Vector potential and gauge theory concept:
a pedagogical introduction
Jun Kishine
Kyushu Institute of Technology, Japan
September 10, 2009
Abstract
@Theoretical Physics Dept. of Ural State University
Contents
1 Introduction
2
2 Kinetic momentum and canonical momentum
2
3 Quantum mechanics
7
4 Vector potential as a phase
11
5 Aharonov-Bhom e¤ect
13
6 Superconductivity and manifestation of the gauge …eld
6.1 Superconductivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6.2 Gauge invariance and Josephson problem . . . . . . . . . . . . . . .
6.3 Superconductivity as Higgs mechanism . . . . . . . . . . . . . . . . .
15
15
20
23
7 Gauging symmetry and gauge …eld theory
26
1
1
Introduction
According to Y. Nambu (Nobel Laureate, 2008, for his description of “spontaneous broken symmetry” in the 1960s), the basic paradigm of theoretical
physics consists of the following four notions:
1. Model building
2. Field theory, accompanying the concept of renormalization
3. Symmetry and its breaking
4. Gauge principle
To work with theoretical physics means you will work out how to materialize the above notions in real nature. In this pedagogical lecture, I will take
examples to demonstrate some of these notions. I will show you how vector
potential in classical electromagnetism is unexpectedly elevated to one of the
most important concept of physics, when it meets quantum mechanics. By
seeking the physical meaning of the vector potential, we reach the concept
of gauge …eld. Furthermore, gauge principle will be introduced as a working
principle of nature. I know that you have just started to learn quantum mechanics. So, I will include very quick survey of the basic concept of quantum
mechanics. These topics may touch 3 and 4 of the above mentioned notions.
Model building (1) and …eld theory (2) require more technical preparation,
that you are supposed to learn in near future.
2
Kinetic momentum and canonical momentum
As all you know, the kinematic momentum is de…ned as
p~kin = m~v :
(1)
I specify the kinematic momentum by subscript "kin." Then, Newtons’equation of motion (EOM) for a moving particle under the potential V is given
by
d~pkin
~
= rV:
(2)
dt
You may also know that the law of mechanics can be constructed Then, we
know that the so-called Lagrangian can be constructed by
1
L (~r; ~v ) = m~v 2 V:
(3)
2
The EOM is derived from the Euler-Lagrange equation,
d
dt
@L
@~v
=
2
@L
;
@~r
which again gives (2).
Then the canonical momentum is de…ned by
p~can =
@L
@~v
(4)
In the case of a free particle, L = 12 m~v 2 gives p~can = m~v , that coincides with
the kinetic momentum p~kin = m~v . The Hamiltonian is given by
p~2can
+ V:
(5)
2m
Please remember that the canonical momentum enters here. The EOM is
derived from the Hamilton’s canonical equations,
H =~v p~can
L=
d~pcan
@H
d~pcan
~
=
=)
= rV;
dt
@~r
dt
d~r
d~r
@H
p~can
=)
=
=
;
dt
@~pcan
dt
m
which again give (2).
Now, let us consider the motion of a charged particle (with charge q) under
~ and the magnetic …eld B.
~ You may know
the electric …eld E
m
d~v
~ + q~v
= qE
dt
~
B:
(6)
Here, it is important to recall that there are two kinds of electric …elds, i.e.,
Coulomb’s static …eld and Faraday’s dynamic …eld. The former is given by
~C =
E
~
rV;
(7)
and the latter is given through the Faraday’s law of electromotive induction,
~
r
~F =
E
~
@B
:
@t
The form of the Faraday’s law motivates us to write
~ =r
~
B
~
A
(8)
~ is called the vector potential. Actually, this form guaranwhere you know A
tees
~ B
~ = 0:
r
3
~ =r
~
By using B
~ you notice
A,
~
r
~F =
E
d ~
r
dt
~ =
A
~
r
~
@A
;
@t
which leads to
~
@A
:
(9)
@t
~ C and E
~ F , we recognize that the electric …eld is generally
Putting together E
given by
~
@A
~
~
(10)
E = rV
@t
Below, we will see that V and A~ are much more fundamental than E~
~ The EOM becomes
and B.
!
~
d~v
@
A
~
~ A
~ :
m =q
rV
+ q~v
r
(11)
dt
@t
~F =
E
Now, let us raise a query: how we can make (11) into the form,
d
~ (
( )= r
):
(12)
dt
Motivation behind this query is "how to construct Lagrangian for a charged
particle in the electromagnetic …eld."
