you use a photon in the spectrum of hydrogen
 
  • #9
30
0
cellphones dont work by means of atomic emission.
they use antennas

I've always heard that an atom takes about 10^-8 sec to emit one photon of energy.

since you are using the bohr radius why dont you use a photon in the spectrum of hydrogen
Thanks for your reminding. The data I used in the previous post may be improper.
 
  • #10
2,257
7
hydrogen spectrum starts at about 1000 angstroms.
there are 3*10^18 angstroms in a light sec
 
  • #11
2,257
7
that gives a frequency of 3*10^15
your highest frequency was 3*10^11

a frequency 10^4 times larger should take 10^12 times less time

500 seconds/10^12= 5*10^-10 seconds
 
  • #12
649
3
I don't have it on me right now, but I think M. Fox treats this specifically in his book "Quantum Optics" by Oxford Master Series in Physics. You should check.
 
  • #13
187
0
[tex]\Delta t = \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{{\left| {\left\langle {f\left| {{\boldsymbol{\hat d}}} \right|i} \right\rangle } \right|}^2}}} \approx \frac{{3\pi {\varepsilon _0}\hbar {c^3}}}{{{\omega ^3}{e^2}{a^2}}} = \frac{{3{c^2}}}{{4\alpha {\omega ^3}{a^2}}}= \frac{{3{c^2}}}{{32\alpha {\pi ^3}{\nu ^3}{a^2}}}\approx \frac{{1.33 \times {{10}^{37}}}}{{{\nu ^3}}}[/tex]
where a is the Bohr radius, alpha is the fine-structure constant, ν is the frequency.

The frequency of the microwave is 300MHz~300GHz. Substitute it into the expression of the Δt, we immediately obtain Δt≈492s~4.92×(10^11)s, which means that if I call you with my cellphone, you will have to wait for Δt before you hear my voice.

Don't you think it a little long?
The actual flux area for an active antenna (an oscilator in sync with incident EM wave, e.g. an atom or cellphone antenna) is determined by the square of the incident EM wavelength, not by the square of Bohr radius. Consequently the photoelectric effect can be fully explained via plain classical EM fields interacting with the electron matter/Dirac field (also via an Schrodinger atom, in dipole approximation). Check the intro section of reference [1] in an https://www.physicsforums.com/showthread.php?t=71297"for literature on this problem. More thorough quantum optics textbooks (such as L. Mandel, E. Wolf "Optical Coherence and Quantum Optics" Cambridge Univ. Press., Cambridge, 1995) also cover such derivation in detail.

A simple physical picture of how such accumulation works is to consider an active dipole which is oscillating in sync with the incident EM wave. The field generated by the dipole is then suporposed with the incident EM field and the resulting Poynting vector is "bent" toward the dipole (a la funnel), energy-momentum being funneled into the dipole from a much wider cross-section area (~ square of the wavelength) than the area of the dipole itself. Any EE textbooks on antennas should cover this type of derivation for the effect (the amplified rates of energy absorption). That's for example how a tiny AM radio with an antenna of just few centimeters in length can receive broadcast at wavelengths in tens or hundreds of meters (its Poynting vector 'collection funnel' is hundreds of meters wide).

Note that the derivations based on Bohr (dipole) radius you suggest (which are typically found in "pedagogical"/handwaving expositions seeking to motivate introduction of photons) belong historically to pre-Schrodinger/Dirac equation QM, the so-called "Old QM'.
 
Last edited by a moderator:

Related Threads on How long does it take an electric dipole to emit a photon?

Replies
7
Views
586
  • Last Post
2
Replies
27
Views
16K
Replies
4
Views
508
Replies
4
Views
5K
Replies
19
Views
3K
  • Last Post
Replies
2
Views
2K
Replies
7
Views
620
Top