Quantum Theory Assignment 3

Transcription

1 Quatum Theory Assigmet 3 Assigmet Cosider a spi-1/ system i a magetic field i the z-directio. The Hamiltoia is give by: ) eb H = S z = ωs z. mc a) Fid the Heiseberg operators S x t), S y t), ad S z t). Let us compute S z t). I this case: Ut) = exp i ωs ) z t. The Heiseberg equatio of motio for S z t) is: where we used ds z t) = 1 i [S zt), H] = 1 i U [S z, H]U = ω i U [S z, S z ]U = S z t) = S z ) = S z, [S z t), H] = [U S z U, H] = U S z UH HU S z U = U S z HU U HS z U = U [S z, H]U, sice [H, U] = [H, U ] =. Usig that [S i, S j ] = iɛ ijk S k, we obtai for S x ad S y : ds x t) = 1 i [S xt), H] = 1 i U [S x, H]U = ω i U [S x, S z ]U = ω S y t)

2 ad ds y t) whose solutio is = ωs x t). Therefore, we also have d S y t) = ω S y t), S y t) = A cosωt) + B siωt), with A ad B operators. Let us determie A ad B: S y ) = S y = A Fially, we have ds y t) ds yt) = ωs x t) = ωa siωt) + ωb cosωt) = ωs x ) = ωs x = ωb. t= S y t) = S y cosωt) + S x siωt) S x t) = S x cosωt) S y siωt) S z t) = S z. b) The system is i the Heiseberg state x, ± : S x x, ± = ± x, ±. The expectatio values of S x = S x ), S y = S y ), S z = S z ) are: x, ± S y x, ± = x, ± S z x, ± = x, ± S x x, ± = ±. Usig S x t), S y t), S z t) obtaied i a): x, ± S x t) x, ± = ± cosωt) x, ± S y t) x, ± = ± siωt) x, ± S z t) x, ± =. c) Fid the Heiseberg basis ket x, ±; t, correspodig to x, ± at t=. x, ±; t = U x, ± = e iωszt/ 1 + ± ) = 1 e iωt/ + ± e iωt/ ).

3 d) Compute x, ±; t x, ±. x, ±; t x, ± = 1 4 = 1 4 e iωt/ + ± e iωt/ ) + ± ) e iωt/ + e iωt/ = 1 cos ωt/) 4 = cos ωt/) = cosωt)). e) Compute x, ; t x, ±. x, ; t x, ± = 1 4 = 1 4 e iωt/ + e iωt/ ) + ± ) e iωt/ e iωt/ = 1 i si ωt/) 4 = si ωt/) = 1 1 cosωt)). f) Calculate the Heiseberg basis kets y, ±; t. We have S y y, ± = ± Compute the probabilities y, ± ad y, ± = 1 + ± i ). Therefore, y, ±; t = U y, ± = e iωs zt/ 1 + ± i ) y, ±; t x, ± = 1 4 = 1 4 y, ; t x, ± = 1 4 = 1 e iωt/ + ± ie iωt/ ) e iωt/ + ie iωt/ ) + ± ) e iωt/ ie iωt/ = siωt)) e iωt/ + ie iωt/ = 1 1 siωt)).

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8 Assigmet Cosider the harmoic oscillator state at t = give by: α = ). a) Calculate α Ht) α. From the Heiseberg equatio of motio, we have: dh = Ht) = H = ω N + 1 ). The expectatio value of the Hamiltoia becomes: α Ht) α = ω α N α + 1 ) [ 1 = ω ) H ) + 1 ] = ω [ ) + + 1) + 1 ) + 1] = ω [ ] = ω + 1). b) Calculate α xt) α. Usig at) = a e iωt ad a t) = a e iωt, we have: x = ) a + a xt) = a e iωt + a e iωt), mω mω α xt) α = To compute this expectatio value, we eed: mω α a e iωt + a e iωt) α a α = ) a α = ).

9 α xt) α = c) Compute α pt) α. = = mω mω [ 1 ) e iωt α ) ] e iωt α e iωt + e iωt) + 1) cosωt). mω Usig dxt) = pt) m ad d xt) = pt), we have m Therefore, we obtai: m d α xt) α = α pt) α α pt) α = m + 1) ω siωt)). mω mω α pt) α = + 1) siωt).

10 Assigmet 3.4 Cosider a oe-dimetioal simple harmoic oscillator H = ω a a + 1 ), E = ω + 1 ). The time-depedet perturbatio is V t) = F x cosωt), so that { H t Ht) = H + V t) t >. For t, the system is i state. At t >, the state i the Schrödiger picture is: α, t S = c t) e ie t/, with c ) = δ. Let us fid a perturbative expasio of α, t S by computig We have with ω = E E. To compute c 1) t), we eed V t): c t) = c ) t) + c 1) t) + c ) t) +... c ) t) = δ c 1) t) = i t e iω t V t ), V t) = F x cosωt) = F cosωt) x = F cosωt) δ 1 mω Therefore, c 1) t) = for all except = 1: c 1) 1 t) = i F mω = i F mω t t e iω 1t cosωt ) e iω t cosωt ),

