Quantum Mechanics
Schrodinger equation
Quantum Mechanics
Propagators : Pg
Quantum Simple Harmonic Oscillator QSHO
Quantum Mechanics
Simulation With GNU Octave
© The scientific sentence. 2010


Fermi's golden rule
1. Transition probability between two states
Given a Hamiltonian
H(t) = H = H_{0} + V(t)
With
V(t) = 0 if t<= t_{0}
V(t) = V(t) if t> t_{0}
If i> are the eigenstates of H_{0}, then:
H_{0} i> = E_{i} i>
Let's consider the system in the state i> = φ(t_{0})>;
at the time t_{0}, so at a later time "t", it is
in the state φ_{s}(t)> = Σ c_{f}(t)f>
The transition probability from i> to f>
is equal to c_{f}(t)^{2}.
We have:
φ_{s}>(t); = U_{s}(t,t_{0})i> so
<fφ>(t); = <fΣ c_{f}(t)f> = c_{f}(t)
Hence:
c_{f}(t) = <fφ_{s}(t)> = <fU_{s}(t,t_{0})i> =
<fexp{ iH_{0}(t  t_{0})/ℏ}U_{i}(t,t_{0})i>
Therefore:
c_{f}(t)^{2} = <fU_{i}(t,t_{0})i>^{2}
We already know:
U_{i}(t,t_{0}) = exp { (i/ℏ) ∫ H_{int}(τ) dτ}
from t_{0} to t.
Then:
c_{f}(t)^{2} = <fexp { (i/ℏ) ∫ V_{i}(τ) dτ}i>^{2}
Therefore, at the first order:
U_{i}(t,t_{0}) = 1  (i/ℏ) ∫ V_{i}(τ) dτ
from t_{0} to t.
Then:
<fU_{i}(t,t_{0})i> =  (i/ℏ) ∫ <fV_{i}(τ)i> dτ
from t_{0} to t.
Recall: V_{i}(t) = exp{ iH_{0}(t  t_{0})/ℏ}V(t)exp{ iH_{0}(t  t_{0})/ℏ}, so:
<fU_{i}(t,t_{0})i> =  (i/ℏ) ∫ dτ <fV(τ)i> exp {i(E_{f}  E_{i})(τ  t_{0})/ℏ}
from t_{0} to t.
P_{i} → _{f} =  (i/ℏ) ∫ dτ <fV(τ)i> exp {i(E_{f}  E_{i})(τ  t_{0})/ℏ}^{2}
(τ: from t_{0} to t)
This is the transition probability of observing
the system in the target state f> prepared in the
state i> at time t, due to V(t).
This formula is known as the
statetostate form Fermi’s Golden Rule
2. Applications:
2.1. Timeindependent perturbation:
V (t) = 0 if t <= 0
V (t) = V if t > 0 (independent of time)
We have in this case:
P_{i} → _{f} =  (i<fVi>/ℏ) ∫ dτ exp {i(E_{f}  E_{i})(τ)/ℏ}^{2} (τ: from t_{0} = 0 to t)
∫ dτ exp {i(E_{f}  E_{i})(τ)/ℏ} = (ℏ/i)(E_{f}  E_{i}) [exp {i(E_{f}  E_{i})t/ℏ}  1]
1  exp{ix} = 1  cos x  i sin x
1  exp{ix}^{2} = (1  cos x  i sin x)(1  cos x + i sin x) =
(1  cos x)^{2} + sin^{2} x = 2  2 cos x = 2 (1  cos x) = 4 sin^{2}(x/2).
Then:
P_{i} → _{f} = [4 <fVi>^{2}/(E_{f}  E_{i})^{2}] sin^{2}[(E_{f}  E_{i})t/2ℏ]
Let's write: E_{f}  E_{i} = ℏ ω, so
P_{i} → _{f} = [4 <fVi>^{2}/ℏ^{2}ω^{2}] sin^{2}[(ωt/2]
Now, let's write:
P_{i} → _{f} = [ <fVi>^{2}/ℏ^{2}] f(ω)
and then set some properties of this function:
f(ω) = (4/ω^{2}) sin^{2}[(ωt/2]
1. lim f(ω) = t^{2}
when fω → 0)
2. f(ω) = 0 if ωt/2 = kπ , that is ω = 2πk/t.
the max of the function is at ωt = 0. The first minimum is
at ω = 2π/t. Therefore the measure of this probability
is appreciable if ω < 2π/t or
ω < 2π/Δt. or ΔE Δt >=0 2πℏ,
E Δt = E_{f}  E_{i}. And we recognize in ΔE >=0 2πℏ/Δt
the incertainty principle.