Note we use the following identity,1
~v
~
r
~ =r
~ ~v A
~
A
~ A:
~
~v r
~ is a function of position ~r and time t.
Next, we note that A
~ (~r; t) @ A
~ (~r; t)
d~r d ~
dA
=
+
A (~r; t)
dt
@t
dt d~r
~ (~r; t)
@A
~ A
~ (~r; t) :
+ ~v r
=
@t
1
Proof:
"ijk "klm vj @l Am = ( il jm
im jl ) vj @l Am
= vj @i Aj vj @j Ai
= @i (vj Aj ) vj @j Ai
4
Using this, we have
d~v
m =q
dt
=q
=q
"
"
~
rV
~
rV
~
rV
~
@A
@t
!
~
r
+ q~v
~
dA
~
+ ~v r
dt
#
~
dA
~ ~v
+ qr
dt
~
A
#
h
~
~ ~v A
~
A +q r
~ A
~
~v r
i
~ :
A
(13)
This gives
d
~ = qr
~ V ~v A
~ ;
m~v + q A
dt
that we exactly desired! Now, we had our potential
~
VEM = qV q~v A
(14)
(15)
and Lagrangian
1
~
L = mv 2 qV + q~v A
2
This Lagrangian gives the canonical momentum
@L
~;
p~can =
= m~v + q A
@~v
that exactly completes our wired EOM (14) in the conventional form
d~pcan
~ EM :
= rV
dt
Or,
~
p~can q A
~v =
m
The corresponding Hamiltonian is
L
1 2
~
= ~v p~can
mv + qV q~v A
2
~
p~can q A
1
~
=
p~can
p~can q A
m
2m
(16)
(17)
(18)
H = ~v p~can
1
~ p~can 1 p~can
p~can q A
m
2
2
1
~ + qV
=
p~can q A
2m
=
5
2
+ qV
~
qA
q
~
qA
p~can
m
!
~
A
~ + qV
qA
(19)
We recognize an important fact that under the in‡uence of the electromagnetic …eld, the canonical momentum does NOT coincides with the kinetic
momentum, i.e.,
p~can 6= p~kin under the EM …eld.
But why this happens? The answer is as follows. Even in the case of electrostatic …eld, once a charged particle moves in the …eld, the particle feels
time-varing magnetic …eld as an outcome of relativity. That is to say, the
moving charge is "dressed" by the surrounding induced magnetic electric …eld.
This induced electric …elds always give additional momentum to the particle.
The momentum that includes (or renormalizes) this additional momentum is
nothing more than the canonical momentum.
The following …gure was adopted from Feynmann’s lecture on physics (Fig.
21-2 in "Quantum Mechanics"). He gave comprehensive way how to understand the di¤erence between the kinetic and canonical momentum.
Now, let us go back to
~ =r
~
B
~ =
E
~
A;
~
rV
(20)
~
@A
:
@t
~ and E
~ do not charge under the transformation
You notice that B
8
~!A
~0 = A
~ r
~ ;
<A
@
: V !V0 =V +
:
@t
6
(21)
(22)
In four-vector notation, we write
~
A = (V; A);
@
~ ;
@ =
; r
@t
A ! A0 = A + @
:
Then, the EM …eld can be all built in the EM …eld tensor,
F
=@ A
@ A :
It is very easy to check
F
! F 0 = @ A0
= @ (A + @
=F :
@ A0
)
@ (A + @
)
The transformation (22) is called gauge transformation and seems nothing
more than mathematical trick at this moment. But you will see this fancy
trick eventually leads to the deepest root of modern physics.
3
Quantum mechanics
Below, I show you the electron di¤raction experiment by A. Tonomura.
Tonomura’s experiment tells us the following two points:
An electron travels as wave, but appears as a particle.
Now, how to describe this wave? The answer is wave function
(r; t) = a(r; t)ei
7
(r;t)
:
(23)
The position where the travelling electron eventually appears is NOT
determined. The accumulation of the electrons occur in probabilistic
manner.
Now, how to determine the probability? The answer is that
P (~r; t) = j (~r; t)j2
(24)
means the probability density of …nding the particle inside the volume element
d = dxdydzat time t. To make this sense, the wave function must be
normalized:
Z
j (~r; t)j2 d = 1:
(25)
Furthermore, the wave function must be
single-valued
continuous
bounded.
The reason why the di¤raction pattern appeared after accumulating huge
number of electrons. An electron that may travel in both sides of the slit is
described by di¤erent wave functions, 1 = A1 ei 1 and 1 = A2 ei 2 . Then,
the superposition gives the interfered wave on the screen,
=
1
+
2
= a1 ei 1 + a2 ei 2 :
This gives the probability density,
j j2 = ja1 ei 1 + a2 ei 2 j2 = a21 + a22 + 2a1 a2 cos(
1
2 ):
The second term gives you the interference pattern!
The next query is how to understand the particle-wave duality. In 1923,
de Broglie suggested an expression for the wavelength associated with the
particle. We summarize his suggestion together with the Einstein relation
which connects the particle energy and wave frequency:
Wave
Particle
Relation
frequency !
energy E
! = E=~ (Einstein)
wave number ~k canonical momentum p~can ~k = p~can =~ (de Broglie)
Free propagation of a particle may be described by a plane wave,
~
(~r; t) = aei(k ~r
!t)
:
Then, Einstein and de Broglie relations give
i
(~r; t) = ae ~ (~p ~r
8
Et)
:
We know that the free particle has energy
p2
E=
:
2m
So, we notice that
(~r; t)satis…es
i~
~2 ~ 2
r (~r; t)
2m
@ (~r; t)
=
@t
as a candidate equation. Furthermore, in the case of motion under the potential V (~r);we know
p2
E=
+ V (~r)
2m
and (~r; t) may satisfy
@ (~r; t)
=
i~
@t
~2 ~ 2
r (~r; t) + V (~r) (~r; t) :
2m
(26)
This is exactly what we seek, called Schrödinger equation which is the most
basic equation in quantum mechanics. In the case where the energy Eis …xed,
(~r; t) =
(~r) e
i E~ t
and the Schrödinger equation reduces to
~2 ~ 2
r (~r) + V (~r) (~r) = E (~r)
2m
(27)
which is called time-independent Schrödinger equation. This heuristic manner of deriving the Schrödinger equation was done by Schrödinger himself.
The Schrödinger equation has a form of eigenvalue problem. Introducing the
di¤erential operator called Hamiltonian
^=
H
~2 ~ 2
r + V (~r)
2m
we have
^ =E
H
In quantum mechanics, this equation has a deep meaning: Acting observation
^
Hon
the state gives you the energy Eas a de…nite measured quantity. To
measure something means...
1. There is a state described by a wave function
9
2. We observe this state: we describe this action as A^
3. and measure this: eigenvalue equation A^ = a gives you the measured
quantity a.
Now, the quantum Hamiltonian is obtained by replacing the momentum
pin the classical Hamiltonian
p~2
H=
+ V (~r)
2m
with he di¤erential operator:
~
i~r
p~ =
Probability current: to interpret the meaning of the wave function, the
concept of the probability current is important. Using (24) and (26), we
obtain the conservation law (or continuity equation),
@P (~r; t) ~ ~
+ r j(~r; t) = 0;
@t
(28)
where the ‡ux is de…ned by
~j(~r; t) = ~
2im
~
r
~
r
(29)
To memorize this result, it is useful to write
~j(~r; t) = ~
2im
~ + c.c.
r
!
@
~
2im
(30)
To check the conservation law is easy. Recalling (26), we have
@
1
=+
@t
i~
@
1
=
@t
i~
~2 ~ 2
r +V
2m
~2 ~ 2
r
+V
2m
10
(31)
;
;
(32)
and then
@P (~r; t)
=
@t
=
=
=
=
4
@
@
+
@t
@t
2
1
~ ~2
r +V
i~
2m
~
~2
~2
r
r
2im
~
~
~
~
r
r
r
2im
~ ~j(~r; t)
r
1
i~
~2 ~ 2
r
2m
+V
Vector potential as a phase
When we translate the classical Hamiltonian into the quantum Hamiltonian,
we must always make replacement
p~can =
~
i~r
(33)
~
Never confused with erroneous relation p~kin = i~r.
But why we need to begin with canonical form? The answer is related to
relativistic symmetry. Compatibility of the canonical EOM with the special
relativity were keenly pointed out by Dirac[P. A. M. Dirac, Rev. Mod. Phys.
21 (1949) 392]. According to Dirac, if a canonical formulation is possible
in one Lorentz frame, it should be possible in every Lorentz frame due to
the principle of relativity. Therefore a Lorentz transformation must be a
canonical transformation of the dynamical variables— for the same reason
that an ordinary spatial rotation is a canonical transformation. This is the
reason why we should begin with the canonical variables to set up canonial
quantization. In summary,
Lagrangian ! Hamiltonian ! Quantum theory
Now, in the case of a particle under the EM …eld, we have
~=
p~can = m~v + q A
11
~
i~r;
(34)
2
1
~ qA
~ + qV
i~r
2m
q~ 2
~2 ~
r i A
+ qV
=
2m
~
~2 ~ 2
=
D + qV;
2m
H=
where
~ =r
~
D
or
D =@
q~
i A
~
(35)
q
i A
~
(36)
is so-called covariant derivative. This is correct Hamiltonian.2
Another observation is the de Broglie’s relation
~ ~k
p~can = m~v + q A=~
This means
(37)
h
i
~
r) = a exp k ~r
~ (~
A
i
~ ~r
m~v + q A
~
i ~
~r
(~r) :
= a exp q A
~
= exp
More generally,
"
i
(~
r
)
=
a
exp
q
~
A
~
Z
~r
~r0
~ d~r
A
#
(~r)
(38)
We see that the presence of the magnetic …eld causes an additional phase
Z ~r
i
~ d~r
A
(39)
P = q
~ ~r0
which is called Peierls phase.
2
Now, let us remind us to pay attention to the electromagnetic energy,
Z
Z
1
~ 2 dV + "0
~ 2 dV;
HEM =
B
E
2 0
2
which exists in reality.
12
5
Aharonov-Bhom e¤ect
The Peierls phase P really comes up in experiment. This is really an amazing
evidence. Let us consider an in…nitely long and narrow solenoid, carrying a
magnetic ‡ux
ZZ
~ dS:
~
B
(40)
=
~
Note that there is NO magnetic …eld Boutside
the solenoid, but nevertheless
~ because
there must be a magnetic vector potential A,
I
~ d~r:
=
A
(41)
C
Now, classical Hamiltonian may be given by
2
1
~
p~can q A
2m
1
1
=
p2r + 2 p
2m
r
H=
q
2
+ p2z :
(42)
Here we used cylindrical coordinates r; ; z;with the zaxis along the solenoid,
~
and if the gauge is appropriately chosen, the vector Ahas
only an azimuthal
component
2 rA = ;
which gives (42). However, the classical canonical transformation
p !p
13
q
2
can totally eliminate the ‡ux term in (42). So we reach the conclusion that
the solenoid does not in‡uence the motion of charges outside it at classical
level. But the situation becomes totally unexpected in quantum world! This
remarkable evidence was enlightened by Aharonov and Bohm[Y. Aharonov
and D. Bohm, Phys. Rev. 115 (1959) 485].
According to (38), we cannot forget the additional phase acquired by an
electron moving along both sides of the solenoid. That is to say, we need to
distinguish two paths connecting the starting point A and the endpoint B,
and write:
rB )
1 (~
= a exp
rB )
2 (~
= a exp
Z
i
~ d~r
e A
~ C1
Z
i
~ d~r
e A
~ C2
(~rA ) ;
(43a)
(~rA ) :
(43b)
Now, the interference may occur at the endpoint:
j
rB )
1 (~
2
=a e
= 2a2
+
R
rB )j2
2 (~
R
2
i
~
e
~ C2 A d~r j
i
~
~ e C1 A d~r
+e
I
e
~ d~r
1 + cos
A
~ C
0
=
I
~ d~r =
A
C
and
0
=
ZZ
h
=2
e
~
r
j (~rA )j2
j (~rA )j2
= 2a2 1 + cos 2
where
(~rA )j2
~
A
~=
dS
2:067833636
(44)
ZZ
~ dS
~
B
1015 Wb
(45)
(46)
is called the ‡ux quanta. The interference was actually observed by (again!)
Tononmura[A. Tonomura, et al., Phys. Rev. Lett. 48, 1443 - 1446 (1982) ].
14
6
6.1
Superconductivity and manifestation of the gauge …eld
Superconductivity
Superconductivity is one of the most amazing phenomena in nature. The
superconducting state appears when the temperature T is lowered below a
critical temperature Tc at which
the resistivity abruptly goes down to zero
15
the electronic heat capacity exhibits a discontinuous jump
the magnetic …eld can penetrate the superconductor only a small distance
100 nm, called the London penetration depth. This is called
the Meissner e¤ect, and is a de…ning characteristic of superconductivity.
By the way, when you think of electric current what you have in mind may
be Ohm’s law:
~
J~O = E;
(47)
where the subscript "o" means the Ohmic current. However, superconducting current is NOT ohmic. That is to say, the superconducting current is
~ Surprisingly, the superconducting current is triggered by
not driven by E.
~ =r
~ A.
~ More surprisingly, the current is not proportional
magnetic …eld B
~ but is proportional to the vector potential as,
to B,
~
J~s = A
(48)
where the subscript "s" means the superconducting current. But where this
strange relation comes from? To see this, let us go back to the relation
16
~v =
~
qA
p~can
:
m
Then, the current carried by the superconducting electron is given by
J~ = e ~v
e
=
p~can e
m
e
~
i~r
=
m
i~e
~
=
r
m
~e
~
=
D
im
(49)
~
A
~
eA
i
e ~
A
~
Recalling (29), more correctly, we have
~e
J~ =
2im
~
D
+ c.c.
But, what’s the meaning of the wave function ? This is the key issue to
understand the superconductivity. Now, I simply say that in the superconducting state, all the electrons get into a single quantum state described by
the so-called macroscopic wave function,
=
p
ei
(50)
p
In the macroscopic wave function, we say the amplitude
and the phase
are decoupled. This decoupling is a consequence of renormalization of electronic degrees freedom to retain only macroscopically surviving degrees of
freedom for our temperature windows.
At this moment , please take as a given wave function which describes
the state as a whole. Then, let us compute the current as follows:
~e
~ + c.c.
J~s =
D
2im
e
~ eA
~
=
i~r
+ c.c.
2m
e p i
~ eA
~ p ei + c.c.
=
e
i~r
2m
17
e p
e i
2m
e p
e i
=
2m
e
~
=
~r
2m
e
~
=
~r
m
=
i~
~
p ~ i
re
p
~p
i~ei r
~ p ei
eA
~
ei r
~ p ei
eA
+ c.c.
+ c.c.
~ + c.c.
eA
~
eA
So, if the phase is "solidi…ed," i.e.,
= constant over the whole region
we have
e2 ~
A
m
J~s =
(51)
This exactly gives (48)! The following story is straightforward. By taking
curl of (51), we have
2
~ J~s = e B
~
r
(52)
m
This is famous London equation.
Next, let us consider physical consequence of (52). For completeness, let
us retrieve Ohmic contribution and write down Ampere’s law,
~
r
~ =
B
J~s + J~O :
0
By taking curl, we have
~
r
~
r
~
~ =
B
0r
J~s +
0
~
r
~
E:
Using (52), we get
e2 ~
B
0
m
where we introduced
~ 2B
~ =
r
0
~
@B
~ 2B
~ =
)r
@t
1
2
1
2
i
0
~
! B;
e2
:
0
m
=
Usually,
109 ( m)
1
;
10 8 m;
5
1
!
0
2
18
0
1012 Hz
=4
10 7 ;
and we can neglect the e¤ect of normal current. So, we have
~ 2B
~ = 1 B:
~
r
2
Let us consider the con…guration shown below:
We easily see that
~
~ 0e z :
B(z)
=B
This is just Meissner e¤ect! We succeeded in explaining this fantastic evidence!
Let us reconsider this result. What is always true is the EOM
m
d~v
~
= e E;
dt
and the physical de…nition of the current density,
J~ = e ~v :
Putting them together, we have
m dJ~
~
= e E:
e dt
By taking curl of this, we have
m d ~
r
e dt
~
J~ = e r
19
~
E:
Now, Faraday’s law
~
r
~
@B
;
@t
~ =
E
also holds any time. So,
m d ~
r
e dt
~
@B
e
:
@t
J~ =
This implies
e2 ~
~
J+
B = const.
m
By putting the constant equal to zero, we return to
~
r
~
r
6.2
e2 ~
B:
m
J~ =
Gauge invariance and Josephson problem
You saw that the superconducting state is described by a single wave function
p i
=
e :
(53)
Now, let us locally "twist" the phase, i.e.,
!
ei
(~r;t)
;
where (~r; t) could be a function of both space and time. Because this twist
causes
~
D
=
~
i~r
D0
=
i~
@
@t
~
eA
eV
~
i~r
!
