Download Q.M3 - Tirgul 9

Q.M3 - Tirgul 9
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
23.12.2015
Contents
1 Parity - Problem 23 from lecture 7
1
2 Finding the evolution operator in interaction picture - Dyson
series
2
3 Time dependant Hamiltonian - different approach
3.1 Time perturbation in harmonic oscillator. . . . . . . . . . . . . .
2
3
4 Class Work
5
1
Parity - Problem 23 from lecture 7
Show that existence of two decay channels of a zero spin K meson into two
pions,K → 2π, and into three pions,K → 3π, indicates the non-conservation of
parity for K decays.
Solution
As you seen in the class the pions have zero spin and negative intrinsic parity.
In the channel of 2π the parity is symmetric and in the channel 3π has negative
parity so, if we have two channel with different parity then the parity is not
conserved.
∗ e-mail:
[email protected]
1
∗
2
Finding the evolution operator in interaction
picture - Dyson series
A 1/2 spin is placed in a constant magnetic field in the z direction (Bz ) - namely,
the unperturbed Hamiltonian H0 is:
1 0
H0 = µBz
(1)
0 −1
where µ is the magnetic moment At the time t = 0 the spin was in the | ↑i state
At the time t = 0 a weak (Bx Bz ) periodic magnetic field is applied at the x
direction, namely the perturbation V (t) is:
0 1
V (t) =
µBx cos(ωt)
(2)
1 0
Find the probability to find the system at | ↓i at time t (in the first order ?).
Solution
In the first order the transition amplitude is
Z t
h↓ |UI (t)| ↑i = −i
dt1 eiΩt1 h↓ |V (t1 )| ↑i
(3)
0
Z
iµBx t
dt1 eiΩt1 cos ωt1
(4)
−
2
0
where we denoted E↑ = −E↓ = Ω/2 = µBz . The transition amplitude now
becomes
µBx e−i(Ω+ω)t − 1 e−i(Ω−ω)t − 1
+
(5)
h↓ |UI (t)| ↑i = −
2
Ω+ω
Ω−ω
Respectively, the rate of the spin-flips is
w↑→↓ =
3
πµ2 Bx2
δ(Ω − ω)
2
(6)
Time dependant Hamiltonian - different approach
Consider time dependant perturbation in Hamiltonian:
b =H
b 0 + Vb (t)
H
(7)
b 0 constant in time and totally soluble H
b 0 |ni = En |ni. Find general way
With H
¬
b
b
b
to solve accurately H = H0 + V (t).
¬ It
is question form of chapter 5.5 in Sakurai
2
Solution
Lets try some wave state form:
|ψ(t)i = Σn cn (t)e−iEn t/~ |ni­
(8)
So now we put the wave function into Schroedinger equation:
b
H|ψ(t)i
∂
b 0 + Vb )|ψ(t)i
= i~ |ψ(t)i = (H
∂t
(9)
So:
∂
(Σn cn (t)e−iEn t/~ |ni)
∂t
b 0 + Vb )Σn cn (t)e−iEn t/~ |ni
= (H
i~
(10)
b 0 e−iEn t/~ |ni = i~ ∂ (e−iEn t/~ |ni) =
By doing the derivative and use the fact that H
∂t
−iEn t/~
e
En |ni we get:
i~Σn ċn e−iEn t/~ |ni = Vb Σn cn (t)e−iEn t/~ |ni
(11)
We multiply from the left by hm| and we get:
i~cm
˙ e−iEm t/~ = Σn cn Vmn e−iEn t/~
(12)
i~cm
˙ = Σn cn Vmn e−(iEn −Em )t/~
(13)
So finely:
We have a set of first order deferential equations for cn .
3.1
Time perturbation in harmonic oscillator.
A harmonic oscillator system with the energy levels En = (n+1/2)~ω is affected
by the electric field applied
(t) =
A −t2 /2τ 2
e
πτ
(14)
Find, in the first order, the probability to find the oscillator in the first exited
state at t → ∞, if at t → ∞ it is in the ground state.
Solution
The potential is:
V (t, x) = −e(t)b
x=−
eA −t2 /2τ 2
e
x
b
πτ
(15)
­ Note that H
b 0 e−iEn t/~ |ni = i~ ∂ (e−iEn t/~ |ni) = e−iEn t/~ En |ni so we look for close
∂t
solution.
3
The transition amplitude in the first order is
Z ∞
eA −t2 /2τ 2
dteiωt
h1|UI (∞, −∞)|0i = i
e
h1|b
x|0i
πτ
−∞
(16)
Since
1
(a+ + a)
2mω
√
h1|a+ |0i = 2
x
b= √
h1|a|0i = 0
(17)
(18)
(19)
one finds
h1|b
x|0i = √
1
mω
The Gaussian integral is easily calculated and finally
√
ieA 2 −ω2 τ 2 /2
h1|UI (∞, −∞)|0i = √
e
mωπ
(20)
(21)
so that the probability of the transition is
P0→1 =
4
2e2 A2 −ω2 τ 2
e
mωπ
Class Work
30 from lecture 7
21, 27, 28, 13 from lecture 8
4
(22)