Dispersion Relation for Schrodinger’s Wave

The dispersion relation tells us how the wave frequency $\omega$ω depends on the wavenumber $k$k.

For a free Schrödinger particle, start with the time-dependent Schrödinger equation:$$i\hbar\frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}$$iℏ∂t∂ψ​=−2mℏ2​∂x2∂2ψ​

Assume a plane-wave solution:$$\psi(x,t) = A e^{i(kx-\omega t)}$$ψ(x,t)=Aei(kx−ωt)

Step 1: Compute the derivatives

Time derivative:$$\frac{\partial \psi}{\partial t} = -i\omega \psi$$∂t∂ψ​=−iωψ

Therefore$$i\hbar\frac{\partial \psi}{\partial t} = \hbar\omega \psi$$iℏ∂t∂ψ​=ℏωψ

Spatial second derivative:$$\frac{\partial^2 \psi}{\partial x^2} = -k^2 \psi$$∂x2∂2ψ​=−k2ψ

Therefore$$-\frac{\hbar^2}{2m} \frac{\partial^2\psi}{\partial x^2} = \frac{\hbar^2k^2}{2m}\psi$$−2mℏ2​∂x2∂2ψ​=2mℏ2k2​ψ

Substituting into Schrödinger’s equation gives$$\hbar\omega\psi = \frac{\hbar^2k^2}{2m}\psi$$ℏωψ=2mℏ2k2​ψ

Cancelling $\psi$ψ,$$\boxed{ \omega=\frac{\hbar k^2}{2m} }$$ω=2mℏk2​​

This is the Schrödinger dispersion relation.

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Visualizing the dispersion relation

The relation is quadratic in $k$k:$$\omega \propto k^2$$ω∝k2

$\omega=\frac{\hbar k^2}{2m}$ω=2mℏk2​

Unlike light waves, where$$\omega = ck$$ω=ck

the Schrödinger particle’s frequency grows as the square of the wavenumber.

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Connection to momentum and energy

Using de Broglie’s relations:$$p=\hbar k$$p=ℏk $$E=\hbar\omega$$E=ℏω

Substituting into the dispersion relation:$$E = \hbar\left(\frac{\hbar k^2}{2m}\right) = \frac{\hbar^2k^2}{2m}$$E=ℏ(2mℏk2​)=2mℏ2k2​

Since $p=\hbar k$p=ℏk,$$\boxed{ E=\frac{p^2}{2m} }$$E=2mp2​​

which is exactly the classical nonrelativistic kinetic energy.

Thus the Schrödinger dispersion relation is simply the wave version of Newtonian mechanics.

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Phase velocity

The phase velocity is$$v_p=\frac{\omega}{k}$$vp​=kω​

Substituting the dispersion relation:$$v_p = \frac{\hbar k}{2m} = \frac{p}{2m}$$vp​=2mℏk​=2mp​

Since$$v=\frac{p}{m}$$v=mp​

we obtain$$\boxed{ v_p=\frac{v}{2} }$$vp​=2v​​

The phase of the wave moves at half the particle velocity.

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Group velocity

A particle is represented by a wave packet, not a single plane wave.

The packet moves at the group velocity:$$v_g = \frac{d\omega}{dk}$$vg​=dkdω​

Differentiating,$$v_g = \frac{\hbar k}{m}$$vg​=mℏk​

Using $p=\hbar k$p=ℏk,$$v_g = \frac{p}{m}$$vg​=mp​

Therefore$$\boxed{ v_g=v }$$vg​=v​

The group velocity equals the particle’s classical velocity.

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Why wave packets spread

Because$$\omega \propto k^2$$ω∝k2

different Fourier components travel at different group velocities:$$v_g=\frac{\hbar k}{m}$$vg​=mℏk​

Large-$k$k components move faster than small-$k$k components.

As time passes, the packet spreads out.

This is called dispersion.

For light in vacuum,$$\omega=ck$$ω=ck

and$$\frac{d\omega}{dk}=c$$dkdω​=c

for every $k$k, so no spreading occurs.

For Schrödinger waves,$$\frac{d^2\omega}{dk^2} = \frac{\hbar}{m} \neq 0$$dk2d2ω​=mℏ​=0

and the packet inevitably disperses.

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Why Lorentz invariance fails

The Schrödinger equation assumes$$E=\frac{p^2}{2m}$$E=2mp2​

which leads directly to$$\omega=\frac{\hbar k^2}{2m}.$$ω=2mℏk2​.

A Lorentz transformation mixes energy and momentum:$$E’=\gamma(E-vp)$$E′=γ(E−vp) $$p’=\gamma\left(p-\frac{vE}{c^2}\right)$$p′=γ(p−c2vE​)

and the transformed quantities no longer satisfy$$E’=\frac{p’^2}{2m}.$$E′=2mp′2​.

Thus the Schrödinger dispersion relation is preserved only under Galilean transformations, not Lorentz transformations.

That is why relativistic quantum theory replaces the Schrödinger relation$$E=\frac{p^2}{2m}$$E=2mp2​

with$$E^2=p^2c^2+m^2c^4,$$E2=p2c2+m2c4,

leading to the relativistic wave equations such as the Klein-Gordon and Dirac equations.