Las reformas electorales - EL POLO DEMOCRÁTICO

2. EL POLO DEMOCRÁTICO

2.6 Las reformas electorales

When the perturbation varies in time, we must use the time-dependent Schr¨ o-dinger equation,

[ ˆH⁽⁰⁾+ ˆH⁽¹⁾(r, t)]Ψ = i~∂Ψ

∂t , (5.35)

where

Hˆ⁽⁰⁾Ψ_i= E_i⁽⁰⁾Ψ_i, and Ψ_i= ψ_ie^−(i/~)Eⁱ⁽⁰⁾^t.

We assume the ψ_ito form a complete orthonormal basis set (hi|ji = δ_ij), and the expansion coefficients will themselves become time-dependent so that

Ψ =X

an(t)Ψn. (5.36)

Using Equation (5.36) in Equation (5.35), we find ( ˆH⁽⁰⁾+ ˆH⁽¹⁾)X

anΨn=X

(i~ ˙aⁿ+ E_n⁽⁰⁾an)Ψn.

We now take the scalar product with hΨk|, obtaining X

[a_nhk| ˆH⁽¹⁾|ni − i~ ˙anδ_kn]e^(i/~)(E^k⁽⁰⁾^−E⁽⁰⁾ⁿ ^)t = 0 .

If we define

ωkn≡ 1

~(E⁽⁰⁾_k − E_n⁽⁰⁾) , then this may be written

i~ ˙a^k=X

anH⁽¹⁾_kne^iω^kn^t. (5.37)

For a large number of terms in the summation, this is difficult (but not im-possible) to solve. We will thus restrict our attention to problems where the initial state is given, so that at t = 0, we know that the system is in state j

and a_j(0) = 1 and all the rest are initially zero. Then for short times, a_j(t) is still close to unity, and the others are still small, so we may approximate

i~ ˙a^k' H⁽¹⁾_kje^iω^kj^t. This equation has solutions

a_j(t) = 1 − i

~ Z t

H⁽¹⁾_jj(t⁰) dt⁰, (5.38)

ak(t) = −i

~ Z t

H⁽¹⁾_kj(t⁰)e^iω^kj^t⁰dt⁰, k 6= j . (5.39) Problem 5.10 Decaying perturbation. A particle in a one-dimensional box (infinite potential well) is initially in the ground state. At time zero, a weak perturbation of the form ˆH⁽¹⁾(x, t) = ˆxe^−t² is turned on. If the box extends from x = 0 to x = a, find the probability that the particle will be found in the first excited state as t → ∞.

Problem 5.11 Perturbed harmonic oscillator. A particle is initially in the ground state of a one-dimensional harmonic oscillator. At time zero, a per-turbation of the form ˆH⁽¹⁾(x, t) = V0(αˆx)³e^−t/τ is turned on (αx = q).

(a) Calculate the probability that, after a sufficiently long time (as t → ∞), the system will have made a transition to an excited state. Consider all possible final states.

(b) Find the ratio of the probability that the system will be in a state higher than n = 1 to the probability that the system will be in the n = 1 state.

Evaluate this ratio for ωτ 1 and for ωτ 1.

(c) For fixed V0/~ω = β, find the probability that the system remains in the ground state, and evaluate P0 for ωτ 1 and for ωτ 1.

5.4.1 Perturbations That Are Constant in Time

If the perturbation ˆH⁽¹⁾ is constant in time after being turned on at t = 0, then the integrals in Equation (5.38) and Equation (5.39) may be evaluated immediately, with the result

aj(t) = 1 − i

H⁽¹⁾_jj t , (5.40)

a_k(t) = H⁽¹⁾_kj

~ωkj

(1 − e^iω^kj^t) , k 6= j . (5.41)

Now the probability that the state Ψkis occupied at time t is given by |ak(t)|², so it amounts to the probability that the system has made a transition from

Approximation Methods 135 state Ψ_j to Ψ_k. We may thus write from Equation (5.41)

P_k(t) = |a_k(t)|²=

2H_kj⁽¹⁾ E_k⁽⁰⁾− E⁽⁰⁾_j

sin²(E_k⁽⁰⁾− E_j⁽⁰⁾)t

2~ , k 6= j . (5.42) We note from this expression that the probability oscillates at frequency ωkj

(since sin²ωkjt/2 = (1 − cos ωkjt)/2), but that due to its resonant denomina-tor, it is large only where E⁽⁰⁾_j is close to E⁽⁰⁾_k . We also note that the assumed smallness of a_k(t) requires either |H⁽¹⁾_kj| |E_k⁽⁰⁾− E⁽⁰⁾_j | or ω_kjt 1.

