schrödinger equation in 3-d - wilfrid laurier university · schrödinger equation in 3-d: ... the...

Schrödinger Equation in 3-D:Time-independent Schrödinger equation in 3-D in sphericalcoordinates: Spherically symmetric potential

! (r,",#) = R(r)Ylm(",#)

Radial equation:

1

R(r)

!!r

r2 !R(r)

!r"#$

%&'(2mr

2

!2

V (r) ( E[ ])*+

,-.= / = l(l +1)

L2W (!,") = #!2W (!,") = l(l +1)!2Y

lm(!,")

Angular equation:

Separable solutions:

Schrödinger Equation in 3-D:Hydrogen atom:

The two-body Hamiltonian for the system is

H =p12

2m1

+p22

2m2

+V(r1 − r2 )

The hydrogen atom is an example of a two-body problem - an electroninteracting with a proton via the Coulomb interaction.

Defining the relative coordinate and relative momentum

r = r1 − r2 , p = m2p1 − m1p2m1 + m2

= −i∂∂r

and the centre-of-mass (CM) coordinate and total momentum

R =m1r1 + m2r2m1 + m2

, P = p1 + p2


The two-body Hamiltonian can be written in CM and relativecoordinates

H =p12

2m1

+p22

2m2

+V(r1 − r2 ) =P2

2M+p2

2µ+V(r)


With total and reduced masses

M = m1 + m2 ≈ mproton , µ =m1m2

M≈ melectron

The potential is only a function of the relative coordinate, so the centreof mass motion is that of a free particleWe now look at the relative coordinate Hamiltonian….


The relative coordinate Hamiltonian for the hydrogen atom is

Hrel =p2

2µ+V (r) = p2

2µ−

e2

4π ∈0

1r


The time-independent Schrodinger equation (energy eigenvalueequation) is

p2

2µ−

e2

4π ∈0

1r

⎧⎨⎩

⎫⎬⎭ψ (r) = Eψ (r)

Schrödinger Equation in 3-D:Hydrogen atom:Rewriting the TISE in spherical coordinates

ψ (r,θ,φ) = R(r)Ylm (θ,φ)

H = −2

2µ1r2

∂∂r

r2∂∂r

⎛⎝⎜

⎞⎠⎟ −

L2

2r2

⎧⎨⎩

⎫⎬⎭−

e2

4π ∈0

1r

⎡

⎣⎢

⎤

⎦⎥ψ (r,θ,φ) = Eψ (r,θ,φ)

Since the Coulomb potential is spherically symmetric, we can applyseparation of variables as before. The solutions are then

The radial wave functions are solutions to the radial equation

1R(r)

∂∂r

r 2∂R(r)∂r

⎛⎝⎜

⎞⎠⎟ −

2µr2

2 −

e2

4π ∈0

1r− E

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= l(l +1)

Schrödinger Equation in 3-D:Hydrogen atom: Solving the energy eigenvalue equation

u(r) = rR(r), κ = −2µE2 , ρ =κ r, ρ0 =

µe2

2π ∈0 2κ

Defining:

The radial equation becomes

1R(r)

∂∂r

r 2∂R(r)∂r

⎛⎝⎜

⎞⎠⎟ −

2µr2

2 −

e2

4π ∈0

1r− E

⎡

⎣⎢

⎤

⎦⎥

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪= l(l +1)

d 2udρ2

= 1− ρ0ρ

+l(l +1)ρ2

⎡

⎣⎢

⎤

⎦⎥u


ρ→ ∞First consider the case of

The above equation becomes

d 2udρ2

= 1− ρ0ρ

+l(l +1)ρ2

⎡

⎣⎢

⎤

⎦⎥u

d 2udρ2

= u

Solutions areu(ρ) = Ae−ρ + Beρ

Require u =0 at infinity, so B=0.


ρ→ 0Next consider the case of

The above equation becomes

d 2udρ2

= 1− ρ0ρ

+l(l +1)ρ2

⎡

⎣⎢

⎤

⎦⎥u

Solutions areu(ρ) = Cρl+1 + Dρ− l

Require u =0 to not blow up at 0, so D=0.

d 2udρ2

=l(l +1)ρ2

⎡

⎣⎢

⎤

⎦⎥u


Taking the asymptotic behaviour into account we guess solutions ofthe form

ρd2vdρ2

+ 2(l +1− ρ) dvdρ

+ [ρ0 − 2(l +1)]v = 0

Then the radial equation becomes

u = ρl+1e−ρv(ρ)

d 2udρ2

= 1− ρ0ρ

+l(l +1)ρ2

⎡

⎣⎢

⎤

⎦⎥u


Assume that v can be expressed as a power series:

ρd2vdρ2

+ 2(l +1− ρ) dvdρ

+ [ρ0 − 2(l +1)]v = 0

Then the radial equation becomes

v = ckρk

k=0

∞

∑

k(k +1)ck+1 + 2(l +1)(k +1)ck+1 − 2kck + [ρ0 − 2(l +1)]ck{ }ρk = 0k=0

∞

∑


Hence the coefficients must satisfy

From which we get a recursion relation:

k(k +1)ck+1 + 2(l +1)(k +1)ck+1 − 2kck + [ρ0 − 2(l +1)]ck{ }ρk = 0k=0

∞

∑

k(k +1)ck+1 + 2(l +1)(k +1)ck+1 − 2kck + [ρ0 − 2(l +1)]ck = 0

ck+1 =2(k + l +1) − ρ0(k +1)(k + 2l + 2)

ck

(Not pretty so far, but almost there…)


