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Amplitude of a photon

  1. Aug 22, 2009 #1
    What's the formula for the amplitude of a photon?

    Thanks.
     
  2. jcsd
  3. Aug 22, 2009 #2

    jtbell

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    What do you mean by the "amplitude of a photon?"
     
  4. Aug 22, 2009 #3
    The EMF energy is the volume integral of (E^2+H^2)/8π. The photon amplitude is determined with the condition that in the whole volume V there is only one photon of the energy ћω.
     
  5. Aug 24, 2009 #4
    :
    I mean the amplitude of probability that has a photon in A at T1 to appear in B at T2.

    Bob_for_short:could you explane me this formula please?(I'm just an incoming freshman)

    Thanks a lot.
     
  6. Aug 25, 2009 #5
    Besides, I read that the probability for a photon in A(x_A,y_A,z_A) at t=T1 to be found in B(x_B,y_B,z_B) at t=T2 was given by a simple formula in which appear (x_A-x_B)^2, (y_A-y_B)^2,(z_A-z_B)^2 and (T1-T2)^2. If anyone has a clue...
     
  7. Aug 25, 2009 #6

    ZapperZ

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    This is what happens when you neglect to provide a clear and complete context to your question.

    Zz.
     
  8. Aug 26, 2009 #7
    Then I should provide a clear context to my question:

    By “amplitude of a photon”, I mean the amplitude of probability for a photon situated in a point A of the space A(xA,yA,zA) at t=T1 to appear in an other point B(xB,yB,zB) at t=T2.
    Indeed, I read (QED: The strange theory of light and matter) that we could not consider that a photon travelled in a straight line and therefore, we must consider all the potential ways the photon could take.
    Besides, a photon has also amplitudes to travel faster or lower than the speed c. These amplitudes reduce each other to zero in long distances, but they must be taken in consideration for short distances.

    I guess this problem is in relation with path integrals, but I also read that the amplitude of a photon had a simple expression that depends on (xA-xB)^2, (yA-yB)^2,(zA-zB)^2 and (T1-T2)^2: I’m looking for this expression.

    Thanks for your help.
     
  9. Aug 26, 2009 #8
    Then it is the amplitude of transition, not just the normalizing factor.
     
  10. Aug 30, 2009 #9
    Assuming thhe journey of the photon from A to B is one-step, and nothing else is involved in the system, then the probability can be described by:

    P(A to B) and is calculated from the inverse of the difference in the squares between the change in position and the change in time (interval)

    The simplification is due to consideration of a photon, becasue photons travel at the speed of light, (B-A) EQUALS (T2 - T1)
    There are probabilities where the speed of the photon does not equal c, but these cancel out accordingly, so the formula is simplified by not having to include the excess of the time difference.

    As you already quoted the cartesian geometrical calculation of the position of A and B, you pretty much had the simplified equation:

    ( (( (BX - AX)^2 ) + ( (BY - AY)^2 ) + ( (BZ - AZ)^2 ) ) ^ (1/2) )

    I don't think it gets any simpler than this, unless perhaps the photon only moves in one dimension?

    ((BX - AX)^2)^1/2)

    Also, this is not including any polarisation.
     
  11. Aug 30, 2009 #10
    The amplitude at a particular point is a complex number whose squared modulus gives you the probability of finding the particle at that point. This number is also called the "wave function" or "state vector" depending on how you like to express it.
    Richard Feynman in some of his books likes to use the term "amplitude". (see his "Lectures on Physics" volume III).
    If you want to calculate the probability a particle known to be in a certain state to be found in a different state , you take the dot product (inner product) of both state vectors.
    The expressions you list at the end are part of the expression for the "interval" in special relativity. (xA-xB)^2+ (yA-yB)^2+(zA-zB)^2 - (T1-T2)^2 does not vary when you look at the events from reference frames moving at different velocities.
     
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