−5π −4π −3π −2π −π 0.00 π 2π 3π 4π 5π

0.2 0.4 0.6 0.8 1.0

...

FIGURE 5.1

Plot of sin²x/x² where x = Et/2~, illustrating the δ function character of sin²x/x².

Now for large t, letting E ≡ E_k⁽⁰⁾− E_j⁽⁰⁾, the oscillating term resembles a delta function δ(E) as illustrated in Figure 5.1 where the only significant con-tribution is near the origin in E. If we assume there exists a near continuum of final states, characterized by the density of final states N (E) = ρ(E) dE, where N (E) is the number of states between E and E + dE, then the area under the graph of |ak(E, t)|²versus E represents the total transition proba-bility. Hence,

P (t) = Z ∞

−∞

|ak(E, t)|²ρ(E) dE = |2H⁽¹⁾_kj|²ρ(E) t 2~

Z ∞

−∞

sin²x x² dx

= 2πt

|H_kj⁽¹⁾|²ρ(E) , (5.43) which indicates that the probability of making a transition is linearly increas-ing with time after the perturbation is turned on. If we look rather at the

transition rate, then the result is R = 2π

|H⁽¹⁾_kj|²ρ(E) , (5.44) which is Fermi’s Golden Rule (Number 2).

Problem 5.12 Two-state problem. If there are only two states, the system oscillates back and forth between them. Assuming only states j (initial state) and k exist (i.e., for which H_kj⁽¹⁾ 6= 0), find the oscillation frequency (or fre-quencies) and aj(t) and ak(t).

5.4.2 Perturbations That Are Harmonic in Time

We now assume that the perturbation oscillates in time, again having been turned on at time t = 0, and we shall assume the form of the perturbation is Hˆ⁽¹⁾(r, t) = 2 ˆH⁽¹⁾(r) cos ωt = ˆH⁽¹⁾(e^iωt+ e^−iωt) . (5.45) Substituting this into Equation (5.39), the integrals are straightforward, so that

ak(t) = −i

~ H⁽¹⁾_kj

Z t 0

[e^i(ω^kj^+ω)t⁰+ e^i(ω^kj^−ω)t⁰] dt⁰

= −H⁽¹⁾_kj e^i(ω^kj^+ω)t− 1

~(ωkj+ ω) +e^i(ω^kj^−ω)t− 1

~(ωkj− ω)

. (5.46)

There are two special cases. When ω ∼ −ωkj, the first term in Equation (5.46) dominates the second term, in which case

~ω ∼ Ej⁽⁰⁾− E_k⁽⁰⁾.

This corresponds to stimulated emission since E_j⁽⁰⁾ > E_k⁽⁰⁾ and the tran-sition is from an initial state with higher energy to a final state with lower energy. The probability is of course maximum when ~ω = Ej⁽⁰⁾− E⁽⁰⁾_k . The other case is where the second term dominates in Equation (5.46), or when ω ∼ ωkj. In this case, we have resonant absorption, since

~ω ∼ Ek⁽⁰⁾− E_j⁽⁰⁾,

and the transition is from a lower energy state to higher energy state. If the perturbation is due to an electromagnetic wave, the perturbating force is an incident photon with frequency ω. Then the two cases can be visualized in figures (a) and (b) of Figure 5.2. Figure (a) shows one photon coming in and two coming out. The second photon is stimulated and is emitted in phase with the incident photon. Figure (b) shows nothing coming out as the perturbing

Approximation Methods 137 Ej

...

•

(a)

... ...