For large k,

ck+1 =2(k + l +1) − ρ0(k +1)(k + 2l + 2)

ck

So

ck+1 ≈2

k +1ck ⇒ ck ≈

2k

k!c0 ⇒ v =

2k

k!c0ρ

k = c0e2ρ

k=0

∞

∑

u = ρl+1e−ρv(ρ) = c0ρl+1eρ

But u does not go to 0 as r goes to infinity !Therefore the series expansion for v cannot be an infinite series.


ck+1 =2(k + l +1) − ρ0(k +1)(k + 2l + 2)

ck

If the series terminates then there is a maximum value kmax such thatckmax +1 = 0

Plugging this into the recurrence relation above,

v = ckρk

k=0

∞

∑

ckmax +1 = 0 =2(kmax + l +1) − ρ0

(kmax +1)(kmax + 2l + 2)ckmax

⇒ 2(kmax + l +1) − ρ0 = 0


Define n (integer):

2n = ρ0 =µe2

2π ∈0 2κ, κ =

−2µE

Thus solving for E we get the famous Bohr formula!

2(kmax + l +1) − ρ0 = 0

n = kmax + l +1

Then

En = −µ22

e2

4π ∈0

⎛⎝⎜

⎞⎠⎟

21n2, n = 1,2,3....(µ ≈ melectron )


κ =−2µE

=1a0

1n, a0 =

4π ∈0 2

me2= 0.0529nm (Bohr radius)

Quantized energies of the hydrogen atom:

En = −µ22

e2

4π ∈0

⎛⎝⎜

⎞⎠⎟

21n2, n = 1,2,3....(µ ≈ melectron )

Degeneracy of energy levels:The energies only depend on the principal quantum number n.For each value of n, there are n-1 possible values of l.For each l there are 2l+1 possible value of m.Thus total number of states with energy En for a given n = n2

Hydrogen emission and absorption spectral lines correspond totransitions between these energy levels.

Schrödinger Equation in 3-D:

Time-independent Schrödinger equation in 3-D: Hydrogen-like

atoms

Energy level diagram for hydrogen (note the degeneracy of excited

states):

Transitions between energy levels

occur by emission or absorption of

a photon.

Photons carry an angular momentum

of .

Thus to conserve angular momentum,

the transition rule must be

l f ! li = ±1

1!

Schrödinger Equation in 3-D:Hydrogen atom: Solving the energy eigenvalue equationEnergy eigenstates of the hydrogen atom:

The energy eigenstates are specified by 3 quantum numbers n, l and m.

ψ (r,θ,φ) = Rnl (r)Ylm (θ,φ)

Rnl (r) =ur=1rρ l+1e−ρv(ρ) = 1

rρ l+1e−ρ ckρ

k

k=0

n− l−1

∑

ck+1 =2(k + l +1) − ρ0(k +1)(k + 2l + 2)

ck

Schrödinger Equation in 3-D:Hydrogen atom: Solving the energy eigenvalue equationEnergy eigenstates of the hydrogen atom:

The function v defined by the above recursion relation is a well knownpolynomial function called the associated Laguerre polynomial.

ψ (r,θ,φ) = Rnl (r)Ylm (θ,φ)

Rnl (r) =ur=1rρ l+1e−ρv(ρ) = 1

rρ l+1e−ρ ckρ

k

k=0

n− l−1

∑

ck+1 =2(k + l +1) − ρ0(k +1)(k + 2l + 2)

ck

Schrödinger Equation in 3-D:Hydrogen atom: Solving the energy eigenvalue equationLaguerre polynomial:

Exact energy eigenstates of the hydrogen atom:

Lq (x) = ex d

q

dxqe! xxq( )

Associated Laguerre polynomial:

Lq! pp(x) = !1

p dp

dxpLq (x)

!nlm(r,",#)

2

na0

$

%&'

()

3

(n * l *1)!2n[n + l)!]

3e*r /na0

2r

na0

$

%&'

()

l

Ln* l*12l+1

2r na0( )+, -.Ylm (",#)

Schrödinger Equation in 3-D:Hydrogen atom: Solving the energy eigenvalue equationLaguerre polynomials:

Lq (x) = ex d

q

dxqe− x xq( )

L0 (x) = 1L1(x) = −x +1L2 (x) = x

2 − 4x + 2L3(x) = −x3 + 9x2 −18x + 6

Schrödinger Equation in 3-D:Time-independent Schrödinger equation in 3-D: Hydrogen-like

atoms

Probability densities:

P(r,!,") = # (r,!,"2

Probability of finding the electron

in a spherical volume:

P(r,!,") = # $#"min

"max

%!min

!max

%rmin

rmax

% r2sin!d&d!dr

Schrödinger Equation in 3-D:Time-independent Schrödinger equation in 3-D: Hydrogen-like

atoms

Radial probability distribution:

P(r)dr = r2R(r)

2dr

This is the probability of finding the electron in a

spherical shell of radius r and thickness dr

Probability that the electron lies between r1 and r2:

P(r1! r ! r

2) = r

2R(r)

2dr

r1

r2

"

Average radius: r = rP(r)dr =0

!

" r3R(r)

2dr

0

!

"


Radial wavefunctions of the hydrogenic atom:

Schrödinger Equation in 3-D:

Time-independent Schrödinger equation in 3-D: Hydrogen-like

atoms

Ground state of the hydrogenic atom:

Although R10(r) peaks at r=0, P(r)=r2 R210(r) peaks at r=a0/Z = a0 (for

hydrogen)


Some excited states of the hydrogenic atom:


Spectroscopic notation:

schrödinger equation in 3-d - wilfrid laurier university · schrödinger equation in 3-d: ... the...

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