MPH-008 Solved Assignment
PART A
Question:-01
a) Write the space translation operator in quantum mechanics for an infinitesimal translation of
ε
y
ε
y
epsi _(y) \varepsilon_y ε y
y
y
y y y
ε
z
ε
z
epsi _(z) \varepsilon_z ε z
z
z
z z z
[
p
^
y
,
p
^
z
]
=
0
p
^
y
,
p
^
z
=
0
[ hat(p)_(y), hat(p)_(z)]=0 \left[\hat{p}_y, \hat{p}_z\right]=0 [ p ^ y , p ^ z ] = 0
Answer:
1. Infinitesimal Translation Operators
The space translation operator for an infinitesimal translation along the
y
y
y y y -direction by
ε
y
ε
y
epsi _(y) \varepsilon_y ε y and along the
z
z
z z z -direction by
ε
z
ε
z
epsi _(z) \varepsilon_z ε z can be written as:
T
^
(
ε
y
,
ε
z
)
=
T
^
(
ε
y
)
T
^
(
ε
z
)
T
^
(
ε
y
,
ε
z
)
=
T
^
(
ε
y
)
T
^
(
ε
z
)
hat(T)(epsi _(y),epsi _(z))= hat(T)(epsi _(y)) hat(T)(epsi _(z)) \hat{T}(\varepsilon_y, \varepsilon_z) = \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_z) T ^ ( ε y , ε z ) = T ^ ( ε y ) T ^ ( ε z ) where
T
^
(
ε
y
)
T
^
(
ε
y
)
hat(T)(epsi _(y)) \hat{T}(\varepsilon_y) T ^ ( ε y ) and
T
^
(
ε
z
)
T
^
(
ε
z
)
hat(T)(epsi _(z)) \hat{T}(\varepsilon_z) T ^ ( ε z ) are the translation operators along the
y
y
y y y – and
z
z
z z z -directions, respectively.
For infinitesimal translations, these operators can be approximated using the exponential form involving the momentum operators
p
^
y
p
^
y
hat(p)_(y) \hat{p}_y p ^ y and
p
^
z
p
^
z
hat(p)_(z) \hat{p}_z p ^ z :
T
^
(
ε
y
)
=
e
−
i
ε
y
p
^
y
/
ℏ
T
^
(
ε
y
)
=
e
−
i
ε
y
p
^
y
/
ℏ
hat(T)(epsi _(y))=e^(-iepsi _(y) hat(p)_(y)//ℏ) \hat{T}(\varepsilon_y) = e^{-i \varepsilon_y \hat{p}_y / \hbar} T ^ ( ε y ) = e − i ε y p ^ y / ℏ
T
^
(
ε
z
)
=
e
−
i
ε
z
p
^
z
/
ℏ
T
^
(
ε
z
)
=
e
−
i
ε
z
p
^
z
/
ℏ
hat(T)(epsi _(z))=e^(-iepsi _(z) hat(p)_(z)//ℏ) \hat{T}(\varepsilon_z) = e^{-i \varepsilon_z \hat{p}_z / \hbar} T ^ ( ε z ) = e − i ε z p ^ z / ℏ 2. Properties of Space Translation Operators
The key properties of the translation operators we will use are:
Translation operators are unitary:
T
^
(
ε
y
)
†
=
T
^
(
−
ε
y
)
T
^
(
ε
y
)
†
=
T
^
(
−
ε
y
)
hat(T)(epsi _(y))^(†)= hat(T)(-epsi _(y)) \hat{T}(\varepsilon_y)^\dagger = \hat{T}(-\varepsilon_y) T ^ ( ε y ) † = T ^ ( − ε y )
The composition of translations is associative:
T
^
(
ε
y
+
ε
y
′
)
=
T
^
(
ε
y
)
T
^
(
ε
y
′
)
T
^
(
ε
y
+
ε
y
′
)
=
T
^
(
ε
y
)
T
^
(
ε
y
′
)
hat(T)(epsi _(y)+epsi_(y)^(‘))= hat(T)(epsi _(y)) hat(T)(epsi_(y)^(‘)) \hat{T}(\varepsilon_y + \varepsilon_y’) = \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_y’) T ^ ( ε y + ε y ′ ) = T ^ ( ε y ) T ^ ( ε y ′ )
