# What does the amplitude of a photon signify in the quantum mechanical model of the universe?

In the wave picture of light, amplitude had it's usual meaning- the maximum distance moved from the mean position by the vibrating medium (the luminiferous ether). The energy and intensity of light was also directly proportional to the square of amplitude. But in the quantum mechanical model of light, the energy is proportional to only the frequency of light, and the intensity was proportional to the number of photons per unit area. However, light is still being treated as a wave even though it's energy is quantized. And any wave which has frequency and wavelength should definitely have an amplitude. So why have we totally forgot about the amplitude of the photon in the quantum mechanical model of light? And what does amplitude signify in the quantum mechanical model?

So why have we totally forgot about the amplitude of the photon in the quantum mechanical model of light? And what does amplitude signify in the quantum mechanical model?

Light is built up by a multitude of photons, but not in the way you imagine. The photon has a wavefunction, its values are complex numbers, i.e. not real numbers as is necessary to be measurable in the laboratory.( This is true in the quantum mechanical framework generally).

From the linked paper above ( also on Arxiv.org ):

Now write the complex wave function as a sum of real and imaginary parts $$\bar E_T(\bar r)$$ and $$\bar B_T(\bar r)$$

$$\bar {\psi}_T(\bar r,t)=2^{-\frac 1 2}\left(\bar E_T(\bar r,t)+i\bar B_T(\bar r,t)\right)$$

The amplitude is connected to probability, not energy as in the classical electromagnetic wave. The real number that can be measured is the complex cojugate square of the wavefunction $$\psi*\psi$$, which is a real number and gives the probability for finding the photon with energy E=h*nu at an $$(x,y,z)$$ spot , as in the image for a single photon at a time double slit experiment.

In quantum mechanics it is the wavefunctions of the photons that are superposed to create the classical beam of light, in a mathematically complex manner , using quantum field theory, described in this blog entry.

• How does this make any sense? I've heard elsewhere that "photons do not have a wave function" and that relativistic particles in general "do not have a wave function". What then is the equation you give above supposed to mean? Or perhaps, what's that other stuff about not having wave functions supposed to mean? Commented Mar 9, 2018 at 9:29
• @The_Sympathizer "heard" is not a reference, I give you a link where one form of a wave function for a photon is seen. There are other formalisms quantizing the form of Maxwell's equations with the A potential ( not E and B as in the reference I gave). Commented Mar 9, 2018 at 12:38
• @The_Sympathizer The statement "relativistic particles in general do not have a wave function" is said by physicist who use quantum field theory, and forget that the "electron fields and photon fields .." on which creation and annihilation operators generate wave packets corresponding to particles, depend on the free particle ( plane wave) quantum mechanical solutions of the corresponding quantum mechanical equation, Dirac or Klein Gordon or Maxwel. QFT is based on quantum mechanics and it allows to calculate measurable quantities for many body systems without solving boundary value problems. Commented Mar 9, 2018 at 12:38