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….