The Hydrogen AtomThe Hydrogen Atom

continued..continued..

Quantum Physics2002

Recommended Recommended Reading:Reading:

Harris Chapter 6, Sections 3,4• Spherical coordinate system

•The Coulomb Potential•Angular Momentum• Normalised Wavefunctions•Energy Levels• Degeneracy

We now have all the components necessary to write down the solutions to the Schrodinger Equation for the Hydrogen Atom. We applied the method of Separation of Variables to wavefunction

Put it all together Put it all together

φθφθψ )r(R,,r

and have found the following solutions for the radial and angular parts.

onwavefuncti AzimuthalimexpA

sPolynomial Legendre Associated P

sPolynomial Laguerre Associated rR)r(R

L

m,l

l,n

L

φφ

θθ

And so the total wavefunction can be written as:

),(YrRNimexpPrRN,,r m,,nm,,nm,,nm,,nm,,n

llllll lllllll φθφθψ

where Nn,l,mL is a normalisation factor which we still have to determine.

Note that the wavefunction depends on three quantum numbers, n, the principal quantum number, l = the total angular momentum and ml the z-component of the angular momentum.

Example: Example: the wavefunction corresponding to the state n = 1, l = 0, ml = 0, what is the explicit form of this wavefunction

φθφθψ 0iexpPrRN,,r 0,00,10,0,10,0,1

writing the explicit forms of the Laguerre and Legendre polynomials gives

0230

0,0,1 ar

expa

2N,,r

0,0,1φθψ

since P0,0() = 1 and exp(i.0.) = 1.

Similary for n = 2, l = 2, mL = 1 we find

φθφθψ iexpPrRN,,r 1,11,21,1,21,1,2

φθφθψ iexpsina2r

expa3r

a2

1N,,r

00230

1,1,21,1,2

The angular part of the wavefunction The angular part of the wavefunction Lets examine the angular part in more detail. We can write the angular part as

imexpP, Lm,l Lφθφθφθψ

as before we can combine these two terms together into a single function Yl,mL(,) the Spherical Harmonics, this function combines the and dependent part of the solution. The Spherical represent the solutions to the Schrodinger equation for a particle confined to move on the surface s a sphere of unit radius. The first few are tabulated on the next page.

Note that the Spherical harmonics depends on two quantum numbers l and ml.

The Spherical Harmonics YThe Spherical Harmonics Yl,mLl,mL((, , ) )

l m L Y l , m L ( , )0 0

π21

1 0θcos

π3

21

1 1 φiesin3

21 θ

2 π

2 0 1θcos35

41 2

π2 1 φieθcosθsin

1521

2 π

2 2 φθπ

i22 esin215

41

l m L Y l , m L ( , )3 0 θcos3cos5

π7

41 3 θ

3 1 φi2 e1cos5θsinπ

2181 θ

3 2 φθθ i22 ecossinπ2

10541

3 3 φθπ

i33 esin35

81

The Spherical Harmonics YThe Spherical Harmonics Yl,mLl,mL((, , ))continued... continued...

to see what these wavefunctions look like see the following websites

http://www.quantum-physics.polytechnique.fr/en

http://www.uniovi.es/~quimica.fisica/qcg/harmonics/harmonics.html

http://wwwvis.informatik.uni-stuttgart.de/~kraus/LiveGraphics3D/ java_script/SphericalHarmonics.html

YY0,00,0

YY1,01,0

YY1,+11,+1

YY1,-11,-1

YY2,02,0

YY2,+22,+2YY2,+12,+1

YY2,-12,-1 YY2,-22,-2

z-Component of Angular momentum z-Component of Angular momentum What happens if we operate on the angular part of the wavefunction with the Lz operator ?

φθφθ

φφ

θφθφ

φθφθφθψ

,YmimexpP m

imexpdd

PiimexpPdd

i

,YL̂imexpPL̂ ,L̂

m,

m,

m,

m,m,

zm,zz

llll

ll

lll

lll

llll

ll

Thus we see that the Spherical Harmonics are eigenfunctions of the Lz-operator with eigenvalues lm

φθφθ ,Ym ,YL̂m,m,z llll l

this shows that the z-component of the angular momentum is quantised in units of

What else can we learn from the angular part of the wavefunction ?

Total Angular Momentum Total Angular Momentum Since the angular momentum is a vector we can write the total angular momentum as

2z

2y

2x

2 LLLL

From this (after some math) we find that

2

2

222

θsin

1θsin

θsin1

L̂φθθ

Let us operate on the angular part of the wavefunction and see what result it gives us, that is

,Y1

,Yθsin

1θsin

θsin1

,YL̂

m,2

m,2

2

22

m,2

φθ

φθφθθ

φθ

l

ll

l

ll

ll

this shows that the the angular part of the wave function is an eigenfunction of the L2 operator, with eigenvalue 21 ll

Total Angular Momentum Total Angular Momentum Since the square of a vector is equal to the square of its magnitude, this means that the magnitude of the angular momentum can take on only the values

2 ...3, 2, 1, 0 where l 1llLL 2

and recall that no physical solution exists unless ml l. So with this restriction we have

l, ...2, 1, 0 where mmLLLZ

The information we now have is the magnitude of the quantised angular momentum Land the magnitude of the z-component of the angular momentum.

