GREEN’S FUNCTION FOR ONE DIMENSIONAL
SCHRÖDINGER EQUATION
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Post date: 16 October 2021.
Griffiths derives the first Born approximation by writing the Schrödinger
equation in integral form as
ˆ
ψ (r) = G (r − r0 ) Q (r0 ) d3 r0
(1)
where
∇2 + k 2 ψ = Q
√
2mE
k≡
h̄
2m
Q≡ 2 Vψ
h̄
and G is the Green’s function satisfying
(2)
(3)
(4)
∇2 + k 2 G (r) = δ 3 (r)
(5)
We can follow through Griffiths’s derivation and adapt it for the one dimensional Schrödinger equation:
2
d
2
+k ψ = Q
(6)
dx2
where k and Q have the same definitions as above. In this case, the Green’s
function satisfies
2
d
2
+ k G (x) = δ (x)
(7)
dx2
and the integral form of the Schrödinger equation is
ˆ
2m
ψ (x) = 2
G (x − x0 ) V (x0 ) ψ (x0 ) dx0
(8)
h̄
We can work with the Fourier transform of G:
1
GREEN’S FUNCTION FOR ONE DIMENSIONAL SCHRÖDINGER EQUATION
ˆ ∞
1
√
G (x) =
eixs g (s) ds
2π −∞
where s has the dimensions of 1/length. Plugging in 7 we get
2
(9)
ˆ ∞ 2
d
1
2
√
+ k eixs g (s) ds = δ (x)
(10)
2
dx
2π −∞
ˆ ∞
1
√
−s2 + k 2 eixs g (s) ds = δ (x)
(11)
2π −∞
We can now use the Fourier transform of the delta function to rewrite the
RHS:
1
√
2π
ˆ
∞
2
−s + k
−∞
2
e
ixs
ˆ ∞
1
g (s) ds =
eixs ds
2π −∞
1
1
g (s) = √
2
2π k − s2
(12)
(13)
Therefore
ˆ ∞
1
eixs
G (x) =
ds
(14)
2π −∞ k 2 − s2
This integral can be done using contour integration in the complex plane,
if we choose a contour that runs along the (horizontal) s axis in the +s
direction from s = −R to s = +R, then arcs back as a semicircle from +R
to −R. The problem is that the horizontal leg of this contour runs through
the two singularities at s = ±k, so we need to change the contour so that
it loops around these singularities with small semicircles of radius ρ (see
Fig. 11.10 in Griffiths). Cauchy’s theorem states that the integral of an
analytic function (that is, a function that doesn’t have any singularities on
the contour) around a closed contour is
˛
f (z)
dz = 2πif (z0 )
(15)
z − z0
if z = z0 is the only singularity within the contour. (More generally, if the
function to be integrated has more than one singularity at locations zi , then
the integral is 2πi ∑i res (zi ) where res (zi ) is the residue of the function at
zi .)
In fact, the derivation in Griffith isn’t really correct, although his Green’s
function is valid. He chooses a contour that runs along the s axis and loops
over the singularity at s = −k and under the singularity at s = +k. He then
states that by Cauchy’s theorem, the integral is just 2πif (+k). That is true
GREEN’S FUNCTION FOR ONE DIMENSIONAL SCHRÖDINGER EQUATION
3
for the contour shown in his book, but what we need is actually the limit
of this integral as the radius of the little semicircles ρ → 0. [To see that
Griffiths’s argument as it stands doesn’t work, suppose we excluded both
singularities by looping over them. In that case, the contour contains no
singularities and the integral 14 is zero, which can’t be true.]
To do it properly, we’ll use Griffiths’s contour but write out the integrals
involved and take the limit as ρ → 0. We start by taking x > 0 and integrating along the s axis from −R to −k − ρ, then over the semicircle at s = −k
to −k + ρ, then from −k + ρ to k − ρ, then under the semicircle at s = +k
to k + ρ, then from ρ to +R and finally around the big semicircle from +R
back to −R. Since the contour contains only the singularity at s = +k, we
have
1
2π
˛
eixz
1
dz =
2
2
k −z
2π
˛
eixz
dz
(k + z) (k − z)
˛
1
eixz
=−
dz
2π (z + k) (z − k)
=−
eikx
2πi eikx
= −i
2π 2k
2k
At this point, Griffiths assumes that this result is the same as
so the Green’s function is
G (x) = −i
(16)
(17)
(18)
´∞
eikx
2k
eixs
−∞ k2 −s2 ds,
(19)
and the integral form is, from 8
ˆ
im
ψ (x) = ψ0 (x) − 2
eik(x−x0 ) V (x0 ) ψ (x0 ) dx0
h̄ k
[Note that this answer cannot be correct, because
ˆ
eixs
ds =
2
2
−∞ k − s
∞
ˆ
∞
cos xs
ds + i
2
2
−∞ k − s
ˆ
(20)
∞
sin xs
ds
2
2
−∞ k − s
(21)
The first integrand on the RHS is an even function of s, and the second
integrand is an odd function. Since we’re integrating over an interval that is
symmetric about s = 0, the second integral is zero, meaning that the overall
integral must be real, which Griffiths’s answer isn’t.]
