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Chapter 4. Harmonic Oscillator

1 Examples of Harmonic Oscillators

Vibration of the nuclei in molecules

r0

r: the distance between two atoms

r

2

00)(

2

1)( rrkVrV +

-V0

Vibration of atoms in solids: phonons Vibration modes of a continuous physical system - application to radiation (photons)

2 Eigenvalues of the Hamiltonian

2.1 Hamiltonian

The Hamiltonian operator is

H = p22m

+ 12

m2x2

Define two operators

a =

m

2h

x +

ip

m

: annihilation operator a =

m

2h

x ip

m

: creation operator

Consider the commutator [a, a] =1

2h(

i[x, p] + i[p, x]) = 1

Using x =

h2m (a + a

) and p = 1i

mh2 (a a),

H =1

2m

1

i

mh

2(a a)

2

+1

2m2

h

2m(a + a)

2

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= h4

a2 aa aa + (a)2

+h

4

a2 + aa + aa + (a)2

=h

2(aa + aa) =

h

2(aa + 1 + aa)

H = h(aa +

1

2

) = h(N +1

2

) where N = aa.

We denote an eigenket of N by its eigenvalue n as N|n = n|n.

H|n = (n + 12

)h|n energy eigenvalue, En = (n + 12)h

2.2 Annihilation and Creation Operator

[N , a] = [aa, a] = a[a, a] + [a, a]a = a[N , a] = [aa, a] = a[a, a] + [a, a]a = a

Thus, Na|n = ([N , a] + aN)|n = (n + 1)a|n: a|n is an eigenstate of N with eigenvalue of (n + 1). a: creation operatorSimilarly, Na|n = ([N , a] + aN)|n = (n 1)a|n: a|n is an eigenstate of N with eigenvalue of (n 1). a: annihilation operator

We write a|n = c|n 1, thenn|aa|n = |c|2n 1|n 1 = |c|2= nn|n

= n

Therefore, |c|2 = n c = n

Thus, a|n = n|n 1 Similarly, a|n = n + 1|n + 1

n is a positive integer.

Proof) n = n|N|n = (n|a)(a|n) 0.If n is not integer, (a)m|n |n m n m is negative when m > n.Thus, n should be positive integer.

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3 Eigenstates of the Hamiltonian

3.1 Construction of the number state |n

The ground state has n = 0: |0 and E0 = 12 h.Successive application of a to the ground state gives rise to

|n =

(a)nn!

|0, En = (n + 1

2)h

3.2 Matrix Elements

Using the number states {|n} as a basis,

n|a|n = nn,n1n|a|n = n + 1n,n+1

n|x|n =

h2m

(nn,n1 + n + 1n,n+1)

n|p|n = i

mh

2(nn,n1 +

n + 1n,n+1)

a =

0

1 0 0 0 0

2 0

0 0 0

3 ...

......

...

0 0 0 0 0 n ...

......

...

a =

0 0 0 0 1 0 0 0

0

2 0 0 ...

......

...

0 0 0 n + 1 0 ...

......

...

3.3 x-representation

x|a|0 =

m

2h

xx + ipm 0

= 0

x + x20

d

dx0(x) = 0 where 0(x)

x

|0

and x0

h

m

d0(x)

0(x)= x

x20dx ln 0(x) = x

2

2x20+ C

0(x) = Ae x2

2x20

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d2

dt2x = 1

m

d

dtp = 2x xt = A cos t + B sin t

At t=0, x0 = A, ddt x = p0m = B

xt = x0 cos t + p0m

sin t, pt = m ddt

xt = p0 cos t mx0 sin t

4 Isotropic 3D Harmonic Oscillator

4.1 Hamiltonian and Eigenvalue Equation

x

y

z

m22

2

2

1

2

rm

m

pH +=

where p2 = p2x + p2y + p

2z and r

2 = x2 + y2 + z2.

Then, H = Hx + Hy + Hz where Hx =p2x

2m+

1

2m2x2

Hx, Hy, Hz: 1D harmonic oscillator Hamiltonian.

Since Hx, Hy, and Hz are commute with H, the eigenvalue equation, H| = E|, can be solvedseeking the eigenvectors of H which are also eigenvectors of Hx, Hy, and Hz .

Eigenvectors and eigenvalues of Hx, Hy, Hz.

Hx|nx = (nx + 12

)h|nx, Hy|ny = (ny + 12

)h|ny, Hz|nz = (nz + 12

)h|nz

The eigenstates common to H, Hx,Hy, and Hz are |nx, ny, nz |nx|ny|nz, in Ex Ey EzH|nx, ny, nz = (nx + ny + nz + 3

2)h|nx, ny, nz

Eigenvectors of H: |nx, ny, nz - tensor products of eigenvectors of Hx,Hy, Hz Eigenvalues ofH: E = (nx+ny+nz+ 32)h - sums of eigenvalues ofHx,Hy, Hz , En = (n+ 32)h

4.2 Degeneracy of the Energy Levels

(1) {Hx, Hy, Hz} is a C.S.C.O. in Er since Hx, Hy, and Hz are C.S.C.O.s in Ex, Ey, and Ez .H does not form a C.S.C.O. since the energy levels En are degenerate.

(2) Degree of degeneracy, gn

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0

1

0

0:

n-1

0

:

n

nz

2

1

n

n+1

0n

0

1n-1

n-1:

0

1

n

:

0

0

ny

nx

+

++==

+++++=

1

)2)(1(2

1

)1(21

n

i

n

nni

nng L

4.3 Annihilation and Creation Operator

[ax, ax] = [ay, a

y] = [az , a

z] = 1

ax|nx, ny, nz =

nx + 1|nx + 1, ny, nzax|nx, ny, nz = nx|nx 1, ny, nz

|nx, ny, nz = 1nx!ny!nz!

