Homework set #11 is due Mon. Dec.2 – after Fall Break....

http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 1

Hydrogen atom

•  Homework set #11 is due Mon. Dec.2 – after Fall Break.

•  Quantization of angular momentum in the Hydrogen Atom!

http://www.colorado.edu/physics/phys2170/

•  Let’s consider N particles whose positions are defined by a set of rectangular coordinates x, y, z. The classical expression for the total energy of the system is

•  Now replace momenta, E with differential operators

•  Multiply both sides by the wavefunction

Physics 2170 – Fall 2013 2

€

12m j

px j2 + py j

2 + pz j2( )

j=1

N

∑ +V (x1,y1,z1,......,xN ,yN ,zN ,t) = E

€

−2

2m j

∂ 2

∂x j2 +

∂ 2

∂y j2 +

∂ 2

∂z j2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

j=1

N

∑ +V (x1,y1,z1,......,xN ,yN ,zN ,t) = i ∂∂t

Multiparticle Schrodinger Eq.


Laplacian •  We use the symbol

which is called the Laplacian in spherical coordinates of the jth particle + write the Schrodinger equation as

So long as V does not depend on t, then we can write

where


€

∇ j2 ≡

∂ 2

∂x j2 +

∂ 2

∂y j2 +

∂ 2

∂z j2

€

−2

2m j

∇ j2Ψ

j=1

N

∑ +VΨ = i∂Ψ∂t

€

Ψ(x1,....,zN ,t) =ψ(x1,....,zN )φ(t)

€

φ(t) = e− iEt /


Schrodinger Eq. for Coulomb •  Get time independent Schrodinger Equation

•  For One-Electron Atom; let Ze be the charge on the Nucleus of mass m1, and –e is charge on electron of

mass m2. The Coulomb potential energy is

where x1, y1, z1 are coordinates of nucleus and x2, y2, z2 are coordinates of the electron.


€

−2

2m j

∇ j2ψ

j=1

N

∑ +Vψ = Eψ

€

V (x1,....,z2) =−Ze2

(x1 − x2)2 + (y1 − y2)

2 + (z1 − z2)2


One Electron Atom


€

−2

2m1∂ 2ψT

∂x12 +

∂ 2ψT

∂y12 +

∂ 2ψT

∂z12

⎛

⎝ ⎜

⎞

⎠ ⎟ −2

2m2

∂ 2ψT

∂x22 +

∂ 2ψT

∂y22 +

∂ 2ψT

∂z22

⎛

⎝ ⎜

⎞

⎠ ⎟ +V (x1,......,z2)ψT = ETψT

Reason for the subscript T will become apparent in a minute. We want to change variables corresponding to the translational motion of the center of mass (x, y, z) and the motion of the two particles relative to each other (r, θ, ϕ).

€

x =m1x1 + m2x2m1 + m2

€

y =m1y1 + m2y2m1 + m2

€

z =m1z1 + m2z2m1 + m2

€

rsin(θ)cos(φ) = x2 − x1

€

rsin(θ)sin(φ) = y2 − y1

€

rcos(θ) = z2 − z1


One – Electron Atom •  Fair amount of algebra- actually tedious, but can get

•  µ is called the reduced mass, but for electron and proton it is about me. Look for solutions of the form

•  Divide top equation by


€

−2

2(m1 + m2)∂ 2ψT

∂x 2+∂ 2ψT

∂y 2+∂ 2ψT

∂z2⎛

⎝ ⎜

⎞

⎠ ⎟ −2

2µ1r2

∂∂r

r2 ∂ψT

∂r⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin(θ)

∂ 2ψT

∂φ 2+

1r2 sin(θ)

∂∂θ(sin(θ)∂ψT

∂θ)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

+V (r)ψT = ETψT

where and

€

µ ≡m1m2

m1 + m2

€

V (r) = −Ze2

r

€

ψT (x,y,z,r,θ,φ) =ψCM (x,y,z)ψ(r,θ,φ)

€

ψT =ψCMψ


Full Schrodinger Equation

•  Note left side is a function only of x,y,z and a function only of r,θ,ϕ . Since sum of two terms is equal to a constant, it is clear that each term must be equal to a constant.

