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Unformatted text preview: IV. Solving the Time
Independent Schrödinger
Equation
Grifﬁths Chapters 2 and 4 1 A. The Operator Approach to the Harmonic Oscillator
Grifﬁths 2.3 • The harmonic oscillator is represented by the force
equation F = kx. • This gives a potential energy of V(x) = kx2/2. • Many potentials will reduce to this form for small
values of x. • Therefore, as a ﬁrst approximation, many physical
phenomena can be understood using the harmonic
potential.
2 A. The Operator Approach to the Harmonic Oscillator • Taylor expand V(x) about
the point x0.
x0 i s a minimum so the first derivative at x0 i s zero 1
V (x) = V (x0 ) + V (x0 )(x − x0 ) + V (x0 )(x − x0 )2 + · · ·
2
drop constant term...
doesn’t change the force k = V’’(x0)
as long as (xx0) is small, higher
o rder terms will be negligible • For our quantum situation, we want to solve the Schrödinger
k
equation for the potential
ω=
m
1
V (X ) = mω 2 X 2
2
3 A. The Operator Approach to the Harmonic Oscillator
1. Wave Function and Energies • The timeindependent Schrödinger equation will be 2 d2 ψ (x) 1
−
+ mω 2 X 2 ψ (x) = E ψ (x)
2m dx2
2 • This has an inﬁnite number of discrete solutions. The ground
state (lowest energy) of this quantum oscillator will be
ψ0 (x) = mω 1 / 4
π ω
− m x2
2 e ground state wave function • 1
E0 = ω
2
ground state energy The higher energy solutions are
ψn (x) = An (a+ ) ψ0 (x)
n En = n+ 1
2 ω
4 A. The Operator Approach to the Harmonic Oscillator • The higher energy solutions are
ψn (x) = An (a+ ) ψ0 (x)
n En = 1
n+
2 ω normalization
c onstant • So if we know what the raising operator a+ is, then we can
ﬁnd all of the higher energy wave functions and their
corresponding energies. • a+ and the related lowering operator a are called ladder
operators.
See board IV.A.1 5 A. The Operator Approach to the Harmonic Oscillator
2. The Number Operator • The number operator is deﬁned as N = a+a.
N ψn (x) = a+ a− ψn (x) = nψn (x)
See board IV.A.2 3. Normalization • The ladder operators can be used to normalize the wave
functions of the harmonic oscillator.
1
ψn (x) = √ (a+ )n ψ0 (x)
n! and En = 1
n+
2 ω normalized See board IV.A.3 6 B. The Hydrogen Atom Grifﬁths 4.1 and 4.2 • The hydrogen atom is one of the few, realistic, systems that
can be solved analytically. • The prescription is to write the timeindependent
Schrödinger equation in spherical coordinates and solve it
using a Coulomb potential. • The math goes on a long time but involves mostly just
separation of variables. • We use special functions such as Legendre polynomials,
spherical harmonics and Laguerre polynomials to get our
answer. • The principal, azimuthal and magnetic quantum numbers
arise naturally in solving the differential equations.
7 B. The Hydrogen Atom 1. The Time Independent Schrödinger Equation in Spherical Coordinates • The time independent Schrödinger equation is
H ψ (r) = E ψ (r) • For a potential V that doesn’t depend on time and is a central
potential (depends only on radial distance r) we get 2 2
−
∇ ψ (r) + V (r)ψ (r) = E ψ (r)
2m • This spherically symmetric potential is most conveniently
solved in spherical coordinates. The Laplacian ∇2 in
spherical coordinates is
2
1∂
1
∂
∂
1
∂
2
2∂
∇=2
r
+2
sin θ
+2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2
8 B. The Hydrogen Atom 2 2
−
∇ ψ (r) + V (r)ψ (r) = E ψ (r)
2m
2
1∂
1
∂
∂
1
∂
2
2∂
∇=2
r
+2
sin θ
+2 2
r ∂r
∂r
r sin θ ∂θ
∂θ
r sin θ ∂φ2 • This partial differential equation can be split into three
ordinary differential equations: one for r, one for θ and one
for ϕ. • This is done using separation of variables where
ψ(r,θ,ϕ) = R(r)Θ(θ)Φ(ϕ)
Angular parts all together, Y(θ, ) = Θ(θ)Φ( ), are called spherical harmonics 9 B. The Hydrogen Atom
Angular parts all together, Y(θ, ) = Θ(θ)Φ( ), are called spherical harmonics • The combined angular equation is
2
1
1∂
∂ Y (θ, φ)
1 ∂ Y (θ, φ)
sin θ
+
= −l(l + 1)
2
2
Y (θ, φ) sin θ ∂θ
∂θ
∂φ
sin θ • The radial equation is
1d
dR(r)
2mr2
r2
−
[V (r) − E ] = l(l + 1)
2
R(r) dr
dr
These are equal to the same separation constant, which
we choose to write in this strange way for later
c onvenience. See board IV.B.1.a
10 B. The Hydrogen Atom • The combined angular equation is
2
1
1∂
∂ Y (θ, φ)
1 ∂ Y (θ, φ)
sin θ
+
= −l(l + 1)
2
2
Y (θ, φ) sin θ ∂θ
∂θ
∂φ
sin θ • The angular equation can be further reduced using separation
of variables again with Y(θ,ϕ) = Θ(θ)Φ(ϕ). We get two
ordinary differential equations.
