Nakahara M. Geometry, Topology and Physics
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where use has been made of the equation of motion. The function E,whichis
often the sum of the kinetic energy and the potential energy, is called the energy.
Example 1.1. (One-dimensional harmonic oscillator)Letx be the coordinate
and suppose the force acting on m is F(x) =−kx, k being a constant. This force
is conservative. In fact, V (x ) =
1
2
kx
2
yields F(x) =−dV (x)/dx =−kx.
In general, any one-dimensional force F(x) which is a function of x only is
conserved and the potential is given by
V (x ) =−
x
F(ξ) dξ.
An example of a force that is not conserved is friction F =−η dx /dt.We
will be concerned only with conserved forces in the following.
1.1.2 Lagrangian formalism
Newtonian mechanics has the following difficulties;
1. This formalism is based on a vector equation (1.1) which is not very easy to
handle unless an orthogonal coordinate system is employed.
2. The equation of motion is a second-order equation and the global properties
of the system cannot be figured out easily.
3. The analysis of symmetries is not easy.
4. Constraints are difficult to take into account.
Furthermore, quantum mechanics cannot be derived directly from
Newtonian mechanics. The Lagrangian formalism is now introduced to overcome
these difficulties.
Let us consider a system whose state (the position of masses for example)
is described by N parameters {q
i
} (1 ≤ i ≤ N). The parameter is an element
of some space M.
2
The space M is called the configuration space and the {q
i
}
are called the generalized coordinates. If one considers a particle on a circle, for
example, the generalized coordinate q is an angle θ and the configuration space
M is a circle. The generalized velocity is defined by ˙q
i
= dq
i
/dt.
The Lagrangian L(q, ˙q) is a function to be defined in Hamilton’s
principle later. We will restrict ourselves mostly to one-dimensional space but
generalization to higher-dimensional space should be obvious. Let us consider
a trajectory q(t)(t ∈[t
i
, t
f
]) of a particle with conditions q(t
i
) = q
i
and
q(t
f
) = q
f
. Consider a functional
3
S[q(t), ˙q(t)]=
t
f
t
i
L(q, ˙q) dt (1.3)
2
A manifold, to be more precise, see chapter 5.
3
A functional is a function of functions. A function f (•) produces a number f (x) for a given number
x. Similarly, a functional F[•] assigns a number F[ f ] to a given function f (x).
called the action. Given a trajectory q(t) and ˙q(t), the action S[q, ˙q] produces
a real number. Hamilton’s principle, also known as the principle of the least
action, claims that the physically realized trajectory corresponds to an extremum
of the action. Now the Lagrangian must be chosen so that Hamilton’s principle is
fulfilled.
It turns out to be convenient to write Hamilton’s principle in a local form
as a differential equation. Suppose q(t) is a path realizing an extremum of S.
Consider a variation δq(t) of the trajectory such that δq(t
i
) = δq(t
f
) = 0. The
action changes under this variation by
δS =
t
f
t
i
L(q + δq, ˙q + δ ˙q) dt −
t
f
t
i
L(q, ˙q) dt
=
t
f
t
i
∂ L
∂q
−
d
dt
∂ L
∂ ˙q
δq dt (1.4)
which must vanish because q yields an extremum of S. Since this is true for any
δq, the integrand of the last line of (1.4) must vanish. Thus, the Euler–Lagrange
equation
∂ L
∂q
−
d
dt
∂ L
∂ ˙q
= 0 (1.5)
has been obtained. If there are N degrees of freedom, one obtains
∂ L
∂q
k
−
d
dt
∂ L
∂ ˙q
k
= 0 (1 ≤ k ≤ N ). (1.6)
If we introduce the generalized momentum conjugate to the coordinate q
k
by
p
k
=
∂ L
∂ ˙q
k
(1.7)
the Euler–Lagrange equation takes the form
d p
k
dt
=
∂ L
∂q
k
. (1.8)
By requiring this equation to reduce to Newton’s equation, one quickly finds the
possible form of the Lagrangian in the ordinary mechanics of a particle. Let us
put L =
1
2
m ˙q
2
− V (q). By substituting this Lagrangian into the Euler–Lagrange
equation, it is easily shown that it reduces to Newton’s equation of motion,
m ¨q
k
+
∂V
∂q
k
= 0. (1.9)
Let us consider the one-dimensional harmonic oscillator for example. The
Lagrangian is
L(x, ˙x ) =
1
2
m ˙x
2
−
1
2
kx
2
(1.10)
from which one finds m ¨x + kx = 0.
