Operator Formalism I - Introduction to Quantum Mechanics I | PHY 4604, Study notes of Physics

Download Operator Formalism I - Introduction to Quantum Mechanics I | PHY 4604 and more Study notes Physics in PDF only on Docsity! 6 Operator Formalism I 6.1 Momentum operator In eqs. 5/(47-48) found ∫ dx|ψ(x)|2 = ∫ dp|f (p)|2 (1) suggesting that if |ψ(x)|2 is the probability density for finding the particle in spatial range [x, x + dx], then |f (p)|2 is prob. density for finding particle’s momentum in range [p, p + dp]. Plane wave ψp = e ipx/h̄ is state of definite momentum. Define momentum operator p̂ = −ih̄ ∂ ∂x , (2) verify p̂ψp = pψp, note difference between operator p̂ and eigenvalue p. Compare with Hψ = Eψ for state of definite energy. 6.2 Expectation values or, how the probabilistic nature of ψ affects physically mea- surable quantities. Consider statistical ensemble of systems with 1 particle de- scribed by ψ. Prob. particle in d3r at r is dP = |ψ(r, t)|2d3r. 1 Calculating averages Calculate avg. value by summing over all possible values of something, weighted with prob. of each. For example, mean value of particle’s position: < r >= ∫ ψ∗rψ d3r (3) if ψ normalized. Called expectation value of r because it is at < r > you would expect to find particle, on avg. Expectation val. of any operator  is just <  >= ∫ ψ∗Â(r)ψ d3r (4) Avg. value of momentum (need to use momentum prob. dens. |f (p)|2: < p > = ∫ d3p f (p)∗pf (p) (5) = ∫ d3pd3r1d 3r2 (2πh̄)3 ψ∗(r1)ψ(r2)pe−ip·(r2−r1)/h̄ (6) Simplify by writing pe−ip·(r2−r1)/h̄ = +ih̄∇2e−ip·(r2−r1)/h̄ (7) where ∇2 grad wrt r2. Integrate by parts in 3D: note for 2 functions g(r), h(r) which fall off suff. rapidly at |r| → ∞, ∫ d3rg∇h = − ∫ d3r(∇g)h (8) since the surface term gh|∞ vanishes. So rewrite (6) as < p >= ∫ d3pd3r1d 3r2 (2πh̄)3 ψ∗(r1)(−ih̄∇2)ψ(r2)e−ip·(r2−r1)/h̄ (9) 2 Examples: 1. α a const. =⇒ α† = α∗ 2. Ô = ∂∂x: ∫ ψ∗ ∂φ ∂x d3r = − ∫ ∂ψ ∗ ∂x φd3r (22) so ( ∂ ∂x )† = − ∂ ∂x (23) 3. (αÔ)† = α∗Ô† so if p̂ = −ih̄ ∂∂x then p̂† = ih̄ ∂∂x † = −ih̄ ∂∂x = p̂. p̂ is said to be self-adjoint or Hermitian. 4. (P̂ + Ô)† = P̂ † + Ô† 5. (P̂ Ô)† = Ô†P̂ † 6. If (ψ, Ôψ) = (Ôψ, ψ)∀ψ, then (ψ, Ôφ) = (Ôψ, φ) so Ô is self-adjoint. Check this! 7. From (16) get (ψ, Ôφ)∗ = (Ôφ, ψ) (24) = (φ, Ô†ψ) (25) so in particular (ψ, Ôψ)∗ = (ψ, Ô†ψ) (26) so that if Ô is self-adjoint, Ô† = Ô, then (ψ, Ôψ)real . All observables wil be represented by self-adjoint operators. 5 6.5 Eigenvalues and eigenfunctions Let Q̂ be a linear operator, and ψ an eigenfunction of Q̂ with eigenvalue q, i.e. Q̂ψ = qψ. If Q̂ is Hermitian, Q̂ = Q̂† then several properties follow: 1. All eigenvalues q are real! Start with (Q̂ψ, ψ) = (ψ, Q̂ψ) (27) Q̂ψ = qψ, (28) then (qψ, ψ) = (ψ, qψ) (29) q∗(ψ, ψ) = (ψ, ψ)q =⇒ (30) q∗ = q (31) 2. Eigenfunctions belonging to different eigenval- ues are orthogonal. If Q̂ψ1 = q1ψ1 (32) Q̂ψ2 = q2ψ2, (33) then (Q̂ψ1, ψ2) = (ψ1, Q̂ψ) (34) q1(ψ1, ψ2) = q2(ψ1, ψ2), (35) so if q1 6= q2, then (ψ1, ψ2) = 0 and the two fctns said to be orthogonal. 6 3. Linearly independent eigenfunctions belonging to the same eigenvalue q can be made orthogo- nal. (Example: recall we studied eigenvalues of 3D SHO in problem set, showed degeneracy—number of eigenfunc- tions corresponding to given eigenvalue– was N 2/2+3N/2+ 1. ) Functions ψ1...ψm are linearly ind. if only soln. to ∑m n=1 ψn = 0 is cn = 0. Suppose Q̂ψn = qψn, n = 1, . . .m. (36) It might be that (ψm, ψn) 6= δmn for this choice of the ψn, but can always find new set of φn obeying Q̂φn = qφn for which (φm, φn) = δmn using Gram-Schmidt orthonor- malization (as in linear algebra!): Let φ1 = ψ1 (37) φ2 = ψ2 + αψ1 (38) with α chosen s.t. (φ1, φ2) = (ψ1, ψ2) + α(ψ1, ψ1) = 0 (39) let φ3 = ψ3 + βφ2 + γψ1 (40) and require (φ1, φ3) = (φ2, φ3) = 0. (41) Last line is two equations determining 2 unknowns β and γ. And so on. . . 4. If P̂ and Q̂ commute, [P̂ , Q̂] = 0, can find set of simultaneous eigenfunctions of P̂ and Q̂. 7