Lecture 21:The Parity Operator
Phy851 Fall 2009
Parity inversion
• Symmetry under parity inversion isknown as mirror symmetry
• Formally, we say that f(x) is symmetricunder parity inversion if f(-x) = f(x)
• We would say that f(x) is antisymmetricunder parity inversion if f(-x)=-f(x)
• The universe is not symmetric underparity inversion (beta decay)– Unless there is mirror matter (and mirror
photons)– Would interact only weakly with matter
via gravity
−
−
−
z
y
x
z
y
x
P a:Parity inversion cannot
be generated byrotations
Parity Operator
• Let us define the parity operator via:
• Parity operator is Hermitian:
• Parity operator is it’s own inverse
• Thus it must be Unitary as well
12 =Π
)(
)(
xxxxxx
xxxxxx
′+=−′=Π′
′+=′−=′Π∗∗
δ
δ
Π=Π†
xxx =−Π=ΠΠ
1
†
−Π=Π
Π=Π
xx −=Π
1† −Π=Π
Properties of the Parity operator
• Parity acting to the left:
• What is the action of the parity operator on ageneric quantum state?– Let:
• Under parity inversion, we would say:
ψψ Π=′
ψψ Π=′ xx
ψψ xx −=′
)()( xx −=′ ψψ
)()( xx ψψ =−′
)()( xx ψψ =′′
xx −=′
Must be true for any physicaltransformation!
( ) xxxx −=−=Π=Π†††
xx −=Π
Eigenstates of Parity Operator
• What are the eigenstates of parity?– What states have well-defined parity?– Answer: even/odd states
• Proof:– Let:
– It follows that:
– But Π2=1, which gives:
πππ =Π
πππ 2=
πππ 22 =Π
12 =π
1±=π
+=+Π xx
+=+− xx
1+=π 1−=π−−=−Π xx
−−=−− xx
Any Even function! Any Odd function!
Parity acting on Momentum states
pxxdxp Π=Π ∫pxxdx −= ∫pxxdx −= ∫
pxepxpxi
−==−−h
hπ2
1
pxxdxp −=Π ∫
pp −=Π
Commutator of X with Π
• First we can compute ΠXΠ :
• So Π and X do not commute
ψψ Π−=ΠΠ XxXx
ψΠ−−= xx
ψxx−=
ψXx−=
XX −=ΠΠ
Π−=ΠΠ XX 2
Π−=Π XX
Π−=Π−Π XXX 2
This is theimportant resultfor calculations
Commutator of X with Π
• Next we can compute [x2,Π]:
• So Π and X2 do commute!
ψψ Π−=ΠΠ 22 XxXx
ψΠ−= xx2
ψxx2=
ψ2Xx=
22 XX =ΠΠ
Π=ΠΠ 222 XX
Π=Π 22 XX
022 =Π−Π XX
This is theimportant resultfor calculations
Commutator with Hamiltonian
• Same results must apply for P and P2, as therelation between Π and P is the same asbetween Π and X.
• Thus
• If Π commutes with X2, then Π commuteswith any even function of X
• Let
• Then
• This means that simultaneous eigenstates ofH and P exist
02,
2
=
Π
M
P
( )[ ] 0, 2 =Π XVeven
)(2
2
XVM
PH even+=
[ ] 0, =Π H
Consequences for a free particle
M
PH
2
2
=
M
kkH
2
22h=
M
kkH
2
22h=−
hMEk
kE2
:1,=
=hMEk
kE2
:2,−=
=
( )2,1,2
1:, EEE +=+
( )2,1,2
1:, EEE −=−
( )xx MEE h
2, cos
1)(
πψ =+
( )xix ME
E h2
, sin)(π
ψ =+
Note that P and Π do not commute, sosimultaneous eigenstates of momentum
and parity cannot exist
• The Hamiltonian of a free particle is:
• Energy eigenstates are doubly-degenerate:
• Note that plane waves, |k〉, areeigenstates of momentum and energy,but NOT parity
• But [H,Π]=0, so eigenstates of energyand parity must exist
Consequences for the SHO
• For the SHO we have:
• Therefore [H,Π]=0, so simultaneouseigenstates of Energy and Parity mustexist
• The energy levels are not-degenerate, sothere is no freedom to mix and matchstates
• Thus the only possibility is that eachenergy level must have definite parity
• The Hermite Polynomials have definiteparity: Hn(-x)=(-1)n Hn(x)
• Thus we have:
222
2
1
2XM
M
PH ω+=
( ) nn n1−=Π
So ground-state (n=0) is even
First excited state (n=1) is odd
(n=2) is even
Etc….