It is convenient for later purposes to introduce the notion of a functional
derivative. Let us consider the case with a single degree of freedom for simplicity.
Define the functional derivative of S with respect to q by
δS[q, ˙q]
δq(s)
≡ lim
ε→0
{S[q(t) + εδ(t − s), ˙q(t) +ε
d
dt
δ(t − s)]−S[q(t), ˙q(t)]}
ε
.
(1.11)
Since
S
q(t) +εδ(t − s), ˙q(t) +ε
d
dt
δ(t − s)
=
dtL
q(t) +εδ(t − s), ˙q(t) +ε
d
dt
δ(t − s)
=
dtL(q, ˙q) + ε
dt
∂ L
∂q
δ(t − s) +
∂ L
∂ ˙q
d
dt
δ(t − s)
+
(ε
2
)
= S[q, ˙q]+ε
∂ L
∂q
(s) −
d
dt
∂ L
∂ ˙q
(s)
+
(ε
2
),
the Euler–Lagrange equation may be written as
δS
δq(s)
=
∂ L
∂q
(s) −
d
dt
∂ L
∂ ˙q
(s) = 0. (1.12)
Let us next consider symmetries in the context of the Lagrangian formalism.
Suppose the Lagrangian L is independent of a certain coordinate q
k
.
4
Such
a coordinate is called cyclic. The momentum which is conjugate to a cyclic
coordinate is conserved. In fact, the condition ∂ L/∂q
k
= 0 leads to
d p
k
dt
=
d
dt
∂ L
∂ ˙q
k
=
∂ L
∂q
k
= 0. (1.13)
This argument can be mathematically elaborated as follows. Suppose the
Lagrangian L has a symmetry, which is continuously parametrized. This means,
more precisely, that the action S =
dtL is invariant under the symmetry
operation on q
k
(t). Let us consider an infinitesimal symmetry operation q
k
(t) →
q
k
(t) + δq
k
(t) on the path q
k
(t).
5
This implies that if q
k
(t) is a path producing
an extremum of the action, then q
k
(t) → q
k
(t) + δq
k
(t) also corresponds to an
extremum. Since S is invariant under this change, it follows that
δS =
t
f
t
i
k
δq
k
∂ L
∂q
k
−
d
dt
∂ L
∂ ˙q
k
+
k
δq
k
∂ L
∂ ˙q
k
t
f
t
i
= 0.
4
Of course, L may depend on ˙q
k
. Otherwise, the coordinate q
k
is not our concern at all.
5
Since the symmetry is continuous, it is always possible to define such an infinitesimal operation.
Needless to say, δq(t
i
) and δq(t
f
) do not, in general, vanish in the present case.
The first term in the middle expression vanishes since q is a solution to the Euler–
Lagrange equation. Accordingly, we obtain
k
δq
k
(t
i
) p
k
(t
i
) =
k
δq
k
(t
f
) p
k
(t
f
) (1.14)
where use has been made of the definition p
k
= ∂ L/∂ ˙q
k
.Sincet
i
and t
f
are arbitrary, this equation shows that the quantity
k
δq
k
(t) p
k
(t) is, in fact,
independent of t and hence conserved.
