Why the Schrödinger Equation Is Not Lorentz Invariant

Key Result

The phase of a Schrödinger plane wave can be rewritten in Lorentz-transformed coordinates. However, the transformed energy and momentum do not satisfy the nonrelativistic Schrödinger dispersion relation.

Therefore, the Schrödinger equation is not Lorentz invariant.

The Free-Particle Plane Wave

A free particle in nonrelativistic quantum mechanics can be represented by the plane wave:

ψ(x,t) = A exp[i(kx − ωt)]

Using the de Broglie relations:

p = ℏk

E = ℏω

the wavefunction can also be written as:

ψ(x,t) = A exp[(i/ℏ)(px − Et)]

For a free particle governed by the Schrödinger equation, energy and momentum are related by:

E = p²/(2m)

Equivalently, frequency and wave number satisfy:

ω = ℏk²/(2m)

This is the nonrelativistic Schrödinger dispersion relation.

Applying a Lorentz Transformation

Consider two inertial reference frames moving relative to one another along the x-axis.

The inverse Lorentz transformation is:

x = γ(x′ + vt′)

t = γ(t′ + vx′/c²)

where:

γ = 1/√(1 − v²/c²)

Substitute these expressions into the phase px − Et:

px − Et = pγ(x′ + vt′) − Eγ(t′ + vx′/c²)

Collecting the coefficients of x′ and t′ gives:

px − Et = γ(p − vE/c²)x′ − γ(E − vp)t′

This can be rewritten as:

px − Et = p′x′ − E′t′

provided that:

p′ = γ(p − vE/c²)

and:

E′ = γ(E − vp)

These are the Lorentz transformation rules for relativistic momentum and energy.

The Transformed Wave

The transformed wave can now be written as:

ψ′(x′,t′) = A exp[(i/ℏ)(p′x′ − E′t′)]

Equivalently:

ψ′(x′,t′) = A exp[i(k′x′ − ω′t′)]

where:

k′ = γ(k − vω/c²)

and:

ω′ = γ(ω − vk)

The phase therefore retains the same form:

kx − ωt = k′x′ − ω′t′

At first glance, this may appear to suggest that the Schrödinger plane wave is compatible with special relativity. The problem becomes apparent when we examine the transformed dispersion relation.

The Schrödinger Dispersion Relation Is Not Preserved

The original wave satisfies:

ω = ℏk²/(2m)

After the Lorentz transformation:

ω′ = γ(ω − vk)

and:

k′ = γ(k − vω/c²)

In general, these transformed quantities do not satisfy:

ω′ = ℏk′²/(2m)

Instead:

ω′ ≠ ℏk′²/(2m)

Therefore, although the transformed expression still looks like a plane wave, it is not generally a solution of the same free-particle Schrödinger equation.

The Central Result

The plane-wave phase px − Et can retain its form under a Lorentz transformation, but the Schrödinger energy–momentum relation is not preserved.

Lorentz-transforming the phase does not make the Schrödinger equation Lorentz invariant.

Why the Conflict Occurs

Special relativity requires energy and momentum to satisfy:

E² = p²c² + m²c⁴

This relation is Lorentz invariant. If it holds in one inertial frame, it holds in every inertial frame.

The Schrödinger equation instead uses the nonrelativistic relation:

E = p²/(2m)

This is only the low-speed approximation for a particle’s kinetic energy. It excludes the rest energy mc² and is not preserved by Lorentz transformations.

The conflict is therefore not with the plane-wave form itself. The conflict arises from combining a Lorentz transformation with a nonrelativistic energy–momentum relation.

Relativistic Quantum Equations

A Lorentz-invariant quantum theory must begin with the relativistic energy–momentum relation:

E² = p²c² + m²c⁴

For spin-0 particles, this leads to the Klein–Gordon equation.

For spin-½ particles, it leads to the Dirac equation.

The Schrödinger equation remains extremely successful in the nonrelativistic domain, where particle speeds are much smaller than the speed of light. Its natural spacetime symmetry is Galilean rather than Lorentzian.

Final Conclusion

The phase:

px − Et

can be rewritten in Lorentz-transformed coordinates by treating energy and momentum as components of a relativistic four-vector.

However, the transformed quantities E′ and p′ do not satisfy the Schrödinger relation:

E′ = p′²/(2m)

Therefore:

The phase can be Lorentz-transformed, but the Schrödinger equation itself is not Lorentz invariant.