(ax)nx(ay)

ny(az)nz |0, 0, 0

r|0, 0, 0 =

m

h

3/4e

m2h

(x2+y2+z2)

5 A Charged Harmonic Oscillator in a Uniform Electric Field

5.1 Hamiltonian

xxqxmxV =

22

2

1)(

m,q

H =

p2

2m +

1

2 m

2

x

2

qx

5.2 Schrodinger Equation

h

2

2m

d2

dx2+

1

2m2x2 qx

(x) = E(x)

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h

2

2m

d2

dx2+

1

2m2

x qm2

2 q

22

2m2

(x) = E(x)

Replace the variable x by u = x qm2 , h

2

2m

d2

du2+

1

2m2u2

(u) = E(u), where E = E +q22

2m2

Eigenfunctions and eigenvalues

n(u) = n

x q

m2

The translation comes from the fact that the electric field exerts a force on the particle.

En = En

q22

2m2= (n +

1

2)h q

22

2m2

x

V(x)

20

m

qx =

2

22

2

m

q

5.3 Electrical Susceptibility

: electric field

+- +-+-

+- +-+-

x

polarized:P=qx Px=qx

Susceptibility: =Px/

(i) When = 0, Px = qn|x|n = 0(i) When = 0,

Px = qn|x|n = q

dxn

x q

m2

xn

x q

m2

= q

dun (u) (u + x0)n (u) , where x = u + x0, x0 =q

m2

= q

dun (u) un (u) + qx0

|n (u) |2du

= qx0 =q2

m2

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Susceptibility, =Px

=

1

q2

m2=

q2

m2

x0 is the mean value of the equilibrium position of the electron

Polarizability Px = qx0 = qx0

5.4 Translation Operator

U() e(aa), is a real constant.

Adjoint U() = e(aa) U()U() = U()U() = 1 U: unitary operator

Under the corresponding unitary transformation, H becomes

H = U()HU() = h

1

2+ U()aaU()

= h

aa +

1

2

where a = U()aU

()

Using eA+B

= eA

eB

e1

2[A,B]

U() = ea+a

= eaea

e2

2 , U() = ea+a = ea

eae2

2

Then,

a = (eaea

e2

2 )a(ea

eae2

2 )

= ea(ea

aea

)ea [ea , a] = ea , eaaea = a = ea(a )ea = a

Similarly, a = a

Thus,

H = h

(a )(a ) + 1

2

= H h(a + a) + 2h

Let = q

1

2mh and use a + a =

2mh x.

H = H q

1

2mhh

2m

hx +

q22

21

2mhh

= H qx + q22

2m2= H() +

q22

2m2

Since H|n = En|n,HU()|n = U()HU()U()|n = U()H|n = EnU()|n

Therefore, |n = U()|n is a eigenstate of H() and eigenvalue is En = (n + 12)h q22

2m2

|n = U()|n = eiphx0|n, x0 = q

m2

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6 Coupled Harmonic Oscillators

6.1 Classical Picture

x1

m m

k1 k2 k1

x2

2

11 mk =

2

22mk =

Equation of motion

md2x1

dt2= k1x1 + k2(x2 x1)

md2x2

dt2= k1x2 + k2(x1 x2)

Introduce xC =12 (x1 + x2) (center of mass motion) and xR = x1 x2 (relative motion).

The equation of motion becomes

d2

dt2xC = 21xC

d2

dt2xR = 21xR 222xR = (21 + 222)xR

xC(t) = x0Ccos(Ct + C), C = 1

xR(t) = x0

R cos(Rt + R), R =

21 + 2

22

: normal vibrational mode

C

R

General motion - linear combination of normal modes.

x1(t) = xC(t) +1

2xR(t)

x2(t) = xC(t) 12

xR(t)

When 2 1, R = C

1 +22

2

21

1/2 C +

2

2

1 beating

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t

x1 Fast oscillaton

Slow oscillaton

1

2

22

CR

C

6.2 Quantum Mechanical Picture

H =1

2m(p21 + p

22) +

1

2m21(x

21 + x

22) +

1

2m22(x1 x2)2

=p2C

2C+

p2R2R

+1

2C

2Cx

2C +

1

2R

2Rx

2R

where pC = p1 + p2, C = 2m, C = 1 and pR =p1p2

2 , R =m2 , R =

21 + 222

H = HC + HR

HC =p2C

2C+ 1

2C

2Cx

2C = (a

CaC +

12

)hC

HR =p2R

2R+

1

2R

2Rx

2R = (a

RaR +

1

2)hR

(1) Eigenvalues and eigenstates

|u = |nC, nR = |nC|nRwhere HC|nC = (nC + 12)hC|nC and HR|nR = (nR + 12 )hR|nR(2) Quantum beats

|(0) = 12

(|0, 1 + |1, 0)

|(t) = 12

(eihHt |0, 1 + e ih Ht |1, 0)

=1

2(ei(C/2+3R/2)t|0, 1 + ei(3C/2+R/2)t|1, 0)

=1

2ei(C+3R)t/2(|0, 1 + ei(CR)t|1, 0)

( )10012

12 = CR

T

=2( )

10012

1

1 +=

Coupling lifts degeneracy

When 2 = 0 (no coupling), two-fold degeneracy exists : |n1, n2 and |n2, n1 has the same energyeigenvalue E = (n1 + n2 + 1)h.

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