1 Harmonic Oscillator

The Hamiltonian of a one-dimensional harmonic oscillator is

Подпись:– 1 a2 mj2

ff = 2mp + ~q •

where the momentum p and coordinate q satisfy the commutation relation

[p, q] = pq — qp = ih.

We introduce a pair of operators, the annihilation operator,

Physics of Solar Energy C. Julian Chen Copyright © 2011 John Wiley & Sons, Inc.

image831

(C.3)

 

a

 

and the creation operator

„ + 1 „ . [шш

a тшш”- V-2ira (C4)

the meanings of these terms well be clarified soon. Using the commutation relation C.2, we have

Подпись: (C.5)H = НшаР a +— Нш,

and the commutation relation

[a, af] = aaP — aPa = 1. (C.6)

To find the eigenstates and energy levels of the Hamiltonian C.1,

Hn) = Enn), (C.7)

it is sufficient to find the eigenstates and eigenvalues of the operator apa,

ap an) = unn), (c.8)

and En = (un + 2)Нш. As a consequence of Eq. C.6, if n) is an eigenstate with eigenvalue un, then an) is also an eigenstate

ap aan) = (un — 1)an) (c.9)

with eigenvalue un — 1. Because {napan) must not be negative, there must be an eigenstate with minimum value, un = 0. For such a state,

afao) =o. (C.10)

On the other hand, if n) is an eigenstate with eigenvalue un, then aPn) is also an eigenstate

aPa(aPn)) = (un + 1)(aPn)) (C.11)

with eigenvalue un + 1. Starting with ), by applying aP many times, we have

aPa (aP) o) = n (aP) o). (C.12)

Therefore, we conclude that the eigenvalues of the operator apa are non-negative integers, and the eigenstates can be constructed from the state with a zero eigenvalue,

The final step is to find the normalization constant Cn. The zeroth-order state is by definition normalized,

(0|0) = 1. (C.14)

By applying a) many times, we have

(0| (a)n (rf) |0) = n!. (C.15)

Therefore, Cn = (n!)-1/2, and finally

in = – Ц (at)”|0). (C.16)

n!

Those are also eigenstates of the harmonic oscillator problem,

Hin) = ^n +hw in). (C.17)

The energy level of the harmonic oscillator is thus quantized, with energy quanta hw. The operator a^ adds an energy quanta to the oscillator, thus the name creation op­erator ; the operator a removes an energy quanta from the oscillator, thus the name annihilation operator. These operators play essential roles in quantum electrodynamics, or the bona fide quantum theroy of radiation.

B. 2 Angular Momentum

The definition of angular momentum in quantum mechanics is similar to that in classical mechanics, except that the order of coordinate r and momentum p is fixed,

m = r x p, (C.18)

or in tensor notation,

mi = eijk Xj pk, (C.19)

where the unit axial tensor eijk is a tensor antisymmetric to all three suffixes, with t123 = 1, and changes sign by exchanging two identical indices, where the value is zero. A sum over j and к is implied.

To simplify notation, a dimensionless version of the angular momentum is defined,

li h mi h 9ijk x jp k. (C.20)

The commutation relations can be obtained from the commutation relations of mo­mentum and coordinate,

which are

Подпись:or in tensor form,

From the components we can form an operator as the square of the modulus of the angular momentum vector,

l2 =PX + /2 + Z (C.24)

As a result of commutation relations C.22, l2 commutes with a component, for example,

[І2л ]=0. (C.25)

Therefore, we can find states 1l, m) which are simultaneously eigenstate of l2 and lz,

l2l, m) = Xl, m),

(C.26)

l z l, m) = fil, m).

In the following, we introduce a pair of operators which are similar to the creation and annihilation operators in the problem of harmonic oscillators,

Using arguments similar to those in the harmonic oscillator problem, we find that /+ l, m) is also an eigenstate of lz with eigenvalue p +1 and /_l, m) is an eigenstate of I z with eigenvalue p — 1. Applying those operators many times, we have

lz l, m) = (p + n) (l+) l, m) (C.31)

and

lz (j_^ l, m) = (p — n) (j_^ l, m). (C.32)