1
d
dΘ(θ)
sin θ
sin θ
+ l(l + 1) sin2 θ = m2
Θ(θ)
dθ
dθ 1 d2 Φ(φ)
= −m2
Φ(φ) dφ2 11 B. The Hydrogen Atom
1
d
dΘ(θ)
sin θ
sin θ
+ l(l + 1) sin2 θ = m2
Θ(θ)
dθ
dθ 1 d2 Φ(φ)
= −m2
Φ(φ) dφ2 • These have unnormalized solutions
Φ(φ) = eimφ and Associated Legendre function Θ(θ) = APlm (cos θ)
Normalization for both
s olutions lumped together in A. 12 ASIDE:
Plm (x) = (1 − x ) 1
Pl (x) = l
2 l! 2 m/2 d
dx l d
dx m Pl (x)
Legendre polynomial (x2 − 1)l See Griffths
4.1.2 for more
L egendre
p roperties and
discussion.
13 B. The Hydrogen Atom
1
d
dΘ(θ)
sin θ
sin θ
+ l(l + 1) sin2 θ = m2
Θ(θ)
dθ
dθ 1 d2 Φ(φ)
= −m2
Φ(φ) dφ2 • These have unnormalized solutions
Φ(φ) = eimφ and Associated Legendre function Θ(θ) = APlm (cos θ)
Normalization for both
s olutions lumped together in A. See board IV.B.1.b
14 B. The Hydrogen Atom • These solutions combined and normalized give the spherical
harmonics.
(2l + 1) (l − m)! imφ m
m
Yl (θ, φ) = ε
e
Pl (cos θ)
4π (l + m)!
where ε = (1)m for m ≥ 0 and ε = 1 for m ≤ 0. • These have unnormalized solutions See
G riffths
4.1.2 for
more
Spherical
H armonic
p roperties
a nd
discussion. 15 B. The Hydrogen Atom • • • The radial equation was
1d
2mr2
2 dR(r )
r
−
[V (r) − E ] = l(l + 1)
2
R(r) dr
dr
We can simplify this by changing variables to u(r) = rR(r).
See board
22
2
d u(r)
l(l + 1)
−
+ V (r) +
u(r) = Eu(r) IV.B.1.c
2
2
2m dr
2m r
This is the same as the one dimensional Schrödinger equation
but with an effective potential Veff(r) given by 2 l(l + 1)
Veﬀ = V (r) +
2m r2
and a normalization condition
∞
u(r)2 dr = 1 Centrifugal term 0 • We can’t go further without specifying the potential V(r). 16 B. The Hydrogen Atom
2. The Hydrogen Atom • The proton of the hydrogen atom is essentially motionless.
We’ll put it at the origin of the spherical coordinate system. • The electron will feel a Coulomb attraction to the proton
given by
e2 1
V (r) = −
4π0 r • The radial equation for the electron will therefore be
2
2
d u(r)
e1
l(l + 1)
−
+−
+
u(r) = Eu(r)
2
2
2m dr
4π0 r 2m r
2 • 2 We want to solve this equation for u(r) and E. 17 B. The Hydrogen Atom
2
2
d u(r)
e1
l(l + 1)
−
+−
+
u(r) = Eu(r)
2
2
2m dr
4π0 r 2m r
2 2 • We want to solve this equation for u(r) and E. • We can simplify the equation by making the substitutions
√
me2
−2mE
ρ0 ≡
κ≡
ρ ≡ κr
2π0 2 κ
We are interested in bound electron
s tates, where E<0, so this argument
o f the square will be positive • This gives
d u(ρ)
ρ0
l(l + 1)
= 1−
+
u(ρ)
2
2
dρ
ρ
ρ
2 See board IV.B.2.a
18 B. The Hydrogen Atom
d u(ρ)
ρ0
l(l + 1)
= 1−
+
u(ρ)
2
2
dρ
ρ
ρ
2 • Next we change the dependent variable too.
u(ρ) = ρl+1 e−ρ v (ρ) • This gives
d2 v (ρ)
dv (ρ)
ρ
+ 2(l + 1 − ρ)
+ [ρ0 − 2(l + 1)]v (ρ) = 0
2
dρ
dρ See board IV.B.2.b • We try a power series solution for this form of the radial
equation.
