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Does the amplitude of the photon oscillations always stay constant and if it is not - what are the physical differences between the photon with higher amplitude in comparison to the one with the less amplitude? Why if the photon is spread on the electron the energy of the photon lose the energy so that the frequency became lower, not the amplitude? Why nobody talk about the intensity of the light in the context of the amplitude value of the electromagnetic waves? Where is the amplitude in the equation E=hv?

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You are thinking classically. A single photon doesn't have a variable amplitude, it only has a frequency (which determines its momentum and energy). When thinking about amplitude for light (luminosity / intensity) what that really means is the number of photons hitting a spot per time interval rather than the amplitude of a single photon. Quantum mechanics can be very confusing at first! –  Brandon Enright May 4 '13 at 3:55
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I think this is actually a duplicate of physics.stackexchange.com/q/47105 –  Ben Crowell Jun 3 '13 at 15:37
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marked as duplicate by Ben Crowell, Brandon Enright, Qmechanic Jun 3 '13 at 19:02

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Does the amplitude of the photon oscillations always stay constant

A photon is an elementary particle and it also does have a frequency associated with it, as you say: E=hv. When we measure one photon, we can only measure the energy transmitted in elementary particle interactions: a photon of appropriate energy kicks an electron to a higher level around the atom and thus is absorbed. One photon differs from the other only through energy and spin orientation. At the photon/particle level there is no frequency to be measured.

The frequency comes in when we have ensembles of photons building up a classical electromagnetic wave, a collective phenomenon. It is the marvelous continuity between classical electrodynamics and quantum ones. Here @LubošMotl gives an exposition of how classical waves emerge from quantum mechanical fields, it needs some physics background though .

Why if the photon is spread on the electron the energy of the photon lose the energy so that the frequency became lower, not the amplitude?

At the particle level only energy can be measured, and the identity E=hv gives a number for v which acquires a meaning in an ensemble of photons, becoming a classical electromagnetic wave. There is only the energy that is lowered in photon particle interactions. No spreads as such.

Why nobody talk about the intensity of the light in the context of the amplitude value of the electromagnetic waves? Where is the amplitude in the equation E=hv?

The intensity of light, classical EM, translates to the number of photons in the ensembles. The energy of the photons gives the frequency of the classical wave to be built up by the photons.

A rough example: have you seen people making waves in a stadium? The wave and the people are independent concepts, even though individual people make up the wave, their kinematics etc are quantized to one person, but a collective phenomenon appears. There is nothing else than an energy in E=hv in the quantum level. The identity allows consistency between classical and quantum.

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"One photon differs from the other only through energy and spin orientation. At the photon/particle level there is no frequency to be measured." This is not true. It's a wave. For example, it can be composed of a wave-train with a certain number of wavelengths. –  Ben Crowell Jun 3 '13 at 15:18
    
"At the photon/particle level there is no frequency to be measured." This is not true. For example, we can observe double-slit interference of a single photon with itself. This interference is related in the expected way to the photon's frequency and wavelength. –  Ben Crowell Jun 3 '13 at 15:31
    
@BenCrowell yes, but it is the probability amplitude frequency, not the electric and magnetic field frequency. An ensemble of photons ends up behaving as expected by classical QED but a photon itself is a single particle, exactly as an electron itself is a single particle, imo. –  anna v Jun 3 '13 at 15:51
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