5.61 Fall 2004
Lecture #7
page 1
SOLUTIONS TO THE SCHRÖDINGER EQUATION
Free particle and the particle in a box
Schrödinger equation is a 2nd-order diff. eq.
()
h2 x
+ V x x = E x
2m x 2
2
() ()
()
()
()
We can find two independent solutions 1 x and 2 x
The general solution is a linear combination
()
()
()
x = A1 x + B 2 x
()
()
A and B are then determined by boundary conditions on x and x .
()
Additionally, for physically reasonable solutions we require that x and
()
x
(I)
be continuous function.
V( x ) = 0
Free particle
()
h2 x
= E x
2m x 2
2
Define
2mE
k = 2
h
2
h2 k 2
or E =
2m
p2
V x = 0, E =
2m
()
de Broglie
p=
h
()
k=
2
p 2 = h2 k 2
p = hk
5.61 Fall 2004
Lecture #7
x
2
2
()
(x)
( )
( )
x = Acos kx + B sin kx
with solutions
( ) = k 2 x
The wave eq. becomes
Free particle
page 2
no boundary conditions
h2 k 2
any A and B values are possible, any E =
possible
2m
So any wavelike solution (traveling wave or standing wave) with any wavelength,
wavevector, momentum, and energy is possible.
(II)
Particle in a box
()
V (x) =
0
( x < 0, x > a )
(0 x a )
V x =
V( x )
0
0
a
x
()
Particle can’t be anywhere with V x = (
)
x < 0, x > a = 0
For 0 x a , Schrödinger equation is like that for free particle.
()
h2 x
= E x
2m x
2
2
( ) = k 2 x
x
2
2
(x)
()
with same definition
k2 =
2mE
h2
or E =
h2 k 2
2m
5.61 Fall 2004
Lecture #7
()
page 3
( )
( )
x = Acos kx + B sin kx
again with solutions
But this time there are boundary conditions!
()
Continuity of x
(i)
()
()
()
0 = a = 0
()
()
0 = Acos 0 + B sin 0 = 0
()
A=0
( )
a = B sin ka = 0
(ii)
Can’t take B = 0 (no particle anywhere!)
Must have
( )
sin ka = 0
ka = n
n = 1, 2, 3,...
k is not continuous but takes on discrete values k =
n
a
Thus integer evolves naturally !!
So solutions to the Schrödinger equation are
n x 0 x a = B sin a (
)
n = 1, 2,3,...
These solutions describe different stable (time-independent or
“stationary”) states with energies
h2 k 2
E=
2m
n2 h2
En =
8ma 2
Energy is quantized!! And the states are labeled by a quantum number n
which is an integer.
Properties of the stationary states
5.61 Fall 2004
(a)
Lecture #7
page 4
The energy spacing between successive states gets progressively
larger as n increases
25EE=
51
2
2
2% h
"
En+1 ! En = n + 1 ! n
#$
&' 8ma2
(
16EE=
41
)
(
)
En+1 ! En = 2n + 1 E1
9EE=
31
4EE=
2218Ehma=
21
(b)
()
The wavefunction ! x is sinusoidal, with the number of nodes
increased by one for each successive state
) (3 nodes)
(
) ( 2 nodes)
! 4 = B sin 4" x a
!3 = 2a 3
! 3 = B sin 3" x a
!n = 2a n
# nodes = n - 1
a!=
2
!1 = 2a
(c)
(
!4 = a 2
(
) (1 node )
= B sin (" x a )
(0 nodes)
! 2 = B sin 2" x a
!1
0
a
x
The energy spacings increase as the box size decreases.
E!
1
a2
5.61 Fall 2004
Lecture #7
page 5
We’ve solved some simple quantum mechanics problems! The P-I-B model is a
good approximation for some important cases, e.g. pi-bonding electrons on
aromatics.
Electronic transitions shift to lower energies as molecular size increases !