3. Commutator of Momentum Operators
We want to show that the commutator of the momentum operators
p
^
y
p
^
y
hat(p)_(y) \hat{p}_y p ^ y and
p
^
z
p
^
z
hat(p)_(z) \hat{p}_z p ^ z is zero:
[
p
^
y
,
p
^
z
]
=
0
p
^
y
,
p
^
z
=
0
[ hat(p)_(y), hat(p)_(z)]=0 \left[\hat{p}_y, \hat{p}_z\right] = 0 [ p ^ y , p ^ z ] = 0 4. Infinitesimal Translation and Commutator
To do this, consider the product of the two translation operators
T
^
(
ε
y
)
T
^
(
ε
y
)
hat(T)(epsi _(y)) \hat{T}(\varepsilon_y) T ^ ( ε y ) and
T
^
(
ε
z
)
T
^
(
ε
z
)
hat(T)(epsi _(z)) \hat{T}(\varepsilon_z) T ^ ( ε z ) :
T
^
(
ε
y
)
T
^
(
ε
z
)
=
e
−
i
ε
y
p
^
y
/
ℏ
e
−
i
ε
z
p
^
z
/
ℏ
T
^
(
ε
y
)
T
^
(
ε
z
)
=
e
−
i
ε
y
p
^
y
/
ℏ
e
−
i
ε
z
p
^
z
/
ℏ
hat(T)(epsi _(y)) hat(T)(epsi _(z))=e^(-iepsi _(y) hat(p)_(y)//ℏ)e^(-iepsi _(z) hat(p)_(z)//ℏ) \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_z) = e^{-i \varepsilon_y \hat{p}_y / \hbar} e^{-i \varepsilon_z \hat{p}_z / \hbar} T ^ ( ε y ) T ^ ( ε z ) = e − i ε y p ^ y / ℏ e − i ε z p ^ z / ℏ 5. Expand the Exponentials
Using the Baker-Campbell-Hausdorff formula for infinitesimal translations, we get:
T
^
(
ε
y
)
T
^
(
ε
z
)
≈
(
1
−
i
ε
y
p
^
y
ℏ
)
(
1
−
i
ε
z
p
^
z
ℏ
)
T
^
(
ε
y
)
T
^
(
ε
z
)
≈
1
−
i
ε
y
p
^
y
ℏ
1
−
i
ε
z
p
^
z
ℏ
hat(T)(epsi _(y)) hat(T)(epsi _(z))~~(1-(iepsi _(y) hat(p)_(y))/(ℏ))(1-(iepsi _(z) hat(p)_(z))/(ℏ)) \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_z) \approx \left( 1 – \frac{i \varepsilon_y \hat{p}_y}{\hbar} \right) \left( 1 – \frac{i \varepsilon_z \hat{p}_z}{\hbar} \right) T ^ ( ε y ) T ^ ( ε z ) ≈ ( 1 − i ε y p ^ y ℏ ) ( 1 − i ε z p ^ z ℏ ) Expanding this product up to the first order in
ε
y
ε
y
epsi _(y) \varepsilon_y ε y and
ε
z
ε
z
epsi _(z) \varepsilon_z ε z :
T
^
(
ε
y
)
T
^
(
ε
z
)
≈
1
−
i
ε
y
p
^
y
ℏ
−
i
ε
z
p
^
z
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
T
^
(
ε
y
)
T
^
(
ε
z
)
≈
1
−
i
ε
y
p
^
y
ℏ
−
i
ε
z
p
^
z
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
hat(T)(epsi _(y)) hat(T)(epsi _(z))~~1-(iepsi _(y) hat(p)_(y))/(ℏ)-(iepsi _(z) hat(p)_(z))/(ℏ)+(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)] \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_z) \approx 1 – \frac{i \varepsilon_y \hat{p}_y}{\hbar} – \frac{i \varepsilon_z \hat{p}_z}{\hbar} + \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] T ^ ( ε y ) T ^ ( ε z ) ≈ 1 − i ε y p ^ y ℏ − i ε z p ^ z ℏ + ε y ε z ℏ 2 [ p ^ y , p ^ z ] Similarly, the product
T
^
(
ε
z
)
T
^
(
ε
y
)
T
^
(
ε
z
)
T
^
(
ε
y
)
