2

I am reading about the coherent states of light (Glauber States), where the starting point is the solution of the stationary Schrödinger equation is solved using analytical and algebraic methods for the 1D-harmonic oscillator. The light is considered as a harmonic oscillator of Hamiltonian $$H = \hbar \omega \left( a^{\dagger}a + \frac{1}{2} \right) \, ,$$ where the operators $a^{\dagger}$ and $a$ are expressed in terms of $p$ and $x$ (momentum and position operators) as $$a=\frac{1}{2m\omega}(x+ip) $$ $$a^{\dagger}=\frac{1}{2m\omega}(x+ip)$$

For a particle of mass, $m$ is ok for me. But for light, the expression of the operators $a^{\dagger}$ and $a$ still working even if light has no mass?

DanielSank
  • 24,439
Malek_Physics
  • 195
  • 2
    I think you should read the accepted answer to this post, keeping in mind that the energy density of an electromagnetic field is $\epsilon E^2 / 2 + B^2 / 2 \mu$. Position and momentum don't make sense for electromagnetism. Instead the two conjugate variables are electric and magnetic field. – DanielSank Apr 17 '20 at 00:24
  • 1
    The x and p for light are oscillating in a notional not real space, and m is an abstract parameter in their dimensional makeup. The immediate variables are dimensionless a s. – Cosmas Zachos Apr 17 '20 at 00:24

0 Answers0