For example, if l = 2 then

6 122L

and the z-components of the allowed angular momenta are

2,1,0,1,2LZ

we can represent this graphically as follows

6

2

-

2-

0

z L

mLZ

1llL θ

Space Quantisation Space Quantisation

The angular momentum vector can only point in certain directions in space Space Quantisation

1ll

mcos

1llm

LL

cos LL 1z θθ

Example Example What is the minimum angle that the angular momentum vector may make with the z-axis in the case where (a) l = 3 and (b) for l = 1?

When l = 3 the magnitude of the angular momentum is

12133L

with 7 possible z-components, mL = -3, -2, -1, 0, 1, 2, 3. The angular momentum vector will make the minimum angle with the z-axis when the z-component is as large as possible, 3

3LZ

12L θ

30

123

cosθ123

LL

θcos 1z

and then

2111L for l = 1:

4521

cos21

LL

cos 1z

θθ

NotationNotationWe can specify the angular momentum states of a single particle as follows

so for example, if the particle has angular momentum l = 2, then it is called a d-state. In this state the magnitude of the angular momentum is

6122L

and its z-component can have the following values,

2,1,0,1,2LZ

l = 0 1 2 3 4 5 6state s p d f g h iNumber of states 1 3 5 7 9 11 13

Note that each angular momentum state has a degeneracy of 2l+1 . Thus for example, a d-state has a degeneracy of 5. this means that there are 5 states with l = 2 which all have the same energy.

Degeneracy Degeneracy We have found that the energy levels are given by

eV

n

6.13

n

1

h8

me

n

1

2

meE

222

4

222

4n

004 επε

so each energy level has a degeneracy of

note that the energy levels are degenerate since they do not depend explicitly on l and ml . Recall that n = 1, 2, 3, 4, …

l = 0, 1, 2, 3, … n-1

ml = - l, -(l -1), ….(l -1), l

21n

0n12

ll

Example: the n = 2 state has a degeneracy of 4:

a single 2-s state with l = 0, ml = 0 and a 3-fold degenerate 2-p state with l = 1, ml = -1, 0, 1

n l mL state Energy1 0 0 1s -13.6 eV2 0 0 2s -3.40 eV

-1 2p -3.40 eV1 0 2p

+1 2p3 0 0 3s -1.51 eV

-1 3p -1.51 eV1 1 3p

+1 3p-2 3d -1.51 eV-1 3d

2 0 3d+1 3d+2 3d

NormalisationNormalisationJust as in the case of one dimensional problems, the total probability of finding the electron somewhere in space must be unity. Using the volume element in polar coordinates, the nomalisation condition becomes

1dV2

spaceall

Lml,n,Ψ

1eesinPdrrR0

imim

0 0

2m,l

22l,n LL

L

2π φφπdφθdθ

The last integral is just 2 so we are left with two integrals.

1drrR0

22l,n

1sinP2

0

2m,l L

π

θdθπand

Using these two integrals we can find the correctly normalised Radial and Angular wavefunctions

ExampleExampleFind the angular normalisation constant for P1,+1().

For l = 1, mL = +1 we have P1,+1() = sin (see table). then

1sinA2sinsinA20

321,1

0

221,1

ππθdθπθdθθπ

π

π83

A1A2 1,121,1

34

Carry out the integral, to find

and the correctly normalised wavefunction is

θπ

θ sin83

P1,1

Probability DensityProbability DensityWhat is the probability of finding the electron within a small volume element dV at position r,, ?

φθθψψ ddrdsinrdVP 22n,l,m

2n,l,m

φθθφθθ ddrdsinYdrrRddrdsinrYR2

m,l22

l,n22

m,ll,n

If we want to find the probability of finding the electron at a distance between r and r + dr (i.e. in a spherical shell of radius r and thickness dr), then we just integrate over and .

φθθπ π

ddsinYdrrRPdr0

2

0

2m,l

22l,n

But, since the spherical harmonics are normalised (by definition), so the integral is equal to 1 and we then have

drrRdrP 22l,nr or 22

l,nr rRP

is the probability of finding the electron at a distance r from the nucleus at any angle . What do these distributions look like ?

1s State1s Staten = 1, l = 0n = 1, l = 0

220,1r rRP 0,1R

2s State2s Staten = 2, l = 0n = 2, l = 0

220,2r rRP 0,2R

2p State2p Staten = 2, l = 1n = 2, l = 1

221,2r rRP

1,2R

3s State3s Staten = 3, l = 0n = 3, l = 0

220,3r rRP

0,3R

3p State3p Staten = 3, l = 1n = 3, l = 1

221,3r rRP 1,3R

3d State3d Staten = 3, l = 2n = 3, l = 2

222,3r rRP 2,3R

Summary of Key PointsSummary of Key Points1) The wavefunctions of the hydrogen atom can be expressed as the product of three one dimensional functions in the variables r, and

2) The mathematical form taken by the wavefunction depends upon three quantum numbers n, l, ml.

φθφθψ )r(R,,r

3) These three quantum numbers can only take integral values subject to the following restrictions:

n = 1, 2, 3, … l < n, ml = 0, 1, 2 … l

orbitals with l = 0, 1 and 2 are known as s, p and d orbitals respectively.The radial solutions Rn,l(r) are given by the Associated Laguerre Polynomials

The angular solutions are given by the Spherical Harmonics Yl,ml( ,) where Yl,ml( ,) =Pl,ml().exp(i ml ), and Pl,ml() are the Associated Legendre Polynomials.