Ph12b Solution Set 4

Transcription

1 Ph Solution Set 4 Feruary 5, 00 The state ψ S un E is transforme to p ψ S un E + p a 0 S γ E + S η E Since γ E an η E aren t mutually orthogonal, we can t use this equation irectly to compute the new ensity operator. Instea, we must change to an orthonormal system, as suggeste in the hint. One way to o this is to keep γ E an to take δ E as the secon asis vector, where δ E is etermine y η E ɛ γ E + ɛ ɛ δ E Sustituting into, we get that the new state is p ψ S un E + pa 0 S + ɛ S γ E + p ɛ ɛ S δ E 3 Rememer that if we have a state ψ i S f i E i with f i E normalize an mutually orthogonal, then we calculate the ensity operator y the formula ˆρ i ψ i S ψ i S Working in the asis 0 S, S of S, we have p ψ S a p a p a 0 S + ɛ S p ɛ p ɛ ɛ S p 0 ɛ ɛ

2 so from 3 we get that the new ensity operator is ˆρ p a a + p a ɛ a ɛ a pɛ a pɛa Comparing this to the original ensity operator ˆρ ψ S ψ S a a a a a lets us rea off that λ pɛ. + pɛ ɛ 0 0 a We egin y analysing the evolution of a single term of the form ψ ψ. If at time t we have ψt ψt, then at time t + t we have ψt + t ψt + t ψt iωt e e ψt ψt + iωt ψt e e ψt ψt iωt e e ψt ψt + iωt ψt ψt e e + ω t e e ψt ψt e e Notice that the right han sie is linear in ψt ψt. Since a ensity operator is a linear comination of ψ i t ψ i t terms, y summing the previous equation over such terms, for an aritrary ensity operator ˆρt we have ˆρt + t ˆρt iωt e e ˆρt + iωtˆρt e e + ω t e e ˆρt e e Now taking the limit of ˆρt+t ˆρt t as t 0 gives ˆρ iω e e ˆρ + iω ˆρ e e 4 t as require. In the asis { g, e }, we have e 0 so e e an Eq. 4 ecomes ρgg ρ iω ge t ρ eg ρ ee 0 ρ eg ρ ee 0 iωρge iωρ eg 0 + iω Componentwise, we get the 4 ifferential equations t ρ gg 0 t ρ ge iωρ ge t ρ eg iωρ eg t ρ ee 0 0 ρ eg ρ ee 0

3 The solutions of these equations, in terms of initial conitions at t 0, are ρ gg t ρ gg 0 ρ ge t e iωt ρ ge 0 ρ eg t e iωt ρ eg 0 ρ ee t ρ ee 0 so in terms of components of ˆρ0, ˆρt is given y ρgg 0 e ˆρt iωt ρ ge 0 e iωt ρ eg 0 ρ ee 0 c Using the same approach as in part a, if at time t we ha a pure state ˆρt ψt ψt, then at time t + t we have the ensity operator ˆρt + t g g ψt + Γt e e ψt ψt g g + Γt ψt e e + Γt g e ψt Γt ψt e g which can e rewritten as ˆρt + t g g ˆρt g g + Γt e e ˆρt g g + Γt g g ˆρt e e + Γt e e ˆρt e e + Γt g e ˆρt e g Moreover, since this equation is linear in ˆρt, it will still hol for an aritrary mixture of states. Since g, e form an orthonormal asis of the atom s Hilert space, we have g g + e e which we use with the previous equation to get ˆρt + t ˆρt t Γt Γt e e ˆρt g g + g g ˆρt e e t t Γ e e ˆρt e e + Γ g e ˆρt e g an taking the limit t 0, this ecomes ρ t Γ e e ˆρ g g Γ g g ˆρ e e Γ e e ˆρ e e + Γ g e ˆρ e g Again using g g + e e, this can e simplifie to ρ t Γ e e ˆρ Γ ˆρ e e + Γ g e ˆρ e g 5 proving the atom s master equation. 3