However, because of Eq. C.24, the eigenvalue ofl z cannot grow indefinitely. Its absolute value must have a maximum. Because of the symmetry, the absolute value of the positive maximum and the absolute value of the negative maximum must beequal. Also, because the difference between the positive maximum and the negative maximum must be an integer, both must be one-half of an integer. Assigning this number as l, the possible eigenvalues of l z, often assigned as m, must be

m = —l, —l + 1, —l + 2, …l — 2, l — 1, l. (C.33)

with

l = 0, 2, 1, 3, 2, 2, 3… (C.34)

It is obvious that l, the maximum absolute value of l z, is also a quantum number for the total angular momentum l2. In the following, we will find the eigenvalue of the operator l 2. Because m = l is the maximum eigenvalue of l z, one must have

l+l, l) =0. (C.35)

In view of Eq. C.30,

l2l, l) = [l_l+ + ll + lz) l, l) = l(l + 1)l, l). (C.36)

Because l and m are independent, finally we have the eigenvalues and eigenstates for the angular momentum operator,

Подпись: (C.37)m2 l, m) = l(l + 1)h l, m), mz l, m) = mh l, m).

C. 3 Hydrogen Atom

In classical physics, if the mass of the proton is large, the Hamiltonian of the hydrogen atom is

2

Подпись:p2 К

2me r ’

where к = e2/4^e0; see Chapter 2. Because of the spherical symmetry of the problem, the angular momentum is conserved. The vector of angular momentum is always perpendicular to the plane of motion,

L = r x p = const. (C.39)

In addition to angular momentum, there is another conserved vector related to the fixed orientation of the long axis of the orbital, which is a result of the Coulomb interaction. It is called the Runge-Lenz vector after its discoverers,

A = ——— — (p x L) = const. (C.40)

r me

Because the angular momentum vector L is perpendicular to the orbital plane and the Runge-Lenz vector is in the plane, the two vectors are perpendicular,

L ■ A = 0. (C.41)

In 1926, a year before Schrodinger discovered his differential equation and solved the hydrogen atom problem, Wolfgang Pauli solved the eigenvalue problem by using the algebraic method of Werner Heisenberg based on the two constants of motion. Pauli’s treatment is as follows:

Подпись: H image837 Подпись: К r Подпись: (C.42)

In quantum mechanics, the Hamiltonian is an operator,

The angular momentum, also an operator,

L = Г x p, (C.43)

satisfies the commutation relation

[Li, Lj ] = iheijk Lk. (C.44)

It also commutes with the Hamiltonian C.42, thus a constant of motion.

Because p x L is not Hermitian, Pauli defined a Hermitian operator equivalent to the classical Runge-Lenz vector

A = ———- — (p x L — L x p) . (C.45)

r 2me

With rather tedious but straightforward algebra, this operator is shown to commute with the quantum-mechanical Hamiltonian, equation C.42. Similar algebra results in the commutation relations,

[Li, Aj] =

ift^ijk Ak,

(C.46)

– n (

—2H –

(C.47)

Aj] = ift

ЄijkLk,

me

image840 Подпись: (C.48)

and the relation

Because A commutes with H, and there are common eigenstates, a reduced vector can be defined,

Подпись:B= I— mi1/2 A.

2H

The commutation relations Eqs C.46 and C.47 are reduced to

[Li:i Bj] rihtijk Bk,

(C.50)

[Bi, B j] ift^ijkLk.

(C.51)

We introduce a pair of operators

J=2 (L + B,

к = 2 (L — B.

(C.52)

The commutation relations are reduced to those of a pair of independent angular mo­menta

Jj ] iftk-ijkJk , (C.53)

[Ki, Kj] = ihkijkKk, (C.54)

Подпись: (C.55)[Ji, Кj] = 0.

Using Eqs. C.49 and C.52, Eq. C.48 becomes

-2-2 = 2 (J2 + K2) + ti2. (C.56)

From Eq. C.52, J2 = K2, and the eigenvalues of J2 and K2 are identical. According to the theory of angular momentum (Eq. C.37), both are j (j + 1). Therefore, the solution is

-1 ^k2-^ = 4j(j + 1)ti2 + ti2 = (2j + 1)2ti2. (C.57)

2 En

Подпись: En Подпись: /ЛК " 2ti2n2 Подпись: (C.58)

From Eq. C.34, n = 2j + 1 can be any positive integer. Finally, the energy eigenvalues of the hydrogen atom are

image847

Leave a reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>