Recall how the Schrödinger equation was motivated by the non-relativistic dispersion relation E=p22m. If we follow the same procedure for the case of a relativistic dispersion relation (E2=p2c2+m2c4), what equation do we arrive at? (For simplicity consider the one-dimensional case)
To derive the relativistic equivalent of the Schrödinger equation, we start with the relativistic dispersion relation: E^2 = p^2c^2 + m^2c^4.
In the non-relativistic case, we take the non-relativistic limit by assuming that the kinetic energy is much smaller than the rest energy, such that E ≈ mc^2 + p^2/2m, which leads to the Schrödinger equation.
To do the same for the relativistic case, we consider the Taylor expansion of the relativistic dispersion relation for small momentum values:
E^2 = p^2c^2 + m^2c^4
Expanding this equation to first order in momentum (p) and simplifying, we get:
E^2 ≈ (mc^2)^2 + 2mc^2p + p^2c^2
Dividing both sides by (mc^2)^2, we obtain:
(E^2)/(m^2c^4) ≈ 1 + 2(E/mc^2)(p/mc) + (p/mc)^2
Since E = mc^2, we can simplify further:
(E^2)/(m^2c^4) ≈ 1 + 2(pc)/(mc^2) + (p/mc)^2
Substituting E = iħ∂/∂t and p = -iħ∂/∂x for the one-dimensional case, where ħ is the reduced Planck constant, we have:
(∂^2ψ)/(∂t^2)(1/(mc^2))^2 ≈ 1 - 2(iħ∂/∂x)/(mc^2) + (-ħ^2/∂^2x^2)/(m^2c^2)^2
Simplifying further and rearranging terms, we arrive at the relativistic counterpart of the Schrödinger equation:
(∂^2ψ)/(∂t^2) - (∇^2ψ)/(c^2) + (m^2c^2/ħ^2)ψ = 0
This equation is known as the Klein-Gordon equation, and it describes the wave function for relativistic particles.
To derive the relativistic equation from the relativistic dispersion relation (E^2 = p^2c^2 + m^2c^4), we need to consider the relativistic energy-momentum relationship.
In the non-relativistic case, the energy of a particle is given as E = p^2 / (2m), where E is the energy, p is the momentum, and m is the mass. The non-relativistic dispersion relation can then be derived from this expression.
In the relativistic case, however, the energy-momentum relationship is different. The correct relationship is E^2 = p^2c^2 + m^2c^4, where E is the energy, p is the momentum, m is the rest mass of the particle, and c is the speed of light in vacuum.
For simplicity, we can consider the one-dimensional case. In this case, we can write the relativistic equation as:
(E^2 - p^2c^2) = m^2c^4
Now, we can rearrange the equation to isolate the energy term:
E^2 = p^2c^2 + m^2c^4
Taking the square root of both sides, we get:
E = ±√(p^2c^2 + m^2c^4)
Therefore, the relativistic equation we arrive at, in the one-dimensional case, is:
E = ±√(p^2c^2 + m^2c^4)
Note that the ± sign indicates that there are two possible solutions (positive and negative energy), which is a characteristic of relativistic equations.