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As $E=hf$ and $\mathrm{amplitude}^2 = \mathrm{energy}$ of an EM wave, if I increase the amplitude does that then affect wavelength and frequency to keep $c$ constant?

I assume so since high energy photons have high-frequency low wavelength but just need some clarification!

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  • $\begingroup$ Amplitude for light is really just the number of photons. Energy is higher though for shorter wavelength light. C is always constant. $\endgroup$ – PhysicsDave Jun 4 at 2:21
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It is very important to differentiate between a single QM photon in this case and classical light.

The two models work together perfectly, as the classical light (EM wave) emerges from the QM photon's herd.

Light as an EM wave does have amplitude, we call it intensity, and it changes with the number of photons it consists of.

A single photon does not have amplitude. Now this is not completely true, because as per QM, the electric field is described by probability amplitudes for each point in space. We can use probability amplitudes for certain field configurations. Whether we can use these probability amplitudes for single photons depends on the definiteness of the state of the single photon (in a defined state, we cannot use probability amplitudes for a single photon).

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  • $\begingroup$ A wave describing a single photon does have an amplitude, just as a Schrödinger wave function has. $\endgroup$ – my2cts Jun 3 at 19:44
  • $\begingroup$ @my2cts only if it (the single photon) is in a well defined state $\endgroup$ – Árpád Szendrei Jun 3 at 19:55
  • $\begingroup$ Also if it is in a mixed state. Just apply superposition. $\endgroup$ – my2cts Jun 3 at 21:38
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The energy of a classical EM wave is proportional to $\omega^2 |A|^2$ where the wave is $A f(t,x)$ and $\int dV |f|^2 = 1 $. The absolute value takes care of the possibility that $A$ and $f$ may be complex valued. For a wave describing a single photon the amplitude is $A = \sqrt{\hbar / \omega}$. If there are N photons and if they are all coherent, the amplitude becomes N times this value. All the time frequency and wavelength of the wave remain the same and related by $\lambda \nu = c/n$.

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