hat(T)(epsi _(z)) hat(T)(epsi _(y)) \hat{T}(\varepsilon_z) \hat{T}(\varepsilon_y) T ^ ( ε z ) T ^ ( ε y ) can be expanded as:
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
(
1
−
i
ε
z
p
^
z
ℏ
)
(
1
−
i
ε
y
p
^
y
ℏ
)
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
1
−
i
ε
z
p
^
z
ℏ
1
−
i
ε
y
p
^
y
ℏ
hat(T)(epsi _(z)) hat(T)(epsi _(y))~~(1-(iepsi _(z) hat(p)_(z))/(ℏ))(1-(iepsi _(y) hat(p)_(y))/(ℏ)) \hat{T}(\varepsilon_z) \hat{T}(\varepsilon_y) \approx \left( 1 – \frac{i \varepsilon_z \hat{p}_z}{\hbar} \right) \left( 1 – \frac{i \varepsilon_y \hat{p}_y}{\hbar} \right) T ^ ( ε z ) T ^ ( ε y ) ≈ ( 1 − i ε z p ^ z ℏ ) ( 1 − i ε y p ^ y ℏ )
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
z
,
p
^
y
]
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
z
,
p
^
y
]
hat(T)(epsi _(z)) hat(T)(epsi _(y))~~1-(iepsi _(z) hat(p)_(z))/(ℏ)-(iepsi _(y) hat(p)_(y))/(ℏ)+(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(z), hat(p)_(y)] \hat{T}(\varepsilon_z) \hat{T}(\varepsilon_y) \approx 1 – \frac{i \varepsilon_z \hat{p}_z}{\hbar} – \frac{i \varepsilon_y \hat{p}_y}{\hbar} + \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_z, \hat{p}_y] T ^ ( ε z ) T ^ ( ε y ) ≈ 1 − i ε z p ^ z ℏ − i ε y p ^ y ℏ + ε y ε z ℏ 2 [ p ^ z , p ^ y ] Since the commutator
[
p
^
z
,
p
^
y
]
=
−
[
p
^
y
,
p
^
z
]
[
p
^
z
,
p
^
y
]
=
−
[
p
^
y
,
p
^
z
]
[ hat(p)_(z), hat(p)_(y)]=-[ hat(p)_(y), hat(p)_(z)] [ \hat{p}_z, \hat{p}_y ] = -[ \hat{p}_y, \hat{p}_z ] [ p ^ z , p ^ y ] = − [ p ^ y , p ^ z ] , the term becomes:
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
T
^
(
ε
z
)
T
^
(
ε
y
)
≈
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
hat(T)(epsi _(z)) hat(T)(epsi _(y))~~1-(iepsi _(z) hat(p)_(z))/(ℏ)-(iepsi _(y) hat(p)_(y))/(ℏ)-(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)] \hat{T}(\varepsilon_z) \hat{T}(\varepsilon_y) \approx 1 – \frac{i \varepsilon_z \hat{p}_z}{\hbar} – \frac{i \varepsilon_y \hat{p}_y}{\hbar} – \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] T ^ ( ε z ) T ^ ( ε y ) ≈ 1 − i ε z p ^ z ℏ − i ε y p ^ y ℏ − ε y ε z ℏ 2 [ p ^ y , p ^ z ] 6. Equate the Two Products
Since translations commute,
T
^
(
ε
y
)
T
^
(
ε
z
)
=
T
^
(
ε
z
)
T
^
(
ε
y
)
T
^
(
ε
y
)
T
^
(
ε
z
)
=
T
^
(
ε
z
)
T
^
(
ε
y
)
hat(T)(epsi _(y)) hat(T)(epsi _(z))= hat(T)(epsi _(z)) hat(T)(epsi _(y)) \hat{T}(\varepsilon_y) \hat{T}(\varepsilon_z) = \hat{T}(\varepsilon_z) \hat{T}(\varepsilon_y) T ^ ( ε y ) T ^ ( ε z ) = T ^ ( ε z ) T ^ ( ε y ) , we have:
1
−
i
ε
y
p
^
y
ℏ
−
i
ε
z
p
^
z
