Download Quantum Mechanics Lecture 2 Dr. Mauro Ferreira

Quantum Mechanics
Lecture 2
Dr. Mauro Ferreira
E-mail: [email protected]
Room 2.49, Lloyd Institute
Postulates of Quantum Mechanics
What’s th
e differenc
e
to classica
P1:
l mechanic
s
?
The state of a dynamic system is completely
specified by a state (wave) function
... whose physical meaning is of a probability density, i.e.,
!
|ψ(x)|2 dx
Probability of finding the particle
in the range between x and x+dx
Mean values and
deviations are direct
consequences of the
probabilistic nature of
the state function
{
!x" =
!
−∞
∞
!
dx ψ ∗ (x) ψ(x) = 1
square-integrable functions
dx ψ ∗ (x) x ψ(x) =
−∞
(∆x)2 =
∞
∞
−∞
!
∞
−∞
dx ψ ∗ (x) (x − "x#)2 ψ(x)
dx x |ψ(x)|2
What does it mean to have a probabilistic description of
the position x of a particle ?
{
!x" =
!
∞
dx ψ ∗ (x) x ψ(x) =
−∞
(∆x)2 =
!
∞
−∞
!
∞
−∞
dx ψ ∗ (x) (x − "x#)2 ψ(x)
dx x |ψ(x)|2
(Mean)
(Mean square deviation)
(∆x)2 =
(∆x)2 =
!
!
∞
−∞
∞
−∞
dx ψ ∗ (x) (x − "x#)2 ψ(x)
dx ψ ∗ (x) [x2 + !x"2 − 2x!x"] ψ(x)
(∆x)2 = !x2 " − !x"2
e
h
t
t
u
o ?
b
a
at tum
h
W men
mo
(Mean square deviation)
!
∆x = !x2 " − !x"2
(Uncertainty in the position)
Following the same formalism, the momentum is given by
!p" =
!
∞
dx ψ ∗ (x) p ψ(x)
−∞
To solve the integral above, we should be able to express the
momentum p in terms of the position x.
However, unlike classical mechanics, p and x are not simultaneously
known, which requires an alternative procedure. Let’s start with a
wave that describes the motion of a free particle:
ψ(x, t) = A ei(kx−ωt)
∂ψ(x, t)
= ikψ(x, t)
∂x
! ∂ψ(x, t)
p ψ(x, t) =
i
∂x
Bearing in mind that p = !k
!
!p" =
i
!
∞
∂ ψ(x)
dx ψ (x)
(Mean)
∂x
−∞
∗
(∆p)2 = !p2 " − !p"2
∆p =
where
!
(Mean square deviation)
!p2 " − !p"2 (Uncertainty in the momentum)
!p2 " = −!2
!
∞
2
∂
ψ(x)
∗
dx ψ (x)
∂x2
−∞
Example:
A particle is described by the following wave function ψ(x) defined in the
region -2<x<2 (see graph below). Calculate the average and the mean
square deviations for both the position and the momentum of the particle.
πx
ψ(x) = A cos( )
4
2.5
2
1.5
1
0.5
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
2
Example:
Normalization condition:
A
2
!
x
+
2
A=
!
πx "+2
sin( 2 )
π
1
2
−2
A
2
!
+2
−2
=1
πx
dx cos ( ) = 1
4
2
Example:
Average position:
Integrand is odd
!x" = A
2
!
+2
πx
dx x cos ( )
4
2
−2
⇒ "x# = 0
The same conclusion could have been obtained from the graph
ψ(x) = A cos(
2.5
πx
)
4
2
1.5
1
0.5
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
x
2
Example:
!
!p" =
i
!
∞
∂ ψ(x)
dx ψ (x)
∂x
−∞
πx
ψ(x) = A cos( )
4
{
∗
∂ψ(x)
Aπ
πx
=−
sin( )
∂x
4
4
−A ! π
!p" =
4i
2
!
Integrand is odd
2
πx
πx
dx cos(
) sin(
)
4
4
−2
⇒ "p# = 0
Example:
!x2 " =
!
!x " = A
2
dx ψ ∗ (x) x2 ψ(x)
−∞
2
!x " = 2 A
2
∞
!
+2
−2
2
πx
dx x cos ( )
4
2
!
+2
0
4
8
!x " = − 2
3 π
2
Integrand is even
?
0
≠
>
2
x
<
d
πx
dx x cos ( )
4
2
2
n
a
0
2
n
a
C
=
>
<x
Example:
!p2 " = −!2
!
∞
2
∂
ψ(x)
∗
dx ψ (x)
∂x2
−∞
∂ 2 ψ(x)
−A π 2
πx
−π 2
=
cos(
)=
ψ(x)
2
∂x
16
4
16
! π
!p " =
16
2
2
2
2 2
!
π
2
!p " =
16
!
2
−2
dx ψ ∗ (x) ψ(x)
Normalization
Example:
!x" = 0
!p" = 0
4
8
2
!x " = − 2
3 π
2 2
!
π
2
!p " =
16
It is instructive to calculate the product Δx.Δp :
∆x ∆p ≈ 0.568 !
... which obviously satisfies the uncertainty principle.
!
∆x ∆p ≥
2