Lecture 21:
The Parity Operator
Phy851 Fall 2009
Parity inversion
• Symmetry under parity inversion is
known as mirror symmetry
 x − x
   
P : y a − y
z − z
   
Parity inversion cannot
be generated by
rotations
• Formally, we say that f(x) is symmetric
under parity inversion if f(-x) = f(x)
• We would say that f(x) is antisymmetric
under parity inversion if f(-x)=-f(x)
• The universe is not symmetric under
parity inversion (beta decay)
– Unless there is mirror matter (and mirror
photons)
– Would interact only weakly with matter
via gravity
Parity Operator
• Let us define the parity operator via:
Π x = −x
• Parity operator is Hermitian:
x Π x′ = x − x′ = δ ( x + x′)
∗
∗
x′ Π x = x′ − x = δ ( x + x′)
Π† = Π
• Parity operator is it’s own inverse
ΠΠ x = Π − x = x
Π2 = 1
• Thus it must be Unitary as well
Π† = Π
Π = Π −1
Π † = Π −1
Properties of the Parity operator
• Parity acting to the left:
x Π = (Π x
†
†
)=
−x
†
= −x
xΠ= −x
• What is the action of the parity operator on a
generic quantum state?
– Let:
ψ′ = Πψ
x ψ′ = x Πψ
x ψ′ = −x ψ
ψ ′( x) = ψ (− x)
ψ ′(− x) = ψ ( x)
x′ = − x
• Under parity inversion, we would say:
ψ ′( x′) = ψ ( x)
Must be true for any physical
transformation!
Eigenstates of Parity Operator
• What are the eigenstates of parity?
– What states have well-defined parity?
– Answer: even/odd states
• Proof:
– Let:
Π π =π π
– It follows that:
2
2
Π π =π π
– But Π2=1, which gives:
π =π2 π
π 2 =1
π = ±1
π = +1
xΠ+ = x +
π = −1
xΠ − =− x −
−x + = x +
−x − =− x −
Any Even function!
Any Odd function!
Parity acting on Momentum states
Π p = ∫ dx Π x x p
= ∫ dx − x x p
= ∫ dx x − x p
i
− px
1
−x p =
e h = x −p
2πh
Π p = ∫ dx x x − p
Π p = −p
Commutator of X with Π
• First we can compute ΠXΠ :
x ΠXΠψ = − x XΠψ
= −x − x Π ψ
= −x x ψ
=− x Xψ
Π X Π = −X
2
Π X Π = −X Π
Π X = −X Π
Π X − X Π = −2 X Π
• So Π and X do not commute
This is the
important result
for calculations
Commutator of X with Π
• Next we can compute [x2,Π]:
x Π X 2 Π ψ = − x X 2Π ψ
= x2 − x Π ψ
= x2 x ψ
= x X2ψ
2
ΠX Π = X
2
2
2
2
ΠX Π = X Π
2
2
ΠX = X Π
ΠX 2 − X 2Π = 0
• So Π and X2 do commute!
This is the
important result
for calculations
Commutator with Hamiltonian
• Same results must apply for P and P2, as the
relation between Π and P is the same as
between Π and X.
• Thus
 P2 
Π ,
=0
 2M 
• If Π commutes with X2, then Π commutes
with any even function of X
[Π,V (X )]= 0
2
even
• Let
• Then
P2
H=
+ Veven ( X )
2M
[Π, H ]= 0
• This means that simultaneous eigenstates of
H and P exist
Consequences for a free particle
• The Hamiltonian of a free particle is:
P2
H=
2M
• Energy eigenstates are doublydegenerate:
h 2k 2
H −k =
2M
h 2k 2
H k =
2M
E ,1 := k
k=
E ,2 := k
2 ME
h
k =−
2 ME
h
• Note that plane waves, |k〉, are
eigenstates of momentum and energy,
but NOT parity Note that P and Π do not commute, so
simultaneous eigenstates of momentum
and parity cannot exist
• But [H,Π]=0, so eigenstates of energy
and parity must exist
1
( E ,1 + E ,2
E ,+ :=
2
1
( E ,1 − E ,2
E ,− :=
2
)
ψ E , + ( x) =
1
cos
π
(
2 ME
h
x
)
)
ψ E , + ( x) =
i
sin
π
(
2 ME
h
x
)
Consequences for the SHO
• For the SHO we have:
P2 1
H=
+ Mω 2 X 2
2M 2
• Therefore [H,Π]=0, so simultaneous
eigenstates of Energy and Parity must
exist
• The energy levels are not-degenerate, so
there is no freedom to mix and match
states
• Thus the only possibility is that each
energy level must have definite parity
• The Hermite Polynomials have definite
parity: Hn(-x)=(-1)n Hn(x)
So ground-state (n=0) is even
• Thus we have:
First excited state (n=1) is odd
n
Π n = (− 1) n
(n=2) is even
Etc….