Seminario

Parity
The parity or space inversion operation converts a right handed coordinate system to left handed: x −→ −x, y −→ −y, z −→ −z. This is a case of a non continuous operation, i.e. the operation cannot be composed of infinitesimal operations. Thus the non continuous operations have no generator. We consider the parity operation, i.e. we let the parity operator π to act on vectors of a Hilbertspace and keep the coordinate system fixed: |α −→ π|α .

are equivalent: πT (dx ) = T (−dx )π. Substituting T (dx ) = 1 − we get the condition {π, p} = 0 or π † pπ = −p, or the momentum changes its sign under the parity operation.
Angular momentum and parity

i dx · p, ¯ h

Like in all symmetry operations we require that π is unitary, i.e. π † π = 1. Furthermore we require: α|π xπ|α = − α|x|α∀|α . So we must have π † xπ = −x, or πx = −xπ. The operators x ja π anti commute. Let |x be a position eigenstate, i.e. x|x = x |x . Then xπ|x = −πx|x = (−x )π|x , and we must have π|x = eiϕ | − x . The phase is usually taken to be ϕ = 0, so π|x = | − x . Applying the parity operator again we get π 2 |x = |x or π = 1. We see that • the eigenvalues of the operator π can be only ±1, • π −1 = π † =π.
Momentum and parity
2 †

In the case of the orbital angular momentum L=x×p one can easily evaluate π † Lπ = π † x × pπ = π † xπ × π † pπ = (−x) × (−p) = L,

so the parity and the angular momentum commute: [π, L] = 0. In R3 the parity operator is the matrix   −1 0 0 P =  0 −1 0  , 0 0 −1 so quite obviously P R = RP, ∀R ∈ O(3). We require that the corresponding operators of the Hilbertspace satisfy the same condition, i.e. πD(R) = D(R)π. Looking at the infinitesimal rotation ˆ ˆ h D( n) = 1 − iJ · n /¯ , we see that [π, J ] = 0 or π † J π = J , which is equivalent to the transformation of the orbital angular momentum. We see that under • rotations x and J transform similarly, that is, like vectors or tensors of rank 1. • space inversions x is odd and J even. We say that underthe parity operation • odd vectors are polar, • even vectors are axial or pseudovectors.

We require that operations • translation followed by space inversion • space inversion followed by translation to the opposite direction

Let us consider such scalar products as p · x and S · x. One can easily see that under rotation these are invariant, scalars. Under the parity operation they transformlike π † p · xπ π † S · xπ = (−p) · (−x) = p · x = S · (−x) = −S · x.

The explicit expression for spherical functions is Ylm (θ, φ) = (−1)m (2l + 1)(l − m)! m Pl (θ)eimφ , 4π(l + m)!

from which as a special case, m = 0, we obtain Yl0 (θ, φ) = 2l + 1 Pl (cos θ). 4π

We say that quantities behaving under rotations like scalars, spherical tensors of rank 0, which under the parity operationare • even, are (ordinary) scalars, • odd, are pseudoscalars.
Wave functions and parity

Depending on the degree l of the Legendre polynomial it is either even or odd: Pl (−z) = (−1)l Pl (z). We see that x |π|α, l0 = (−1)l x |α, l0 , so the state vectors obey π|α, l0 = (−1)l |α, l0 . Now [π, L± ] = 0 and Lr |α, l0 ∝ |α, l, ±r , ± so the orbital angular momentum states satisfy the relation π|α, lm= (−1)l |α, lm . Theorem 1 If [H, π] = 0, and |n is an eigenstate of the Hamiltonian H belonging to the nondegenerate eigenvalue En , i.e. H|n = En |n , then |n is also an eigenstate of the parity. Proof: Using the property π 2 = 1 one can easily see that the state 1 (1 ± π)|n 2 is a parity eigenstate belonging to the eigenvalue ±1. On the other hand, this is also an eigenstate of the HamiltoniaH with the energy En : 1 1 H( (1 ± π)|n ) = En (1 ± π)|n . 2 2 Since we supposed the state |n to be non degenerate the 1 states |n and 2 (1 ± π)|n must be the same excluding a phase factor, 1 (1 ± π)|n = eiϕ |n , 2 so the state |n is a parity eigen state belonging to the eigenvalue ±1 Example The energy states of a one dimensional harmonic oscillator are non degenerate and the Hamiltonian even,...