qm_ch4
-
Upload
sidhartha-samtani -
Category
Documents
-
view
222 -
download
0
Transcript of qm_ch4
-
8/3/2019 qm_ch4
1/10
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
1
-
8/3/2019 qm_ch4
2/10
= 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.
2
-
8/3/2019 qm_ch4
3/10
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
3
-
8/3/2019 qm_ch4
4/10
Normalization: A = 11/4
x0
0(x) = 11/4
x0
e x2
2x20
In general, n(x) = x|n = x|1
n! (a)n|0 = 1
1/42nn! 1
xn+1/20
x x20
d
dxn
ex2
2
x2
0
3.4 Expectation Values
n|x|n =
h
2m(
nn,n1 +
n + 1n,n+1) = 0
n|p|n = imh2 (nn,n1 + n + 1n,n+1) = 0n|x2|n = h
2mn|
(a)2 + aa + aa + a2
|n = h2m
n|
aa + aa
|n
=h
2mn|
2aa + 1
|n = h2m
(2n + 1) = (n +1
2)
h
m
n|p2|n = mh2
n|
(a)2 aa aa + a2
|n = mh2
n|
aa + aa
|n
=mh
2n|
2aa + 1
|n = (n + 12
)mh
xp = (x2 x2)1/2(p2 p2)1/2 = (n +1
2)
h
m1/2
(n +1
2)mh
1/2
= (n +1
2)h
Note:
(i) T = p22m = 12 h(n + 12) = 12H, V = 12m2x2 = 12 h(n + 12) = 12H: Virial theorem
(ii) When n = 0, xp = h2 : minimum uncertainty - wave function 0(x): Gaussian
3.5 Time Evolution ofx
and
p
d
dtx = 1
ih[x, H] = p
md
dtp = 1
ih[p, H] = m2x
4
-
8/3/2019 qm_ch4
5/10
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
5
-
8/3/2019 qm_ch4
6/10
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)
6
-
8/3/2019 qm_ch4
7/10
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
7
-
8/3/2019 qm_ch4
8/10
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
8
-
8/3/2019 qm_ch4
9/10
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
9
-
8/3/2019 qm_ch4
10/10
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.
10