4 Since g 0 an e 0, we can rewrite Eq. 5 in the matrix form as t ρ eg ρ ee Γ Γ 0 ρ eg ρ ee 0 ρgg ρ + Γ ge ρ eg ρ ee Γρee Γ ρ ge Γ ρ eg Γρ ee 0 Componentwise, we get the 4 ifferential equations t ρ gg Γρ ee t ρ ge Γ ρ ge t ρ eg Γ ρ eg t ρ ee Γρ ee ρ eg ρ ee 0 The one for ρ gg is couple with ρ ee, so we start y solving the other three, ρ ge t e Γt/ ρ ge 0 ρ eg t e Γt/ ρ eg 0 ρ ee t e Γt ρ ee 0 Now integrating the equation for ρ gg /t from 0 to t we get ρ gg t ρ gg 0 + e Γt ρ ee 0 so in terms of components of ˆρ0, ˆρt is given y ρgg 0 + e ˆρt Γt ρ ee 0 e Γt/ ρ ge 0 e Γt/ ρ eg 0 e Γt ρ ee 0 e Comining the two ifferential equations gives a new ifferential equation, an we can comine them at the componentwise level, getting t ρ gg Γρ ee t ρ ge Γ + iω ρ ge t ρ eg Γ iω ρ eg t ρ ee Γρ ee 4

5 As in part, we start y solving the uncouple ifferential equations, getting ρ ge t e Γ +iωt ρ ge 0 ρ eg t e Γ iωt ρ eg 0 ρ ee t e Γt ρ ee 0 after which we integrate the equation for ρ gg, getting ρ gg t ρ gg 0 + e Γt ρ ee 0 To summarize, in terms of components of ˆρ0, ˆρt is given y ρ ˆρt gg 0 + e Γt ρ ee 0 e Γ +iωt ρ ge 0 e Γ iωt ρ eg 0 e Γt ρ ee 0 3 a Since we have / proaility for each of the states ψ H, ψ T, the ensity operator is ˆρ ψ H ψ H + ψ T ψ T 0 + / / / 0 / 3/4 /4 /4 /4 Eigenvalues λ of ˆρ satisfy the characteristic equation which has the solutions 3 4 λ 4 λ λ ± ± 8 6 c From the equation 3/4 λ± /4 u± 0 /4 /4 λ ± v ± we get u ± + v ± 0. If we focus on normalize eigenvectors, u ± + v±, an comining these two equations gives, after some algera, u± ± / v ± ± / 5

6 By comparing these with the values of cosπ/8, sinπ/8 which you can calculate using the half-angle formulas, you may notice that u+ cosπ/8 v + sinπ/8 u sinπ/8 v cosπ/8 Another reason to suspect that the angle π/8 woul have something to o with the solution is just y noticing that ψ H, ψ T are vectors with phases 0 an π/4, an since they contriute equally to the ensity operator, it is to e expecte that the ensity operator has some sort of symmetry aroun the axis with angle π/8. 4 a Let M e the matrix ψ a N a,. By the singular value ecomposition, there exist unitary matrices U u ai N a,i an V v j N j, such that M UDV with D ij N i,j a iagonal real N N matrix with nonnegative eigenvalues. In componentwise notation, M UDV ecomes ψ a u ai ij v j i,j which we can use to express ψ as ψ a, a,i,j, ψ a e a f u ai ij v j e a f N N ij u ai e a v j f i,j a Now we use the conition that U an V are unitary. Rememer that unitary matrices correspon to transformations from one orthonormal asis to another. Treating U as a unitary matrix acting on H A, we see that e i u ai e a a efines a new orthonormal asis of H A, an treating V as a unitary matrix acting on H B, we see that f j 6 v j f

7 efines a new orthonormal asis of H B note that here the summation is in the secon inex of v j, so this transform actually correspons to V T, ut that is irrelevant ecause the transpose of a unitary matrix is again unitary. With these new ases for H A an H B, the state ψ is represente as ψ ij e i f j i,j But rememer that D is a iagonal real matrix with nonnegative eigenvalues! Thus, ij 0 for i j. Denote also p i ii. Then the previous equation ecomes ψ pi e i f i 7 i Since we know that the state is normalize, ψ ψ, which can, y the formula given in the prolem, e expresse componentwise as p i i Thus, N i p i. From Eq. 7, since f i are orthonormal vectors, we get ˆρ A p i e i e i 8 i Since { e i } form an orthonormal asis, in that asis ˆρ a is a iagonal matrix, p p 0 ˆρ A p N Analogously, the ensity operator for system B can e written in asis { f i } as p 0 0 ˆρ B p i f i f i 0 p 0. i p N c Since from part we have representations of ˆρ A an ˆρ B as iagonal matrices, their eigenvalues are just the elements on the iagonal. Thus, reaing off from Eqs. 9 an 0 we see that ˆρ A an ˆρ B have the same eigenvalues p, p,..., p N. 7