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VARIANCE, ROOT-MEAN SQUARE, OPERATORS,
EIGENFUNCTIONS, EIGENVALUES
≡ Deviation ofi
th x measurement from average value i
− x
x i
− x ≡ Average deviation from average value
But for particle in a box, x i
− x = 0
2
≡ Square of deviation ofi
th measurement from average
value
x i
− x
2
x i
− x ≡ σ
2 ≡ the Variance inx x
Note
(^2 ) 2 2
x i
− x = x − x = σ x
The Root Mean Square (rms) or Standard Deviation is then
σ
2
1 2
2 = x x x ⎢ ⎥
The uncertainty in the measurement ofx, Δx, is then defined as
Δ x = σ x
σ x
for particle in a box
a ∞
σ x
2
=
ψ * x
2
ψ x dx −
( ) x^ ( ) ψ^ *^ ( x^ ) x^ ψ^ ( x dx )
0 −∞
2
⎛ 2 ⎞ a^ ⎛ (^) n π x ⎞ ⎡⎛ 2 ⎞ a^ ⎛ (^) n π x ⎞ ⎤
a (^) ⎠
0
x
2 sin
2
a (^) ⎠
dx −
a (^) ⎠
0
x sin
2
a (^) ⎠
dx
Evaluate integral by parts
2 2
2 ⎤
⇒ σ
2
a
a
)
2
a
x
( n π ⎦
a
2
( n^ π^ )
2 ⎤
σ
2
= − 2
x
( n π )
2
1 2
Δ x = σ = x
2 (
a
n π )
( n π
)
2
Note that deviation increases witha, and depends weakly onn.
Now suppose we want to test the Heisenberg Uncertainty Principle for the
particle in a box.
2 ⎤
1 2
2 2
We need and p to get Δ p = σ =
p p − p p ⎣
∞
But do we write (^) p = ψ * x x dx? ∫ −∞
( ) p^ ψ^ ( )
what do we put in here??
We need the concept of an OPERATOR
Af
x x ( ) = g ( )
operator acts on function to get a new function
ˆ A
a
Special Case
If Af x x
( ) = a f ( )
number (constant)
then f (^) ( x ) is called an eigenfunction of the operator
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 (^) ) x = E ψ x ⎥ ψ (^) ( ) ( )
2 m (^) dx
2
H operator Eigenfunction constant
(Hamiltonian)
or H
ψ = E ψ with H
x
2 d
2
( )
2 m dx
2
The Hamiltonian operator, acting on an eigenfunction, gives the energy.
i.e. the Hamiltonian is the energy operator
2
If V x then ( )
= 0 , E = K.E. =
p
2 m
( )
2
∴ H
p ˆ
( )
(^2) d
2
⇒ p ˆ^ = −!
2
2 m (^) dx
2
means p ( ) ( p ˆ^ ) i.e. the operator acts sequentially on the function
2
( )
ˆ (^) p ˆ
σ
p ˆ^ f x ( )
ˆ (^) p x ˆ ˆ (^) x ⎤
⎦
= p g ( ) p ( ) f
( )
( )
2
( )
= ˆ^
( ) = p pf ˆ^ x
⇒ p ˆ^ p ˆ^
− i!
dx ⎠
− i!
dx ⎠
2
dx
2
( ) ( )
⎛ (^) d ⎞ ⎛ (^) d ⎞ (^) d
2
d
∴ p ˆ^ = − i! Momentum operator (in 1D)
dx
p
for Particle in a Box
2 ∞^ ⎛ (^) d ⎞
2
⎡ ∞ (^) ⎛ d ⎞
ψ ( ) x dx
2
2 σ
2 = p − p = ∫ −∞
ψ * ( )
− i!
dx ⎠
ψ x dx −
∫
ψ * x
dx ⎦
x ( ) −∞
( ) − i! p
Noteorder is now very important! Operator acts only on the function to its
right.
∞ (^) ⎛ d ⎞
p = ∫ −∞
ψ * (^) ( )
x − i!
ψ (^) ( x dx )
dx
a
1 2
⎛ (^) n π x ⎞
⎛ (^) d ⎞
1 2
⎛ (^) n π x ⎞
∫ 0
a ⎠
sin
a ⎠
− i!
dx ⎠
a ⎠
sin
a ⎠
⎥ dx
a
⎛ 2 ⎞ ⎛ (^) n π x ⎞
⎛ (^) d ⎞ ⎛ (^) d ⎞
1 2
⎛ (^) n π x ⎞
1 2
2 p = ⎢ ∫ 0 ⎢⎝
a (^) ⎠
sin
a (^) ⎠
− i!
dx (^) ⎠
− i!
dx (^) ⎠
a (^) ⎠
sin
a (^) ⎠
⎥ dx
2 a
2 2
a
⎛ (^) n π
a
x
⎛ (^) n
a
π ⎞^ ⎛^ n π
a
x ⎞^2
a
2
⎛ (^) n
a
π
∫ 0
a
sin
2
⎛ (^) n π
a
x ⎞
∫ 0
sin
sin
dx =
dx
n
2
π
2
!
2
2
a
