Download Q.M3 - Tirgul 10

Q.M3 - Tirgul 10
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
31.12.2014
Contents
1 Sakurai 6.1 - Lippmann-Schwinger formalism
1
2 Scattering Amplitude - Important equations
3
3 First order - simple example
4
4 Translation invariance potential
5
5 Scattering of negative potential sphere
5
1
Sakurai 6.1 - Lippmann-Schwinger formalism
The Lippmann-Schwinger formalism can also be applied to a one-dimensional
transmission-reflection problem with a finite-range potential V (x) 6= 0 for 0 <
|x| < a.
√
a) Suppose we have an incident wave coming from the left: hx|φi = eikx / 2π.
How must we handle the singular 1/(E − H0 ) operator if we are to have a
transmitted wave only for x > a and a reflected wave and the original wave for
x < −a? Is the E → E + i prescription still correct? Obtain an expression for
the appropriate Green’s function and write an integral equation for hx|ψ (+) i.
b) Consider the special case of an attractive δ-function potential.
V (x) = −
γ~2
δ(x) γ > 0
2m
Solve the integral equation to obtain the transmission and reflection amplitudes.
c) The one-dimensional δ-function potential with γ > 0 admits one (and only
one) bound state for any value of γ . Show that the transmission and reflection
amplitudes you computed have bound-state poles at the expected positions when
∗ e-mail:
[email protected]
1
∗
k is regarded as a complex variable.
Answer
The Lippmann-Schwinger equation:
|ψ ± i = |φi +
1
V |ψ ± i
E − H0 ± i
(1)
So:
hx|ψ
(±)
Z
i = hx|φi +
dx0 hx|
1
V |x0 ihx0 |ψ ± i
E − H0 ± i
(2)
We will use |ψ + i for the transmitted wave eikx , and will use |ψ − i for the reflected
wave e−ikx . So:
1
|x0 i
G± (x0 , x) = hx|
E − H0 ± i
ZZ
1
=
dp00 dp0 hx|p0 ihp0 |
|p00 ihp00 |x0 i
E − H0 ± i
Z
0
0 0
1
1
dp0 eip x/~ e−ip x /~
=
02
p
2π
E − 2m ± i
Now we use the fact that p0 = ~q and E = ~2 k 2 /(2m) so we get:
Z
0
0 0
1
1
dp0 eip x/~ e−ip x /~
G± (x0 , x) =
02
p
2π
E − 2m ± i
Z ∞
0
2m
1
= 2
dq eiq(x−x ) 2
2~ π ∞
k − q 2 ± i
√
The poles are at q = ± k 2 ± i ' ±k ± i0 so:
Z ∞
0
2m
2mi ±ik|x−x0 |
1
G± (x0 , x) = 2
dq eiq(x−x ) 2
=
e
2
2~ π ∞
k − q ± i
2k~
(3)
(4)
(5)
So:
eikx
im
hx|ψ + i = √ −
k~
2π
Z
a
0
eik|x−x | V (x0 )hx|ψ + idx0
(6)
−a
It is ok to replace |x − x0 | with x − x0 because in these boundary condition x ≥ a
and −a ≤ x0 ≤ a.
2
b) when V = − γ~
2m δ(x) then we take equation 6 and we get:
hx|ψ + i = hx|φi +
iγ ik|x|
e
h0|ψ + i
2k
(7)
We can substitute now boundary condition x = 0 and get:
h0|ψ + i = h0|φi +
2
iγ
h0|ψ + i
2k
(8)
So we get:
1
1
h0|ψ + i = √
iγ
2π 1 − 2k
(9)
So equation 7 will be read
√
h0|ψ + i = eikx / 2π + √
iγ
eik|x|
2π(2k − iγ)
(10)
The Transition matrix T is:
T =1+ √
iγ
2π(2k − iγ)
(11)
In the same way the reflation matrix R is:
R= √
iγ
2π(2k − iγ)
(12)
c) For positive γ we get one pole which correspond to bounded state with the δ
function so:
2k − iγ → k = −
iγ
2
(13)
which gives a decaying result agreeing with Schroedinger equation.
2
Scattering Amplitude - Important equations
Important equations for proceeding are:
0
G(x, x0 ) = −
2m
hx|ψ i = hx|φi − 2
~
±
Z
1 e±ik|x−x |
4π |x − x0 |
(14)
0
e±k|x−x |
d x
V (x0 )hx0 |ψ ± i
4π|x − x0 |
3 0
(15)
so
1
eikx
hx|ψ + i = √ [eikx +
f (k 0 , k)]
r
2π
(16)
where:
f (k 0 , k) = −
(2π)3 2m 0
hk |V |ψ + i
4π ~2
(17)
the deferential scattering amplitude is:
dσ
= |f (k 0 , k)|2
dΩ
3
(18)
and :
Im(f (θ = 0)) =
kσt
4π
It is important to note that for first order (Born Approximation):
Z
m
~ ~0 ~0
f (1) (k 0 , k) = −
d3 x0 ei(k−k )x V (x0 )
2π~2
(19)
(20)
By taking that |~k − k~0 | = q = 2ksin( θ2 ) and the fact ~k~x = r|k|cos(θ) so after
integration we get:
Z ∞
Z
m
2m ∞
f (1) (θ) = − 2
rV (r)(eiqr − e−iqr )dr = − 2
rV (r) sin(qr)dr (21)
~ iq 0
~ q 0
3
First order - simple example
Find the scattering amplitude in first order of the following potential:
V0
r≤a
V (r) =
0
r>a
Solution
The integral from equation 21 will give:
f (1) (θ) = −
2m V0 a3 sin(qa)
[
− cos(qa)]
~2 (qa)2
qa
4
(22)
4
Translation invariance potential
5
Scattering of negative potential sphere
At a certain sphere in space there is a constant negative potential −V0 . By
measuring the quantities of particles scattering on this potential, suggest a way
to measure its volume.
Solution
5
We will expect to find a characteristic differential cross section for this potential,
so lets find it and see if we can extract some volume observable.
Z
−1 ∞
f (θ) α
rV (r) sin(qr)dr
q 0
Z
V0 ∞
rθ(r − R) sin(qr)dr
=
q 0
Z
V0 R
V0
=
r sin(qr)dr = 3 sin(qR) − qR cos(qR)
(23)
q 0
q
Hence:
dσ
sin(x) − x cos(x)
α
dΩ
x3
(24)
where x = qR = 2kR sin( θ2 ). The deferential cross section will behave like
If we
know the minimum points, we can find the radius of the potential. x = 1.43π
so
R=
1.43π
2k sin( θ21 )
(25)
So by knowing the angle of the minimum deferential cross section and knowing
the momentum we can conclude the radius of the potential.
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