Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absorption And Emission Of Light According To Quantum Mechanics

  1. Jan 26, 2014 #1
    Hi everyone,

    I just wanted to double check on a fundamental concept of quantum physics.

    Let us consider a monochromatic beam of photons,incident on a material(of a finite thickness,say 15 cm.) which reflects, transmits, and also absorb photons.

    Now, in QED by RP Feynman it is stated that there is no way to tell surely whether a photon interacting with matter would be reflected or transmitted. All we can count is the probability of how much photons that will be transmitted or reflected by looking at the intensity of transmitted and reflected electromagnetic raditions.

    So,according to this concept, even when a photon has energy to electronically excite electron to higher energy state and it falls on the surface of such a material, whether a photon will be absorbed or pass away without causing excitation depends just depends on probability and it is not compulsory for a photon to be absorbed if it has energy to cause electronic excitation.

    And also when the electron falls back to ground state whether it will release energy by photo-emission(reverse of photon absorption) or by emission of thermal phonons(Thermal de-excitation) also depends on probability and we cannot directly tell that a particular event(here, photon-emission or thermal de-excitation) is 100% sure to happen.

    Am I correct or is there a fundamental flaw in my concept. If so, please help by proper reasoning.

    Thanks For Your Kind Attention:confused:
  2. jcsd
  3. Jan 26, 2014 #2


    Staff: Mentor

    Yes - in QM all we can predict is probabilities.

    In fact QM can be viewed as a generalization of standard probability theory:

  4. Jan 26, 2014 #3
    Thank You very much Mr. Bhobba...!!! for your attention and the fantastic link..!!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook