Quantum Mechanics and its mathematical tools
· adek
Henryk Zo÷
¾
1. Inadequacy of the classical physics and the old quantum theory
In the beginning of the twentieth century there appeared some phenomena which
were non-explainable using the classical physics. Let me recall some of them.
1. Radiation of the black body. The black body is an isolated matter which
only emits radiational energy in form of electromagnetic waves. The dependence of
the energy of the emitted waves on the wave length could not been explained on the
ground of classical physics. Max Planck made calculations which suggested that
the shape of the wavelength–energy curve could be explained when one assumes
that the radiation is emitted only in quanta. It means that the admitted energies
are
E = n }! = n h ; n = 1; 2; : : : ;
where ! = 2 is the angular velocity, is the frequency (see (2.4) and } = 2h is an
universal constant such that the Planck constant
h
6:62 10
27
erg s:
(Recall that erg is a CGS unit of energy, erg = g cm2 =s2 = 10 7 J s = 10 7 N m s;
where Joule J and the Newton N are SI unit of energy and force.)
Therefore these portions of the electromagnetic radiation could be interpreted
as particles, namely the photons, with the mass m = 0: According to the Einstein
equation (3.3), E = c2 p2 ; also the absolute momentum jpj should be quantized,
jpj = Ec = }!
c = } jkj : The principal quanta of energy and momenta should satisfy
the relation
(1.1)
E = }!;
p = }k
where k is the wave vector.
2. The Compton scattering and the photoelectric phenomenon. The
Compton’s experiment of scattering of the light (electromagnetic waves) on electrons
revealed that the photons of light behave like particles, equipped with momentum,
energy and with zero mass.
The photoelectric phenomenon appears when a monochromatic light with …xed
frequency ! is directed to a metal’s surface. If the energy }! of the quanta of the
photons is greater that some threshold energy W then the surface emits electrons
with kinetic energy T = }! W: This kinetic energy does not depend on the
radiation intensity. This also con…rms the corpuscular nature of the light.
3. The Davisson–Germer experiment. In the textbook version of this experiment a ‡ow of electrons passes through two near and small holed. On a screen
behind the holes a di¤raction picture formed by electrons appears. In the original
experiment a bundle of electrons was re‡ected from a nickel crystal surface and the
e¤ect was the same, even when the bundle’s intensity was so low that the electrons
were passing separately. This suggests that sometimes electrons behave rather like
waves (not like particles).
4. The Stern–Gerlach experiment. In this experiment a bundle of electrons
was divided into two subbundles when passing through a constant magnetic …eld.
Since the interaction of charged particles with the magnetic …eld is realized through
1
2
the component Mz in the direction of the magnetic …eld of the angular momentum
of the particle, the experiment suggests that the angular momentum of electrons is
also quantized. Moreover, as we shall see it below, the electron besides the angular
momentum has a kind of ‘internal angular momentum’called the spin.
5. The Franck–Hertz experiment and the Balmer–Ritz classi…cation
of the spectral lines. In the Franck–Hertz experiment it was con…rmed that
electrons loose discrete portions of energy in collisions with photons.
The Balmer–Ritz classi…cation says that the frequencies of the emitted or absorbed photons in interaction with the hydrogen atom satisfy the following rule
(1.2)
!=N
1
m2
1
n2
;
; m; n = 1; 2; : : : ;
where N is a normalizing constant. For m = 2 it is called the Balmer series and
was discovered in 1885 by Balmer.
6. The atoms’ stability. When one assumes that the electron in an atom
moves along a circular (or elliptical orbit) around its kernel then one arrives to a
contradiction, because then the electron should loose its energy. Indeed, by the
Maxwell equations it should generate a varying magnetic …eld B which, on order,
should generate a varying electric …eld E; hence an electromagnetic wave should be
emitted. After a very short time ( 10 9 s) the electron should fall into the kernel.
This evidently contradicts with the observed atoms’stability.
The old quantum theory was created by de Broglie, Bohr and others in order
to explain the above facts using some simple assumptions and rules.
One of these assumptions stated that with a particle, like the electron, one should
associate a wave, called the de Broglie wave, of the form
(1.3)
const e
i(Et px)=}
where E and p are the energy and the momentum of the particle.
The rules are the following Bohr rules:
1. The angular momentum of the electron is an integer multiple of };
(1.4)
M = n};
n = 1; 2; : : :
2. The electron in an atom has only discrete energies En and a jump from
an energy level En to an energy level Em is associated with an electromagnetic
radiation with the angular velocity ! nm de…ned by the formula
(1.5)
}! nm = jEn
Em j :
Theorem 1.1. Assuming the above rules formula (1.2) follows.
Proof. For an electron, with the mass m and the electric charge e; which moves
along a circular trajectory of radius r (with center at the proton with the charge +e)
with an angular velocity ! (and the linear velocity v = r!) we have the following
formulas (the Coulomb’s rule and the third component of the angular momentum):
e2
= mr! 2 ; Mz = m!r2 :
r2
Since Mz = n} (by rule 1) we get the possible values of the radius
r = rn = r 0 n 2 ;
r0 = }2 =me2 :
3
2
The energy of the electron consists of the kinetic energy T = 12 m (r!) and of the
2
potential energy U = er . We …nd
E = T + U = En =
e2
2r0
1
:
n2
Application of rule 2 gives the thesis of the theorem.
The above calculations were successful, but there remained too many open questions. So this theory did not developed further. The old quantum theory was
subsequently replaced by mathematically more rigorous theory called the Quantum Mechanics.
2. Basics of Quantum Mechanics
We begin with underlining the fact that Quantum Mechanics is a completely
mathematical theory. Physicists, like Heisenberg, Planck, Schrödinger and others,
constructed a mathematically rigorous theory whose theorems turned out to be in
agreement with the physical observations. The latter fact decided that the theory
survived.
The main mathematical tools were: the probability theory and the linear algebra of eigenvalues and eigenvectors of linear operators (including elements of the
representation theory). The functional analysis came into the game later, when a
rigorous treatment of unbounded di¤erential operators was needed.
Let us present the main postulates of Quantum Mechanics.
P0 The states of the physical system are de…ned via so called the wave functions (or the state functions) which are complex valued functions on the con…guration space
: Q ! C:
The physical quantities, like energy, momentum, position, are de…ned via certain
^ @x ), called the observables, which act on
linear di¤erential operators A^ = A(x;
the wave functions.
For example, if d = dim Q = 1 then the position operator x
^ acts as the multiplication by the argument
(^
x )(x) = x (x)
and the ‘momentum operator’ A^ = i@x acts as the di¤erentiation
A^ (x) =
id =dx:
Important assumption is that:
the observables are symmetric operators, i.e.
Z
Z
^
(A')(x)
(x)dx =
'(x)A^ (x)dx:
Q
Q
Then one can associate with such observable A^ its collection of eigenvalues n and
eigenfunctions
un (x): By the symmetry the eigenfunctions
are pairwise orthogonal,
R
R
um um = 0 for m 6= n: We assume also that un un = 1: (Often the eigenvalues
are not discrete, but at this stage we skip such subtleties).
4
P1 The only possible results of measurement of a physical quantity represented
by an observable A^ are the eigenvalues n of this operator. After the measurement the system switches to the state
= un represented by the corresponding
eigenfunction.
P2 If the physical system is in a state represented by an eigenfunction (x) =
un (x) then the observation result is the corresponding eigenvalue n ; i.e. A^ =
n :
P3 The average of many observations of the physical observable A^ in the same
state equals
R
D E
(A^ )(x) (x)dx
^
(2.1)
A = R
:
(x) (x)dx
P4 The probability of obtaining
quantity A^ in a state equals
(2.2)
P(
n)
=
n
Z
as a result of measurement of the physical
2
(x)un (x)dx :
It is the same as the probability P (un ) that we are in the eigenstate un :
R
Remark 2.1. Introducing the Hermitian product ('; ) = Q '(x) (x)dx we
equip the space of states with the structure of a pre-Hilbert space, which is completed to the Hilbert space
(2.3)
H = L2 (Q):
^
Then the symmetry condition means (A';
) = ('; A^ ): Later usually D
we Eshall
2
assume that the state functions are normalized, k k = ( ; ) = 1: Then A^ =
A^ ;
2
: Moreover, P ( n ) = j( ; un )j :
P
P
2
2
an un (x); an 2 C; then
jan j = 1; P ( n ) = P (un ) = jan j and
D If
E (x)P=
2
A^ =
n jan j : Here we see the probabilistic aspect of the above postulates.
For example, h^
xi can be interpreted as Ex, the expectation value of the ‘random
2
variable’x with the probability distribution de…ned by j (x)j dx:
Of course, there appears the problem of de…nition of the operators associated
with the very concrete physical quantities. Here we have two rules.
The …rst rule is called the correspondence rule which stated the following:
relations satis…ed by physical quantities in the classical mechanics hold true after
replacing the corresponding quantities by corresponding quantum operators.
Example 2.1. If x
^; y^; z^ are the position operators and p^x ; p^y ; p^z are the momentum operators, the the angular momentum operators are following
^ x = y^p^z z^p^x ; M
^ y = z^p^x x
^z = x
(2.4)
M
^p^y ; M
^p^y y^p^x :
The energy operator of a particle, i.e. the Hamiltonian, equals
^ =E
^ = 1 p^2 + U (^
x):
(2.5)
H
2m
Remark 2.1. For the physicists of the beginning of the twentieth century
the correspondence rule was quite natural. But from the mathematical point of
5
view it is not that obvious. The problem is that the algebra of operators is not a
commutative algebra. So the problem of ordering of product of observables appears.
One could de…ne the z–th component of the angular momentum operators as p^y x
^
p^x y^; which is wrong.
The problem of a proper de…nition of a ‘homomorphism’from the algebra C 1 (T Q)
of smooth functions on the phase space T Q to the algebra of di¤erential operators is called the quantization problem. It is one of most important problem in the
modern mathematical physics (see [Ber], [BS]).
The another rule is the famous indeterminacy principle which is crucial for
the Quantum Mechanics’philosophy. Suppose that we are dealing with two observations A and B (of some physical quantities). After applying any of the observations
the state of the physical system changes. But the changes infected by these observations may be not harmonious. It may turn out that AB 6= BA, the result of two
observations may depend on the order in which the observations are made.
This non-commutativity of the observations is easily translated into the mathematical language. It means the non-commutativity of the corresponding observable
operators, the commutator
h
i
^ B
^ := A^B
^ B
^ A^
A;
is nonzero. For example, if eigenfunctions un of the operator A^ are no longer
^ and
eigenfunctions of the operator B
is a given state then application of the
observation BA (…rstly B and next A) switches the system to some state um (with
2
^ ; un ) ), whereas the observation AB switches the system to
the probability (B
^
some eigenfunction of B:
Moreover, on the basis of the old quantum theory we can even estimate quali^
tatively the value of the above commutator for some speci…c operators A^ and B:
It should be either zero, if the operators commute, or of the order }; if the corresponding physical quantities are complementary.
By de…nition, two classical quantities are complementary when their product
is measured in g m2 =s: For example, the pairs: x and px ; t and E; ' and Mz are
complementary; here ' = arg(x + iy) is the angle in the xy coordinates.
The physicists express the non-commutativity in the form of the following Heisenberg’s indeterminacy principle
(2.6)
x
px
};
t
E
};
'
Mz
}:
(To be correct, one should write x y const }, etc. See Theorem 2.1 below. But
the above form looks simpler and is traditional.) The meaning of the inequalities
(2.6) is that the product of discrepancies of quantum measurement of, say the
position x and the corresponding momentum px ; is always separated from zero.
If one knows an exact position of an electron then one can say nothing about its
momentum and vice versa.
Therefore we postulate [^
x; p^x ] = }; where is a constant to be found. Since
[^
x; @x ] = x 0x (x )0 =
it should be p^x =
}@x : The operator
}@x has
the eigenfunctions of the form
up = const e
px= }
6
with the eigenvalues p: But the de Broglie wave of a particle with momentum p
equals eipx=} (see (1.3)). It follows that should equal i and hence
(2.7)
p^x =
i}@x ;
p^y =
i}@y ;
p^z =
i}@z :
Knowing this we get the Hamiltonian operator, called also the Schrödinger
operator,
}2
2m
^ =E
^=
H
(2.8)
+ U (^
x)
(see (2.5)) where x = (x1 ; : : : ; xd ) and the Laplacian
= @x21 + : : : + @x2d :
Using and the fact that the quantities t (the time) and E (the energy) are
^ = const i}@t : But the de Broglie wave e i(Et px)=} has
complementary, we get E
^ So
the energy E, which should be the eigenvalue of E:
^ = i}@t :
E
(2.9)
Moreover, we obtain the following Schrödinger equation
@
^
=H
@t
which describes the evolution of the wave function
(2.10)
i}
(x) = (x; t) in time.
Example 2.2. For a free particle, i.e. with U (x)
0; the de Broglie wave
2
@
p2
}2 @
i(Et px)=}
^
=e
satis…es i} @t = E and H = 2m @x2 = 2m
: The classical
formula E =
p2
2m
follows.
The following theorem provides a probabilistic interpretation of the Heisenberg’s
indeterminacy principle. For the operator x
^ in a normalized state we de…ne its
mean value as h^
xi = h^
xi = (^
x ; ) and its standard deviation as
1=2
x
^ = h^
x
Analogously we de…ne
h^
xii
p^x =
= ((^
x
1=2
h^
xi) ; )
:
p^:
Theorem 2.1. We have
1
}
2
and the equality take place when the wave function can be reduced to the Gaussian
function
x
^
p^
(x) = p
(2.11)
1
2
e
x2 =2
2
by a translation of x and multiplication by the oscillating factor ei x :
Proof. By shifting of the variable x we can assume that h^
xi = 0: The identity
i}@x ei
x
= ei x ( }
i}@x )
shows that the shifting of the variable p corresponds to the multiplication of the
state function by an oscillating factor. Therefore assume that also h^
pi = 0: Thus
2
x
^= x
^2 ;
and 2 p^ = p^2 ; :
By the Schwartz inequality we get
2
x
^
2
2
p^ = k^
x k
2
k^
p k
2
2
j(^
x ; p^ )j = j(^
px
^ ; )j :
7
The number in the right-hand side of the latter inequality is estimated from below
by
1
1
2
2
2
jIm(^
px
^ ; )j = j(^
px
^ ; ) ( ; p^x
^ )j = j([^
p; x
^] ; )j :
4
4
As [^
p; x
^] = i} this equals 14 } 2 = 14 }:
The Schwartz inequality becomes an equality when p^ = x
^ for a constant
: The modulus of a complex number equals it imaginary part when the real part
vanishes. So the inequality from the thesis of Theorem 2.1 is an equality when
0 = 2 Re(^
x ; p^ ) = ( + )(^
x ; p^ ):
Thus
= i is imaginary and the equation
i}
gives
= const e
2
x =}
0
=i
:
Remark 2.2. The Gaussian probability distribution p21 exp( x2 =2 2 )dx has
the standard deviation ; which is interpreted as x appearing in inequalities (2.6).
Let the wave function be (x) = (2 x2 ) 1=4 exp( x2 =4 x2 ):
Recall that the eigenfunction of the momentum operator p^x are up = const
eipx=} : Recall also that the probability of falling into the state up after measuring
2
the momentum is P (p) = j'(p)j ; where
Z
(2.12)
'(p) = ( ; up ) = const
(x)e ipx=} dx:
Standard calculations show that '(p) = const exp( p2 x2 =}2 ) and hence
2
j'(p)j = const exp( p2 =2(}=2 x)2 = const exp( p2 =2 p2x ):
We see that x
px = }2 :
The transformation (2.12) is the well known Fourier transformation, normalized
accordingly to the quantum mechanics aims. Therefore the indeterminacy principle
has pure mathematical meaning that, if a function is well localized in the position
space then its Fourier transform is weakly localized in the momentum space, and
vice versa.
Remark 2.3. (The Newtonian equations) The Schrödinger equation (2.10) with
^
the initial condition jt=0 = 0 has the solution t = e iHt=} 0 , where U (t) =
^
e iHt=} is the evolution operator. This is called the Schrödinger image. In the
further sections we show that its is a unitary operator and that U (t) = U ( t) =
U (t) 1 .
^
Recall that we are interested in the quantities (A';
) which are the amplitudes
of transition from a state ' to a state : When the states evolve we deal with the
quantities
^ t ; t ) = (AU
^ (t)'; U (t) ) = (U ( t)AU
^ (t)'; ):
(A'
^
But the latter quantity can be interpreted as (A(t)';
); where
(2.13)
^ = U ( t)AU
^ (t):
A(t)
It means that the physical observable evolves, while the states remain constant.
This interpretation of evolution is called the Heisenberg image. If course, the
Schrödinger image and the Heisenberg image are equivalent.
8
^ satisfy the equations
It is easy to check that the evolving operators A(t)
d ^
^
^
^
^
(2.14)
i} A(t)
= [A(t);
H];
A(0)
= A;
dt
^ =
(at least at the formal level). Taking the operators x
^ and p^x as A^ and H
1 2
^x + U (^
x); we obtain the equations
2m p
d^
px
=
dt
d^
x
= p^x ;
dt
These are the Newtonian equations.
@U
(^
x):
@x
3. Bounded states and free states
This section is devoted to study the eigenvalues, denoted by E (the energy), and
the eigenfunctions uE = uE (x) of the Schrödinger operator (2.8). We begin with
the following observation.
Proposition 3.1. If the eigenfunction uE is in the Hilbert space H = L2 (Q)
2
and kuE k = 1; then the probability distribution juE (x)j dx is stationary, i.e. does
not change with time.
^
^ E = EuE ; hence t = U (t)uE = e iHt=}
Proof. We have Hu
uE = e iEt=} uE : It
follows
t (x)
=e
iEt=}
uE (x): It is then clear that j
2
t (x)j
=j
2
0 (x)j
:
The eigenfunctions uE which belong to the Hilbert space H are called the
bounded states. It means that the particle is localized, or bounded, in a …xed
region of the phase space. The amplitude of the wave function is …xed, only its
phase (i.e. the argument) changes.
The equation
^ =E
(3.1)
H
is called the stationary Schrödinger equation.
Example 3.1. (In…nite potential well ) This example can be found in any textbook on Quantum Mechanics.
The potential energy function U (x) is 0 for jxj < 1 and +1 otherwise. The
stationary Schrödinger equation (3.1) with = u means the following
u00 (x) =
2
k 2 u(x);
jxj < 1;
2mE
}2
u(x)
0;
jxj
a;
where k =
> 0: (It is easy to see that the case E < 0 is not realized.) So we
get the following boundary value problem
u00 + k 2 u = 0;
u(1) = u( 1) = 0:
Is solutions are
u2l (x) = A sin(l x);
u2l+1 (x) = B cos((l + 1=2) x):
The corresponding values kn of k are k2l = 2l 2 and k2l+1 = (2l + 1) 2 : Thus
kn = n =2 and the corresponding eigenvalues (energies of bounded states) are
2 2
En =
} 2
n :
8m
This example suggests that the eigenvalues corresponding to the bounded states
are discrete in the spectrum of the Hamiltonian operator.
9
But we know at least one operator, namely p^x = i}@x ; for which the eigenvalues
are not discrete and the eigenfunctions are not quadratically integrable. Here the
eigenfunctions are up = eipx=} (and do not belong to H) and the corresponding
eigenvalues p …ll the whole real line. The states up (x) correspond to a state of a
particle which moves with constant velocity v = p=m; its momentum is …xed but
its position is completely unknown.
^ The
Similar situation may occur (and occurs) for the Schrödinger operator H:
eigenstates uE (x) such that
uE (x)
const eipx=}
as
jxj ! 1
are called the free states. Also linear combination of such states is called a free
state.
Example 3.2. (Finite potential well ) The potential energy equals
U (x) = 0 for jxj < 1;
U (x) = V for jxj > 1:
The stationary Schrödinger equation (3.1) leads to the equations
2Em
u(x) = 0; jxj < 1;
}2
2(E V )
(3.3)
u00 (x) +
u(x) = 0; jxj > 1:
}2
The bounded states should satisfy ju(x)j ! 0 as jxj ! 1: Therefore it should
be 2(E}2 V ) = K 2 < 0: Then
u00 (x) +
(3.2)
u(x) = C1 e
Kx
for x > 1;
u(x) = C2 eKx for x <
1:
The potential is an even function of x and hence the eigenfunction can be divided
into even and odd. For even solutions we have C1 = C2 = C and for odd solutions
we have C1 = C2 = C:
Consider the domain jxj < 1: The case E < 0 is not physical and it can be
checked that it is not realized. So 0 < E < V; e.g. for bounded states. Denoting
k02 = 2Em=}2 we get the even solutions D cos k0 x and the odd solutions D sin k0 x;
below we put D = 1:
The eigenfunctions should be of class C 1 which implies the gluing conditions
cos k0 = Ce
K
;
k0 sin k0 = KCe
K
;
e.g. in the even case. The resulting equation k0 tan k0 = K, where k0 and K
depend on E; has …nitely many solutions E = En : In the odd case the corresponding
equation takes the form k0 cot k0 = K and also has only …nitely many solutions.
Consider the case E > V: Then u(x) = C1 eikx + C2 e ikx ; k 2 = 2(E V )m=}2 ;
for jxj > 1 in the even case. The gluing conditions
cos k0 = C1 eik + C2 e
ik
;
k0 sin k0 = ikC1 eik
iC2 e
ik
are satis…ed by some constants C1;2 : Analogously one …nds odd free states.
It follows that the (physical) spectrum of the Hamiltonian operator consists
of …nitely many points in the interval (0; V ) (the discrete spectrum) and of the
in…nite interval [V; 1) (the continuous spectrum). Moreover, for each E > V the
corresponding eigenspace is two dimensional; we say that the continuous spectrum
is of multiplicity 2:
10
4. Bounded states for the hydrogen atom
The classical Hamiltonian of the electron in the hydrogen atom takes the form
H=
p2
+ U (r);
2m
e2
;
r
U (r) =
p
where r = x2 + y 2 + z 2 = jqj :
Writing q = rer (where er is a unit vector) we get q_ = re
_ r + re_ r = re
_ r +r_e
(where is the polar angle in the plane spanned by q and q_ and e is a unit vector
in this plane orthogonal to er ): Therefore p2 = mq_2 = pr er + pt e ; where pr = mr_
j
is the radial part of the momentum and pt = mr _ = jM
r is the tangential par of
the momentum; here M = q p is the angular momentum. Therefore
p2 = p2r + p2t ;
p2t = M 2 =r2 :
The corresponding quantum mechanical operator takes the form
2
^2
e2
^ = p^r + M
(4.1)
H
;
2m 2mr2
r
^ 2 =r2 is a corresponding splitting of the operator ~2 : The following
where p^2r + M
result is proved in the next subsection.
Theorem 4.1. In the spherical coordinates r; ';
y = r sin sin '; z = r cos ) we have
(4.2)
p^2r =
~2 r
2
@r r2 @r =
(such that x = r sin cos ';
~2 (@r2 + 2r
1
@r );
1 @
@
1 @2
sin
+
sin @
@
sin2 @'2
^ 2 commutes with the Schrödinger operator H
^ and with r^:
and M
2
2
^
Moreover, the operator M , acting on L2 (S ; sin d'd ); has the eigenvalues
(4.3)
^2 =
M
~2
~2 l(l + 1);
(4.4)
l = 0; 1; 2; : : :
each of multiplicity 2l + 1 and with eigenfunctions
(4.5) Yl;m ('; ) = const e
im'
jmj
(sin )
zl;jmj (cos ); m =
where zl;jmj (w) are polynomials such that the functions
are the so-called Legendre functions of the …rst kind.
l; l + 1; : : : ; l
jmj
Pl (w)
1; l;
= (1 w2 )jmj=2 zl;jmj (w)
Since d3 x = r2 dr sin d'd we can represent the Hilbert space H = L2 (R3 ) as
the following tensor product
(4.6)
H = L2 (R+ ; r2 dr)
L2 (S2 ; sin d'd ):
The space L2 (S2 ) is split into the subspaces C Yl;m and the space H is split into
^
subspaces L2 (R+ ; r2 dr) Yl;m which are invariant with respect to the operator H
2
^
^
^
(by the commutativity of H and M ): Therefore the eigenfunctions for H should
be of the form
= u(r)Yl;m ('; );
where the function u satis…es the following 1-dimensional Schrödinger equation
(4.7)
~2 1
~2 l(l + 1)
2
@
r
@
+
r
r
2m r2
2mr2
e2
r
u(r) = Eu(r):
11
Remark 4.1. When one puts u(r) = (r)=r then the above equation takes the
2
l(l+1)
~2 00
form 2m
+ ~ 2mr
+ U (r) = E ; which recalls the classical Hamiltonian.
2
The factor 1=r is ‘quantum mechanical’.
We seek solutions u(r) to equation (4.7), with E = jEn j < 0; which belong to
L2 (R+ ; r2 dr): It means that ju(r)j r2 and ju0 (r)j r2 are integrable.
The physicists use the following normalizations
=
n r;
jEn j = me4 =(2~2
u(r) = un;l ( );
2
n );
where
2
n
They …nd the equation
1
(4.8)
2
2
@
@
= 8m jEn j =~2 :
l(l + 1)
1
+
4
2
n
un;l ( ) = 0:
The following lemma is proved by direct calculations.
Lemma 4.1. As ! 1 equation (4.8) has two independent solutions of the
form
n 1
n 1
u1 ( )
e =2 ; u2 ( )
e =2 ;
and as
! 0 it has two meromorphic independent solutions of the form
l
u3 ( )
;
l 1
u4 ( )
Moreover, the solutions are analytic functions of
:
in C n 0:
Since we look for bounded states, we must choose the solution u2 near in…nity.
Also the solution u4 near zero is ‘bad’because its derivative is not square integrable
with respect to the measure 2 d : Therefore the solution un;l ( ) should be an integer
function. Moreover, the function un;l ( ) l e =2 should be integer and of polynomial
growth; so, it should be a polynomial denoted by L( ) = Ln;l ( ); called the Laguerre
polynomial. Thus
un;l ( ) = const l e =2 Ln;l ( );
where the polynomial L satis…es the Laguerre equation
Assuming L( ) =
P
L00 + (2l + 2
ak
k
) L0 + (
n
l
1) = 0:
we …nd the recurrent relations
(k(k + 1) + 2(l + 1)(k + 1)) ak+1 = (k + l + 1
n )ak :
Since aj = 0 for large j we get that
n
=n
is integer and
k+l+1
n:
It is a condition for the eigenvalues of the Schrödinger operator, because
me4 1
; n = 1; 2; : : : :
2~2 n2
We can formulate the following result, which explains the Balmer–Ritz classi…cation of the spectral lines in formula (2.1) in Subsection 2.
(4.9)
En =
12
Theorem 4.2. The point spectrum of the hydrogen atom Hamiltonian consists
of the points (4.9). Each such eigenvalue has multiplicity n2 and the corresponding
eigenfunctions take the form
jmj
un;l;m (r; '; ) = const eim' (sin )
=
n r;
n
zl;jmj (cos ) l e
=2
Ln;l ( );
= 2me2 =(~2 n):
Proof. It remains to calculate the multiplicity. For …xed n the number l can take
he values 0; 1; : : : ; l 1 and for
n and l the number m can take 2l + 1 values.
Pn…xed
1
Thus the multiplicity equals l=0 (2l + 1) = n2 :
The continuous spectrum of he hydrogen atom operator is discussed in Chapter
3 (Section 3) below.
5. The angular momentum and the representation theory of so(3)
The components of the angular momentum operator can be written as linear
vector …elds:
^ x = i~(y@z z@y )
^y
^z
M
i~Ax q; M
i~Ay q; M
i~Az q;
where q = (x; y; z) and
2
0 0
(5.1)
Ax = 4 0 0
0 1
3
2
0
0
1 5 ; Ax = 4 0
1
0
2
3
0
0 1
0 0 5 ; Ax = 4 1
0
0 0
1
0
0
3
0
0 5
0
are antisymmetric matrices, generating the Lie algebra so(3): The algebra so(3) is
the Lie algebra of the group SO(3) if rotations in R3 and is isomorphic with R3
equipped with the vector product as the commutator.
^ and of A are following
The commutation relations for the components of M
h
i
h
i
h
i
^ x; M
^y
^z; M
^y; M
^ z = i~M
^ x; M
^z; M
^ x = i~M
^y;
(5.2)
M
= i~M
[Ax ; Ay ]
=
Az ;
[Ay ; Az ] =
Ax ;
[Az ; Ax ] =
Ay :
Proposition 5.1. The angular momentum operator realizes a representation of
the Lie algebra so(3) in the Hilbert space H = L2 (R3 ); e.g. modulo the factor i~:
Interesting question is whether this representation of the Lie algebra arises from
a representation of the Lie group SO(3) (or maybe O(3)) in H by means of unitary
operators. As we shall see, this is true for the model of the hydrogen atom presented
in the previous subsection, but in a more adequate model of the electron there exist
so-called spin representations which do not arise from any representation of the Lie
group SO(3):
5.1. Representation of so(3) in L2 (S2 ). Let us calculate the Laplacian in the
spherical variables (following the V. Arnold’s book [Arn]). Denoting the spherical coordinates (x1 ; x2 ; x3 ) = (r; '; ) and the orthogonal frame (e1 ; e2 ; e3 ) =
(er ; e' ; e ); we write the metric as follows
ds2 = dr2 + r2 sin2 d'2 + r2 d 2 = E1 dx21 + E2 dx22 + E32 dx23 :
p
Since hdxi ; ei i = dxi =ds = 1= Ei we have
dr(er ) = 1;
d'(e' ) = 1=(r sin );
d (e ) = 1=r:
13
With a vector …eld A = A1 (x)e1 + A2 (x)e2 + A3 (x)e3 one associates two di¤erential forms
! 1A = (A; ); ! 2A = det (A; ; ) :
In fact, we have
X p
! 1A =
Ai Ei dxi ;
p
p
p
! 2A = A1 E2 E3 dx2 ^ dx3 + A2 E3 E1 dx3 ^ dx1 + A3 E1 E2 dx1 ^ dx2 :
We have the following formulas: df = (Of; ) = ! 1Of ; which de…nes the gradient
of a function
X 1 @f
@f
1 @f
1 @f
p
ei =
er +
e' +
e ;
grad f =
@r
r sin @'
r@
Ei @xi
p
and d! 2A = div A V OL = div A
E1 E2 E3 dx1 ^ dx2 ^ dx3 ; which de…nes the
divergence of a vector …eld
p
p
p
1
@
@
@
div A = p
A1 E2 E3 +
A2 E3 E1 +
A3 E1 E2
:
@x2
@x3
E1 E2 E3 @x1
Since the Laplacian of a function f equals f = div grad f the previous formula,
1=2
with Aj = Ej
@f =@xj ; gives
!
!
r
r
1
@
E2 E3 @f
@
E3 E1 @f
f = p
f
+
E1 @x1
@x2
E2 @x2
E1 E2 E3 @x1
!
r
@
E1 E2 @f
+
g
@x3
E3 @x3
=
Therefore
equal
r
1
r2 sin
=
=
1 @
r2 @r
r
@
@r
+
r2
1
r2
@
@r
r2 sin
t;
;
@f
@r
+
@
@'
where the radial
t
=
1
r2
1 @f
sin @'
r
+
@
@
sin
and the tangential
1 @2
1 @
2 @'2 + sin @
sin
sin
@
@
@f
@
:
t
parts are
:
The operators r and r2 t commute and it follows that the tangential part should
equal the square of the angular momentum divided by ~2 r2 :
^2 = M
^ x2 + M
^ y2 + Mz2 = }2 r2 t
(5.3)
M
Hence we have proved …rst part of Theorem 4.1.
We note also the expression of the third component in the spherical coordinates:
^ z = i~ @
(5.4)
M
@'
which follows directly from the formula tan ' = y=x: The other components have
^ x = i~ sin ' @ + cot cos ' @
more complicated expressions:
M
and
@
@'
^ y = i~
M
@
:
cos ' @@ + cot sin ' @'
^ 2 commutes with the components M
^ x; M
^y; M
^z:
Lemma 5.1. The operator M
^ 2 with M
^ z follows directly from their expressions
Proof. The commutation of M
in formulas (5.3) and (5.4).
14
It follows also from the commutation relations (5.2). Indeed, we have
^ 2; M
^z]
[M
^ x2 ; M
^ z ] + [M
^ 2; M
^z]
= [M
y
^ x [M
^ x; M
^ z ] + [M
^ x; M
^ z ]M
^x + M
^ y [M
^y; M
^ z ] + [M
^y; M
^ z ]M
^y
= M
^ xM
^y
= i~( M
^yM
^x + M
^yM
^x + M
^ xM
^ y ) = 0:
M
^ 2 is also called the Casimir operator of the representation.)
(The operator M
^ 2 and M
^ z commute and are symmetric, they have can be
Since the operators M
simultaneously diagonalized, e.g. in the space L2 (S2 ): The eigenfunctions of the
^ z with eigenvalues = ~m are
operator M
um ('; ) = const( ) eim'
^ 2 with
and it is clear that m should be integer. The common eigenfunctions of M
^ z should take the form
eigenvalues ~2 and M
Y ;m ('; ) = eim' Pe ;m ( )
n
o
m2
where Pe = Pe ;m satis…es the equations sin1 @@ sin @@
Pe =
Pe: The
sin
substitution
w = cos ;
i.e. Pe( ) = P (w), leads to the Legendre equation
d
dw
(5.5)
w2 )
(1
d
dw
m2
1 w2
P+
P = 0:
Of course, P (w) should be …nite in the closed interval [ 1; 1]:
The only singularities of the function P can appear at the endpoints of this
interval.
Lemma 5.2. Near w = 1 we have two independent solutions to (5.5) of the
form
P (1) = (1 w)jmj=2 (a0 + a1 (1 w) + : : :); Q(1) = (1 w)
and near w =
P
( 1)
jmj=2
(a0 + a1 (1 w) + : : :)
1 we have two independent solutions of the form
jmj=2
= (1+w)
(a0 +a1 (1+w)+: : :); Q(
If m = 0 then the solutions Q(1) and Q(
1)
Of course, our solution should equal P
is a condition for the eigenvalue ~2 : So,
(1)
P
;m (w)
= (1
1)
= (1+w)
jmj=2
(a0 +a1 (1+w)+: : :):
take another form (with logarithm).
near w = 1 and P (
1)
near w =
1; it
w2 )jmj=2 z(w)
where the function z(w) satis…es the equation
(1
w2 )z 00
2((1 + jmj)wz 0 + (
jmj (jmj + 1)w = 0
P
and is an integer function of w 2 C: The Taylor expansion z = ak wk implies the
recurrent relations
(k + 1)(k + 2)ak+2 = f(k + jmj)(k + jmj + 1)
g ak :
If all ak 6= 0; then ak+2
ak and the function z(w) were not integer. So some
coe¢ cient before ak in the above formula must vanish. We have
= l(l + 1)
15
for a nonnegative integer
l
jmj :
jmj
The obtained solutions zl;jmj (w) are polynomials such that the functions Pl
=
2 jmj=2
(1 w )
zl;jmj are the so-called Legendre functions of the …rst kind (see [BE1,
BE2]). In fact, we have zl;jmj (w) =
djmj
P (w)
dwjmj l
where Pl (w) are the standard LeP1
1=2
gendre polynomials with the generating function l=0 Pl (w) l = 1 2 w + 2
(see [Sch]).
If k = l jmj is odd then we have a0 = 0 and the Legendre polynomial is an
odd function of w: If k is even, then a0 6= 0; a1 = 0 and we have an eve Legendre
polynomial.
Therefore we have proved the second statement of Theorem 4.1.
Remark 5.1. For …xed integer l
0 the spherical functions Yl;m = const
jmj
e
Pl (cos ); m = l; l+1; : : : ; l; generate a subspace of L2 (S2 ) which supports
an irreducible representation of the Lie algebra so(3) (see also Subsection 3 below).
But these representations of the Lie algebra can be integrated to representations of
the Lie group SO(3):
Indeed, the rotation by the angle
around the OZ axis corresponds to the
^ z =~) which acts on the wave function Yl;m by multiplication by
operator exp(i M
the number eim :
im'
5.2. Interaction with the magnetic …eld. The energy levels of the bounded
states of an electron in the hydrogen atom depend only on the number l; but not
on m: In order to see that the states with di¤erent m’s are physically di¤erent
one must embedded the hydrogen atom in a constant magnetic …eld. Then the
energy levels En become separated. Therefore we need to have a formula for the
Hamiltonian operator of the electron in presence of a magnetic …eld.
Proposition 5.2. The classical Hamiltonian in this situation is following
e 2
1
(p
A) + e';
2m
c
where A is the vector potential and ' is the scalar potential de…ned by B = O
E = 1c @A
O':
@t
(5.6)
H=
A;
Proof. The equations of motion with the Hamilton function (5.6) take the form
q_ = Hp0 =
1
(p
m
e
A);
c
p_ =
Hq0 =
e
e
(p
A; OA) eO':
mc
c
P
P
@Aj
i
= i q_i ( @A
i
@qj
@qi ) +
P
@Aj
i
So p = mq_ + ec A: Since (q;
_ OA)j = i q_i @A
@qj
@qi q_i =
P @Aj
P @Aj
dAj
@Aj
@Aj
(q_ B)j + i @qi q_i and dt = i @qi q_i + @t ; we get mq_ + @t = ec q_ B eO':
This means exactly that m•
q = eE + ec q_ B; which agrees with the classical notions
of the electromotoric and Lorentz forces (see formula (2.3) in Section 1).
Lemma 5.3. When B = const the vector potential can be taken in the form A =
q: In particular, for an electron in the hydrogen atom in presence of a constant
1
e
e2
magnetic …eld B the Hamiltonian (5.6) takes the form H = 2m
(p 2c
B q)2 jqj
:
1
2B
16
Then for small e=c the approximate contribution to the energy arising from the
interaction with the magnetic …eld equals
e
e
e
e
(q;
_ A) = (q;
_ q B) =
(q mq;
_ B) =
(M; B)
c
2c
2mc
2mc
where M is the angular momentum of the electron. Assuming B is directed along
the OZ axis (with length B) we have the the quantum mechanical Hamiltonian
e2
e
~2
^z
BM
2me
r
2me c
where me is the electron’s mass (in order to avoid confusion with the index m in the
^ z ). Since the operator M
^ z commutes with
eigenvalues of M
and 1r the operator
(5.7) has the same eigenfunctions as the pure hydrogen Hamiltonian operator (4.1),
i.e. the functions un;l;m de…ned in Theorem 4.2. But the eigenvalues corresponding
to these eigenfunctions do not equal En (de…ned in Theorem 4.2). The following
result is known as the Zeeman e¤ ect.
(5.7)
^ =
H
Theorem 5.1. The energies of the bounded states for the operator (5.7) equal
(5.8)
En;m = En + m E;
E=
e~B
;
2me c
jmj < n:
Remark 5.2. If the electron is moving with constant velocity v along a circular
orbit of radius r then we can associate with it the current j = v; where is the
charge’s density equal 2 e r ; thus jjj = 2ejvjr : Here the classical contribution to
R the
e
energy is the average of the contribution along the orbit and equals 2c
(q
v; B) = ( ; B); where
e
jjj
area ~n =
M:
c
2me c
The vector is the magnetic moment (of the current generated by the electron)
which is parallel to the unit vector ~n orthogonal to the orbit’s plane and area is the
area bounded by the circuit. This is a physical interpretation of the Hamiltonian
(5.7).
e~
The value j j = 2m
is called the Bohr magneton.
ec
=
5.3. Spin. Theorem 5.1 implies, in particular, that the …rst level E1 is not split
(because then l = 0 and m = 0): But observations did not con…rm this. This level
was split into two sublevels. Explanation of this phenomenon lies in considering
new representations of the Lie algebra so(3): We begin with some algebraic stu¤
and then apply it to the quantum physics.
Introduce the operators
^ =M
^x
M
^y;
iM
A = Ax
iAy :
^ z (respectively, with Az ) they satisfy the following commutation
Together with M
relations
h
i
h
i
^z; M
^
^ ;
^ +; M
^
^z
(5.9)
M
= ~M
M
= 2~M
(and similar relations are satis…ed by A’s). We can say that the matrices Az ; A+ ; A
generate the complex Lie algebra sl(2; C), which is isomorphic with the complex Lie
^z; M
^ +; M
^ de…ne a representation
algebra so(3; C) = so(3) C; and the operators M
17
of sl(2; C): Recall that the Lie algebra sl(2; C) consists of 2
trace.
2 matrices with zero
Theorem 5.2. Any …nite dimensional irreducible representation of the Lie algebra sl(2; C) is equivalent to C2l+1 ; l = 0; 21 ; 1; 32 ; : : : ; with a basis f m gm= l; l+1;:::;l
such that
^ z m = m~ m
M
^
^ 2 = ~2 l(l + 1)I.
and M are de…ned below. Moreover here M
Proof. Let j be a basis of a space of irreducible representation generated by
^ z and such that M
^z j = j j:
and M
^ + j : Then (5.9) implies that
Fix an element j and consider the vector ' = M
^
^
^
^
^ z with the
Mz ' = M+ Mz j + ~M+ j = ( j + ~)'. Thus ' is an eigenvector of M
^ j is an eigenvector of
eigenvalue j + ~: Analogously we …nd that the vector M
^
Mz with the eigenvalue j ~:
Since we consider an irreducible representation we can numerate the vector j
such that
^ + j = const j+1 (if M
^ + j 6= 0) and M
^ j = const j 1 (if it is nonzero):
M
^
M
^ z is generated by in…nitesimal rotations around the OZ
Since the operator M
^ z and M
^ z are the same. It follows
axis in the negative direction, the spectra of M
^
that the spectrum of Mz equals fm~ : m = l; l + 1; : : : ; l 1; lg ; where l~ is the
maximal eigenvalue. We see also that l 2 21 Z is a half-integer or integer.
^ 2 we …rstly note that it is proportional to the identity,
To compute the operator M
^2 =
M
I (as commuting with the whole representation). By (5.9) we have
^2 = M
^ M
^+ + M
^ z2 + ~M
^ z . Since M
^ + l = 0 we get M
^ 2 l = 0 + ~2 l 2 l + ~2 l l :
M
Therefore = ~2 l(l + 1):
De…nition 5.1. The representation described in Theorem 5.2 are called the
spin l representations.
Example 5.1. The integer spin representations are the representations obtained
in Subsection 2 above. They are subrepresentations of so(3) in L2 (S2 ) and also
in H = L2 (R3 ) = L2 (S2 ) L2 (R+ ; r2 dr). Recall that they arise from unitary
representations of the group SO(3):
The spin 0 representation is the trivial 1 dimensional and all the operators
A ; Az are equal zero.
The spin 1 representation is the representation in R3 such that (A ) = A
and (Az ) = Az :
The spin 21 representation is realized in C2 : We have
0
0
^+ = ~
M
^x =
and M
~
2
x;
x
^y =
M
=
~
2
y;
0 1
1 0
^z =
M
;
y
1
0
~
2
=
^ =~
; M
z;
0
1
0
0
where
0
i
i
0
;
z
=
1
0
0
1
are the so-called Pauli matrices.
The latter representation does not arise from a representation of the Lie group
@
^ z and the
SO(3): Indeed, the matrix Az corresponds to the vector …eld @'
= ~i M
1;l
18
matrix e Ax is a matrix of rotation by the angle
e2 Ax = I (identity matrix). But
e
^ z =~
iM
ei
=
around the OZ axis. In particular,
=2
0
0
e
i =2
^ z =~) = id:
and exp(2 M
In fact, the spin 21 representation of the algebra so(3) arises from a genuine
representation of a larger group than SO(3): It is the group
SU (2) = fA : AA = 1; det A = 1g =
a
c
c
a
2
2
: jaj + jcj = 1
of unitary 2 2 matrices. The group SU (2) covers two times the group SO(3);
hence the both groups have the same Lie algebra. Topologically SU (2) is the 3dimensional sphere S3 and SO(3) is the real 3-dimensional projective space RP3
and the covering SU (2) ! SO(3) is the same as the antipodal points identi…cation.
One can say that the spin 21 representation arises from ‘2–valued representation’
of the group SO(3):
As we mentioned in the beginning of this subsection physical experiments demonstrate realization of the spin 12 representation for the electron in the hydrogen atom.
On the other hand, we do not know any natural way to include this representation
into our standard Hilbert space L2 (R3 ): The physicists found the following solution
to this problem.
Instead of acting in the space L2 (R3 ) we should act in the Hilbert space
H := L2 (R3 )
(5.10)
C2 ' L2 (R3 )
which consists of vector-valued functions
ator should take the form
(5.11)
^ =
H
~2
2me
e2
r
id
=
1
2
L2 (R3 );
: Moreover, the Hamilton oper-
e
^z
B(M
2me c
id + ~ id
z ):
~2
e2
^
It means that the operators 2m
r and Mz act on the both components 1
e
and 2 in the same way, while the operator z multiplies 1 by 1 and 2 by 1:
It means that the electron has some ‘internal’ discrete degrees of freedom and
the operator S^ = ~2 z is the spin operator acting only on these internal discrete
variables. One would like to interpret this classically as a kind of rotation of the
electron around some (undetermined) axis. But this is not accurate interpretation
~
because the coe¢ cient before id
z is ~ (not 2 ), as it follows from physical
experiments.
Anyway, the operator (5.11) acts diagonally and has the eigenfunctions un;l;m
1
0
0 and un;l;m
1 with the eigenvalues which di¤er from the eigenvalues from
e~
B: Therefore we have the following
Theorem 5.1 by 2m
ec
Theorem 5.3. The energy levels of the bounded states of the operator (5.11)
are following
En;m; = En + (m
where En and
E were de…ned before.
1) E;
jmj < n;
19
The phenomenon of having internal spin by elementary particles is common. If
a particle has spin l; then the corresponding Hilbert space is
H = L2 (R3 )
C2l+1
^ + S;
^ where
and the angular momentum operator is replaced by the operator M
^
^
^
^
M ' M id acts in the usual (di¤erential) way and S ' id S acts in algebraic
way.
Remark 5.3. The above Zeeman e¤ ect observed in the discrete spectrum of the
hydrogen atom is not end of the story. There exists the so-called Lamb shift which
says that the energy levels take the form
2
Enj = En 1 +
where j +
1
2
1
j + 1=2
n
3
4n
;
^ + S^ and
is he maximal eigenvalue of the operator M
e2
1
~c
137
is the so-called …ne structure constant. This shift arises when one takes into account
the e¤ect of interaction of the electron with the electromagnetic radiation caused
by his movement along the orbit.
Very precise calculation of the Lamb shift and its agreement with experiments
in the 1950’s was a great success of the quantum electrodynamics.
(5.12)
=
Remark 5.4. (The group Spin(n)) There exists a generalization of the twofold
covering of the group SO(3) to higher dimensions.
Firstly we note that the Pauli matrices x ; y ; z satisfy the following anticommuatation relations
f
x;
xg
=f
y;
yg
=f
z;
zg
= 2; f
x;
yg
=f
y;
zg
=f
z;
xg
= 0;
where fA; Bg = AB + BA is the anticommutator. Mathematicians say that they
generate the Cli¤ord algebra Cl(R3 ):
By de…nition, for a real vector space V of dimension n equipped with a scalar
product Q(x; y) (which we assume standard Q(ei ; ej ) = ij ); the Cli¤ ord algebra
Cl(V ) is an algebra generated by 1 and the basic vectors ei with the relations
fei ; ej g = 2Q(ei ; ej ):
In the Cli¤ord algebra Cl(V ) we de…ne the ‘conjugation’x ! x by
De…ne the group
ei1 : : : eik ! ( 1)k eik : : : ei1 :
P in(V ) = g 2 Cl(V ) : gg = 1; gV g
1
=V :
This group acts in natural way on the space V by means of orthogonal transformations; so we have a homomorphism : P in(V ) ! O(V ): The group Spin(V ) =
1
(SO(V )) is called the spinor group and is twofold covered over the group
SO(V ) :
2:1
Spin(V ) ! SO(V ):
Indeed, the transformations x ! gxg 1 and x ! g 0 xg 0 1 of V are the same when
g 0 = g (because they induce the same automorphisms of the Cli¤ord algebra). But
the second condition in the de…nition of the group P in(V ) implies that 2 = 1:
20
When V is the standard Rn we use the notation Spin(n):
The 2–valued mapping from SO(V ) to Spin(V ) can be de…ned directly. For
example, if ex ! cos ' ex sin ' ex ; ey ! sin ' ex + cos ' ey is a rotation in
the OXY plane in V = R3 , then we associate with it the element
g = cos('=2) 1
sin('=2)
x y
=
cos '2 + i sin '2
0
0
cos '2
i sin '2
from the Cli¤ord algebra Cl(R3 ):
We have Spin(2) = SU (1) ' S1 (with twofold covering over SO(2) ' S1 );
Spin(3) ' SU (2) and Spin(4) ' SU (2) SU (2):
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