Q.M3 - Tirgul 12 Contents 1 Spherical well potential

Q.M3 - Tirgul 12
Roee Steiner
Physics Department, Ben Gurion University of the Negev, BeerSheva 84105, Israel
14.1.2015
Contents
1 Spherical well potential
1
2 Attractive delta shell potential
7
3 Eikonal approximation
9
1
Spherical well potential
Consider the spherical well potential:
−V0
V (r) =
0
r<a
r>a
(1)
Answer the following questions for l = 0 partial wave.
(a) Work out the conditions for the bound states. Plot the bound state energies as a function of V0 .
(b) Calculate the phase shift δ0 and the S-matrix e2iδ0 .
(c) Show that the poles of the S-matrix on the upper half plane of k corresponds precisely to the bound states.
(d) Plot the cross section σ0 as a function of ka for a several values of V0 to
show its behavior far away from, just below, exactly on, and just above the
threshold bound state, and compare them to the geometric cross section
4πa2 .
1
Solution
a
Let’s first solve the eigenvalue problem:
Generally:
−
~2 2
∇ ψ − V0 ψ + V0 Θ(r − a)ψ = Eψ
2m
(2)
Where Θ(r − a) is simply the Heaviside step function ...
If we stick to the l = 0 mode, the nabla operator can be taken simply to
be:
∇2 ψ =
∂2
(rψ)
r∂r2
(3)
Where here, we could simplify things by stating the obvious: l = 0 ⇒ ψ = R(r),
thus we get (inside the well):
~2
(rR)00 − V0 R = ER
(4)
2mr
2m(V0 + E)
(rR) = 0
(5)
⇒ (rR)00 +
~2
U 00
2m(V0 + E)
(rR) ≡ U ⇒
=−
(6)
U
~2
Where, in order to get a bound state we need the particle’s energy to be smaller
than the ”walls” of the well, satisfying E < 0.
−
We have two cases here: E > −V0 , and E < −V0 . For the second case:
2m(V0 + E)
2m(V0 + E)
>0⇒−
≡ κ2
(7)
2
~
~2
A sinh (κr)
ψin =
(8)
r
And that’s not so interesting though, since we want states which are classically
bound thus: −V0 < E < 0
The solution for which is:
E < −V0 ⇒ −
sin(kr)
kr
eq(a−r) sin(ka)
=A
kr
ψin = A
ψout
(9)
Where k, and q are given by:
k2 =
2m(V0 + E)
~2
2mE
q2 = − 2
~
2
(10)
(11)
As for the energies, by the derivative continuity condition we get an equation
for k, and q:
−
k
= tan(ka)
q
(12)
Or more explicitly with constraints on the energy E we get:
!
r
r
(V0 + E)
2ma2 (V0 + E)
= tan
−
−E
~2
(13)
This has an obvious solution at E = −V0 , but that’s not in our interval for E
so we need another (stinkier) solution...
This could be solved graphically:
2
What is seen here is a qualitative graph for V0 = 10, and b = 2ma
~2
We can also perform a graph of the resonance has the well dip.
3
We can see that has the well is dipper there is more resonance energy
b
Before we get the exact solution we should mention that
lim k cot(δ0 ) → −
k→0
1
a
(14)
where a is the ”scattering length”. Since k is very small that limit is possible if
cot(δ0 ) is very large, namely δ0 = 0, π, 2π, · · · .
But at the transition (the bound state) we have that δ0 = 0, π/2, 3π/2, · · · .