•  Let the first term = ECM and the second E, such that •  ET = ECM + E


€

−2

2(m1 + m2)∂ 2ψT

∂x 2+∂ 2ψT

∂y 2+∂ 2ψT

∂z2⎛

⎝ ⎜

⎞

⎠ ⎟ −2

2µ1r2

∂∂r

r2 ∂ψT

∂r⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin(θ)

∂ 2ψT

∂φ 2+

1r2 sin(θ)

∂∂θ(sin(θ)∂ψT

∂θ)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

+V (r)ψT = ETψT

€

−2

2(m1 + m2)1

ψCM

∂ 2ψCM

∂x 2+∂ 2ψCM

∂y 2+∂ 2ψCM

∂z2⎛

⎝ ⎜

⎞

⎠ ⎟ + [−

2

2µ1ψ

1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin(θ)

∂ 2ψ∂φ 2

+1

r2 sin(θ)∂∂θ(sin(θ)∂ψ

∂θ)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

+V (r)] = ET


Final Solutions


€

−2

2(m1 + m2)1

ψCM

∂ 2ψCM

∂x 2+∂ 2ψCM

∂y 2+∂ 2ψCM

∂z2⎛

⎝ ⎜

⎞

⎠ ⎟ + [−

2

2µ1ψ

1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin(θ)

∂ 2ψ∂φ 2

+1


∂θ)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

+V (r)] = ET

€

−2

2(m1 + m2)1

ψCM

∂ 2ψCM

∂x 2+∂ 2ψCM

∂y 2+∂ 2ψCM

∂z2⎛

⎝ ⎜

⎞

⎠ ⎟ = ECM

€

−2

2µ1ψ

1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin(θ)

∂ 2ψ∂φ 2

+1


∂θ)

⎧ ⎨ ⎩

⎫ ⎬ ⎭

+V (r) = E

We are going to neglect the Center of Mass Motion and concentrate on the interaction and energy levels between the proton and the electron. Let’s start to look at the hydrogen atom and third equation.


Clicker Question 1 Set frequency to AD

Q. The potential seen by the electron in a hydrogen atom is… A. Independent of distance B. Spherically symmetric C. An example of a central force potential D. Constant E. More than one of the above

The potential seen by the electron is


Clicker Question 1 Set frequency to AD

Q. The potential seen by the electron in a hydrogen atom is… A. Independent of distance B. Spherically symmetric C. An example of a central force potential D. Constant E. More than one of the above

The potential seen by the electron is

Spherically symmetric (doesn’t depend on direction). It depends only on distance from proton so it is a central force potential.


3-D central force problems The hydrogen atom is an example of a 3D central force problem. The potential energy depends only on the distance from a point (spherically symmetric)

x y

z

θ

φ

r

Spherical coordinates is the natural coordinate system for this problem.

Engineering & math types sometimes swap θ and φ. General potential: V(r,θ,φ)

Central force potential: V(r)

€

−2

2µ1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sinθ

∂∂θ

sinθ ∂ψ∂θ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin2θ

∂2ψ∂φ 2

⎡

⎣ ⎢

⎤

⎦ ⎥ +V (r)ψ = Eψ

The Time Independent Schrödinger Equation (TISE) becomes:

We can use separation of variables so

CM of mass motion is ignored!


Volume Element in Spherical Coordinates


€

dτ = (rsinθdφ)(rdθ)dr = r2 sinθ dr dθ dφ

€

ψ2

0

2π

∫0

π

∫0

∞

∫ r2 sinθ dr dθ dφ


Separation of Variables


€

∂ψ∂r

=ΘΦ∂R∂r

=ΘΦdRdr

∂ψ∂θ

= RΦ∂Θ∂θ

= RΦ dΘdθ

∂ 2ψ∂φ 2

= RΘ∂2Φ∂φ 2

= RΘ d2Φdφ 2

€

−2

2µ1r2

∂∂r

r2 ∂ψ∂r

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sinθ

∂∂θ

sinθ ∂ψ∂θ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1r2 sin2θ

∂2ψ∂φ 2

⎡

⎣ ⎢

⎤

⎦ ⎥ +V (r)ψ = Eψ

€

Now divide by ψ = RΘΦ to get

€

sin2θR

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +sinθΘ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1Φ∂2Φ∂φ 2

+2µr2 sin2θ2

[E −V (r)] = 0

We will find that the solution to the wavefunction depends on 3 quantum numbers,

€

n, , m .


Separation of Variables cont.


