Module 35: Algebra of Operators Lecture 35: Algebra of

Module 35: Algebra of Operators
Lecture 35: Algebra of Operators
In the previous chapter it was mentioned that in quantum mechanics the
physical observables are replaced with Hermitian operators. In this chapter
we define Hermitian operators and show that they possess real eigenvalues.
We will restrict ourselves to one space dimension just for simplicity. The
calculations can easily be extended for higher dimensions. We call a function,
ψ(x), if it is not singular( does not become infinitely large) for finite x and
it vanishes for |x| → ∞. For example the normal curve, exp(−x2 ), is a well
behaved function whereas the sin(x) is not.
35.1
Hermitian Operators
ˆ is termed as Hermitian if it satisfies the following property:
An operator O
Z
∞
−∞
ˆ
φ (x)Oψ(x)dx
=
∗
Z
∞
−∞
ˆ ∗ φ∗ (x)dx,
ψ(x)O
(35.1)
ˆ∗
where φ(x) and ψ(x) are two well behaved functions and the operator O
ˆ and φ(x) respectively by complex
and the function φ(x)∗ are obtained from O
conjugation (i.e. by interchanging i with −i). In the left hand side of the
ˆ acts on the function ψ(x) whereas in the right
equation (35.2) the operator O
∗
ˆ acts on the function φ∗ .
hand side the operators O
35.1.1
Examples of Hermitian operators
The position operator x is Hermitian as
Z
∞
−∞
∗
φ (x)xψ(x)dx =
Z
∞
−∞
∗
ψ(x)xφ (x)dx =
Z
∞
−∞
ψ(x)x∗ φ∗ (x)dx,
(35.2)
where we used x = x∗ for x is postition and it is always real. Functions ψ(x)
and φ(x) are scalar functions and are commutative.
201
202
CHAPTER 35. ALGEBRA OF OPERATORS
∂
The momentum operator, px = −i¯
h ∂x
, is again Hermitian. It can be shown
as follows.
Z ∞
∂ψ(x)
∂
h
dx
φ∗ (x)
φ (x)pˆx ψ(x)dx =
φ (x)(−i¯
h )ψ(x)dx = −i¯
∂x
∂x
−∞
−∞
−∞
Z ∞
∂φ∗ (x)
= −i¯
hφ∗ (x)ψ(x) |∞
+i¯
h
ψ(x)dx,
−∞
∂x
−∞
Z
∞
Z
∗
∞
∗
where we used integration by parts to arrive at the last step. Now, ψ(±∞) →
0 and φ(±∞) → 0, since the functions are well behaved. So,
Ã
Z ∞
∂
∂φ∗ (x)
dx =
ψ(x) −i¯
h
φ (x)pˆx ψ(x)dx = i¯
h
ψ(x)
∂x
∂x
−∞
−∞
−∞
Z
∞
∗
=
Z
Z
∞
∞
−∞
!∗
φ∗ (x)dx
ψ(x)(ˆ
px )∗ φ∗ (x)dx
(35.3)
ˆ † = O.
ˆ Moreover (AˆB)
ˆ †=B
ˆ † Aˆ† .
For Hermitian operators one writes O
35.1.2
Eigenvalues of Hermitian operators
Now we show that Hermitian operators possess real eigenvalues. In the above
let us take a special case where , φ(x) = ψ(x) = χ(x), and χ(x) is an eigenfuncˆ with an eigenvalue λ (i.e. Oχ(x)
ˆ
tion of the Hermitian operator, O
= λχ(x)).