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
1
−
i
ε
y
p
^
y
ℏ
−
i
ε
z
p
^
z
ℏ
+
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
1
−
i
ε
z
p
^
z
ℏ
−
i
ε
y
p
^
y
ℏ
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
1-(iepsi _(y) hat(p)_(y))/(ℏ)-(iepsi _(z) hat(p)_(z))/(ℏ)+(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)]=1-(iepsi _(z) hat(p)_(z))/(ℏ)-(iepsi _(y) hat(p)_(y))/(ℏ)-(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)] 1 – \frac{i \varepsilon_y \hat{p}_y}{\hbar} – \frac{i \varepsilon_z \hat{p}_z}{\hbar} + \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] = 1 – \frac{i \varepsilon_z \hat{p}_z}{\hbar} – \frac{i \varepsilon_y \hat{p}_y}{\hbar} – \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] 1 − i ε y p ^ y ℏ − i ε z p ^ z ℏ + ε y ε z ℏ 2 [ p ^ y , p ^ z ] = 1 − i ε z p ^ z ℏ − i ε y p ^ y ℏ − ε y ε z ℏ 2 [ p ^ y , p ^ z ] 7. Solve for the Commutator
Comparing the terms on both sides, we get:
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
−
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)]=-(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)] \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] = – \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] ε y ε z ℏ 2 [ p ^ y , p ^ z ] = − ε y ε z ℏ 2 [ p ^ y , p ^ z ]
2
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
0
2
ε
y
ε
z
ℏ
2
[
p
^
y
,
p
^
z
]
=
0
2(epsi _(y)epsi _(z))/(ℏ^(2))[ hat(p)_(y), hat(p)_(z)]=0 2 \frac{\varepsilon_y \varepsilon_z}{\hbar^2} [\hat{p}_y, \hat{p}_z] = 0 2 ε y ε z ℏ 2 [ p ^ y , p ^ z ] = 0 For this equality to hold for arbitrary
ε
y
ε
y
epsi _(y) \varepsilon_y ε y and
ε
z
ε
z
epsi _(z) \varepsilon_z ε z :
[
p
^
y
,
p
^
z
]
=
0
[
p
^
y
,
p
^
z
]
=
0
[ hat(p)_(y), hat(p)_(z)]=0 [\hat{p}_y, \hat{p}_z] = 0 [ p ^ y , p ^ z ] = 0 Conclusion
We have shown that the commutator of the momentum operators
p
^
y
p
^
y
hat(p)_(y) \hat{p}_y p ^ y and
p
^
z
p
^
z
hat(p)_(z) \hat{p}_z p ^ z is zero:
[
p
^
y
,
p
^
z
]
=
0
p
^
y
,
p
^
z
=
0
[ hat(p)_(y), hat(p)_(z)]=0 \left[\hat{p}_y, \hat{p}_z\right] = 0 [ p ^ y , p ^ z ] = 0
b) Show that the angular momentum operator
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y is even under a parity transformation.
Answer:
To show that the angular momentum operator
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y is even under a parity transformation, we need to understand how parity transformations affect the position and momentum operators, and subsequently the angular momentum operator.