∞
v (ρ) =
cj ρj
j =0 19 B. The Hydrogen Atom v (ρ) = • •
• ∞
cj ρj j =0 This is a solution provided the coefﬁcients obey the recursion
relation
2(j + l + 1) − ρ0
See board
cj +1 =
cj
IV.B.2.c
(j + 1)(j + 2l + 2)
In order for this solution to be ﬁnite, the series must
terminate at some maximum value of j, call it jmax.
jmax must obey the relationship See board IV.B.2.d
2(jmax + l + 1) − ρ0 = 0
• Deﬁne the principal quantum number n = jmax + l + 1. The
allowed energies are therefore
2 2
Bohr
E1
m
e
1
En = 2 = −
,
n = 1, 2, 3, · · ·
2
2
Formula
n
2
4π0
n
20 B. The Hydrogen Atom E1
m
En = 2 = −
n
2 2 e2
4π0 2 1
n2 , 4π0 2
Bohr
−10
a≡
= 0.529 × 10 m
2
Radius
me Bohr
n = 1, 2, 3, · · ·
Formula See board IV.B.2.e • We have the allowed energy levels for the hydrogen atom.
Now we need the corresponding wave functions. • The radial part of the wave function is
1 l+1 −ρ
Rnl (r) = ρ e v (ρ)
r where v(ρ) is a polynomial of degree jmax = n  l  1with
coefﬁcients given by the previous recursion formula.
cj +1 = 2(j + l + 1) − ρ0
(j + 1)(j + 2l + 2) cj 21 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre
polynomials.
+1
v (ρ) = L2l−l−1 (2ρ)
n 22 ASIDE:
Associated Laguerre
p olynomial Lp−p (x)
q Lq (x) ≡ e x d
dx ≡ (−1) q p d
dx p Lq (x) Laguerre polynomial (e−x xq ) See Griffths pages
152 and 153 for more
L aguerre properties
a nd discussion.
23 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre
polynomials.
+1
v (ρ) = L2l−l−1 (2ρ)
n • The hydrogen wave functions are the combination of the
radial and angular solutions.
ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ) • Substituting in all our results we get the wave functions ψnlm (r, θ, φ) = 2
na 3 (n − l − 1)! −r/na
e
2n[(n + l)!]3 2r
na l 2l+1
m
Ln−l−1 (2r/na) Yl (θ, φ) Hydrogen atom wave functions
See board IV.B.2.f 24 B. The Hydrogen Atom • v(ρ) is expressed in terms of the associated Laguerre
polynomials.
+1
v (ρ) = L2l−l−1 (2ρ)
n • The hydrogen wave functions are the combination of the
radial and angular solutions.
ψnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ) • Substituting in all our results we get the wave functions ψnlm (r, θ, φ) = 2
na 3 (n − l − 1)! −r/na
e
2n[(n + l)!]3 2r
na l 2l+1
m
Ln−l−1 (2r/na) Yl (θ, φ) Hydrogen atom wave functions
See board IV.B.2.f 25 B. The Hydrogen Atom 26 B. The Hydrogen Atom
3. The Spectrum of Hydrogen • The hydrogen wave functions we have found are stationary
states. In principle, the hydrogen electron, if left alone, will
never change states. • In reality, there are always collisions between atoms or with
photons (light). These interactions are beyond our present
timeindependent discussion. • The result of such interactions however, is that the hydrogen
electron may undergo a transistion, that is, change states. It
will absorb or emit energy in the process. • The energy the hydrogen atom releases (or absorbs) is equal
to the energy difference between the bound states. 27 B. The Hydrogen Atom ground state
e nergy Eγ = Ei − Ef = −13.6eV
energy of
e mitted
photon energy of
i nitial state energy of
f inal state
f re • 1
1
2 − n2
ni
f Recall that Eν = hν and λ = c/ν.
Planck
Formula q ue principal
q uantum
n umbers
n cy wave length 1
=R
λ
t
an
con st 1
1
2 − n2
nf
i
g
e2
be r R = m
d
Ry
4π c3 4π0 2 = 1.097 × 107 m−1 28 B. The Hydrogen Atom ground state
e nergy Eγ = Ei − Ef = −13.6eV
energy of
e mitted
photon energy of
i nitial state energy of
f inal state
f re • Recall that Eν = hν and λ = c/ν.
Planck
Formula q ue principal
q uantum
n umbers
n cy wave length 1
=R
λ
t
an
con st 1
1
2 − n2
nf
i
g
e2
be r R = m
d
Ry
4π c3 4π0 • 1
1
2 − n2
ni
f 2 = 1.097 × 107 m−1 Bohr’s theory of the hydrogen atom was able to explain these
experimental results.
29 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.
 Spring '08
 RESCH
 Energy, Force, Potential Energy

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