Example 1.2. Let us consider a particle m moving under a force produced by a
spherically symmetric potential V (r),wherer,θ,φ are three-dimensional polar
coordinates. The Lagrangian is given by
L =
1
2
m[˙r
2
+r
2
(
˙
θ
2
+ sin
2
θ
˙
φ
2
)]−V (r).
Note that q
k
= φ is cyclic, which leads to the conservation law
δφ
∂ L
∂
˙
φ
∝ mr
2
sin
2
θ
˙
φ = constant.
This is nothing but the angular momentum around the z axis. Similar arguments
can be employed to show that the angular momenta around the x and y axes are
also conserved.
A few remarks are in order:
• Let Q(q) be an arbitrary function of q. Then the Lagrangians L and
L + dQ/dt yield the same Euler–Lagrange equation. In fact,
∂
∂q
k
L +
dQ
dt
−
d
dt
∂
∂ ˙q
k
L +
dQ
dt
=
∂ L
∂q
k
+
∂
∂q
k
dQ
dt
−
d
dt
∂ L
∂ ˙q
k
−
d
dt
∂
∂ ˙q
k
j
∂ Q
∂q
j
˙q
j
=
∂
∂q
k
dQ
dt
−
d
dt
∂ Q
∂q
k
= 0.
• An interesting observation is that Newtonian mechanics is realized as an
extremum of the action but the action itself is defined for any trajectory. This
fact plays an important role in path integral formation of quantum theory.
1.1.3 Hamiltonian formalism
The Lagrangian formalism yields a second-order ordinary differencial equation
(ODE). In contrast, the Hamiltonian formalism gives equations of motion which
are first order in the time derivative and, hence, we may introduce flows in the
phase space defined later. What is more important, however, is that we can make
the symplectic structure manifest in the Hamiltonian formalism, which will be
shown in example 5.12 later.
Suppose a Lagrangian L is given. Then the corresponding Hamiltonian is
introduced via Legendre transformation of variables as
H (q, p) ≡
k
p
k
˙q
k
− L(q, ˙q), (1.15)
where ˙q is eliminated in the left-hand side (LHS) in favour of p by making use of
the definition of the momentum p
k
= ∂ L(q, ˙q)/∂ ˙q
k
. For this transformation to
be defined, the Jacobian must satisfy
det
∂p
i
∂ ˙q
j
= det
∂
2
L
∂ ˙q
i
˙q
j
= 0.
The space with coordinates (q
k
, p
k
) is called the phase space.
Let us consider an infinitesimal change in the Hamiltonian induced by δq
k
and δp
k
,
δ H =
k
δp
k
˙q
k
+ p
k
δ ˙q
k
−
∂ L
∂q
k
δq
k
−
∂ L
∂ ˙q
k
δ ˙q
k
=
k
δp
k
˙q
k
−
∂ L
∂q
k
δq
k
.
It follows from this relation that
∂ H
∂p
k
=˙q
k
,
∂ H
∂q
k
=−
∂ L
∂q
k
(1.16)
which are nothing more than the replacements of independent variables.
