solving the hydrogen atom problem

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7.3 The hydrogen atom

Slides: Video 7.3.3 Solving the

hydrogen atom problemText reference: Quantum Mechanics

for Scientists and EngineersSections 10.2 – 10.3 (up to “Bohr

radius and Rydberg energy”)



The hydrogen atom

Solving the hydrogen atom pr

Quantum mechanics for scientists and engineers Da



Hamiltonian for the hydrogen atom

The electron and proton each have a mass

me and m p respectively

We expect

kinetic energy operators

associated with each of these massespotential energy

from the electrostatic attraction ofelectron and proton




Hence, the Hamiltonian becomes

where we mean

and similarly forand

is the position vector of the electron co

and similarly for

2 22 2ˆ

2 2e p e p

e p H V m m r r

2 2 22

2 2 2e

e e e x y z

2

p

e e e e x y z r i j k

pr




The Coulomb potential energy

depends on the distance

between the electron and proton coo

which is important in simplifying the solu

The Schrödinger equation can now be written ex

4

e pV

r r

e hr r

2 2

2 2 , , , , ,2 2

, , ,

e p e p e e e p p p

e p

e e e

V x y z x y zm m

E x y z x

r r



Center of mass coordinates

The potential here is only a function of

the separation of the electron and proton

We could choose a new set of six coordinatin which three are the relative positions

i.e., a relative position vector

from which we obtain

What should we choose for the other threecoordinates?

e r r

e p x x x e p y y y e p z z z x y r i j

2 2 2

e pr x y z r r




The position R of the center of mass of twomasses is the same as

the balance point of a light-weight beamwith the two masses at opposite ends

and so isthe weighted average of the positio

of the two individual masses

where M is the total mass

e e p pm m

r r

R

e m




Now we construct the differential operatorsneed

in terms of these coordinatesWith

then for the new coordinates in the x dire

we have

and similarly for the y and z directio

X Y Z

R i j k

e e p pm x m x

X

e p x x x




So

dropping the explicit statement of variables he

constant

where is the so-called reduced mass

2 2 2

2 2 2 2

1 1 1 1e h

e e p p e p

m m

m x m x M X m m

m

2 2

2 2

1 1

X x




The same kinds of relations can be written for eaother Cartesian directions

so if we define

and

we can write the Hamiltonian in a new fo

with center of mass coordinates

which now allows us to separate the

2 2 22

2 2 2

X Y Z

R

2 2 2

2 2

x y

r

2 2

2 2ˆ2 2

H V M

R r r




To separate the six-dimensional differential equa

using these coordinates

next, presume the wavefunction can be writ

Substituting this form in the Schrödinger equatio

the Hamiltonian

we obtain

, S U R r R r

2 2

2 2ˆ

2 2

H V

M

R r r

2 22 2

2 2U S S V U ES M

R rr R R r r




With

then dividing by and moving some

The left hand side depends only on Rand the right hand side depends only on r

so both must equal a “separation” constantwhich we call E CoM

2 2

2 2

2 2

U S S V U

M

R rr R R r r

S U R r

2 22 21 1

2 2S E V U

S M U

R rR r R r



f



Center of mass motion

is the Schrödinger equation for a free particle of

with wavefunction solutions

and eigenenergies

This is the motion of the entire hydrogen atomas a particle of mass M

2

2

2 CoM

S E S R R R

expS i R K R

2 2

2CoM K E

l i i i



Relative motion equation

The other equation

corresponds to the “internal” relativemotion of the electron and proton

and will give us the internal statesi.e., the orbitals and energies

of the hydrogen atom

2

2

2 H V U E U

r r r r

solving the hydrogen atom problem

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