A parity transformation, denoted by
P
^
P
^
hat(P) \hat{P} P ^ , is an operation that inverts the spatial coordinates. For a vector
r
→
=
(
x
,
y
,
z
)
r
→
=
(
x
,
y
,
z
)
vec(r)=(x,y,z) \vec{r} = (x, y, z) r → = ( x , y , z ) , the parity transformation results in:
P
^
r
→
P
^
−
1
=
−
r
→
P
^
r
→
P
^
−
1
=
−
r
→
hat(P) vec(r) hat(P)^(-1)=- vec(r) \hat{P} \vec{r} \hat{P}^{-1} = -\vec{r} P ^ r → P ^ − 1 = − r → This implies the following transformations for the components of the position vector:
P
^
x
P
^
−
1
=
−
x
,
P
^
y
P
^
−
1
=
−
y
,
P
^
z
P
^
−
1
=
−
z
P
^
x
P
^
−
1
=
−
x
,
P
^
y
P
^
−
1
=
−
y
,
P
^
z
P
^
−
1
=
−
z
hat(P)x hat(P)^(-1)=-x,quad hat(P)y hat(P)^(-1)=-y,quad hat(P)z hat(P)^(-1)=-z \hat{P} x \hat{P}^{-1} = -x, \quad \hat{P} y \hat{P}^{-1} = -y, \quad \hat{P} z \hat{P}^{-1} = -z P ^ x P ^ − 1 = − x , P ^ y P ^ − 1 = − y , P ^ z P ^ − 1 = − z Similarly, the momentum operator
p
→
=
(
p
x
,
p
y
,
p
z
)
p
→
=
(
p
x
,
p
y
,
p
z
)
vec(p)=(p_(x),p_(y),p_(z)) \vec{p} = (p_x, p_y, p_z) p → = ( p x , p y , p z ) transforms under parity as:
P
^
p
→
P
^
−
1
=
−
p
→
P
^
p
→
P
^
−
1
=
−
p
→
hat(P) vec(p) hat(P)^(-1)=- vec(p) \hat{P} \vec{p} \hat{P}^{-1} = -\vec{p} P ^ p → P ^ − 1 = − p → This implies:
P
^
p
x
P
^
−
1
=
−
p
x
,
P
^
p
y
P
^
−
1
=
−
p
y
,
P
^
p
z
P
^
−
1
=
−
p
z
P
^
p
x
P
^
−
1
=
−
p
x
,
P
^
p
y
P
^
−
1
=
−
p
y
,
P
^
p
z
P
^
−
1
=
−
p
z
hat(P)p_(x) hat(P)^(-1)=-p_(x),quad hat(P)p_(y) hat(P)^(-1)=-p_(y),quad hat(P)p_(z) hat(P)^(-1)=-p_(z) \hat{P} p_x \hat{P}^{-1} = -p_x, \quad \hat{P} p_y \hat{P}^{-1} = -p_y, \quad \hat{P} p_z \hat{P}^{-1} = -p_z P ^ p x P ^ − 1 = − p x , P ^ p y P ^ − 1 = − p y , P ^ p z P ^ − 1 = − p z 2. Angular Momentum Operator
The angular momentum operator
L
→
^
L
→
^
hat(vec(L)) \hat{\vec{L}} L → ^ is given by:
L
→
^
=
r
→
×
p
→
L
→
^
=
r
→
×
p
→
hat(vec(L))= vec(r)xx vec(p) \hat{\vec{L}} = \vec{r} \times \vec{p} L → ^ = r → × p → For the
y
y
y y y -component of the angular momentum operator,
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y , we have:
L
^
y
=
x
p
z
−
z
p
x
L
^
y
=
x
p
z
−
z
p
x
hat(L)_(y)=xp_(z)-zp_(x) \hat{L}_y = x p_z – z p_x L ^ y = x p z − z p x We now apply the parity transformation to
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y :
P
^
L
^
y
P
^
−
1
=
P
^
(
x
p
z
−
z
p
