In analytical mechanics and quantum field theory, minimal coupling refers to a coupling between fields which involves only the charge distribution and not higher multipole moments of the charge distribution. This minimal coupling is in contrast to, for example, Pauli coupling, which includes the magnetic moment of an electron directly in the Lagrangian.[1]
In electrodynamics, minimal coupling is adequate to account for all electromagnetic interactions. Higher moments of particles are consequences of minimal coupling and non-zero spin.
In Cartesian coordinates, the Lagrangian of a non-relativistic classical particle in an electromagnetic field is (in SI Units):
l{L}=\sumi\tfrac{1}{2}m
| x |
| 2 | |
| i |
+\sumiq
| x |
iAi-q\varphi
where is the electric charge of the particle, is the electric scalar potential, and the,, are the components of the magnetic vector potential that may all explicitly depend on
xi
t
This Lagrangian, combined with Euler–Lagrange equation, produces the Lorentz force law
m\ddot{x
and is called minimal coupling.
Note that the values of scalar potential and vector potential would change during a gauge transformation,[2] and the Lagrangian itself will pick up extra terms as well, but the extra terms in the Lagrangian add up to a total time derivative of a scalar function, and therefore still produce the same Euler–Lagrange equation.
The canonical momenta are given by
pi=
| \partiall{L | |
Note that canonical momenta are not gauge invariant, and are not physically measurable. However, the kinetic momenta
Pi\equivm
| x |
i=pi-qAi
are gauge invariant and physically measurable.
The Hamiltonian, as the Legendre transformation of the Lagrangian, is therefore
l{H}=\left\{\sumi
| x |
ipi\right\}-l{L}=\sumi
| |||||||||
| 2m |
+q\varphi
This equation is used frequently in quantum mechanics.
Under a gauge transformation,
A → A+\nablaf, \varphi → \varphi-
| f |
,
where f(r,t) is any scalar function of space and time, the aforementioned Lagrangian, canonical momenta and Hamiltonian transform like
L → L'=L+q
| df | |
| dt |
, p → p'=p+q\nablaf, H → H'=H-q
| \partialf | |
| \partialt |
,
which still produces the same Hamilton's equation:
| \begin{align} \left. | \partialH' |
| \partial{xi |
In quantum mechanics, the wave function will also undergo a local U(1) group transformation[3] during the gauge transformation, which implies that all physical results must be invariant under local U(1) transformations.
The relativistic Lagrangian for a particle (rest mass and charge) is given by:
l{L}(t)=-mc2\sqrt{1-
| |||||
| 2}{c |
2}}+q
| x |
(t) ⋅ A\left(x(t),t\right)-q\varphi\left(x(t),t\right)
Thus the particle's canonical momentum is
p(t)=
| \partiall{L | |
that is, the sum of the kinetic momentum and the potential momentum.
Solving for the velocity, we get
| x |
(t)=
| p-qA | |||||
|
2}}
So the Hamiltonian is
l{H}(t)=
| x |
⋅ p-l{L}=c\sqrt{m2c2+{\left(p-qA\right)}2}+q\varphi
This results in the force equation (equivalent to the Euler–Lagrange equation)
| p |
=-
| \partiall{H | |
from which one can derive
| \begin{align} | d{d | |||||||||||||||
|
The above derivation makes use of the vector calculus identity:
\tfrac{1}{2}\nabla\left(A ⋅ A\right) = A ⋅ JA = A ⋅ (\nablaA) = (A{ ⋅ }\nabla)A+A{ x }(\nabla{ x }A).
An equivalent expression for the Hamiltonian as function of the relativistic (kinetic) momentum,, is
l{H}(t)=
| x |
(t) ⋅ P(t)+
| mc2 | |
| \gamma |
+q\varphi(x(t),t)=\gammamc2+q\varphi(x(t),t)=E+V
This has the advantage that kinetic momentum can be measured experimentally whereas canonical momentum cannot. Notice that the Hamiltonian (total energy) can be viewed as the sum of the relativistic energy (kinetic+rest),, plus the potential energy, .
\varphi
\sqrt{g}d4x
S=\intd4x\sqrt{g}\left(-
| 1 | |
| 2 |
R+
| 1 | |
| 2 |
\nabla\mu\varphi\nabla\mu\varphi-V(\varphi)\right)
where
g:=\detg\mu\nu