5.61 Fall 2007
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.
()
!2 ∂ ψ x
−
+ V x ψ x = Eψ x
2m ∂x 2
2
() ()
()
()
()
We can find two independent solutions φ1 x and φ2 x
The general solution is a linear combination
()
()
()
ψ x = Aφ1 x + Bφ2 x
()
()
A and B are then determined by boundary conditions on ψ x and ψ ′ x .
()
Additionally, for physically reasonable solutions we require that ψ x and
()
ψ′ x
(I)
be continuous function.
V( x ) = 0
Free particle
()
!2 ∂ ψ x
−
= Eψ x
2m ∂x 2
2
Define
2mE
k = 2
!
2
!2 k 2
or E =
2m
p2
V x = 0, E =
2m
()
de Broglie
p=
h
λ
⇒
()
k=
⇒
2π
λ
p 2 = !2 k 2
⇒
p = !k
5.61 Fall 2007
Lecture #7
∂x
2
2
()
( x)
( )
( )
ψ x = Acos kx + B sin kx
with solutions
⇒
( ) = −k ψ
∂ 2ψ x
The wave eq. becomes
Free particle
page 2
⇒
no boundary conditions
!2 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 = ∞
(
)
ψ x < 0, x > a = 0
⇒
For 0 ≤ x ≤ a , Schrödinger equation is like that for free particle.
()
!2 ∂ ψ x
−
= Eψ x
2m ∂x 2
2
( ) = −k ψ
∂ 2ψ x
∂x
2
2
( x)
()
with same definition
k2 =
2mE
!2
or E =
!2 k 2
2m
5.61 Fall 2007
Lecture #7
()
page 3
( )
( )
ψ x = Acos kx + B sin kx
again with solutions
But this time there are boundary conditions!
()
Continuity of ψ x
(i)
()
()
()
ψ 0 = ψ a = 0
⇒
()
()
ψ 0 = Acos 0 + B sin 0 = 0
()
⇒
A=0
( )
ψ a = B sin ka = 0
(ii)
Can’t take B = 0 (no particle anywhere!)
Must have
⇒
( )
sin ka = 0
⇒
ka = nπ
n = 1, 2, 3,...
k is not continuous but takes on discrete values k =
Thus integer evolves naturally !!
nπ
a
So solutions to the Schrödinger equation are
⎛ nπ x ⎞
ψ 0 ≤ x ≤ a = B sin ⎜
⎝ a ⎟⎠
(
)
n = 1,2,3,...
These solutions describe different stable (time-independent or
“stationary”) states with energies
!2 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 2007
(a)
Lecture #7
page 4
The energy spacing between successive states gets progressively
larger as n increases
!
!
E5 = 25E1
2
2
2⎤ h
⎡
En+1 − En = n + 1 − n
⎢⎣
⎦⎥ 8ma 2
(
E4 = 16E1
)
(
)
En+1 − En = 2n + 1 E1
E3 = 9E1
E2 = 4E1
E1 = h2 8ma 2
(b)
()
The wavefunction ψ x is sinusoidal, with the number of nodes
increased by one for each successive state
) (3 nodes)
(
) ( 2 nodes)
ψ 4 = B sin 4π x a
λ3 = 2a 3
ψ 3 = B sin 3π x a
λn = 2a n
# nodes = n - 1
aλ=
2
λ1 = 2a
(c)
(
λ4 = a 2
(
) (1 node)
= B sin (π x a )
(0 nodes)
ψ 2 = B sin 2π x a
ψ1
0
a
x
The energy spacings increase as the box size decreases.
E∝
1
a2
5.61 Fall 2007
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 !