x
)
P
^
−
1
P
^
L
^
y
P
^
−
1
=
P
^
(
x
p
z
−
z
p
x
)
P
^
−
1
hat(P) hat(L)_(y) hat(P)^(-1)= hat(P)(xp_(z)-zp_(x)) hat(P)^(-1) \hat{P} \hat{L}_y \hat{P}^{-1} = \hat{P} (x p_z – z p_x) \hat{P}^{-1} P ^ L ^ y P ^ − 1 = P ^ ( x p z − z p x ) P ^ − 1 Using the transformations of the position and momentum operators under parity:
P
^
x
P
^
−
1
=
−
x
,
P
^
p
z
P
^
−
1
=
−
p
z
P
^
x
P
^
−
1
=
−
x
,
P
^
p
z
P
^
−
1
=
−
p
z
hat(P)x hat(P)^(-1)=-x,quad hat(P)p_(z) hat(P)^(-1)=-p_(z) \hat{P} x \hat{P}^{-1} = -x, \quad \hat{P} p_z \hat{P}^{-1} = -p_z P ^ x P ^ − 1 = − x , P ^ p z P ^ − 1 = − p z
P
^
z
P
^
−
1
=
−
z
,
P
^
p
x
P
^
−
1
=
−
p
x
P
^
z
P
^
−
1
=
−
z
,
P
^
p
x
P
^
−
1
=
−
p
x
hat(P)z hat(P)^(-1)=-z,quad hat(P)p_(x) hat(P)^(-1)=-p_(x) \hat{P} z \hat{P}^{-1} = -z, \quad \hat{P} p_x \hat{P}^{-1} = -p_x P ^ z P ^ − 1 = − z , P ^ p x P ^ − 1 = − p x Substituting these into the expression for
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y :
P
^
(
x
p
z
−
z
p
x
)
P
^
−
1
=
(
P
^
x
P
^
−
1
)
(
P
^
p
z
P
^
−
1
)
−
(
P
^
z
P
^
−
1
)
(
P
^
p
x
P
^
−
1
)
P
^
(
x
p
z
−
z
p
x
)
P
^
−
1
=
(
P
^
x
P
^
−
1
)
(
P
^
p
z
P
^
−
1
)
−
(
P
^
z
P
^
−
1
)
(
P
^
p
x
P
^
−
1
)
hat(P)(xp_(z)-zp_(x)) hat(P)^(-1)=( hat(P)x hat(P)^(-1))( hat(P)p_(z) hat(P)^(-1))-( hat(P)z hat(P)^(-1))( hat(P)p_(x) hat(P)^(-1)) \hat{P} (x p_z – z p_x) \hat{P}^{-1} = (\hat{P} x \hat{P}^{-1})(\hat{P} p_z \hat{P}^{-1}) – (\hat{P} z \hat{P}^{-1})(\hat{P} p_x \hat{P}^{-1}) P ^ ( x p z − z p x ) P ^ − 1 = ( P ^ x P ^ − 1 ) ( P ^ p z P ^ − 1 ) − ( P ^ z P ^ − 1 ) ( P ^ p x P ^ − 1 )
=
(
−
x
)
(
−
p
z
)
−
(
−
z
)
(
−
p
x
)
=
(
−
x
)
(
−
p
z
)
−
(
−
z
)
(
−
p
x
)
=(-x)(-p_(z))-(-z)(-p_(x)) = (-x)(-p_z) – (-z)(-p_x) = ( − x ) ( − p z ) − ( − z ) ( − p x )
=
x
p
z
−
z
p
x
=
x
p
z
−
z
p
x
=xp_(z)-zp_(x) = x p_z – z p_x = x p z − z p x 4. Conclusion
We find that:
P
^
L
^
y
P
^
−
1
=
L
^
y
P
^
L
^
y
P
^
−
1
=
L
^
y
hat(P) hat(L)_(y) hat(P)^(-1)= hat(L)_(y) \hat{P} \hat{L}_y \hat{P}^{-1} = \hat{L}_y P ^ L ^ y P ^ − 1 = L ^ y This shows that the angular momentum operator
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y is invariant under the parity transformation, which means it is even under parity. Therefore,
L
^
y
L
^
y
hat(L)_(y) \hat{L}_y L ^ y is an even operator under parity transformation.