gnuplot> set xrange [ 0 : 7 ]
gnuplot> set yrange [ 0 : 1.5 ]
gnuplot> plot (sin(x))**2/x**2
2.2. Timeindependent perturbation at an infinite time
Now we want to set the formula for t that tends touward ∞,
that is:
lim P_{i} → _{f} = [4 <fVi>^{2}/ℏ^{2}ω^{2}] sin^{2}[(ωt/2]
t → ∞
We know
∫ dx sin^{2}(x)/x^{2} = π
x:  ∞ → +∞
Let's write:
∫ dx sin^{2}(x)/x^{2} g(0) = πg(0)
 ∞ → + ∞
g is a function that we are going to determine.
Let x = ωt, so:
πg(0) = ∫ dx sin^{2}(x)/x^{2} g(0) =
=
lim ∫ dx sin^{2}(x)/x^{2} g(ω)
x: ∞ → + &infin
t → ∞ or ω → 0
=
lim ∫ (1/t)dω sin^{2}(ωt)/ω^{2} g(ω)
x: ∞ → + &infin
t → ∞ or ω → 0
=
∫ lim (1/t) sin^{2}(ωt)/ω^{2} g(ω) dω
x: ∞ → + &infin
t → ∞ or ω → 0
Rewriting the equation gives:
∫ lim (1/t) sin^{2}(ωt)/ω^{2} g(ω) dω = πg(0)
x: ∞ → + &infin
t → ∞ or ω → 0
This equality is valid only if:
lim (1/t) sin^{2}(ωt)/ω^{2} = πδ(ω)
according to the property of Dirac function wich is:
∫f(x) δ(x  a) = h(a)
x: ∞ → + ∞
Thus:
lim (1/t) sin^{2}(ωt)/ω^{2} = πδ(ω)
as we see:
∫ πδ(ω) g(ω) dω = πg(0)
x: ∞ → + ∞
Our probability, at t → + ∞ becomes:
lim P_{i} → _{f} = [4 <fVi>^{2}/ℏ^{2}] sin^{2}[(ωt/2] / ω^{2}
t → ∞
= [4 <fVi>^{2}/ℏ^{2}] (t/2) π ^{2}/ℏ δ(E_{f}  E_{i})
(We have replaced t by t/2 and use a property of δ function:
δ(ax) = (1/a) δ(x)
Therefore:
The probability of transition from i> to f>, by
unit of time (transition rate) is:
P_{i} → _{f} = (2 π/ℏ) <fVi>^{2} δ(E_{f}  E_{i})
2.3. Timedependent perturbation
If V(t) is given by:
V(t) = 0 if t<= 0
V(t) = V exp{iωt} + V^{+}exp{iωt} if t> 0
Called armonic perturbations
we will have:
P i → f =  i/ℏ ∫ dτ exp{i(E_{f}  E_{i})τ/ℏ} [<fVi> exp{iωτ} +
<fV^{+}i> exp{iωτ}] ^{2}
from t_{o} to t
= [1  exp{i((E_{f}  E_{i})/ℏ + ω)t}/ ((E_{f}  E_{i} + ℏω) fVi> +
[1  exp{i((E_{f}  E_{i})/ℏ  ω)t}/ ((E_{f}  E_{i}  ℏω) fV^{+}i>^{2}
That leads to:
P_{i} → _{f} = (2 π/ℏ) <fVi>^{2} δ(E_{f}  E_{i} + ℏω) +
(2 π/ℏ) <fV^{+}i>^{2} δ(E_{f}  E_{i}  ℏω)
If E_{f} > E_{i} there is absorption (energy uptake),
then only the 2nd term contributes
If E_{f} < E_{i} there is emission (energy loss),
then only the 1st term contributes
3. Distribution of final states
The firstorder term that we have used in
U_{i}(t,t_{0}) = exp { (i/ℏ) ∫ H_{int}(τ) dτ}
from t_{0} to t.
allows only direct transitions between
i> and f> . The secondorder term
accounts for transitions occuring through
all possible intermediate states of f>
We don’t have strictly real monochromatic light, but a frequency
spectrum of ω.Therefore, we use the radiation density.
We use also the term Density of states &ro;(E_{f}).
The last version of the Fermi's golden rule becomes:
P(i→f) = Σ(f) P_{i} → _{f} = (2 π/ℏ) <fVi>^{2} δ(E_{f}  E_{i})
= ∫ ρ(E_{f}) dE_{f} [P_{i} → _{f}]
P(i→f) = ∫ρ(E_{f}) dE_{f} [P_{i} → _{f}]