...

Ek ...

• (b) ...

E_k

...

•

(c)

...

FIGURE 5.2

Three kinds of interaction between photons and atoms. (a) Stimulated emis-sion. (b) Absorption. (c) Spontaneous emisemis-sion. The first two are resonant interactions while the last is nonresonant

photon is resonantly absorbed. Figure (c) represents spontaneous emission and is a nonresonant process. The relationships between these emission and absorption processes will be treated in more detail in Chapter 9 where their relation to masers and lasers will be discussed.

When neither term is near resonance, a_k is small, so the general condition for a significant transition probability is that ω = ±ω_kj. Since only one of the terms is important at a time, we may write the transition probability from Equation (5.46) as

|ak(t)|²=

2H_kj⁽¹⁾

~(ω ± ωkj)

sin² (ω ± ωkj)t 2

, (5.47)

where the minus sign corresponds to absorption and the plus sign to emission.

Doing the same integral over energy as in the previous case, the transition rate from state j to state k is given by

Rk= 2π

|H⁽¹⁾_kj|²ρ(Ek)δ(Ekj∓ E) , (5.48) where E is the photon energy if the perturbing force is an electromagnetic field.

Problem 5.13 Dipole matrix elements. Calculate all of the dipole matrix elements for the transition n = 2 to n = 1 for the one-electron atom, i.e., find all of the nonzero elements of the form hdi ≡ hψ2`m|qr|ψ100i for hdxi, hdyi, and hdzi.

5.4.3 Adiabatic Approximation

When the perturbation is turned on slowly, so that the characteristic fre-quency changes little over a period, then we may estimate the probability of a transition from Equation (5.37), where we again assume the initial state to be j and that for nominally short times, only transitions from the initial state need be considered, then

i~ ˙a^k' H⁽¹⁾_kje^iω^kj^t.

Assuming now that H⁽¹⁾_kj depends on time, we may integrate by parts to obtain a_k(t) = −i

~ Z t

H_kj⁽¹⁾(t⁰)e^iω^kj^t⁰dt⁰

= − 1

~ωkj

H⁽¹⁾_kj(t⁰)e^iω^kj^t⁰

+ 1

~ωkj

Z t 0

dt⁰H⁽¹⁾_kj(t⁰)e^iω^kj^t⁰dt⁰

= − 1

~ωkj

H_kj⁽¹⁾(t)e^iω^kj^t+ 1

~ωkj

Z t 0

dt⁰H⁽¹⁾_kj(t⁰)

e^iω^kj^t⁰dt⁰ (5.49) since H⁽¹⁾_kj(0) = 0 in the adiabatic approximation. Since H⁽¹⁾_kj(t) is slowly varying, the remaining integral may be made very small, so the adiabatic approximation normally includes only the leading term.

Example 5.3

Equilibrium shift. We consider a harmonic oscillator that undergoes a linear shift of the equilibrium position from 0 to x1, where x1 is a constant. The equilibrium position is given by

x₀(t) = x₁t/t₀0 ≤ t ≤ t₀ x₁ t ≥ t₀ We may write the Hamiltonian then as

Hˆ⁽⁰⁾= pˆ²

2m+¹₂kx², t ≤ 0 H =ˆ pˆ²

2m+¹₂k[x − x₀(t)]², t > 0 Hˆ⁽¹⁾= ˆH − ˆH⁽⁰⁾= ¹₂k [x0(t)]²− kxx0(t) .

We take the system to be initially in the ground state, so ψj = ψ0(x). We wish to find the probability of finding the system in state ψ1, so k = 1. Then the matrix element is given by

H⁽¹⁾₁₀ = hψ1|¹₂kx0(t)²|ψ0i − hψ1|kxx0(t)|ψ0i

= 0 −kx₀(t)

α hψ1|q|ψ0i

= −kx0(t)

√2α ,

Approximation Methods 139 where q = αx. Then we have ω₁₀= (³₂~ω −¹₂~ω)/~ = ω so

a1(t) = kx0(t)

~ω r

2mωe^iωt=αx0(t)

√

2 e^iωt.

and the probability of finding the system in state ψ1 is |a1(t)|² = q²₁t²/2t²₀ where q1= αx1for 0 ≤ t ≤ t0 and |a1|²= q₁²/2 for t > t0.