!
i~
~
A
e
@
@t
e
~~
r
e
~@
V
e @t
;
;
we see that the gauge transformations become
~~
r ;
e
~@
:
e @t
~!A
~
A
V !V
20
(54)
(55)
Let us consider a superconducting hollow cylinder under the magnetic
‡ux [see below]. Eq.(54) leads to the following consequence:
I
C
~ d~r !
A
=
I
~
e
~ d~r
A
C
~
e
I
C
~
r
d~r
;
where we took the contour C inside the superconductor. Now, to absorb the
change of the vector potential, we conclude
~
e
= 0:
But
should be equal to 2 n because of single-valuedness of the wave
function. So, we reach the conclusion
=
h
n;
e
(56)
where we introduced the ‡ux quantum
0
=
h
h
=
= 2:067833636
e
2e
1015 Wb
(57)
Notice that this ‡ux quantum is a bit di¤erent from the previous one (46),
where the charge e is replaced with
e = 2e:
21
The reasoning of this replacement needs microscopic theory, so called BardeenCooper-Schrie¤er (BCS) theory, established in 1957. This theory is regarded
as the most sophisticated theory in modern physics (Prof. Y. Nambu declared this). Of course, to say much about this theory is far beyond the
present scope. Please wait until you will get familiar with more advanced
quantum theory in near future.
Next, let us see Eq.(55) which leads to the equation of motion for the
macroscopic phase
e
@
= V
(58)
@t
~
where V means external voltage by the electric …eld. Eq. (58) describes
“phase slippage”by the external voltage and directly gives the AC Josephson
e¤ect. The Josephson current is given by
j = j0 sin
12 ;
where 12 denotes the phases di¤erence between the two superconductors
sandwiched with an insulator barrier. When the external voltage is zero,
we observe the DC Josephson current and the Fraunhofer pattern (RowellAnderson oscillation) appears.
Rowell-Anderson oscillation
of the Josephson current
22
Now we consider the r.f. voltage
V = V0 + V1 cos(!t)
is applied.3 Eq. (58) gives
(t) = (0) +
e
e
V0 t +
V1 sin(!t);
~
~!
and
e
e
Vt+
V1 sin(!t)
~
~!
j0 sin [V (0) + !J t + sin(!t)]
1
X
e
Jn
= j0
V1 sin [V (0) + !J t + n!t] ;
~!
n= 1
j = j0 sin V (0) +
(59)
(60)
(61)
with Jn denoting the Bessel function. Then, when the matching condition
!J + n! = 0
is satis…ed,the phase di¤erence becomes time-independent and therefore socalled Shapiro step appears.
6.3
Superconductivity as Higgs mechanism
Let us go back to the Lagrangian of a particle under the EM …eld,
1
L = mv 2
2
~
qV + q~v A
(62)
This Lagrangian gives the canonical momentum
p~can =
@L
~:
= m~v + q A
@~v
(63)
~
Now, let us pay attention to the A-term
~
LA = q~v A
q
~ A
~
=
p~can q A
m
q
q2 ~ 2
~
A:
= p~can A
m
m
3
Note that this situation is exactly similar to the ESR problem in the chiral magnet, I guess.
23
(64)
We will apply this to the case of superconducting state, where q =
is to say, we take an expectation value with respect to
p i
=
e :
e . That
That is to say, we handle with
e 2 ~2
q
~
p~can A +
A
m
m
:
But recall
= constant over the whole region.
We also regard as constant. Then, So, the manipulation becomes quite
simple to obtain
2
e2
~2 = e A
~ 2:
j j2 A
(65)
LHiggs =
m
m
The reason why I put the subscript "Higgs" becomes clear shortly. Now,
~ ? To see
what does this mean? What happed about the vector potential A
this, we do not forget the energy stored in the …eld itself, i.e.,
ZZZ
1 ~2
"0 ~ 2
B d :
(66)
E +
EEM =
2
2 0
Maybe, you know the corresponding Lagrangian density is
LEM =
=
"0 ~ 2
E
2
"0
2
~
@A
@t
1 ~2
B
2 0
!2
1 ~
r
2 0
Now, putting HHiggs together, we obtain
!2
~
"0 @ A
e 2 ~2
L = LEM + LHiggs =
+
A
2 @t
m
~
A
2
1 ~
r
2 0
~
A
2
:
(67)
~
The second term has a profound meaning. To see this, Fourier transform A
as
~ / ei(~k ~r !t) :
A
(68)
24
Then, we have an expression like
"0 2 e 2
1 2 ~2
L!