Z
∞
ˆ
φ∗ (x)Oψ(x)dx
−
−∞
Z ∞
−∞
Z ∞
λ
Z
ˆ
χ∗ (x)Oχ(x)dx
−
∗
χ (x)λχ(x)dx −
−∞
Z ∞
−∞
∗
∞
ˆ ∗ φ∗ (x)dx = 0
ψ(x)O
−∞
Z ∞
Z
χ (x)χ(x)dx − λ
−∞
∞
−∞
∗
(λ − λ∗ )
Z
ˆ ∗ χ∗ (x)d = 0
χ(x)O
χ(x)λ∗ χ∗ (x)dx = 0
∞
−∞
Z ∞
−∞
χ(x)χ∗ (x)dx = 0
χ∗ (x)χ(x)dx = 0
(35.4)
Since χ(x) is a well behaved function so it would be square integrable (i.e.
2
∗
−∞ |χ(x)| > 0 and finite), and hence from (35.4) λ = λ , showing eigenvalue
λ is real.
R∞
35.1.3
Orthogonality of eigenfunctions of Hermitian Operators
ˆ has two different eigenfunctions, χ(x) and
Suppose a Hermitian operator O
ˆ
ξ(x) with distinct eigenvalues λ and η respectively, i.e. Oχ(x)
= λχ(x) and
ˆ
Oξ(x) = ηξ(x). Now with the identification ψ(x) = χ(x) and φ(x) = ξ(x) we
35.2. COMMUTATORS AND UNCERTAINTY RELATION
have
Z
∞
ˆ
φ (x)Oψ(x)dx
−
−∞
Z ∞
Z
λ
∗
−∞
∞
∞
∗
ξ (x)λχ(x)dx −
−∞
∗
ˆ ∗ φ∗ (x)dx = 0
ψ(x)O
−∞
Z ∞
ˆ
ξ (x)Oχ(x)dx
−
∗
−∞
∞
Z
Z
Z
ξ (x)χ(x)dx − η
−∞
∞
−∞
∗
(λ − η)
203
Z
ˆ ∗ ξ ∗ (x)d = 0
χ(x)O
χ(x)η ∗ ξ ∗ (x)dx = 0
∞
−∞
Z ∞
−∞
χ(x)ξ ∗ (x)dx = 0
ξ ∗ (x)χ(x)dx = 0
(35.5)
where we used the reality of the eigenvalues, η = η ∗ . Since λ 6= η (initial assumption) the expression (35.5) shows that the eigenfunctions χ(x) and ξ(x),
ˆ are orwhich correspond to distinct eigenvalues of the Hermitian operator O
thogonal, viz.
Z ∞
ξ ∗ (x)χ(x)dx = 0
(35.6)
−∞
35.2
Commutators and Uncertainty Relation
ˆ B],
ˆ of two operators, Aˆ and B,
ˆ is defined as
Commutator, [A,
ˆ B]
ˆ = AˆB
ˆ −B
ˆ A.
ˆ
[A,
(35.7)
A commutator is also an operator. Commutators are useful in quantum mechanics. If two observables do not commute then we do not have simultaneous
eigenfunctions of these two observables, i.e there does not exist a state waveˆ = aψ and Bψ
ˆ = bψ. Hence it is not meaningful
function ψ such that Aψ
to talk about the eigenvalues of these two observables simultaneously. For
∂
ˆ ≡ pˆx = −i¯
example, take Aˆ ≡ xˆ and B
h ∂x
, then
[ˆ
x, pˆx ] = [ˆ
x, −i¯
h
∂ψ(x)
∂
∂
]ψ(x) = −i¯
hx
+ i¯
h (xψ(x))
(35.8)
∂x
∂x
∂x
∂ψ(x)
∂ψ(x)
= −i¯
hx
+ i¯
hx
+ i¯
hψ(x) = i¯
hψ(x).
