In classical mechanics, if you assume the speed of light is constant, it's easy using the mirror and photon thought experiment to conclude that given observers $U,V$ with $U,V$ travelling at speeds $u,v$ from some stationary frame $s$, that $u$ will see $v$ to travel at speed

$$ v' = \frac{v- u}{1 - \frac{uv}{c^2}} $$

Assuming this addition law for velocities its easy to see why the speed of light is constant for all observers since

$$ \forall u , \frac{ c - u}{ 1 - \frac{uc}{c^2}} = c \frac{c-u}{c-u} = c$$

Now assuming the same addition law for velocities theres a very intuitive "quantum" version of the effect by saying if $U$ has velocity wave-function $\psi_U$ and $V$ has velocity wave function $\psi_V$ that the wave function $V$ from $U's$ perspective can be computed to be

$$ \psi_{V-U}(v') = \int_{v \in \text{dom} \ \psi_V}\psi_U \left( \frac{v-v'}{1+ \frac{v}{c^2}} \right) \psi_V \left(v \right) $$

By obeying this "convolution" rule, one notices that the restriction of a velocity distribution to just its domain moving at the speed of light, transforms to another distribution with its domain still moving at the speed of light, regardless of what distribution its "viewed" from. Thats in some sense the same as "speed of light is the same for all quantum observers". Formally put

$$\forall \psi_U , \forall v' \in \left( \text{dom} \ \int_{v \in \lbrace {v} \ s.t. |v| = c \rbrace }\psi_U \left( \frac{v-v'}{1+ \frac{v}{c^2}} \right) \psi_V \left(v \right) \right), |v'| = c $$

Now I recently learned QFT in the form of building a location-space and attaching to each element of the location-space a hilbert space of "vibrational modes", and then using a hamiltonian to describe the evolution of the system.

But given no well defined concept of an "observer". How does one make the statement "the speed of light is the same in all reference frames" in QFT?

I'm looking for (in spirit) something like: Uncertainty principle in quantum field theory where the answerer relates subjectively to the form of the equation that the asker presents.


1 Answer 1


You don't need an observer to define a reference frame in ordinary special relativity. A reference frame is just a particular way of assigning coordinates to spacetime. Causally-connected inertial reference frames are linked to each other by Lorentz transforms, and so you might expect Lorentz invariance to be the fundamental statement of a constant speed of light.

Similarly, in QFT, the statement of a constant speed of light follows from the Lorentz-invariance of the Lagrangian. The Lagrangian qualitatively defines the physics that is allowed in a particular field theory, including the propagation of photons, and so saying "The speed of light is constant in every reference frame" is the same as saying "The Lagrangian is Lorentz-invariant."

(In fact, in some extended Standard Model theories, we assume that the Lagrangian has some small Lorentz-violating component. This immediately leads to observable vacuum dispersion, in which the speed of light is very slightly different coming from different directions. Thus, without saying that the Lagrangian is Lorentz-invariant, you explicitly don't have constant light speed. Here's a review of one of those theories: https://arxiv.org/abs/hep-th/0312310.)


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