Hamilton’s equations of motion are obtained from these equations if the Euler–
Lagrange equation is employed to replace the LHS of the second equation,
˙q
k
=
∂ H
∂p
k
˙p
k
=−
∂ H
∂q
k
. (1.17)
Example 1.3. Let us consider a one-dimensional harmonic oscillator with the
Lagrangian L =
1
2
m ˙q
2
−
1
2
mω
2
q
2
,whereω
2
= k/m. The momentum conjugate
to q is p = ∂ L/∂ ˙q = m ˙q, which can be solved for ˙q to yield ˙q = p/m.The
Hamiltonian is
H (q, p) = p ˙q − L(q, ˙q) =
p
2
2m
+
1
2
mω
2
q
2
. (1.18)
Hamilton’s equations of motion are:
d p
dt
=−mω
2
q
dq
dt
=
p
m
. (1.19)
Let us take two functions A(q, p) and B(q, p) defined on the phase space of
a Hamiltonian H . Then the Poisson bracket [A, B] is defined by
6
[A, B]=
k
∂ A
∂q
k
∂ B
∂p
k
−
∂ A
∂p
k
∂ B
∂q
k
. (1.20)
Exercise 1.1. Show that the Poisson bracket is a Lie bracket, namely it satisfies
[A, c
1
B
1
+ c
2
B
2
]=c
1
[A, B
1
]+c
2
[A, B
2
] linearity (1.21a)
[A, B]=−[B, A] skew-symmetry (1.21b)
[[A, B], C]+[[C, A], B]+[[B, C], A]=0 Jacobi identity. (1.21c)
The fundamental Poisson brackets are
[p
i
, p
j
]=[q
i
, q
j
]=0 [q
i
, p
j
]=δ
ij
. (1.22)
It is important to notice that the time development of a physical quantity
A(q, p) is expressed in terms of the Poisson bracket as
dA
dt
=
k
dA
dq
k
dq
k
dt
+
dA
d p
k
d p
k
dt
=
k
dA
dq
k
∂ H
∂p
k
−
dA
d p
k
∂ H
∂q
k
=[A, H ]. (1.23)
If it happens that [A, H ]=0, the quantity A is conserved, namely d A/dt = 0.
The Hamilton equations of motion themselves are written as
d p
k
dt
=[p
k
, H ]
dq
k
dt
=[q
k
, H ]. (1.24)
Theorem 1.1. (Noether’s theorem)LetH (q
k
, p
k
) be a Hamiltonian which is
invariant under an infinitesimal coordinate transformation q
k
→ q
k
= q
k
+
ε f
k
(q).Then
Q =
k
p
k
f
k
(q) (1.25)
is conserved.
Proof. One has H (q
k
, p
k
) = H (q
k
, p
k
) by definition. It follows from q
k
=
q
k
+ ε f
k
(q) that the Jacobian associated with the coordinate change is
ij
=
∂q
i
∂q
j
δ
ij
+ ε
∂ f
i
(q)
∂q
j
6
When the commutation relation [A, B] of operators is introduced later, the Poisson bracket will be
denoted as [A, B]
PB
to avoid confusion.
up to (ε). The momentum transforms under this coordinate change as
p
i
→
j
p
j
−1
ji
p
i
− ε
j
p
j
∂ f
j
∂q
i
.
Then, it follows that
0 = H (q
k
, p
k
) − H (q
k
, p
k
)
=
∂ H
∂q
k
ε f (q) −
∂ H
∂p
j
εp
i
∂ f
i
∂q
j
= ε
∂ H
∂q
k
f
k
(q) −
∂ H
∂p
j
p
i
∂ f
i
∂q
j
= ε[H, Q]=ε
dQ
dt
,
which shows that Q is conserved.
This theorem shows that to find a conserved quantity is equivalent to finding
a transformation which leaves the Hamiltonian invariant.
A conserved quantity Q is the ‘generator’ of the transformation under
discussion. In fact,
[q
i
, Q]=
k
∂q
i
∂q
k
∂ Q
∂p
k
−
∂q
i
∂p
k
∂ Q
∂q
k
=
k
δ
ik
f
k
(q) = f
i
(q)
which shows that δq
i
= ε f
i
(q) = ε[q
i
, Q].
A few examples are in order. Let H = p
2
/2m be the Hamiltonian of a free
particle. Since H does not depend on q, it is invariant under q → q +ε·1, p → p.
Therefore, Q = p · 1 = p is conserved. The conserved quantity Q is identified
with the linear momentum.
Example 1.4. Let us consider a paticle m moving in a two-dimensional plane with
the axial symmetric potential V (r). The Lagrangian is
L(r,θ) =
1
2
m(˙r
2
+r
2
˙
φ
2
) − V (r).
The canonical conjugate momenta are:
p
r
= m ˙rp
θ
= mr
2
˙
θ.
