Parity are equivalent: The parity or space inversion operation converts a right 0 0 handed coordinate system to left handed: πT (dx ) = T (−dx )π. x −→ −x, y −→ −y, z −→ −z. Substituting i T (dx0) = 1 − dx0 · p, This is a case of a non continuous operation, i.e. the ¯h operation cannot be composed of infinitesimal operations. we get the condition Thus the non continuous operations have no generator. We consider the parity operation, i.e. we let the parity {π, p} = 0 or π†pπ = −p, operator π to act on vectors of a Hilbert space and keep the coordinate system fixed: or the momentum changes its sign under the parity operation. |αi −→ π|αi. Angular momentum and parity Like in all symmetry operations we require that π is In the case of the orbital angular momentum unitary, i.e. π†π = 1. L = x × p Furthermore we require: one can easily evaluate † hα|π xπ|αi = −hα|x|αi ∀|αi. π†Lπ = π†x × pπ = π†xπ × π†pπ = (−x) × (−p) So we must have = L, π†xπ = −x, so the parity and the angular momentum commute: or πx = −xπ. [π, L] = 0. The operators x ja π anticommute. In R3 the parity operator is the matrix Let |x0i be a position eigenstate, i.e. −1 0 0 x|x0i = x0|x0i. P = 0 −1 0 , 0 0 −1 Then 0 0 0 0 xπ|x i = −πx|x i = (−x )π|x i, so quite obviously and we must have PR = RP, ∀R ∈ O(3). π|x0i = eiϕ| − x0i. We require that the corresponding operators of the The phase is usually taken to be ϕ = 0, so Hilbert space satisfy the same condition, i.e. π|x0i = | − x0i. πD(R) = D(R)π. Applying the parity operator again we get Looking at the infinitesimal rotation π2|x0i = |x0i D(nˆ) = 1 − iJ · nˆ/¯h, or we see that π2 = 1. [π, J] = 0 or π†Jπ = J, We see that which is equivalent to the transformation of the orbital angular momentum. • the eigenvalues of the operator π can be only ±1, We see that under −1 † • π = π = π. • rotations x and J transform similarly, that is, like vectors or tensors of rank 1. Momentum and parity • space inversions x is odd and J even. We require that operations We say that under the parity operation • translation followed by space inversion • odd vectors are polar, • space inversion followed by translation to the opposite direction • even vectors are axial or pseudovectors. Let us consider such scalar products as p · x and S · x. The explicit expression for spherical functions is One can easily see that under rotation these are invariant, s scalars. Under the parity operation they transform like (2l + 1)(l − m)! Y m(θ, φ) = (−1)m P m(θ)eimφ, l 4π(l + m)! l π†p · xπ = (−p) · (−x) = p · x π†S · xπ = S · (−x) = −S · x. from which as a special case, m = 0, we obtain r We say that quantities behaving under rotations like 2l + 1 Y 0(θ, φ) = P (cos θ). scalars, spherical tensors of rank 0, which under the l 4π l parity operation are Depending on the degree l of the Legendre polynomial it • even, are (ordinary) scalars, is either even or odd: • odd, are pseudoscalars. l Pl(−z) = (−1) Pl(z). We see that Wave functions and parity Let ψ be the wave function of a spinles particle in the hx0|π|α, l0i = (−1)lhx0|α, l0i, state |αi, i.e. ψ(x0) = hx0|αi. so the state vectors obey Since the position eigenstates satisfy π|α, l0i = (−1)l|α, l0i. 0 0 π|x i = | − x i, Now [π, L ] = 0 the wave function of the space inverted state is ± and hx0|π|αi = h−x0|αi = ψ(−x0). r L±|α, l0i ∝ |α, l, ±ri, Suppose that |αi is a parity eigenstate, i.e. so the orbital angular momentum states satisfy the relation π|αi = ±|αi. π|α, lmi = (−1)l|α, lmi. The corresponding wave function obeys the the relation Theorem 1 If [H, π] = 0, ψ(−x0) = hx0|π|αi = ±hx0|αi = ±ψ(x0), and |ni is an eigenstate of the Hamiltonian H belonging i.e. it is an even or odd function of its argument. to the nondegenerate eigenvalue En, i.e. Note Not all physically relevant wave function have H|ni = E |ni, parity. For example, n then |ni is also an eigenstate of the parity. [p, π] 6= 0, Proof: Using the property π2 = 1 one can easily see that the state so a momentum eigenstate is not an eigenstate of the 1 parity. The wave function corresponding to an eigenstate (1 ± π)|ni 2 of the momentum is the plane wave is a parity eigenstate belonging to the eigenvalue ±1. On 0 ip0·x0/h¯ ψp0 (x ) = e , the other hand, this is also an eigenstate of the Hamiltonia H with the energy En: which is neither even nor odd. Because 1 1 H( (1 ± π)|ni) = En (1 ± π)|ni. [π, L] = 0, 2 2 the eigenstate |α, lmi of the orbital angular momentum Since we supposed the state |ni to be non degenerate the 2 1 (L ,Lz) is also an eigenstate of the parity. Now states |ni and 2 (1 ± π)|ni must be the same excluding a phase factor, m 0 1 Rα(r)Yl (θ, φ) = hx |α, lmi. (1 ± π)|ni = eiϕ|ni, 2 0 0 In spherical coordinates the transformation x −→ −x so the state |ni is a parity eigen state belonging to the maps to eigenvalue ±1 Example The energy states of a one dimensional r −→ r harmonic oscillator are non degenerate and the θ −→ π − θ (cos θ −→ − cos θ) Hamiltonian even, so the wave functions are either even φ −→ φ + π (eimφ −→ (−1)meimφ). or odd. Note The nondegeneracy condition is essential. For Let us suppose that at the moment t0 = 0 the state of the p2 system is |Li. At a later moment, t, the system is example, the Hamiltonian of a free particle, H = 2m, is even but the energy states descibed by the state vector |L, t = 0; ti p02 0 H|p0i = |p0i 1 2m = √ (e−iES t/h¯ |Si + e−iEAt/h¯ |Ai) 2 are not eigenstates of the parity because 1 = √ e−iES t/h¯ (|Si + e−i(EA−ES )t/h¯ |Ai), 2 π|p0i = | − p0i. because now the time evolution operator is simply The condition of the theorem is not valid because the 0 0 −iHt/h¯ states |p i and | − p i are degenerate. We can form parity U(t, t0 = 0) = e . eigenstates √ 1/ 2(|p0i ± | − p0i), At the moment t = T/2 = 2π¯h/2(EA − ES) the system is in the pure |Ri state and at the moment t = T again in which are also degenerate energy (but not momentum) its pure initial state |Li. The system oscillates between eigenstates. The corresponding wave functions the states |Li and |Ri at the angular velocity 0 0 0 ±ip0·x0/h¯ ψ 0 (x ) = hx | ± p i = e EA − ES ±p ω = . ¯h are neither even nor odd, whereas When V → ∞, then EA → ES. Then the states |Li and hx0|(|p0i + | − p0i) ∝ cos p0 · x0/¯h |Ri are degenerate energy eigenstates but not parity eigenstates. A particle which is localized in one of the hx0|(|p0i − | − p0i) ∝ sin p0 · x0/¯h wells will remain there forever. Its wave function does are. not, however, obey the same symmetry as the Example One dimensional symmetric double well Hamiltonian: we are dealing with a broken symmetry. Selection rules Suppose that the states |αi and |βi are parity eigenstates: π|αi = α|αi π|βi = β|βi, where α and β are the parities (±1) of the states. Now S A † † Sym m etric ( ) A ntisym m etric ( ) hβ|x|αi = hβ|π πxπ π|αi = −αβhβ|x|αi, The ground state is the symmetric state |Si and the first so excited state the antisymmetric state |Ai: hβ|x|αi = 0 unless α = −β. H|Si = ES|Si Example The intensity of the dipole transition is π|Si = |Si proportional to the matrix element of the operator x between the initial and final states. Dipole transitions are H|Ai = EA|Si thus possible between states which have opposite parity. π|Ai = −|Ai, Example Dipole moment. If where E < E . When the potential barrier V between S A [H, π] = 0, the wells increases the energy difference between the states decreases: then no non degenerate state has dipole moment: lim (EA − ES) → 0. hn|x|ni = 0. V →∞ The same holds for any quantity if the corresponding We form the superpositions operator o is odd: 1 π†oπ = −o. |Li = √ (|Si + |Ai) 2 1 |Ri = √ (|Si − |Ai), 2 which are neither energy nor parity eigenstates..