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

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MIT Courses

This course presents an introduction to quantum mechanics. It begins with an examination of the historical development of quantum theory, properties of particles and waves, wave mechanics and applications to simple systems — the particle in a box, the harmonic oscillator, the rigid rotor and the ...

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Uploaded on April 25, 2023
Number of pages 9
Written in 2007/2008
Type Class notes
Professor(s) Prof. robert guy griffin
Contains All classes

solutions to the schrodinger equation

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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

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