The Hamiltonian is
H = p
r
˙r + p
θ
˙
θ − L =
p
2
r
2m
+
p
2
θ
2mr
2
+ V (r).
This Hamiltonian is clearly independent of θ and, hence, invariant under the
transformation
θ → θ + ε · 1, p
θ
→ p
θ
.
The corresponding conserved quantity is
Q = p
θ
· 1 = mr
2
˙
θ
that is the angular momentum.
1.2 Canonical quantization
It was known by the end of the 19th century that classical physics,
namely Newtonian mechanics and classical electromagnetism, contains serious
inconsistencies. Later at the beginning of the 20th century, these were resolved by
the discoveries of special and general relativities and quantum mechanics. So far,
there is no single experiment which contradicts quantum theory. It is surprising,
however, that there is no proof for quantum theory. What one can say is that
quantum theory is not in contradiction to Nature. Accordingly, we do not prove
quantum mechanics here but will be satisfied with outlining some ‘rules’ on which
quantum theory is based.
1.2.1 Hilbert space, bras and kets
Let us consider a complex Hilbert space
7
={|φ, |ψ,...}. (1.26)
An element of
is called a ket or a ket vector.
A linear function α :
→ is defined by
α(c
1
|ψ
1
+c
2
|ψ
2
) = c
1
α(|ψ
1
) +c
2
α(|ψ
2
) ∀c
i
∈ , |ψ
i
∈ .
We employ a special notation introduced by Dirac and write the linear function
as α| and the action as α|ψ∈
. The set of linear functions is itself a vector
space called the dual vector space of
, denoted
∗
. An element of is called
a bra or a bra vector.
Let {|e
1
, |e
2
,...} be a basis of .
8
Any vector |ψ∈ is then expanded
as |ψ=
k
ψ
k
|e
k
,whereψ
k
∈ is called the kth component of |ψ.Nowlet
us introduce a basis {ε
1
|, ε
2
|,...} in
∗
. We require that this basis be a dual
basis of {|e
k
},thatis
ε
i
|e
j
=δ
ij
. (1.27)
7
In quantum mechanics, a Hilbert space often means the space of square integrable functions L
2
(M)
on a space (manifold) M. In the following, however, we need to deal with such functions as δ(x) and
e
ikx
with infinite norm. An extended Hilbert space which contains such functions is called the rigged
Hilbert space. The treatment of Hilbert spaces here is not mathematically rigorous but it will not cause
any inconvenience.
8
We assume is separable and there are, at most, a countably infinite number of vectors in the basis.
Note that we cannot impose an orthonormal condition since we have not defined the norm of a vector.
Then an arbitrary linear function α| is expanded as α|=
k
α
k
ε
k
|,where
α
k
∈ is the kth component of α|. The action of α|∈
∗
on |ψ∈ is now
expressed in terms of their components as
α|ψ=
ij
α
i
ψ
j
ε
i
|e
j
=
ij
α
i
ψ
j
δ
ij
=
i
α
i
ψ
i
. (1.28)
One may consider |ψ as a column vector and α| as a row vector so that α|ψ
is regarded as just a matrix multiplication of a row vector and a column vector,
yielding a scalar.
It is possible to introduce a one-to-one correspondence between elements in
and
∗
. Let us fix a basis {|e
k
} of and {ε
k
|} of
∗
. Then corresponding to
|ψ=
k
ψ
k
|e
k
, there exists an element ψ|=
k
ψ
∗
k
ε
k
|∈
∗
. The reason
for the complex conjugation of ψ
k
becomes clear shortly. Then it is possible to
introduce an inner product between two elements of
.Let|φ, |ψ∈ .Their
inner product is defined by
(|φ, |ψ) ≡φ|ψ=
k
φ
∗
k
ψ
k
. (1.29)
We customarily use the same letter to denote corresponding bras and kets. The
norm of a vector |ψ is naturally defined by the inner product. Let |ψ =
√
ψ|ψ. It is easy to show that this definition satisfies all the axioms of the norm.
