Download chapter 7 part 1

Chapter 7: Quantum Mechanics in Three
Dimensions / Theory of Hydrogen Atom/
Quantum-classical hybrid” interpretation of
quantum numbers
time independent, we used to have
d 2 ( x ) 2m
 2 {E  U ( x )} ( x )  0
2
dx

if wave function is 3 dimensional
 2 ( x, y, z )  2 ( x, y, z )  2 ( x, y, z ) 2m


 2 {E  U ( x, y, z )} ( x, y, z )  0
dx 2
dy 2
dz 2

2m
  ( x, y, z )  2 {E  U ( x, y, z )} ( x, y, z )  0

2
possible to consider particle in 3D box with infinite
high (and thick) potential walls – impenetrable – just a
model that does not really correspond to reality at wave function
(evanescent wave) always leaks into barrier – chapter 6
tunneling- and may be picked up at the other side, become a
traveling wave again, i.e. a particle that can move
so U(x,y,z) = U(r) = 0 inside the box, 0 < x,y,z < L
then 3D wave function is separable into product of three 1D
wave functions
1
 ( r )   1 ( x ) 2 ( y ) 3 ( z )
putting back into rearranged time independent Schrödinger
equation
d 2 1 2 3
d 2 1 2 3
d 2 1 2 3
2
2



E
2m 1 2 3
dx 2
2m 1 2 3
dy 2
2m 1 2 3
dz 2
2
 2 d 2 1 ( x )
 2 d 2 2 ( y )
 2 d 2 3 ( z )



E
2
2
2
2m 1 dx
2m 2 dy
2m 3 dz
for a infinite square well we could write (with U(x,y,z) = 0
inside the box)
 2 d 2 1 ( x )

 E1
2
2m 1 dx
(so all the things we had
for 1D are now here three times)
 2 d 2 2 ( y )
 2 d 2 3 ( z )

 E 2 and 
 E3
2m 2 dy 2
2m 3 dz 2
so E1 + E2 + E3 = E
solutions for this differential equation in x are: A sin (k1x) and
B cos (k1x) and linear combinations of the two,
2
where k1=
2mE1
2
only sin function can vanish at wall x = 0
as wave function has to vanish at x = L also we have
sin (k1L) = sin (n1 π)
where n1 is a positive integer
(quantum number)
an integral number of half-wavelength must fit into the box
same applies to y and z, so
 ( r )   1 ( x ) 2 ( y ) 3 ( z ) 
A sin( k1 x ) sin( k 2 y ) sin( k 3 z )
let’s see what we get if we differentiate
 2 d 2 ( x, y, z )
 2 d 2 ( x, y, z )
 2 d 2 ( x, y, z )



E
2
2
2
2m
dx
2m
dy
2m
dz
remember the rules
d sin( Ax 2 )
 2 Ax cos( Ax 2 ) and
dx
d cos( Ax 2 )
 2 Ax sin( Ax 2 )
dx
we get:
3
2
2
2
2
 {
k1 A sin( k1 x ) sin( k 2 y ) sin( k3 z ) 
k2 A sin( k1 x ) sin( k 2 y ) sin( k3 z )
2m
2m
2
2

k3 A sin( k1 x ) sin( k 2 y ) sin( k3 z )}  E
2m
2

2
2
2
E
(k1  k 2  k 3 )
2m
which is equivalent to
1
 22 2
2
2
2
2
2
(
p

p

p
)

(
n

n

n
y
z
1
2
3 ) = KE
2
E = 2m x
2mL
with relation between wave number k1, wavelength and
momentum follows

px = k1  n1 L
analogous relations for y and z


k

n
2
2
py =
L


k

n
3
3
pz =
L
where ni are three quantum numbers
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which we can summarize as
2 2
1


2
2
2
2
2
2
E
( p x  p y  pz ) 
(
n

n

n
1
2
3 )  KE
2
2m
2mL
ni ≠ 0
so 1, 2, 3, ….
ni = 0 would mean the wave function would vanish everywhere
2
and   0 , means there is no particle in the 3D box
so our energy states are a function of three
quantum numbers
E n1n2n3 
 22
2
2mL
( n1  n2  n3 )
2
2
2
two cases to be distinguished either L is the same for all
three dimensions or its different for in each direction x,y,z
E211 
 22
2mL2
( 2 2  12  12 )  E121  E112
three states that have the same energy – called
degenerate (strong language what is really meant
it’s quite unusual in nature, it’s due to the so called
5
symmetry L1 = L2 = L3, it’s usually due to the
symmetry of the potential (which was in our
special case 0 in the box everywhere)
will n1 = n2 = n3 = 1 be a degenerate energy level?
no it is the ground state
in idealized hydrogen atom with (3 quantum numbers) all
energy levels are degenerate, in real hydrogen atom with 4
quantum numbers, there are tiny differences in energy
between the levels lifting this degeneracy
usually in a man made quantum system, quantum
box: L1 ≠ L2 ≠ L3, so no degeneracy here, models and
idealizations may be degenerate more often than the real things
for probability to find particle in an infinitesimal
small volume element (as a coefficient  1 or in %),
P(x,y,z) dv =(x,y,x,t)2 dv
assuming that wave function has been normalized
P   ( x, y, z, t ) 2 dxdydz  1 (by the usual procedure),
……. just kind of everything is there three times
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