Problem 5.14 The neglected term in the adiabatic approximation. For the example above for the shifting equilibrium position of a harmonic oscillator,

(a) Denoting the normally kept term as calculated above as a⁽⁰⁾₁ (t) and the normally neglected term in Equation (5.49) as a⁽¹⁾₁ (t), find a⁽¹⁾₁ (t).

(b) Show that for short times (ωt 1), |a⁽¹⁾₁ (t)/a⁽⁰⁾₁ (t)| ∼ 1, while for longer times (ωt0 1), |a⁽¹⁾₁ (t0)/a⁽⁰⁾₁ (t0)| 1.

5.4.4 Sudden Approximation

For this case, the perturbation is turned on suddenly at t = 0, and then held constant. This approximation suits the example above better, so we will compare the results of the two approaches to the same problem. In this case we assume the eigenvalues and eigenfunctions for the initial state are known and the general solution may be written

Ψ⁽ⁱ⁾=X

aiψ_i⁽ⁱ⁾e^−iEⁱ⁽ⁱ⁾^t/~, t < 0 ,

and for the final state, we likewise know the eigenvalues and eigenfunctions, with general solution

Ψ^{(f )} =X

b_jψ^{(f )}_j e^−iE^{(f )}^j ^t/~, t ≥ 0 .

The wave function must be continuous at t = 0 (the equilibrium may not be continuous, but the particle probability density must be), so we have at t = 0

aiψ⁽ⁱ⁾_i =X

bjψ_j^{(f )},

so we take the scalar product with hψ^{(f )}_k |, and obtain bk =X

aihψ^{(f )}_k |ψ_i⁽ⁱ⁾i .

For the example above, the initial and final states are harmonic oscillator states, but the initial state has x = 0 as the equilibrium position and the final

state has x₁ as the equilibrium position. Given the initial state is the ground state and that we wish to have the probability of finding the system in the first excited state of the new system, we calculate

b1= hψ₁^{(f )}|ψ⁽ⁱ⁾₀ i , with

ψ⁽ⁱ⁾₀ (q) = 1

π^1/4e^−q²^/2 ψ₁^{(f )}(q) =

√2

π^1/4e^−(q−q¹⁾²^/2(q − q1) . Thus,

b1= r2

π Z ∞

−∞

e^−q²^/2−(q−q¹⁾²^/2(q − q1) dq

= r2

πe^−q²¹^/4 Z ∞

−∞

e^−(q−q¹^/2)²(q − q₁) dq

= r2

πe^−q²¹^/4 Z ∞

−∞

e^−u²(u −¹₂q1) du

= −q1

√2e^−q¹²^/4. From this we obtain

|b1|²= q₁² 2e^−q²¹^/2.

For small q1, this result is the same as we obtained for the adiabatic approx-imation.

Problem 5.15 Spring break. If one spring of a pair of springs breaks at t = 0 such that k → ¹₂k, find the probability that a system that is initially in the first excited state, ψ1, will remain in that state, and then calculate the probability that it will be found in the third excited state after the break.

Problem 5.16 Tritium decay. Use the sudden approximation to calculate the probability that the 1s electron of tritium will remain in the 1s state (ground state) of³He⁺when tritium decays by β emission so that Z changes from 1 to 2.

Problem 5.17 Moving wall. A particle of mass m is in the ground state of an infinitely deep, one-dimensional, square well.

(a) If the wall separation is suddenly doubled, what is the probability that the particle will remain in the ground state?

(b) If the wall separation is suddenly halved, what is the probability that the particle will remain in the ground state?

Approximation Methods 141

In document 9 EL FRENTE AMPLIO Y EL POLO DEMOCRÁTICO ALTERNATIVO: UNA COMPARACIÓN Facultad de: Ciencias Políticas y Relaciones Internacionales Claudia Lucía Rodríguez Quiñones (página 94-100)