! +
+
k A
2
m
2 0
"0 2
1 2 e2
~ 2:
A
=
!
k +
2
"0 0
m
This indicates the dispersion relation may be given by
s
1 2
e2
!=
k +
:
"0 0
"0 m
As you know,
c=p
1
"0
(69)
(70)
(71)
(72)
;
0
gives the velocity of light in vacuum. So, by putting
e2
M =
;
"0 m
2
we have …nal result
!=
(73)
p
c2 k 2 + M 2 :
(74)
This means that the "mass" M appears in the excitation spectrum of photon!
The photon acquires the mass! This is exactly so called Higgs mechanism,
which gives the mass to the massless particle. In the superconducting state,
the …led (Cooper pair …eld) plays a role of Higgs …eld. Superconductivity
is only one known way to give the mass to photon. You see that the mass is
directly related to the London’s penetration depth
1
2
e2
= 0
= "0
m
0M
2
:
(75)
Or we notice a marvelous relation
M=
25
c
:
(76)
We see that …nite mass gives …nite penetration depth. If the mass of
photon is so heavy that M ! 1,
! 0, i.e., the magnetic …eld can never
enter the superconducting material. This correspondence will persuade you
to under stand the reason why we say the photon got mass. It is really
amazing that this kind of fancy mechanism realized in laboratory, by just
cooling the sample. The Higgs mechanism in the superconductor is drawn
by picture, so called Feynmann diagram shown below.
7
Gauging symmetry and gauge …eld theory
Let us …nalize our story about vector potential. In the previous sections,
we have seen that the vector potential plays unexpected roles in quantum
physics. But, now let us consider the whole story in a reverse way. I mean
that we demand the following principle of local gauge invariance:
physical laws must be invariant under
where
!e
i ~q
(77)
depends on both space and time. Now, the evidence,
~ !
i~r
~ e
i~r
i ~q
=e
i ~q
~
i~r
~
qr
forces us to introduce something to compensate the term
be done by modifying the derivative like
~ !
i~r
~
i~r
26
~
qA
~
i~D;
(78)
~ . This can
qr
(79)
where the covariant derivative (35),
q~
i A;
(80)
~
again appeared. But, notice that time, the logic is reverse. We demanded
the principle (77) and the covariant derivative appeared as a consequence of
this principle. Then, we see
~
D
~
i~r
~
qA
e
i ~q
~
r
=e
i ~q
~
i~r
~
qA
~
qr
=e
i ~q
~
i~r
~0
qA
;
(81)
where the transformation
~!A
~0 = A
~+r
~ ;
A
completely absorbs everything. q
has a geometrical meaning, …rst enlightThe transformation ! e i ~
ened by Herman Weyl. Recall is generally complex. So, we have real and
imaginary axis to "measure" this. But, how about changing this measure
from space to space and time to time. That is to say, there may be freedom to choose the local frame depending on x and t. However, this freedom
apparently causes some ambiguity.
How to compare (x) and (x + dx)?
To make the story simple, I will omit t for the time being and consider only
one spatial coordinate See the …gure below. The covariant derivative plays
a role to rescale (or gauge) the measure of the frame. If we use exactly the
same frame, we should have
@ (x)
;
(82)
@x
as you learned in your calculus class. But, if we took the freedom of arbitrary
choice of the frame, we may have something like
(x + dx) =
(x + dx) =
(x) + dx
(x) + dx
@ (x)
+ g (x) ;
@x
(83)
where the term g (x) connects the local frames at x and x + dx. Now, you
notice that this form is exactly consistent with the covariant derivative
(x + dx) =
(x) + dx
27
@ (x)
@x
q
i A (x) :
~
(84)
The vector potential plays a role of "connection" that bridges nearby local
frames. This is exactly the geometrical meaning behind the gauge transformation. The principle (77) is now fully understood from physical and
geometrical view points. This principle, the local gauge principle, eventually
is regarded as one of the most fundamental principle that govern the natural
low of forces.
The quantum electrodynamics (QED), the Glashow-Weinberg-Salam theory of electroweak processes among guarks and leptons, and quantum chromodynamics (QCD) on quarks are all founded based on this principle. But,
please keep in mind that the physical meaning of the gauge theory is most
evidently manifested in superconductivity. That’s why elementary particle
theory after 1960s have borrowed its physical foundation on the theory of superconductivity (phenomenological Ginzburg-Landau theory and microscopic
BCS theory).
28