quantum mechanics for scientists and engineers...for this simple model differentiation shows that...

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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


2 2

22 4total kinetic potentialo o o

eE E Ea a

211

2 0.529 5.4 29 10 moo Åa

ex


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


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

quantum mechanics for scientists and engineers...for this simple model differentiation shows that...

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