Post on 21-Mar-2020
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Quantum Mechanics for Scientists and Engineers
David Miller
The hydrogen atom
The hydrogen atom
Multiple particle wavefunctions
Multiple particle systems
How should we tackle this problem of two particles, electron and proton?
We start by generalizing the Schrödinger equationwriting generally for time-independent problems
where now we mean that the Hamiltonian is the operator representing the energy of the entire system
and is the wavefunction representing the state of the entire system
H E
H
Multiple particle wavefunctions
For the hydrogen atomthere are two particles
the electron and the proton Each of these has a set of coordinates
associated with it xe, ye, and ze for the electron and xp, yp, and zp for the proton
The wavefunction will therefore in general be a function of all six of these coordinates
The hydrogen atom
Solving the hydrogen atom problem
Hamiltonian for the hydrogen atom
The electron and proton each have a massme and mp respectively
We expect kinetic energy operators
associated with each of these massespotential energy
from the electrostatic attraction of electron and proton
Hamiltonian for the hydrogen atom
Hence, the Hamiltonian becomes
where we mean
and similarly for and
is the position vector of the electron coordinatesand similarly for
2 2
2 2ˆ2 2e p e p
e p
H Vm m
r r
2 2 22
2 2 2ee e ex y z
2p
e e e ex y z r i j k
pr
Hamiltonian for the hydrogen atom
The Coulomb potential energy depends on the distance
between the electron and proton coordinateswhich is important in simplifying the solution
The Schrödinger equation can now be written explicitly
2
4e po e p
eV
r rr r
e hr r
2 22 2 , , , , ,
2 2
, , , , ,
e p e p e e e p p pe p
e e e p p p
V x y z x y zm m
E x y z x y z
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 coordinates in which three are the relative positions
i.e., a relative position vectorfrom which we obtain
What should we choose for the other three coordinates?
e pr r
e px x x e py y y e pz z z x y z r i j k
2 2 2e pr x y z r r
Center of mass coordinates
The position R of the center of mass of two masses is the same as
the balance point of a light-weight beam with the two masses at opposite ends
and so is the weighted average of the positions
of the two individual masses
where M is the total mass
e e p pm mM
r r
R
e pM m m
Center of mass coordinates
Now we construct the differential operators we need
in terms of these coordinatesWith
then for the new coordinates in the x directionwe have
and similarly for the y and z directions
X Y Z R i j k
e e p pm x m xX
M
e px x x
Center of mass coordinates
Using the standard method of changing partial derivatives to new coordinates
and fully notating the variables held constantthe first derivatives in the x direction become
and similarlyp p p
e
x X x Xe e ex x x
mX xx x X x x M X x
e e e
p
x X x Xp p px x x
mX xx x X x x M X x
Center of mass coordinates
The second derivatives become
and similarly
2
2
p p pe e ex x x
x x x
22 2 2
2 2 2
e
p p
X x x Xp x Xx
m mx M X x M x X X x
2 2 2
2 2e e
X x x Xx X
m mM X x M x X X x
p p
e
x Xe ex x
mM x X x x
Center of mass coordinates
Sodropping the explicit statement of variables held constant
where is the so-called reduced mass
2 2 2 2
2 2 2 2 2
1 1 1 1e h
e e p p e p
m mm x m x M X m m x
e p
e p
m mm m
2 2
2 2
1 1M X x
Center of mass coordinates
The same kinds of relations can be written for each of the other Cartesian directions
so if we define
and
we can write the Hamiltonian in a new form with center of mass coordinates
which now allows us to separate the problem
2 2 22
2 2 2X Y Z
R
2 2 22
2 2 2x y z
r
2 2
2 2ˆ2 2
H VM
R r r
Center of mass coordinates
To separate the six-dimensional differential equationusing these coordinates
next, presume the wavefunction can be written
Substituting this form in the Schrödinger equation with
the Hamiltonian we obtain
, S U R r R r
2 2
2 2ˆ2 2
H VM
R r r
2 2
2 2
2 2U S S V U ES U
M
R rr R R r r R r
Center of mass coordinates
With
then dividing by and moving some terms
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 ECoM
2 2
2 2
2 2U S S V U ES U
M
R rr R R r r R r
S UR r
2 2
2 21 12 2
S E V US M U
R rR r r
R r
CoME
Relative motion
Center of mass motion
Center of mass coordinates
Hence we have two separated equations
where
We can now solve these separately
2
2
2 CoMS E SM
R R R
2
2
2 HV U E U
r r r r
H CoME E E
Center of mass motion
is the Schrödinger equation for a free particle of mass Mwith wavefunction solutions
and eigenenergies
This is the motion of the entire hydrogen atomas a particle of mass M
2
2
2 CoMS E SM
R R R
expS i R K R
2 2
2CoMKEM
Relative motion equation
The other equation
corresponds to the “internal” relative motion of the electron and proton
and will give us the internal states i.e., the orbitals and energies
of the hydrogen atom
2
2
2 HV U E U
r r r r
The hydrogen atom
Informal solution for the relative motion
Bohr radius and Rydberg energy
We presume that the hydrogen atom will have some characteristic size
which is called the Bohr radius aoWe expect that the “average” potential energy
strictly, its expectation value will therefore be
2
4potentialo o
eEa
Bohr radius and Rydberg energy
For a reasonable smooth wavefunction of size ~ aothe second spatial derivative will be
Note this is only meant to a rough estimateonly within some moderate factor
r
aoao
0
0~
slope
oa
0
~
slope
oa
0 / 0 /~
2o o
o
a aa
20 / oa
Bohr radius and Rydberg energy
Remembering that for a mass the kinetic energy operator is
The “average” kinetic energy will therefore be
Now, in the spirit of a “variational” calculationwe adjust the parameter ao to get the lowest value of the total energy
Such variational approaches can be justified rigorously as approximations for the lowest energy
2 2/ 2
2
22kinetico
Ea
Bohr radius and Rydberg energy
With our very simple model, the total energy is
The total energy is a balance between the potential energy
which is made lower (more negative) by choosing ao smaller
and the kinetic energy which is made lower (less positive) by making ao
larger
2 2
22 4total kinetic potentialo o o
eE E Ea a
For this simple model
differentiation shows that the choice of ao that minimizes the energy overall is
which is the standard definition of the Bohr radius We therefore see that the hydrogen atom
is approximately 1 Å in diameter
Bohr radius and Rydberg energy
2 2
22 4total kinetic potentialo o o
eE E Ea a
211
2 0.529 5.4 29 10 moo Åa
ex
Bohr radius and Rydberg energy
With this choice of aothe corresponding total energy of the state is
We can usefully define the “Rydberg” energy unit
in which case
22 2
22 2 4totalo o
eEa
22 2
2 13.62 2 4
eVo o
eRya
totalE Ry
Bohr radius and Rydberg energy
Though we have produced
the Bohr radius
and the Rydberg
by informal argumentsthey will turn out to be rigorously meaningful
The energy of the lowest hydrogen atom state is -Ry
211
2 0.529 5.4 29 10 moo Åa
ex
22 2
2 13.62 2 4
eVo o
eRya