∂x
∂x
The expression (35.9) shows [x, px ] = i¯
h. This commutation relation is known
as fundamental commutation relation and it precipitates in position-momentum
uncertainty relation. The Cauchy-Schwartz inequality shows
ˆ 2 ih(∆B)
ˆ 2 i ≥ 1 | h[A,
ˆ B]i
ˆ |2 ,
h(∆A)
4
(35.9)
where the notation is explained in eqn. (20.39) of the previous chapter . By
ˆ ≡ pˆx we get the famous Hesenberg ‘uncertainty relation’,
taking Aˆ ≡ xˆ and B
h∆ˆ
xih∆ˆ
px i ≥ h
¯ /2, since
h(∆ˆ
x)2 ih(∆ˆ
px )2 i ≥
h
¯2
1
| h[ˆ
x, pˆx ]i |2 = .
4
4
(35.10)
204
CHAPTER 35. ALGEBRA OF OPERATORS
We write the composite observables in quantum mechanics in the following
way. First we write down the observables classically and then change the
observables to corresponding operators. The following example would make
the process clear. Suppose we want to find the operator for the x−component
of the angular momentum, lx , we write,
lx = ypz − zpy ;
(35.11)
∂
∂
ˆlx = yˆpˆz − zˆpˆy = −i¯
hyˆ + i¯
hzˆ .
∂z
∂y
Note that classically the order of the variables does not matter, that is, we
first put the position and then the momentum or first the momentum and
then the position. But in quantum mechanics this may be crucial to get the
right operator. In the above example of angular momentum, since [ˆ
y , pˆz ] =
0 = [ˆ
z , pˆy ] we could have as well written the operator as ˆlx = pˆz yˆ − pˆy zˆ like the
classical case without any trouble but this is not always the case. For example,
if the clasical quantity is, say x2 p2x then we have the following choices for the
corresponding quantum mechanical operator, xˆpˆ2x xˆ or pˆx xˆ2 pˆx . The right choice
will be picked up by the experiment. Note that the other possibilities, viz.
xˆ2 pˆ2x , pˆx xˆpˆx xˆ etc. are eliminated by the Hermiticity requirement.
Example: Calculate [ˆlz , xˆ].
[ˆlz , xˆ] = [ˆ
xpˆy − yˆpˆx , xˆ] = [ˆ
xpˆy , xˆ] − [ˆ
y pˆx , xˆ]
= xˆ[ˆ
py , xˆ] + [ˆ
x, xˆ]ˆ
py − yˆ[ˆ
px , xˆ] − [ˆ
y , xˆ]ˆ
px = i¯
hyˆ
Problems
1. Identify Hermitian operators among the following:
∂2
∂3
∂3
∂
∂
ˆ
, ii) ∂x
ˆpˆ, vii) i ∂t
.
i) ∂x
2 , iii) i ∂x3 , iv) − ∂z 3 , v) lz , vi) x
Ans. ii), iii), v), vii).
2. Show the following identities:
ˆ C]
ˆ = [A,
ˆ C]
ˆB
ˆ + A[
ˆ B,
ˆ C]
ˆ
i) [AˆB,
ˆ Cˆ D]
ˆ = C[
ˆ A,
ˆ D]
ˆB
ˆ + [A,
ˆ C]
ˆD
ˆB
ˆ + A[
ˆ B,
ˆ C]
ˆD
ˆ + AˆC[
ˆ B,
ˆ D]
ˆ
ii)[AˆB,
ˆ [B,
ˆ C]]
ˆ + [B,
ˆ [C,
ˆ A]]
ˆ + [C,
ˆ [A,
ˆ B]]
ˆ =0
iii) [A,
3. Calculate the following commutators:
i) [ˆ
x, pˆy ], ii) [ˆ
y , zˆ], iii) [ˆ
pz , pˆy ], iv) [ˆly , ˆlz ], v) [ˆlx , pˆz ], vi) [ˆ
py pˆz , xˆyˆ]
2
2
2
2
2
vi) [ˆl , ˆlz ], where ˆl = ˆlx + ˆly + ˆlz
Ans. i) 0, ii) 0, iii) 0, iv) i¯
hˆlx , v) −i¯
hpˆy , vi) −i¯
hxˆpˆz vii) 0.
(35.12)

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