Physical Chemistry - Solutions to The Schrodinger Equation & Quantum Mechanical Particle in a Box_lecture7-8

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
2
()
−
2m ∂x 2
+ V x ψ x = Eψ x () () ()
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 () be continuous function.


(I) Free particle V( x ) = 0


!2 ∂ ψ x
2
()
−
2m ∂x 2
= Eψ x ()
2mE !2 k 2
Define k = 2
2
or E =
! 2m

p2
()
V x = 0, E =
2m
⇒ p 2 = !2 k 2 ⇒ p = !k


h 2π
de Broglie p= ⇒ k=
λ λ

, 5.61 Fall 2007 Lecture #7 page 2



∂ 2ψ x ( ) = −k ψ
The wave eq. becomes
∂x 2
2
( x)
with solutions () ( )
ψ x = Acos kx + B sin kx ( )
Free particle ⇒ 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 =∞ ( x < 0, x > a )
V ( x) = 0 (0 ≤ x ≤ a ) V( x )
0
0 x a



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
2
()
−
2m ∂x 2
= Eψ x ()
∂ 2ψ x ( ) = −k ψ
∂x 2
2
( x) with same definition

2mE !2 k 2
k2 = or E =
!2 2m