graphically it is:
4
In order to calculate the exact phase shift we must assume, at large distances the
eigenfunctions go like the Hankel function set we all know and love (or spherical
Bessel, whichever you fancy better), thus:
ψl=0,r>>a =
sin(kr)
cos(kr)
cos(δ0 ) −
sin(δ0 )
kr
kr
(15)
And we need to ”tape it” at the boundary to our own bound state of:
ψbound =
sin(κr)
κr
(16)
So after some algebra we get the expression:
cos(ka − δ0 ) = cos(κa)
k
sin(ka − δ0 ) = sin(κa)
κ
k
⇒ tan(ka − δ0 ) = tan(κa)
κ
k tan(κa) − κ tan(ka)
⇒ δ0 = − arctan
k tan(ka) tan(κa) + κ
(17)
(18)
(19)
(20)
Now, S0 (k) = e2iδ0 , so in theory we have a complete expression for S0 (k), but
in practice we want a little more legible one by:
tan(x) =
eix − e−ix
e2ix − 1
sin(x)
= −i ix
=
−i
cos(x)
e + e−ix
e2ix + 1
(21)
Thus:
S−1
tan(δ0 ) − i
⇒S=−
S+1
tan(δ0 ) + i
k tan(κa) − κ tan(ka) − i (k tan(ka) tan(κa) + κ)
⇒S=−
k tan(κa) − κ tan(ka) + i (k tan(ka) tan(κa) + κ)
− tan(δ0 ) = i
5
(22)
Which satisfy the three ”rules of thumb” for the S matrix (existence of abound
state, unitary, threshold behavior i.e Sl=0 = 1 at k = 0).
c
As for poles of the S-matrix, they are at (look at the denominator of S in
equation 22):
k tan(κa) − κ tan(ka) = −i(k tan(ka) tan(κa) + κ)
tan(κa)
κ
κ
and − = tan(ka) tan(κa)
⇒ =
k
tan(ka)
k
tan(κa)
⇒ − tan(ka) tan(κa) =
⇒ tan2 (ka) = −1
tan(ka)
(23)
(24)
(25)
This is solvable only by employing complex numbers, by assuming k is complex:
k → ±iq ⇒ tanh2 (qa) = 1 ⇒ q → ∞
(26)
But the real value of this calculation is showing that there is a pole in S for the
transformation k → ±iκ. and so where the outgoing wave function is
eikr
: k → iκ
r
e−κr
⇒
r
(27)
Which is exactly the bound state wave at large distances.
d
We know that
σ=
4π X 2
sin (δl )
k2
(28)
4π
sin2 (δ0 )
k2
(29)
l
for l = 0 we have
σ=
We also know that
tan(δ0 ) = −
k tan(κa) − κ tan(ka)
k tan(ka) tan(κa) + κ
6
(30)
We can use the trigonometric relationship
k2
σ
4π
k2
σ
cos2 (δ0 ) = 1 −
4π
sin2 (δ0 ) =
(31)
(32)
(33)
so:
s
tan(δ0 ) =
sin2 (δ0 )
=
cos2 (δ0 )
s
1
k2
4π σ
k2
− 4π
σ
(34)
so
σ=
2
4π tan2 (δ)
k 2 1 + tan2 (δ)
(35)
Attractive delta shell potential
Consider the attractive delta shell potential λ > 0
V (r) = −
~2 λ
δ(r − a)
2µ
(36)
(a) Calculate the phase shift δl (k) , where l is the angular momentum quantum
number.
(b) In the case of l = 0, investigate the existence of bound states by examining
the analytic properties of the partial scattering amplitude. Are there any
resonances?
Solution
a
Again, the eigenvalue problem:
λ~2
~2 2
∇ ψ−
δ(r − a)ψ = Eψ
2m
2µ
1
l(l + 1)
2mE
⇒ r 6= a
(rR)00 −
R=− 2 R
r
r2
~
2mE
l(l
+
1)
⇒ U 00 −
U + k2 U = 0
k 2= ~2
r2
⇒ R = Al jl (kr) + Bl nl (kr)
−
⇒ψ=
∞
X
l
X
Yml (θ, φ) [Al,m jl (kr) + Bl,m nl (kr)]
l=0 m=−l
7
(37)
(38)
(39)
(40)
(41)
As always, the Bessel functions of the second kind all explode at zero so we
might as well write the inner function as simply:
ψin =
∞ X
l
X
Al,m Yml (θ, φ)jl (kr)
(42)
l=0 m=−l
so let’s ”tape” the inner and outer functions:
ψin (a) = ψout (a) (43)
⇒
∞
X
l
X
Al,m Yml (θ, φ)jl (ka) =
∞
X
l0
X
0
Yml 0 (θ, φ) [Cl0 ,m0 jl (ka) + Dl0 ,m0 nl (ka)] (44)
l0 =0 m0 =−l0
l=0 m=−l
Linear independence of Yml functions select the modes and so we get:
Al jl (ka) = Cl jl (ka) + Dl nl (ka)
(45)
the m index is suppresses since the scatterer is spherically symmetric, all m −
modes vanish except the m = 0 mode.
