5.61 Fall 2007
Lecture #10
page 1
THE POSTULATES OF QUANTUM MECHANICS
(time-independent)
Postulate 1:
The state of a system is completely described by a
( )
wavefunction ψ r,t .
Postulate 2:
All measurable quantities (observables) are described by
Hermitian linear operators.
Postulate 3:
The only values that are obtained in a measurement of an
observable “A” are the eigenvalues “an” of the corresponding operator “ Â ”. The
measurement changes the state of the system to the eigenfunction of  with
eigenvalue an.
Postulate 4:
If a system is described by a normalized wavefunction
ψ,
then the average value of an observable corresponding to  is
a = ∫ ψ *Âψ dτ
Implications and elaborations on Postulates
#1]
(a)
The physically relevant quantity is ψ
( ) ( )
( )
ψ * r,t ψ r,t = ψ r,t
2
≡
2
probability density at time t
and position r
(b)
( )
ψ r,t must be normalized
∫ ψ *ψ dτ = 1
(c)
( )
ψ r,t must be well behaved
5.61 Fall 2007
#2]
(a)
Lecture #10
(i)
(ii)
Single valued
ψ and ψ ′ continuous
(iii)
Finite
Example:
() ()
Particle in a box eigenfunctions of Ĥ
()
Ĥ x ψ n x = Enψ n x
But if
page 2
ψn
⎛ 2⎞
x =⎜ ⎟
⎝ a⎠
()
12
⎛ nπ x ⎞
sin ⎜
⎝ a ⎟⎠
ψ is not an eigenfunction of the operator, then the statement is not
true.
e.g.
()
ψ n x above with momentum operator
12
⎛ nπ x ⎞ ⎤
d
d ⎡⎛ 2 ⎞
⎥
p̂nψ n x = −i! ψ n x = −i! ⎢⎜ ⎟ sin ⎜
dx
dx ⎢⎝ a ⎠
⎝ a ⎟⎠ ⎥
⎣
⎦
⎡⎛ 2 ⎞ 1 2 ⎛ nπ x ⎞ ⎤
≠ pn ⎢⎜ ⎟ sin ⎜
⎟⎠ ⎥
a
a
⎝
⎠
⎝
⎢⎣
⎥⎦
()
(b)
In order to create a Q.M. operator from a classical observable, use
x̂ = x and
e.g.
()
p̂x = −i!
d
and replace in classical expression.
dx
1 2
1
!2 d 2
K.E. =
p̂ =
p̂ p̂ = −
(1D)
2m
2m
2m dx 2
!2 ⎛ ∂2
∂2
∂2 ⎞
=−
+
+
(3D)
2m ⎜⎝ ∂x 2 ∂y 2 ∂z 2 ⎟⎠
( )( )
Another 3D example: Angular momentum
L=r×p
5.61 Fall 2007
Lecture #10
page 3
⎛ d
d⎞
lx = ypz − zp y = −i! ⎜ y − z ⎟
dy ⎠
⎝ dz
⎛ d
d⎞
l y = zpx − xpz = −i! ⎜ z
−x ⎟
dz ⎠
⎝ dx
⎛ d
d⎞
lz = xp y − ypx = −i! ⎜ x − y ⎟
dx ⎠
⎝ dy
(c)
Linear means
() ()
()
()
()
()
ˆ x + Ag
ˆ x
 ⎡⎣ f x + g x ⎤⎦ = Af
 ⎡⎣ cf x ⎤⎦ = c ⎡⎣ f x ⎤⎦
(d)
and
Hermitian means that
( )
∗
∗
∫ ψ 1 Âψ 2 dτ = ∫ ψ 2 Âψ 1 dτ
and implies that the eigenvalues of  are real. This is important!!
Observables should be represented as real numbers.
Take Âψ = aψ
Proof:
( )
( ) dτ
∗
∫ ψ Âψ dτ = ∫ ψ Âψ
∫ψ
⇒
∗
( )
aψ dτ = ∫ ψ aψ
∗
∗
dτ
a = a*
true only if a is real
(e)
Eigenfunctions of Hermitian operators are orthogonal
i.e.
if
Âψ m = amψ m
and Âψ n = anψ n
5.61 Fall 2007
Lecture #10
then
∫ψ
ψ n dτ = 0
∗
m
page 4
if m ≠ n
Proof:
(
)
∗
∗
∫ ψ m Âψ n dτ = ∫ ψ n Âψ m dτ
an ∫ ψ m∗ ψ n dτ = am∗ ∫ ψ nψ m∗ dτ
(a
⇒
n
(a
n
− am∗
) ∫ψ
∗
m
− am∗
) ∫ψ
∗
m
ψ n dτ = 0
= 0 if
n=m
Example:
ψ n dτ = 0
= 0 if n ≠ m
Particle in a box
ψ4
As much + as - area
ψ 1∗ψ 4
000aaadxdxdxψψψψψψ
140130120
ψ3
Eigenfunctions
ˆH
of
ψ 1∗ψ 3
⇒
ψ2
ψ 1∗ψ 2
ψ1
0
x
a
0
x
a
In addition, if eigenfunctions of  are normalized, then they are orthonormal
∫ψ
ψ n dτ = δ mn
∗
m
Krönecker delta
5.61 Fall 2007
Lecture #10
page 5
⎧1 if m = n (normalization)
δ mn = ⎨
⎩0 if m ≠ n (orthogonality)
#3]
If
ψ is an eigenfunction of the operator, then it’s easy, e.g.
Ĥψ n = Enψ n
But what if
e.g.
measurement of energy yields value
E
n
ψ is not an eigenfunction of the operator?
ψ could be a superposition of eigenfunctions
ψ = c1φ1 + c2φ2
Âφ1 = a1φ1
where
and Âφ2 = a2φ2
2
2
Then a measurement of A returns either a1 or a2 , with probability c1 or c2
respectively, and making the measurement changes the state to either φ1 or φ2 .
a1
a
1
ψ
measure
#4]
a
2a2
φ1 (probability c12 )
φ2 (probability c22 )
This connects to the expectation value
(i)
If ψ n is an eigenfunction of  , then Âψ n = anψ n
a = ∫ ψ n∗ Âψ n dτ = an ∫ ψ n∗ψ n dτ = an
a = an
only value possible
5.61 Fall 2007
(ii)
Lecture #10
page 6
If ψ = c1φ1 + c2φ2 as above
a = ∫ ψ ∗ Aˆψ dτ =
∫ (c φ
1 1
) (
)
∗
+ c2φ2 Aˆ c1φ1 + c2φ2 dτ = c12 a1 + c22 a2
c12 is the probability of measuring a1
<a > = average of possible values weighted by their probabilities