# The Schrodinger equation for quantisation of light? Why not with the Dirac equation?

Light is a born relativistic. However, when we quantise the electromagnetic wave, we start with the time-independent Schrodinger equation, which is a non-relativistic equation. Why is this fine?

Wouldn't it more make sense if we use the Dirac equation for the quantisation of light? or do I misunderstand the physics of light with respect to the quantisation?

• Are you sure that you start with Schrodingers equation? Can you give a reference? – Fabian Oct 21 '16 at 17:21
• "However, when we quantise the electromagnetic wave, we start with the time-independent Schrodinger equation" Can't say I've seen any treatments that do that. Starting with a massive particle and Schrödinger's equation is common. Townsend starts with light and with ad hoc probability amplitudes in an homage to Feynman's pop sci QED book. – dmckee --- ex-moderator kitten Oct 21 '16 at 17:29
• @dmckee Sorry for misleading. I meant when we quantise the EM wave, we find the EM mode first, and then plug into the expression of classical EM energy. This energy is used as the Hamiltonian, which will have the same form of the harmonic oscillator, upon which we quantise the Hamiltonian in the same way as the harmonic oscillator. Once we get the final form, we plug into the time-independent Schrodinger equation to find the energy eigenstate and also into the time-dependent Schrodinger equation to find the dynamics of a given state of light. This is what I saw in many text books. – Veteran Oct 21 '16 at 18:51
• If you are talking about, for instance, the particle in a box (infinite potential well problem), then you may have been mislead by an analogy meant to help you understand. The standing wave modes of the particle in box are essentially the same as the modes of a string or closed pipe or of an electromagnetic field in a conducting box so those other systems are often used as analogies when talking about the quantum problem. – dmckee --- ex-moderator kitten Oct 21 '16 at 22:48
• @dmckee Many thanks for your reply. It helps me get closer to what I want to know. May I ask you one more? For the EM modes in free space, we also assume there is a 3D periodic structure which also supports the 3D standing waves. For this problem, we also use the same mathematical structures of the harmonic oscillator, and finally we take L->infinity, where L is the spatial period. Is this still safe for quantisation of light? Can such quantised light (or photon) also satisfy the Dirac equation? So sorry if this is so stupid question, but I hope you understand where I got lost. – Veteran Oct 23 '16 at 17:53

Light obeys Maxwell's equations. So when we quantize light, the typical starting point is the Maxwell Lagrangian:

$$\mathscr{L} = \frac{1}{4}F^{ab}F_{ab}$$

where

$$F_{ab} = \partial_{a}A_{b} - \partial_{b}A_{a}$$

where $A_{a}$ is the 4-dimensional vector potential $(\phi, {\vec A})$ such that ${\vec E} = -{\vec \nabla}\phi$ and ${\vec B} = {\vec \nabla} \times {\vec A}$.

Taking the variation of the action with respect to $A_{a}$ gives the equation of motion:

$$0 = \nabla_{b}F^{ba}$$

which can be shown to be equivalent to Maxwell's equations in vacuum.

Doing a proper job of quantizing the above lagrangian requires several non-trivial tricks in quantum field theory to handle the fact that the Maxwell theory is a gauge theory and its evolution contains a constraint (this means that the naïve Hamiltonian is zero, and a "normal" Legendre transform won't just simply work). These technical issues CAN be resolved, but typically, students are expected to have the quantization of the Klein-Gordon and Dirac fields under their belt before they do so.

Neither the Klein-Gordon nor the Dirac equations are appropriate, because photons have spin 1.