3
$\begingroup$

In classical physics, we have second-order equations like Newton's laws, so we need to specify both position (zeroth order) and velocity (first order) of a particle as initial conditions, in order to pick out a unique solution.

In non-relativistic quantum mechanics, we have Schrödinger's equation, which is first-order. As initial data we can therefore choose only the wavefunction's value at each point in space, but not its time derivative. This seems to dovetail with the uncertainty principle, which says a quantum particle does not have independent position and momentum degrees of freedom. We can choose the wavefunction to specify either a position or a momentum, but not both. (Or we can also choose a wavefunction that has uncertain position and momentum, within the bounds of the uncertainty principle.)

In quantum field theory of fermions, we have the Dirac equation which is again first-order like Schrödinger's. But not every quantum particle is a fermion, and, AFAIK relativistic non-fermion particles obey the Klein-Gordon equation, which is second-order!

So with the Klein-Gordon equation we can apparently choose both the wavefunction and its time derivative at each point in space, giving more freedom than the Schrödinger equation. Why do we have this extra freedom, and how can it be reconciled with the uncertainty principle as applied to relativistic bosons?

$\endgroup$
1
$\begingroup$

Klein-Gordon (and actually also Dirac) equation is usually considered a classical field equation. To obtain a quantum field theory, you have to quantize it to become an operator on the symmetric (for bosons) Fock space. Then you have again a Schrödinger-type equation for the wavefunction, with an Hamiltonian that embodies the Klein-Gordon dynamics. For a free scalar KG quantum field, the Hamiltonian would read (in the momentum representation): $$H=\int_{\mathbb{R}^d} \sqrt{k^2+m^2}a^*(k)a(k)dk\; ,$$ where $a^{\#}(k)$ are the bosonic annihilation/creation operators that satisfy the canonical commutation relations $[a(k),a^*(l)]=\delta(k-l)$. Given a wavefunction $\Psi$ in the Fock space, you have the usual Schrödinger equation $$i\partial_t \Psi= H\Psi\; ;$$ which is first order in time (and consistent with the square root in the Klein-Gordon free Hamiltonian).

$\endgroup$
  • $\begingroup$ the Dirac equation is a relativistic wave equation for spin-1/2 fermions. If I remember correctly, it's this equation that enable to link spin with relativistics effects . And, for me (and wikipedia), Klein-Gordon is also a relativistic equation for scalar bosons.Why do you say it's non-relativistic? $\endgroup$ – sailx Jul 20 '14 at 9:52
  • $\begingroup$ I am saying it is not quantized not non-relativistic: if you interpret them as equations for the wavefunction, you have some paradoxical or unrealistic behaviour (as also explained in the wikipedia article, at least for Dirac equation: In quantum field theories such as quantum electrodynamics, the Dirac field is subject to a process of second quantization, which resolves some of the paradoxical features of the equation.). Dirac equation is nevertheless used sometimes in applications, but as far as I know not the non-quantized Klein-Goordon. $\endgroup$ – yuggib Jul 20 '14 at 10:01
  • $\begingroup$ My mistake, for me, classical means non-relativistic... $\endgroup$ – sailx Jul 20 '14 at 13:03
1
$\begingroup$

For a connection between Schr. eq. and complex Klein-Gordon eq., see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.

The complex Klein-Gordon eq. in QFT describes both particle and anti-particle excitations. When scaling to the appropriate non-relativistic one-particle sector to derive the Schr. eq. for the complex wavefunction then degrees of freedom are lost.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.