Now, let’s try for the jump in derivative:
The whole spiel, with integrating the T.I.S.E. comes out to be simply:
a+
mλ 2
∂(rR) =−
· r
a R(a)
(46)
∂r
µ
a−
mλ
a+
aR(a)
(47)
⇒ (R + rR0 )|a− = −
µ
mλ
⇒ R0+ (a) − R0− (a) = −
R(a)
(48)
µ
mλ
⇒ Cl jl0 (ka) + Dl n0l (ka) − Al jl0 (ka) = −
Al jl (ka)
(49)
µ
Now, since ate very large distances we have (method of partial waves):
ψf ar = eiδl [cos(δl )jl (kr) − sin(δl )nl (kr)]
(50)
Thus we can identify:
Dl = − sin(δl )
(51)
Cl = cos(δl )
Dl
= tan(δl )
⇒−
Cl
So all that’s left is to find the coefficients Cl , Dl ...
mλ jl0 (ka)
mλ
0
−Dl nl (ka) + nl (ka)
−
= Cl
jl (ka)
µ
jl (ka)
µ
⇒ tan(δl ) = −
mλ
Dl
µ jl (ka)
=h
Cl
0
nl (ka) + nl (ka) mλ
µ −
8
jl0 (ka)
jl (ka)
i
(52)
(53)
And it’s very easy to show that in the limit λ → ±∞ we get the known factor
for hard sphere scattering:
jl (ka)
nl (ka)
tan(δl ) =
(54)
b
Now, in the case of l = 0 let’s get the phase shift δ0 :
tan(δ0 ) =
amλ sin(ka)
akµ sin(ka) + (2µ − amλ + aµk cot(ka)) cos(ka)
(55)
Now, in order to find resonant states, all we have to do is demand a vanishing
denominator:
akµ sin(ka) + (2µ − amλ + aµk cot(ka)) cos(ka) = 0
(56)
λ
ka + (2 − am ) cot(ka) + ka cot2 (ka) = 0
µ
(57)
Which is:
which can be written has:
sin(2ka) =
2ka
(am µλ − 2)
(58)
this can be solved graphically, lets say x = ka and b = (am µλ − 2) the solution
is the blue line in the graph:
3
Eikonal approximation
f (k 0 , k) = −ik
Z
∞
db b J0 (kbθ)[e2i∆(b) − 1]
(59)
0
where
∆(b) =
−m
2k~2
Z
∞
−∞
9
p
V ( b2 + z 2 )dz
(60)
Eikonal approximation is valid only for high energies where l cannot be neglected. From classical mechanics we know that lmax = kR and in this case
(high energies) δl → ∆(b)|b=l/k .
Example
Use Eikonal approximation and find the phase shift of the Gaussian potential
2
2
V = V0 e−r /a
Solution
We can write
V = V0 e−r
2
/a2
= V0 e−(b
2
+z 2 )/a2
(61)
So
Z
p
2
2
2
−m ∞
V ( b2 + z 2 )dz =
V0 e−(b +z )/a dz
2
2k~
−∞
−∞
√
Z ∞
2
2
π −ma
−m
−b2 /a2
−z 2 /a2
V0 e
V e−b /a
=
e
dz =
2 0
2k~2
2
2k~
−∞
∆(b) =
−m
2k~2
Z
∞
(62)
So
√
2
2 2
π −ma
π −ma
−b2 /a2
V0 e
|b=l/k =
V0 e−l /k a
δl =
2
2
2 2k~
2 2k~
√
10
(63)