Coherent states of the harmonic oscillator

Transcription

1 Coherent states of the harmonic oscillator In these notes I will assume knowlege about the operator metho for the harmonic oscillator corresponing to sect..3 i Moern Quantum Mechanics by J.J. Sakurai. At a couple of places I refefer to this book, an I also use the same notation, notably x an p are operators, while the corresponig eigenkets are x etc. 1. What is a coherent state? Remember that the groun state 0, being a gaussian, is a minimum uncertainty wavepacket: Proof: Since it follows that x = h mω (a + a ) p = mω h (a a ) 0 (a + a )(a + a ) 0 = 0 aa 0 = 1 (1) 0 (a a )(a a ) 0 = 0 aa 0 = 1 () x 0 p 0 = h h 1( 1) = 4 4 an finally since x 0 = p 0 = 0, it follows that ( x) 0 ( p) 0 = h 4 We can now ask whether n is also a minimum uncertainty wave packet. Corresponing to (1) an () we have n (a + a )(a + a ) n = n aa + a a n = n a a + [a, a ] n = n + 1 an similarly which implies n (a a )(a a ) n = (n + 1) (3) ( x) n ( p) n = h 4 (n + 1) (4) 1

2 so n is not minimal! Clearly a crucial part in 0 being a minimal wave packet was a 0 = 0 0 a a 0 = 0 this correspons to a sharp eigenvalue for the non-hermitian operator mωx+ip even though, as we saw, there was (minimal) ispersion in both x an p. It is natural to expect other minimal wave-packets with non zero expectation values for x an p but still eigenfunctons of a, i.e. a α = α α (5) which implies α a a α = α. It is trivial to check that this inee efines a minimal wave-packet α (a + a ) α = (α + α ) α (a a ) α = (α α ) α (a + a )(a + a ) α = (α + α ) + 1 α (a a )(a a ) α = (α α ) 1 from which follows an accoringly ( x) α = x α x α = h mω ( p) α = p α p α = hmω ( x) α ( p) α = h 4 (6) So the states α, efine by (6), satisfies the minimum uncertainty relation. They are calle coherent states an we shall now procee to stuy them in etail.

3 . Coherent states in the n-representation In the n base the coherent state look like: α = n c n n = n n n α (7) Since n = (a ) n 0 (8) we have an thus n α = αn 0 α (9) α = 0 α n (10) The constant 0 α is etermine by normalization as follows: 1 = n solving for 0 α we get: α n n α = 0 α m=0 α m m! = 0 α e α 0 α = e 1 α (11) up to a phase factor. Substituting into (10) we obtain the final form: α = e 1 α n (1) Obviously α are not stationary states of the harmonic oscillator, but we shall see that they are the appropriate states for taking the classical limit. A very convenient expression can be erive by using the explicit expression (8) for n : which implies n = (a ) n 0 = e αa 0 α = e 1 α +αa 0 = e αa α a 0 (13) 3

4 3. Orthogonality an completeness relations We procee to calculate the overlap between the coherent states using (1). an similarly so or α β = n α n β = e 1 α 1 (α β) n β = exp ( 1 α 1 β + α β) (14) β α = exp ( 1 α 1 β + β α) (15) α β = α β β α = exp ( α β + α β + αβ ) α β = e α β. (16) Since α β 0 for α β, we say that the set { α } is overcomplete. There is still, however, a closure relation: α α α = (α ) n α m α e α m n (17) m! where the measure α means summing over all complex values of α, i.e. integrating over the whole complex plane. Now, writing α in polar form: we get α e α (α ) n α m = m,n α = re iφ α = φ r r (18) 0 = πδ m,n 1 π r re r r m+n φ e i(m n)φ Using this we finally get: α α α = π n 0 0 r (r ) m e r = πm!δ m,n n n = π n or equivalently α π α α = 1 (19) 4

5 4. Coherent states in the x-representation Remember that x 0 is a minimal gaussian wave packet with x = p = 0. Since x α is also a minimal wave packet an or R(α) = α a + a α = I(α) = α a a α = i x α = h mω R(α) p α = mω h I(α) mω α x α h (0) 1 α p α mω h (1) () it is natural to expect that x α is a isplace gaussian moving with velocity v = p/m. We now procee to show this. Using the previously erive form (13) of the coherent states expresse in terms of the groun state of the harmonic oscillator we get x α = x e 1 α +αa 0 = e 1 α x e α mω ip (x h mω ) 0 Acting from the left with x gives us 1 : x α = e 1 α e α mω h (x i mω ( i h x )) x 0 = Ne 1 α e α (x x x 0 0 x ) e 1 ( x x 0 ) (3) where we have use the explicit form of x 0 an introuce the constants: x 0 = h mω N = 4 mω π h (4) For notational simplicity we also put y = x /x 0. With these substitutions equation (3) will look like: Using the commutator relation 1 c.f. Sakurai p.93 x α = Ne 1 α e α (y y ) e 1 y (5) e A+B = e 1 [A,B] e A e B (6) 5

6 which is vali if both A an B commute with [A, B], we get e α (y y ) e 1 y = e α 4 + α y e α y e 1 y (7) an thus, noting that e a y, is a translation operator x α = Ne 1 α 1 y 1 α + αy = N exp ( 1 (y R(α)) + i I(α)y ii(α)r(α)) (8) Using () the resulting expression for the wavefunction of the coherent state is x α = Ne mω h (x x α) + ī h p αx i h p α x α (9) an since the last term is a constant phase it can be ignore an we finally get which is the promisse result. ψ α (x ) = ( mω π h )1/4 e ī h p αx mω h (x x α), (30) 5. Time evolution of coherent states The time evolution of a state is given by the time evolution operator U(t). Using what we know about this operator an what we have learne so far about the coherent states we can write: α, t = U(t, 0) α(0) = e ī h Ht α(0) = e ī h Ht e 1 α(0) But the n :s are eigenstates of the hamiltonian so: which is the same as α, t = e 1 α(0) n n (α(0)) n n (31) (α(0)) n e ī h ω h(n+ 1 )t (a ) n 0 (3) α, t = e 1 α(0) e i ωt n (α(0)e iωt a ) n 0 = c.f. Sakurai chapter.1 exp ( 1 α(0) i ωt + α(0)e iωt a ) 0 (33) 6

7 Comparing this expression with (13), it is obvious that the first an the thir term in the exponent, operating on the groun state, will give us a coherent state with the time epenent eigenvalue e iωt α(0) while the secon term only will contribute with a phase factor. Thus we have: α, t = e i ωt e iωt α(0) = α(t) (34) So the coherent state remains coherent uner time evolution. Furthermore, or in components Defining the expectation values, α(t) = e iωt α(0) α(t) = iωα(t) (35) t { R(α) = ωi(α) t I(α) = ωr(α) (36) t we get { x(t) = α(t) x α, t p(t) = α(t) p α, t x(t) = h R(α) = h p(t) ωi(α) = t mω t mω m p(t) = i m hω ( i) I(α) = m hω ωr(α) = t t mω x(t) or in a more familiar form { p(t) = m x(t) = mv(t) t p(t) = t mω x(t) (37) (38) (39) i.e. x(t) an p(t) satisfies the classical equations of motion, as expecte from Ehrenfest s theorem. In summary, we have seen that the coherent states are minimal uncertainty wavepackets which remains minimal uner time evolution. Furthermore, the time epenant expectation values of x an p satifies the classical equations of motion. From this point of view, the coherent states are very natural for stuying the classical limit of quantum mechanics. This will be explore in the next part. 7