5.61 Fall 2007
Lecture #9
page 1
VARIANCE, ROOT-MEAN SQUARE, OPERATORS,
EIGENFUNCTIONS, EIGENVALUES
xi − x
xi − x
≡
Deviation of ith measurement from average value <x >
≡
Average deviation from average value <x >
xi − x
But for particle in a box,
(x − x )
2
(x − x )
Square of deviation of ith measurement from average
≡
i
=0
value <x >
2
≡ σ x2
i
(x − x )
Note
i
the Variance in x
≡
2
= x2 − x
2
= σ x2
The Root Mean Square (rms) or Standard Deviation is then
12
2
σ x = ⎡ x2 − x ⎤
⎢⎣
⎦⎥
The uncertainty in the measurement of x, Δx , is then defined as
Δx = σ x
σ x for particle in a box
σ x2 =
∫
a
0
()
()
∞
() ()
ψ * x x 2ψ x dx − ∫ ψ * x xψ x dx
−∞
⎡⎛ 2 ⎞ a
⎤
⎛ 2⎞ a
⎛ nπ x ⎞
2 ⎛ nπ x ⎞
= ⎜ ⎟ ∫ x 2 sin 2 ⎜
dx
−
x
sin
dx
⎢
⎥
⎜ ⎟ ∫0
⎜⎝ a ⎟⎠
⎝ a⎠ 0
⎝ a ⎟⎠
⎣⎝ a ⎠
⎦
2
5.61 Fall 2007
Lecture #9
page 2
Evaluate integral by parts
⎡ 2
a
a2
⎢
σ =
−
⎢ 3 2 nπ
⎣
2
x
⇒
σ =
2
x
( )
a
( )
2
( )
4 nπ
2
⎡ nπ
⎢
⎢ 3
⎣
2
⎤
− 2⎥
⎥
⎦
( )
⎡
a ⎢ nπ
Δx = σ x =
2 nπ ⎢ 3
⎣
( )
⎤ ⎡ 2⎤
⎥ − ⎢a ⎥
2
⎥ ⎣4⎦
⎦
2
12
⎤
− 2⎥
⎥
⎦
Note that deviation increases with a, and depends weakly on n.
Now suppose we want to test the Heisenberg Uncertainty Principle for the
particle in a box.
We need
12
2
p and p 2 to get Δp = σ p = ⎡ p 2 − p ⎤
⎥⎦
⎢⎣
But do we write
p =
∫
∞
−∞
() ()
ψ * x pψ x dx ?
what do we put in here??
We need the concept of an OPERATOR
()
()
ˆ x =g x
Af
operator acts on function to get a new function
5.61 Fall 2007
e.g.
Lecture #9
⎛ 2x ⎞
d ⎛ x2 ⎞
= ⎜ ⎟
⎜
⎟
dx ⎝ 3 ⎠
⎝ 3⎠
operator function
new function
page 3
5.61 Fall 2007
Lecture #9
page 4
Special Case
()
()
ˆ x =af x
Af
If
number (constant)
()
then f x
is called an eigenfunction of the operator
a
ˆA
and
is the eigenvalue.
This is called an eigenvalue problem (as in linear algebra).
Quantum mechanics is full of operators and eigenvalue problems!!
e.g. Schrödinger’s equation:
⎡ !2 d 2
⎤
+
V
x
⎢−
⎥ψ x
2
⎣ 2m dx
⎦
()
Ĥ operator
(Hamiltonian)
Ĥψ = Eψ
or
()
()
= Eψ x
Eigenfunction
constant
!2 d 2
Ĥ x = −
+V x
2m dx 2
()
with
()
(in 1D)
The Hamiltonian operator, acting on an eigenfunction, gives the energy.
i.e. the Hamiltonian is the energy operator
p2
If V x = 0 , then E = K.E. =
2m
()
∴
( p̂ )
2
means
( pˆ ) ( p̂ )
( p̂ )
Ĥ =
2
2m
⇒
( p̂ )
2
d2
= −!
