Download One dimensional quantum mechanics: the free particle

One Dimensional Quantum
Mechanics: The Free Particle
LUKE CORCOS
WITH JACKY CHONG
DIRECTED READING PROGRAM
Books
Lectures on Quantum
Mechanics for
Mathematics Students
by Faddev and Yakubovskii
Modern Quantum Mechanics
by Sakurai
Introduction
Quantum mechanical behavior is described by
the Schrödinger equation:
d (t ) ˆ
 H (t )
Time Dependent: ih
dt
Time Independent:
Hˆ  (t )  E (t )
E=Energy; h=Plancks constant
Introduction
ˆ2
P
 V (Qˆ )
Hamiltonian: Hˆ 
2m
Coordinate Representation
Q
Qˆ  ( x)  x ( x)
P
h 
ˆ
P ( x) 
 ( x)
i x
Q= Position Operator
P= Momentum Operator
Momentum Representation

ˆ
Q ( p)  ih  ( p)
p
Pˆ  ( p )  p ( p )
Free Particle (V=0)
When V=0:
ˆ2
P
Hˆ 
2m
Let’s first look at solution of the Time-Independent
Schrödinger Equation and solve for the spectrum of
the Hamiltonian Operator
2
ˆ
P
ˆ
H  E 
  E
2m
Spectrum of Free Particle
h 
ˆ
In the coordinate representation: P ( x) 
 ( x)
i x
Pˆ 2
h 2 d 2
So
  E  
 E
2
2m
2m dx
For simplicity, define units such that
h=1 and m=1/2
Define the wave number k>0 s.t. k 2  E
Spectrum of Free Particle
So  ' ' k 2  0
which has solutions:
  k ( x)  Ce ikx

Now Normalize:
*

 k ' ( x) k ( x)dx  C


2
ikx ik ' x
e
 e dx

Using the property of Fourier Transforms

C
2
e

ikx ik ' x
e
1
dx  2 C  (k  k ' )  C 
2
2
Free Particle Dynamics
To understand how the Free Particle develops with
time, we now shift to the Time-Dependent
Schrödinger Equation
d (t ) ˆ
ih
 H (t )
dt
For simplicity, let’s start with the momentum
representation:
Pˆ  ( p )  p ( p )
Free Particle Dynamics
d ( p, t )
2
i
 p  ( p, t )
dt
Where
 ( p,0)   ( p)

and
  ( p)

The Solution:  ( p, t )  Ce
ip2t
C   ( p,0)   ( p)
 ( p, t )   ( p)e
ip2t
2
dp  1
Free Particle Dynamics
Now we put the solution into the coordinate
representation using the Inverse Fourier Transform
 ( x, t )  F [ ( p)e
1
F [ ( p)]   ( x)
1
ip2t
]  F [ ( p)]  F [e
1
1
F [e
1
ip 2t
ip2t
1
]
e
2it
x2
i
4t
]
The General Solution
1
 ( x, t ) 
2it


(
y
)
e



x  y 2

4it
dy
Approximation of Long Term Dynamics
 ( x, t )  F [ ( p)e
1
1
 ( x, t ) 
2

  ( p )e
ip2t
]
i ( px p 2t )
dp

1
  ( x, t ) 
2

  ( p )e
it (
px 2
p )
t
dp

Let’s now use the stationary phase approximation
to look at the dynamics as t  
Stationary Phase Approximation
This method approximates integrals of the form
b
I ( N )   g ( x)eiNf ( x ) dx
N 
a
This has the solution
1
2
 2

i
1
~
~
~


I (N )  
g ( x ) exp[ iNf ( x )  sgn f ' ' ( x )]  O 
~

4
N
 N f ''(x ) 
~
Where x is the point where f ' ( x)  0
Approximation of Long Term Dynamics
1
 ( x, t ) 
2
N t

  ( p )e
it (
px 2
p )
t
dp

px
f ( p) 
 p2
t
1
x
 ( x, t ) 
 ( )e
2t
2t
x2 
i(  )
4t 4
x
~
p
2t
 1  as t  
 O 
t 
Consequences
1
x
 ( x, t ) 
 ( )e
2t
2t
x2 
i(  )
4t 4
1
 O 
t 
x
1. The point of stationary phase occurs at p 
2t
So recalling that m=1/2, the classical
momentum equation
p  mv is realized.
1
C

(
x
,
t
)

O
(
)
2.
so  ( x, t ) 
t
t
2
This implies  ( x, t )  0 as t  
Source: https://en.wikipedia.org/wiki/Wave_packet#Free_propagator
Thank you for your attention