Phys460.nb Back to our example. on the same quantum state. i.e., if we have initial condition (5.241) ψ(t = 0) = χ n (t = 0)

Transcription

1 Phys46.nb 89 on the same quantum state. i.e., if we have initial condition ψ(t ) χ n (t ) (5.41) then at later time ψ(t) e i ϕ(t) χ n (t) (5.4) This phase ϕ contains two parts ϕ(t) - E n(t) t + ϕ B (t) (5.43) The first part is the dynamic phase factor. The second part is the Berry phase t ϕ B (t) χn (τ) i d χ n (τ) (5.44) For a time-dependent Hamiltonian, we change the Hamiltonian by making some of the control parameters time-dependent (e.g B x and B y ). Let s use R 1 (τ), R (τ) R n (τ) to represent these time-dependent parameters. d χ n (τ) χ n (τ) d R 1 (τ) R 1 + χ n (τ) R d R (τ) + + χ n (τ) R n dr n (τ) t dr 1 (τ) t dr (τ) t dr n (τ) ϕ B (t) χ n (τ) i R1 χ n (τ) + χ n (τ) i R χ n (τ) + + χ n (τ) i Rn χ n (τ) Ri dr 1 (τ) χ n (τ) i R1 χ n (τ) + Ri d R (τ) χ n (τ) i R χ n (τ) + + Ri dr n (τ) χ n (τ) i Rn χ n (τ) Ri dr χn (τ) i R χ n (τ) (5.45) (5.46) Here, all possible values of control parameters form a n-dimensional space, with coordinates R (R 1, R,, R n ). Each point in this space gives a set of values for my control parameters. For a time-dependent H, it corresponds to a curve in this space, R (τ). For each time τ, the value of my control parameters is one point in the space with coordinate R (τ). R is the n-dimensional gradient in this space and Ri d R is a line integral in this space alone the curve of R (τ). If the Hamiltonian at time t T goes back to the initial Hamiltonian at t, this integral becomes a loop integral before for t T, R i ϕ B (T) dr χ n (τ) i R χ n (τ) (5.47) This is really similar to the relation between magnetic flux and the vector potential Φ B d r A (5.48) Remember, Φ B also introduces a phase factor for wavefunctions (the AB effect), there is in fact a very strong analogy between Berry phase and magnetic flux. In some sense, the Berry phase is a magnetic flux in the parameter space. Comment #1: The general proof can be found in the textbook. Comment #: only works for adiabatic procedures. As we will show below, if we are not slow, the wavefunction will NOT satisfy ψ(t) e i ϕ(t) χ n (t). Thus, it makes no sense to take about phase factor Back to our example In our example, there are two control parameters B x and B y, i.e. we have a D parameter space. As time changes, we get a curve in this space (B x, B y ) (B sin α cos ω t, B sin α sin ω t) (5.49) which is a circle around the origin with radius B sin α. This circle defines my loop integral dr My wavefunction

2 9 Phys46.nb χ n (τ) χ + (τ) e i ω τ sin α (cos ω t + i sin ω t) sin α Bx + i By sin α B sin α B sin α (5.5) HEre, we used the fact that B sin α B x + B y Thus χ n (τ) (5.51) χ n (τ) i R χ n (τ) ( χ n (τ) i Bx χ n (τ), χ n (τ) i By χ n (τ) ) (5.5) χ n (τ) i R χ n (τ) is a two component vector. The first component is χ n (τ) i Bx χ n (τ) i Bx i 1 Bx +By () Bx - sin α (Bx +By ) 3/ i By -i Bx By (Bx +By ) 3/ sin α (5.53) B x - i B y B x + B y sin α i B y - i B x B y B x + B y 3/ sin α (B x - i B y ) (B x + i B y ) α B y sin B x + B y B y sin α B x + B y The second component χ n (τ) i By χ n (τ) B x - i B y B x + B y i By sin α i i B x - B x B y i i Bx +i By -() By (Bx +By ) 3/ sin α B x + B y 3/ sin i i Bx +By () By - sin α (Bx +By ) 3/ α - (B x - i B y ) (B x + i B y ) α B x sin B x + B y - B x sin α B x + B y (5.54) So sin α χ n (τ) i R χ n (τ) (B y, -B x ) - B x + B y Here, we convert to the polar coordinate r and ϕ. And sin α B x + B y ϕ (5.55) ϕ -B y B x + B y, B x B x + B y (5.56) is the unit vector along the circular direction ϕ B (T) dr χ n (τ) i R χ n (τ) π B x + B y d ϕ - sin α B x + B y -sin α π dϕ - π sin α -π(1 - cos α) π (cos α - 1) (5.57) Same as what we found above from directly solving for ψ(t).