dx 2
2
i.e. the operator acts sequentially on the function
5.61 Fall 2007
Lecture #9
page 5
( p̂ ) f ( x ) = ( pˆ ) ( pˆ ) f ( x ) = pˆ ⎡⎣ pfˆ ( x )⎤⎦ = pˆ ⎡⎣ g ( x )⎤⎦
2
⎛
d ⎞⎛
d⎞
d2
p̂ p̂ = ⎜ −i! ⎟ ⎜ −i! ⎟ = −!2 2
dx ⎠ ⎝
dx ⎠
⎝
dx
( )( )
⇒
p̂ = −i!
∴
σ
p
d
dx
Momentum operator (in 1D)
for Particle in a Box
σ 2p = p 2 − p
2
2
⎡ ∞
⎤
⎛
⎛
d⎞
d⎞
= ∫ ψ * x ⎜ −i! ⎟ ψ x dx − ⎢ ∫ ψ * x ⎜ −i! ⎟ ψ x dx ⎥
−∞
−∞
dx ⎠
dx ⎠
⎝
⎝
⎣
⎦
∞
()
()
()
()
Note order is now very important! Operator acts only on the function to its
right.
p =
=
∫
a
0
⎛
d⎞
ψ
*
x
−i!
ψ x dx
∫−∞
⎜⎝
dx ⎟⎠
∞
()
()
12
⎡⎛ 2 ⎞ 1 2 ⎛ nπ x ⎞ ⎤ ⎛
⎛ nπ x ⎞ ⎤
d ⎞ ⎡⎛ 2 ⎞
⎢⎜ ⎟ sin ⎜
⎥
⎢
−i!
sin
⎟⎠ ⎜⎝
⎟⎠ ⎜⎝ a ⎟⎠
⎜⎝ a ⎟⎠ ⎥dx
a
a
dx
⎝
⎠
⎝
⎢⎣
⎥⎦
⎢⎣
⎥⎦
=0
p
2
=
2!2
=
a
=
∫
a
0
12
⎡⎛ 2 ⎞ 1 2 ⎛ nπ x ⎞ ⎤ ⎛
⎛ nπ x ⎞ ⎤
d ⎞⎛
d ⎞ ⎡⎛ 2 ⎞
⎥
⎢
⎢⎜ ⎟ sin ⎜
⎟⎠ ⎜⎝ −i! dx ⎟⎠ ⎜⎝ −i! dx ⎟⎠ ⎜⎝ a ⎟⎠ sin ⎜⎝ a ⎟⎠ ⎥dx
a
a
⎝
⎠
⎝
⎢⎣
⎥⎦
⎢⎣
⎥⎦
2
⎛ nπ x ⎞ ⎛ nπ ⎞
⎛ nπ x ⎞
2!2 ⎛ nπ ⎞
∫0 sin ⎜⎝ a ⎟⎠ ⎜⎝ a ⎟⎠ sin ⎜⎝ a ⎟⎠ dx = a ⎜⎝ a ⎟⎠
a
n 2π 2 !2
a2
2
2 ⎛ nπ x ⎞
sin
∫0 ⎜⎝ a ⎟⎠ dx
a
2
5.61 Fall 2007
Note
Lecture #9
p
2
n2 h2
n2 h2
=
= 2m
= 2mE
4a 2
8ma 2
page 6
as expected
E = K.E. since V(x) = 0
n2π 2 !2
σ =
= Δp
a2
( )
2
p
⇒
( )
⎡
a ⎢ nπ
ΔxΔp =
2 nπ ⎢ 3
⎣
( )
2
12
⎤
− 2⎥
⎥
⎦
2
( )
⎡
nπ ! ! ⎢ nπ
=
a
2⎢ 3
⎣
2
12
⎤
− 2⎥
⎥
⎦
always > 1
∴
ΔxΔp ≥
!
2
as expected from Heisenberg Uncertainty Principle