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….