11 sice ω 1 = E 1 E We have that: Fially, = ω = ω. t To first order, we have e iω t cosωt ) = 1 c 1) 1 t) = F mω = 1 i t e iω +ω)t + e iω ω)t ) [ e iω +ω)t 1 ω + ω [ e iω +ω)t 1 ω + ω ] + eiω ω)t 1 ω ω ] + eiω ω)t 1. ω ω α, t S = e iω t/ + e i3 ω t/ c 1) 1 t) 1. Let us compute x to lowest o-vaishig order: ) x = x + e iω t c 1) 1 t) x 1 + e iω t c 1) 1 t) 1 x = Re e iω t c 1) 1 t) x 1, sice x = ad 1 x = x 1 = x 1. Note: The term c 1) 1 t) 1 x 1 is here equal to zero. However, it would oly cotribute to secod order so that, to compute x to lowest o-vaishig order, we would eglect this term eve if ot zero. To first order, x is ) x = Re e iω t c 1) 1 t) mω = F e iωt e iω t Re mω ω + ω = F cosωt) cosω t) m ω ω = F m si ω+ω t) si ω ω t) ω ω = F siω + t) siω t), m Ω + Ω ) + e iωt e iω t ω ω where Ω + = ω + ω ad Ω = ω ω. Let us check the validity of the expasio of x for ω ω. Rewrite x as: x = F siω +t) siω t) m t. Ω + t Ω t

12 For ω ω, Ω ad siω t) Ω t 1. Cosequetly, sice Ω + ω, we have x F mω t siω t). At t =, x = which is cosistet with α, t = S =. For t >, x grows while oscillatig, cosistet with a classical harmoic oscillator drive at resoace frequecy. Assigmet 3.5 Cosider a system of two spi-1/ particles. The Hamiltoia is give by t < Ht) = ) 4 S 1 S t, where S 1 ad S are the spi operators of particles 1 ad. a) Note that S 1 S = 1 Therefore, H = ) S S 1 S. ) S S 1 S ) ad commutes with S, S z, S 1, S. The eigestates of H are the states s, m : s = 1, m = ±1, E = ) 3 ) = s =, m = E = 3 ) = 3. The eigekets ad eigevalues are s = 1, m = ±1,, E = s =, m =, E = 3. The states +, +, +,,, +,, are eigekets of the operators S 1, S 1z, S, S z. They ca be writte i terms of the kets s, m as +, + = 1, 1, = 1, 1 +, = 1 1, +, ), + = 1 1,, ).

13 At t =, the system is i the state At t >, the state α, t is give by α, t = = +,. α, t = e iht/ α, t = = e iht/ + = 1 e i t/ 1 + e i3 t/ ) = 1 [ e i t/ +, +, + ) + e i3 t/ +,, + ) ]. Let us compute the probability of fidig the system i the states +, +, +,,, +,, : Note that the probabilities sum to 1. P ++) = + + α, t = P ) = α, t = P + ) = + α, t = 1 e i t/ + e i3 t/ 4 = 1 e i t/ e i t/ + e i t/) 4 ) t = cos P +) = + α, t = 1 e i t/ e i3 t/ 4 = 1 e i t/ e i t/ e i t/) 4 ) t = si. b) We ow solve the problem usig perturbatio theory. The state at time t is give by 4 α, t = c t)e iet/, =1 where { } = { +, +, +,,, +,, }. At t =, α, t = i = +,. We have: c ) c 1) = δ i = i t e iω it V i t ),

14 where ω i = E E i = sice Ht < ) =. Let us compute V i t ) = V t ) i = H i for t > : V i = H + = H ) = ). Therefore, we have: ++ V + = V + = + V + = 1 3 ) = + V + = ) =. The probability of fidig the system i a state { } = { +, +, +,,, +,, } is P ) = c t) = c ) t) + c 1) t) + c ) t) +... = c ) t) + λc 1) t) + λ c ) t) +... = c ) t) [ c + λrec ) t) c 1) t)) + λ 1) t) ] + Rec ) t) c ) t)) + Oλ 3 ), where we itroduced a parameter λ to idicate the order of the perturbatio we will set λ = 1 at the ed of the calculatio). For a state differet tha the iitial state + i), we have c ) = δ i = P i) = λ c 1). Therefore, the leadig order for the probability is secod order ad P i) = P ++) = P ) = P +) = i t ) t =. c 1) To compute P + ), ote that P ) = 1 ad this idetity must hold at ay order: P ) = P ) ) + λ P 1) ) + λ P ) ) +... = 1 P ) ) = 1 ad P k ) ) =. :

15 Here, we computed P ++), P ), P +) to secod order so that P +) to secod order is give by: P + ) = 1 P ++) P ) P +) ) t = 1. You could have tried to solve the problem by directly computig P = i) = P + ). The probability of fidig the system i the iitial state is [ c ] P = i) = c =i t) = 1 + λrec 1) i t)) + λ 1) i t) + Rec ) i t)) + Oλ 3 ), where we used c ) =i = δ ii = 1. Therefore, to compute P = i) to secod order, we also eed c ) t): c ) =i = 1 c 1) =i c ) =i = t = i = i t i ) 4 t t V im t )V mi t ) m=1 = 1 4 t V im V mi = 1 t m=1 4 V im m=1 = 1 t + 4 ) = 5 ) t. where we used that ω mi =, V im is time-idepedet, ad V mi = Vim. Fially, to secod order, we have [ c ] P + ) = 1 + Rec 1) 1) i t)) + i t) + Rec ) i t)) [ c ] 1) = 1 + i t) + c ) i t) ) t = ) t ) t = 1, as obtaied above.

16 These perturbative results are valid for t the probabilities should be less tha 1). For t, the results foud i a) ca be expaded as: ) t P + ) = cos [ 1 1 ) ] t ) ) ) 4 t t = 1 + O ) t P +) = si ) t.