3 Phys46.nb Resonance and transition rate Non-adiabatic regime If we change B fast enough (not slowly), then the system could jump/tunnel to a different quantum state P + > (5.58) Here, we can ask the question, which ω maximizes the probability for the system to change from one state to another? Here, for simplicity, we will only focus on the case of small α. B (B sin α cos ω t, B sin α sin ω t, B cosα) (5.59) At small α, B (α B cos ω t, α B sin ω t, B ) (5.6) The x and y component are much smaller than the z component. Notice that the x and y components depends on time, where the z component is independent of time, this small α regime is the case that time independent part dominate and the time-dependent part is small and can be treat as a small perturbation. H e m Bx By -i i + Bz 1-1 e m Bz e m Bx By -i i e B m α e B cos ω t 1 + sin ω t -i m 1 i We can rewrite the Hamiltonian as (5.61) H(t) H + α H ' (t) (5.6) where H e B m 1-1 ω (5.63) is the unperturbed Hamiltonian and it is independent of time. H ' (t) e B m cos ω t 1 + sin ω t -i 1 i ω 1 e -i ω t e i ω t (5.64) depends on time. Because α is small. This part is a small time-dependent perturbation. At small α, we have a time-independent system with Hamiltonian H. And we apply a small perturbation, which depends on time t, α H ' (t). H has two eigenstates, spin up and down, with eigenenergy + ω1 for the system to be in the other state is P - ω sin α ω + ω 1 - ω ω 1 cos α sin λ t and - ω1. If we start from the spin up state at t, at time t the probability (5.65) For small α, sinα α and cosα 1. λ ω + ω 1 - ω ω 1 cos α ω + ω 1 - ω ω 1 ω - ω 1 P - α ω ω - ω sin 1 t ω + ω 1 - ω ω 1 α ω ω - ω 1 t (ω - ω 1 ) sin sin (ω-ω1) t α ω ω - ω 1 (5.66) For any fixed t, the maximum of this function is at ω ω 1. P - α ω t. Remember, H has two eigenstates. And we are considering the transition between them. The energy difference between there two states is ΔE E + - E - ω 1 /. Thus, the maximum transition probability happens at

4 9 Phys46.nb ω ΔE/ (5.67) This conclusion turns out to be universal. If we have a time-independent system with multiple energy levels. We prepare the system in one eigenstate with energy E i, and then we apply a time-dependent perturbation with frequency ω. If we want the system to turn into a different quantum state with energy E f, the best frequency that we shall apply is ω E f - E i /. This is what we called a resonance. A resonance in quantum mechanics is not something really surprising. In classical mechanics, we have similar phenomenon. If a system has some intrinsic oscillation frequency ω (say a harmonic oscillator), then the system will response very strongly to external perturbations, if the frequency of the perturbation is the same as ω. In quantum mechanics, the intrinsic frequency ω is the energy difference between to eigenstates of H divided by Transition rate If we apply the perturbation for longer time, the probability for the transition increases. So to be fair, we shall define the transition rate R(t) dp -(t) dt α ω sin[(ω - ω 1) t] (ω - ω 1 ) d sin (ω-ω1) t d t α ω ω - ω 1 sin (ω-ω1) t cos (ω-ω1) t α ω (ω - ω 1 ) (ω - ω 1 ) (5.68) In the large t limit (i.e., if we wait for long enough time) lim t + R(t) α ω lim t + sin[(ω - ω 1 ) t] (ω - ω 1 ) In the long time limit, the transition rate is a constant (independent of time) α ω π δ(ω - ω 1 ) (5.69) R α ω π δ(ω - ω 1 ) α ω 1 π δ(ω - ω 1 ) (5.7) Here, the delta function tells us that if we want the transition to happen, when we wait for long enough time, we will find that the frequency of the perturbation have to match E f - E i /. Comment A: the infinite that comes with the δ function is not a problem, for two reasons 1. This is transition rate not probability. The probability is the integral of the transition rate P d t R(t). As long as the integral is finite (and probability smaller than 1), it is safe.. The transition rate above only consider perturbation with a unique frequency. In reality, we cannot really have a unique frequency. The energy must be distributed in some energy range. For example, if we shine light on a system, the light cannot really has one single frequency, the intensity must have some distribution I dω ρ(ω) (5.71) where I is the total intensity of the light, and ρ(ω) is the intensity density at some frequency. There, the total transition rate is R dω R(ω) ρ(ω) d ω ρ(ω) α ω π δ(ω - ω 1 ) α ω π ρ(ω 1 ) (5.7) After the frequency integral, R is no longer infinity. Comment B: the coefficient α ω1 π has one clear physical meaning f α H ' i ( 1 ) α ω 1 e -i ω t e i ω t 1 ω 1 ei ω t (5.73) Here i (1, ) is the initial state (spin up state at t ) and t >. We find that f (, 1) is the final state that they system will transit to at some later time

5 Phys46.nb 93 f α H ' i α ω 1 (5.74) In other words, R i f α ω 1 π δ(ω - ω 1 ) π f α H ' i π f α H ' i δ( E f - E i -ω ) δ(ω - ω 1 ) π f α H ' i δ ω - E f - E i (5.75) The transition rate from an initial state i to a final state is R i f π f α H ' i δ( E f - E i -ω ) (5.76) If we wait enough time, the transition will happen and can only happen when the frequency of the perturbation matches the energy difference between the final and initial states. If our perturbation do not have a unique frequency (in any real experiment, the frequency cannot be really unique), we shall need to use a function ρ(ω) to describe the intensity distribution. Here, R i f dω ρ(ω) R(ω) d ω ρ(ω) π π f α H ' i ρ E f - E i f α H ' i δ( E f - E i -ω ) (5.77)