CHM 532 Notes on Creation and Annihilation Operators

Transcription

1 CHM 53 Notes on Creation an Annihilation Operators These notes provie the etails concerning the solution to the quantum harmonic oscillator problem using the algebraic metho iscusse in class. The operators we introuce are calle creation an annihilation operators, names that are taken from the quantum treatment of light (i.e. photons). The quantum treatment of electromagnetic raiation has similarities with the harmonic oscillator problem. In the stuy of photons, creation operators create photons an annihilation operators annihilate photons. As iscusse below, creation operators create one quantum of energy in the harmonic oscillator an annihilation operators annihilate one quantum of energy. We begin with the Hamiltonian operator for the harmonic oscillator expresse in terms of momentum an position operators taken to be inepenent of any particular representation Ĥ = ˆp µ + 1 µωˆx. (1) We next introuce the imensionless operators ˆQ an ˆP, relate to ˆx an ˆp by the equations an Using it can easily be shown that ˆx = Substituting Eqs. () an (3) into Eq.(1) we obtain ˆQ () µω ˆp = (µω) 1/ ˆP. (3) [ˆx, ˆp] = i (4) [ ˆQ, ˆP ] = i. (5) Ĥ = 1 ω( ˆP + ˆQ ). (6) We next efine Ĥ 1 = Ĥ ω = 1 ( ˆP + ˆQ ). (7) 1

2 It is easy to see that if ψ is an eigenfunction of Ĥ1 with the imensionless eigenvalue ɛ, then ψ is also an eigenfunction of Ĥ with eigenvalue E = ωɛ. We next efine an annihilation operator by The ajoint of the annihilation operator â = 1 ( ˆQ + i ˆP ). (8) â = 1 ( ˆQ i ˆP ) (9) is calle a creation operator. Clearly, â is not Hermitian. Using Eq.(5), it is easy to show that the commutator between creation an annihilation operators is given by We next solve Eqs.(8) an (9) for ˆQ an ˆP [â, â ] = 1. (10) ˆQ = 1 (â + â ) (11) ˆP = 1 i (â â ) (1) an substitute into Eq.(7) to obtain Ĥ 1 = 1 ( 1 (â + â + ââ + â â) 1 ) (â + â ââ â â) (13) Using the commutator Eq.(10), the expression for Ĥ1 becomes where = 1 (ââ + â â). (14) Ĥ 1 = 1 (â â + 1) = ˆN + 1, (15) ˆN = â â (16) is calle the number operator. We see that if ψ n is an eigenfunction function of ˆN with eigenvalue n; i.e. ˆNψ n = nψ n (17) then ψ n is also an eigenfunction of Ĥ1 with eigenvalue n + 1/ as well as an eigenfunction of Ĥ with eigenvalue (n + 1/)ω. We seek then the eigenfunctions an eigenvalues of ˆN. We begin with Eq.(17), an we assume the eigenfunctions have been normalize so that ψ n ψ m = δ n,m. (18) We erive the eigenvalues of ˆN by proving a series of lemmas. The proofs of the lemmas make use of Eq.(10).

3 Lemma 1 Lemma Lemma 3 Lemma 4 where c is a constant. n 0. (19) n = ψ n ˆN ψ n (0) = ψ n â â ψ n (1) = âψ n âψ n 0 () ˆNâ = â( ˆN 1). (3) ˆNâ = â ââ (4) = (ââ 1)â (5) = â(â â 1) = â( ˆN 1). (6) ˆNâ = â ( ˆN + 1). (7) ˆNâ = â ââ (8) = â (â â + 1) = â ( ˆN + 1). (9) âψ n = c ψ n 1 (30) We must show that âψ n is an eigenfunction of ˆN with eigenvalue n 1. ˆN(âψ n ) = â( ˆN 1)ψ n = â(n 1)ψ n (31) Lemma 5 where c + is a constant. = (n 1)(âψ n ). (3) â ψ n = c + ψ n+1 (33) 3

4 We must show that â ψ n is an eigenfunction of ˆN with eigenvalue n + 1. ˆN(â ψ n ) = â ( ˆN + 1)ψ n = â (n + 1)ψ n (34) = (n + 1)(â ψ n ). (35) Lemma 6 an c = n (36) c + = n + 1. (37) n = ψ n ˆN ψ n = ψ n â â ψ n (38) = âψ n âψ n = c ψ n ψ n = c (39) or Next c = n. (40) n + 1 = ψ n ˆN + 1 ψ n = ψ n â â + 1 ψ n (41) = ψ n ââ ψ n (4) = â ψ n â ψ n = c + (43) or c + = n + 1. (44) Lemma 7 n is an integer. We use the metho of proof by contraiction. We assume n is not an integer, an show this leas to a result that contraicts one of our lemmas. We consier the process âψ n = nψ n 1 (45) â ψ n = n(n 1)ψ n (46) until â m ψ n = n(n 1)... (n m + 1)ψ n m. (47) If n is not an integer, there exits a series of values of m such that n m < 0. However, from the first lemma, the eigenvalue of ˆN must be positive or zero. Consequently, we have a contraiction, an n must be an integer. 4

5 Lemma 8 The smallest value of n is 0. Notice that an âψ 1 = ψ 0 (48) âψ 0 = 0. (49) Then n = 0 is the smallest eigenvalue of ˆN. We have then prove that the eigenvalues of ˆN are n = 0, 1,,..., an as a consequence, the eigenvalues of Ĥ are E n = (n + 1/)ω with n = 0, 1,,..., in agreement with the solution by ifferential equation. We next show that all matrix elements an expectation values of observables with respect to harmonic oscillator eigenfunctions can be evaluate using creation an annihilation operators. From Eqs.() an (11), we can write ˆx = (â + â ). (50) µω Then ψ n ˆx ψ m = ψ n â + â ψ m (51) µω = ψ n ( m ψ m 1 + m + 1 ψ m+1 ) µω (5) = ( mδ n,m 1 + m + 1δ n,m+1). µω (53) The expectation value of x (i.e. n = m) vanishes for any value of n. To evaluate matrix elements of ˆx, we use the operator ˆx = µω (â + â ) (54) = µω (â + â + ˆN + 1). (55) We en by showing that creation an annihilation operators can be use to construct the eigenfunctions of the Hamiltonian. We begin with the expression âψ 0 = 0. (56) From Eq.(8), we have 1 ( ˆQ + i ˆP )ψ 0 = 0. (57) 5

6 Now 1 ˆP = ˆp (58) µω 1 = µω i x = ( 1 µω ) 1/ i Q x Q (59) Then from Eq.(57) which is easily solve to give ( 1 ) 1/ µω i ( ) µω 1/ Q = 1 i Q. (60) ψ 0 Q + Qψ 0 = 0 (61) ψ 0 = Ae Q / where A is a constant. Using Eq.(), ψ 0 can be expresse in terms of the coorinate x an subsequently normalize to etermine A. To etermine ψ 1, we use â ψ 0 = ψ 1 = 1 ( Q ) ψ 0 = A Qe Q / Q where A is a constant. Again, using Eq.(), ψ 1 can be expresse in terms of x an normalize. Finally, it is easy to show that ψ n = 1 n! â n ψ 0. (64) (6) (63) 6