€

sin2θR

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +sinθΘ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ +

1Φ∂2Φ∂φ 2

+2µr2 sin2θ2

[E −V (r)] = 0

€

sin2θR

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +sinθΘ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ +2µr2 sin2θ2

[E −V (r)] = −1Φ∂2Φ∂φ 2

This equation can be correct only if both sides of it are equal to the same constant.

€

−1Φ∂2Φ∂φ 2 = ml

2 Φ(φ) = Aeim lφ

€

ml = 0, ±1, ± 2,.....Called the magnetic quantum number and implies that z component of angular momentum is quantized!


The Φ(φ) part The variable φ only appears in the TISE as

€

∂2ψ∂φ 2

∝ −Cψ

So we should not be surprised that the solution is Note that m is a separation variable and not the electron mass. We use me for the electron mass.

x y

z

θ

φ

r

Are there any constraints on m? What can we say about and ?

They have to be the same!

Since cosine and sine have periods of 2π, as long as m is an integer (positive, negative, or 0) the constraint is satisfied.


The Theta Term


€

−1Rddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ − E −V (r)[ ] 2µr2

2−

1Θsinθ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ = −

m2

sin2θ

Transposing and rewriting

€

1Rddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +2µr2

2E −V (r)[ ] =

m2

sin2θ−

1Θsinθ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟

One side contains only r dependence; the other only θ; set both equal to a constant, called α.

€

1r2

ddr

r2 dRdr

⎛

⎝ ⎜

⎞

⎠ ⎟ +2µ2

E −V (r)[ ]R =αRr2

€

m2Θsin2θ

−1sinθ

ddθ

sinθ dΘdθ

⎛

⎝ ⎜

⎞

⎠ ⎟ =αΘ


The Θ(θ) part

x y

z

θ

φ

r

The solution to the Θ(θ) part is more complicated so we skip solving the differential equation + just show results.

The end result is that there is another quantum variable ℓ which must be a non-negative integer and ℓ ≥ |m|.

The ℓ variable quantizes the total angular momentum:

Note, for large ℓ, so ℓ is basically the total angular momentum and m is the z-component of the angular momentum.

Since the z-component cannot be larger than the total, |m| ≤ ℓ.

Solving for Θ, and requiring behavior is finite everywhere, forces

€

α = ( +1)

€

= m ,m +1,m + 2,m + 3.....where

€

= orbital quantum number


Clicker question 2 Set frequency to DA Total angular momentum is ℓ can be 0, 1, 2, 3, …

The z-component of the angular momentum is where m can be 0, ±1, ±2, … ±ℓ

Just like with any vector, the total angular momentum can be written

Q. Given the rules above, can Lx=Ly=0? That is, can L=Lz? A. Yes, in more than one case B. Yes, but only in one case C. Never

http://www.colorado.edu/physics/phys2170/ Physics 2170 – Fall 2013 19 so (except for m=ℓ=0)

and there is no integer between ℓ and ℓ+1

Clicker question 2 Set frequency to DA Total angular momentum is ℓ can be 0, 1, 2, 3, …

The z-component of the angular momentum is where m can be 0, ±1, ±2, … ±ℓ

Just like with any vector, the total angular momentum can be written

Q. Given the rules above, can Lx=Ly=0? That is, can L=Lz? A. Yes, in more than one case B. Yes, but only in one case C. Never

It is possible for Lx=Ly=Lz=0 in which case L = 0 so ℓ=0 and m=0

In general, if Lx=Ly=0 then simplifies to which means or . But


Angular Momentum


If hydrogen atom were classical, the electron’s orbit, like a planet would be an elipse, but it is not!


Spherical harmonics We have determined the angular part of the wave function so

has become with the quantum numbers ℓ and m specifying the total angular momentum and the z-component of angular momentum.

This angular solution works for any central force problem.

The combination are the spherical harmonics

Spherical harmonics are 3-D and complex (real and imaginary terms), making it very difficult to display.


Spherical harmonics We have determined the angular part of the wave function so

has become with the quantum numbers ℓ and m specifying the total angular momentum and the z-component of angular momentum.

This angular solution works for any central force problem.

The combination are the spherical harmonics


A few of the spherical harmonics

http://oak.ucc.nau.edu/jws8/dpgraph/Yellm.html

Real part only Colors give phase Red = +1

Cyan = -1 Purple = +i

Green = -i

Homework set #11 is due Mon. Dec.2 – after Fall Break....

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