Note that the norm is real and non-negative thanks to the complex conjugation in
the components of the bra vector.
By using the inner product between two ket vectors, it becomes possible
to construct an orthonormal basis {|e
k
} such that (|e
i
, |e
j
) =e
i
|e
j
=δ
ij
.
Suppose |ψ=
k
ψ
k
|e
k
. By multiplying e
k
| from the left, one obtains
e
k
|ψ=ψ
k
.Then|ψ is expressed as |ψ=
k
e
k
|ψ|e
k
=
k
|e
k
e
k
|ψ.
Since this is true for any |ψ, we have obtained the completeness relation
k
|e
k
e
k
|=I, (1.30)
I being the identity operator in
(the unit matrix when is finite dimensional).
1.2.2 Axioms of canonical quantization
Given an isolated classical dynamical system such as a harmonic oscillator, we
can construct a corresponding quantum system following a set of axioms.
A1. There exists a Hilbert space
for a quantum system and the state of the
system is required to be described by a vector |ψ∈
.Inthissense,
|ψ is also called the state or a state vector. Moreover, two states |ψ and
c|ψ (c ∈
, c = 0) describe the same state. The state can also be described
as a ray representation of
.
A2. A physical quantity A in classical mechanics is replaced by a Hermitian
operator
ˆ
A acting on
.
9
The operator
ˆ
A is often called an observable.
The result obtained when A is measured is one of the eigenvalues of
ˆ
A.(The
Hermiticity of
ˆ
A has been assumed to guarantee real eigenvalues.)
A3. The Poisson bracket in classical mechanics is replaced by the commutator
[
ˆ
A,
ˆ
B]≡
ˆ
A
ˆ
B −
ˆ
B
ˆ
A (1.31)
multiplied by −i/
¯
h. The unit in which
¯
h = 1 will be employed hereafter
unless otherwise stated explicitly. The fundamental commutation relations
are (cf (1.22))
[ˆq
i
, ˆq
j
]=[ˆp
i
, ˆp
j
]=0 [ˆq
i
, ˆp
j
]=iδ
ij
. (1.32)
Under this replacement, Hamilton’s equations of motion become
d ˆq
i
dt
=
1
i
[ˆq
i
, H ]
d ˆp
i
dt
=
1
i
[ˆp
i
, H ]. (1.33)
When a classical quantity A is independent of t explicitly, A satisifies the
same equation as Hamilton’s equation. By analogy, for
ˆ
A which does not
depend on t explicitly, one has Heisenberg’s equation of motion:
d
ˆ
A
dt
=
1
i
[
ˆ
A,
ˆ
H]. (1.34)
A4. Let |ψ∈
be an arbitrary state. Suppose one prepares many systems, each
of which is in this state. Then, observation of A in these systems at time t
yields random results in general. Then the expectation value of the results is
given by
A
t
=
ψ|
ˆ
A(t)|ψ
ψ|ψ
. (1.35)
A5. For any physical state |ψ∈
, there exists an operator for which |ψ is one
of the eigenstates.
10
These five axioms are adopted as the rules of the game. A few comments
are in order. Let us examine axiom A4 more carefully. Let us assume that |ψ is
normalized as |ψ
2
=ψ|ψ=1 for simplicity. Suppose
ˆ
A(t) has the set of
discrete eigenvalues {a
n
}with the corresponding normalized eigenvectors {|n}:
11
ˆ
A(t)|n=a
n
|nn|n=1.
9
An operator on is denoted by ˆ. This symbol will be dropped later unless this may cause
confusion.
10
This axiom is often ignored in the literature. The raison d’etre of this axiom will be clarified later.
11
Since
ˆ
A(t) is Hermitian, it is always possible to choose {|n} to be orthonormal.