The Hamiltonian of a charged particle in an EM field
Prof. R. S. Sorbello, Dept of Physics, UWM
Consider a particle of mass m and charge q that moves in a magnetic field B(r)
and an electric field E(r), both of which are independent of time. (Time dependent fields
are discussed later.) In the Hamiltonian formulation of classical mechanics, one writes
the "Hamiltonian" H for the motion of a particle in Cartesian coordinates as
H= H(x,y,z, px, py, pz) where p is the so-called "canonical momentum," which we will see
does not equal the "mechanical momentum" mv when a B-field is present.
Now in the Hamiltonian formulation, we don't explicitly use Newton's 2nd Law,
but instead, we use Hamilton's equations of motion, which are derived from Newton's
2nd Law. Hamilton's equations are:
x =
∂H
,
∂px
∂H
p x = −
,
∂x
y =
∂H
,
∂py
∂H
p y = −
,
∂y
z =
∂H
,
∂pz
∂H
p z = −
.
∂z
(1)
In these expressions, H is regarded as a function of the six variables x, y, z, px, py, and pz ,
each of which depends only on time, and whose only connection with one another is
through the set of equations (1). The over-dot symbol in Hamilton's equations denotes
the derivative with respect to time.
We wish to show that for this problem the Hamiltonian is given by
(p − qA(r))2
H=
+ qV (r) ,
2m
where
and
(2)
B=∇×A
(3)
E = – ∇V.
(4)
In Eq. (2), the "square" of a vector is defined, as usual, as the vector dotted into itself.
We can prove that Eq (2) is the correct Hamiltonian by showing that using it in
the set of equations (1) gives the correct equation of motion m dv/dt = qE + q(v × B). It
2
is simplest to consider only one component of the equation of motion, say the
x-component, noting that if we can show that is true for the x-component, it must be true
for the y- and z- components also. We can proceed in two steps (details are left to the
reader):
a) Write A in Cartesian components as A(x, y, z) = (Ax (x, y, z), Ay (x, y, z), Az (x, y, z)) ,
and use the Hamilton equation x =
∂H
to immediately deduce that mvx = px – qAx ,
∂px
where vx ≡ x . (Note that this result implies that mv = p – qA.)
b) Use the relation px = mvx +qAx from part (a) into the LHS of the Hamilton equation
p x = −
∂H
. Then use Eqs. (3) and (4) plus vector algebra to show that the resulting
∂x
Hamilton equation gives the x-component of the eq. of motion
∂A
∂A
∂A
m dv/dt = qE + q(v × B). In working this out, use A x (x, y, z) = x x + y x + z x
∂x
∂y
∂z
by the chain rule.
Comment #1: An important application of H is in quantum mechanics, where the timeindependent Schroedinger equation is Hψ = Eψ , and in the expression for H we make
the substitution p → –i∇, according to the rules of quantum mechanics. Thus, if a B
field is present, it enters H only in the term (p–qA)2/2m = (–i∇ – qA)2/2m. (This
assumes that the particle has no intrinsic magnetic moment due to spin, otherwise one
must include an additional interaction term –µ⋅B, where µ is the magnetic moment due to
spin.)
Comment #2: In the case of time dependent fields, we are no longer in the regime of
electrostatics and Eq (4) is therefore no longer valid, and it must be replaced by
E = – ∇V – ∂A/∂t,
(5)
3
where now A=A(r,t) and V= V(r,t). (See Griffiths pages 416-417.) However,
Hamilton's equations (1) are still valid with H= H(x,y,z, px, py, pz , t), where H now has
explicit time dependence in addition to the implicit time dependence in the variables
x,y,z, px, py, pz . The Hamiltonian turns out to be again given by Eq. (2) except that A(r)
and V(r) are replaced by A(r,t) and V(r,t), respectively. With these forms for A and V,
Hamilton's equations lead to m dv/dt = qE + q(v × B), as can be verified by the same
method used above for time-independent fields except that we use Eq. (5) instead of
Eq. (4) in the derivation.
Comment #3: For time-dependent EM fields in quantum mechanics, the time-dependent
Schroedinger equation is H Ψ = i
∂Ψ
, where H is the Hamiltonian described in
∂t
Comment #2, and we again make the substitution p → –i∇ in H. Thus, if a B field is
present, it enters H only in the term (p–qA)2/2m = (–i∇ – qA)2/2m, as found previously
for the case of time-independent fields discussed in Comment #1.
Comment #4: Sometimes it is desirable to describe a particle in an EM field by the
"Lagrangian" rather than by the Hamiltonian. The Lagrangian, L, for a particle can be
found from the Hamiltonian by the Legendre-transformation rule
 y, z ,t) = x px + y py + z pz − H (x, y, z, px , py , pz ,t) . Using Hamilton's equations
L(x, y, z, x,
 y, z and t, one can then derive
to replace px, py, pz by the appropriate functions of x, y, z, x,
the result L = mv2/2 + qA⋅v –qV. When this expression for L is used in the so-called
"Lagrange equations", namely,
d ⎛ ∂L ⎞ ∂L
= 0 , and similarly for y and z,
⎜ ⎟−
dt ⎝ ∂x ⎠ ∂x
one obtains after some algebra, the correct equation of motion m dv/dt = qE + q(v × B).
The fact that the expression L = mv2/2 + qA⋅v –qV gives the correct equation of motion is
enough to establish that this expression is the correct Lagrangian, and one does not need
to first find the Hamiltonian as we have done here.