plane wave schrodinger wave solution – Transformation under a Lorentz Transform

Key Result – The phase can be Lorentz-transformed, but the Schrödinger dispersion relation is not Lorentz invariant.

Start with the free-particle Schrödinger plane wave:$$\psi(x,t)=A e^{i(kx-\omega t)}$$ψ(x,t)=Aei(kx−ωt)

Using de Broglie relations,$$p=\hbar k,\qquad E=\hbar \omega$$p=ℏk,E=ℏω

so$$\psi(x,t)=A e^{\frac{i}{\hbar}(px-Et)}$$ψ(x,t)=Aeℏi​(px−Et)

For the nonrelativistic Schrödinger equation,$$E=\frac{p^2}{2m}$$E=2mp2​

so$$\omega=\frac{\hbar k^2}{2m}$$ω=2mℏk2​

Now apply a Lorentz transformation along the $x$x-axis:$$x=\gamma(x’+vt’)$$x=γ(x′+vt′)$$t=\gamma\left(t’+\frac{v x’}{c^2}\right)$$t=γ(t′+c2vx′​)

where$$\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$$γ=1−v2/c2​1​

Substitute these into the phase:$$px-Et = p\gamma(x’+vt’) – E\gamma\left(t’+\frac{vx’}{c^2}\right)$$px−Et=pγ(x′+vt′)−Eγ(t′+c2vx′​)

Collect terms in $x’$x′ and $t’$t′:$$px-Et = \gamma\left(p-\frac{vE}{c^2}\right)x’ + \gamma(pv-E)t’$$px−Et=γ(p−c2vE​)x′+γ(pv−E)t′

Rewrite it as$$px-Et = p’x’-E’t’$$px−Et=p′x′−E′t′

so we identify$$\boxed{ p’=\gamma\left(p-\frac{vE}{c^2}\right) }$$p′=γ(p−c2vE​)​

and$$\boxed{ E’=\gamma(E-vp) }$$E′=γ(E−vp)​

Therefore the transformed wave is$$\boxed{ \psi'(x’,t’) = A e^{\frac{i}{\hbar}(p’x’-E’t’)} }$$ψ′(x′,t′)=Aeℏi​(p′x′−E′t′)​

or$$\boxed{ \psi'(x’,t’) = A e^{i(k’x’-\omega’t’)} }$$ψ′(x′,t′)=Aei(k′x′−ω′t′)​

with$$\boxed{ k’=\gamma\left(k-\frac{v\omega}{c^2}\right) }$$k′=γ(k−c2vω​)​

and$$\boxed{ \omega’=\gamma(\omega-vk) }$$ω′=γ(ω−vk)​

So the phase transforms nicely:$$kx-\omega t = k’x’-\omega’t’$$kx−ωt=k′x′−ω′t′

However, here is the important point.

For the Schrödinger wave,$$\omega=\frac{\hbar k^2}{2m}$$ω=2mℏk2​

But after Lorentz transformation,$$\omega’=\gamma(\omega-vk)$$ω′=γ(ω−vk)$$k’=\gamma\left(k-\frac{v\omega}{c^2}\right)$$k′=γ(k−c2vω​)

In general,$$\boxed{ \omega’ \neq \frac{\hbar k’^2}{2m} }$$ω′=2mℏk′2​​

So the transformed wave is not generally another valid Schrödinger plane wave with the same nonrelativistic dispersion relation.

That is the core result:$$\boxed{ \text{The phase } px-Et \text{ can be Lorentz transformed, but the Schrödinger equation itself is not Lorentz invariant.} }$$The phase px−Et can be Lorentz transformed, but the Schro¨dinger equation itself is not Lorentz invariant.​

For Lorentz invariance, the energy relation must be relativistic:$$\boxed{ E^2=p^2c^2+m^2c^4 }$$E2=p2c2+m2c4​

which leads to the Klein-Gordon equation for scalar particles or the Dirac equation for spin-$\frac12$21​ particles.