−ikx
A similar analysis for x < 0 gives G (x) = −i e 2k so the overall integral
form is
GREEN’S FUNCTION FOR ONE DIMENSIONAL SCHRÖDINGER EQUATION
4
ˆ
im
ψ (x) = ψ0 (x) − 2
eik|x−x0 | V (x0 ) ψ (x0 ) dx0
(22)
h̄ k
However, this isn’t really the correct way to do this. Let’s look at it in more
detail.
Since x > 0, we can let the big semicircle go to infinity since xz has a
positive imaginary part for all points on this semicircle, so eixz → 0 as the
semicircle goes to infinity, and the integral over that part of the contour also
goes to zero. So we’re left with
"ˆ
−k−ρ
ˆ
k−ρ
ˆ
#
ˆ
eixz
dz +
k2 − z 2
ˆ
ˆ
eixz
dz +
2
2
−k k − z
ˆ
ikx eixz
e
+
+
dz = −2π i
2
2
2k
−∞
−k+ρ
k+ρ
+k k − z
(23)
where the two integrals on the right are integrals around the little semicircles at s = ±k. Note that the semicircle at −k is traversed in a clockwise
direction, while that at +k is counterclockwise.
The integral around the semicircle at s = −k is over z as it takes on
the values z = −k + ρeiθ for π ≥ θ ≥ 0 (that is, we start the integration at
θ = πand proceed to θ = 0), and for s = +k, z = k + ρeiθ for −π ≤ θ ≤ 0.
The only quantity that is being integrated over in these two integrals is θ, so
we can transform to θ using dz = iρeiθ dθ. We have
∞
iθ
eix[−k+ρe ]
iρeiθ dθ
(24)
iθ + ρ2 e2iθ
2kρe
π
ˆ 0 ixρeiθ
e
−ikx
dθ
(25)
= ie
iθ
π 2k + ρe
Taking the limit, and reversing the limits of integration and hence the
sign, we get
eixz
dz =
2
2
−k k − z
0
ˆ
eixz
lim
dz = −ie−ikx
ρ→0 −k k 2 − z 2
ˆ
0
π
dθ
2k
(26)
πie−ikx
(27)
2k
We can do the same calculation at the other singularity at s = +k and we
get
ˆ
eixz
πieikx
lim
dz
=
−
(28)
ρ→0 +k k 2 − z 2
2k
Putting these results into 23 and taking the limit there, we get
=−
GREEN’S FUNCTION FOR ONE DIMENSIONAL SCHRÖDINGER EQUATION
ˆ
eixs
πi eikx − e−ikx
ds
=
−
2
2
k
2
−∞ k − s
π
= sin (kx)
k
5
∞
(29)
(30)
(note that this answer is real, as required) so from 14 we get the Green’s
function
G (x) =
1
sin (kx)
2k
(31)
and the integral form is, from 8
m
ψ (x) = ψ0 (x) + 2
2h̄ k
ˆ
sin (k (x − x0 )) V (x0 ) ψ (x0 ) dx0
(32)
2
d
2 ψ = 0.
+
k
where ψ0 is a free particle solution of dx
2
If x < 0, we need to use a large semicircle that is below the s axis, so that
xz > 0. The contour now includes the singularity at s = −k and excludes
the one at s = +k, and the contour is traversed clockwise, so we pick up a
minus sign when using Cauchy’s theorem:
1
2π
˛
eixz
1
dz =
2
2
k −z
2π
˛
eixz
dz
(k + z) (k − z)
˛
1
eixz
=−
dz
2π (z + k) (z − k)
=+
2πi e−ikx
e−ikx
= −i
2π (−2k)
2k
(33)
(34)
(35)
Equation 23 now becomes
"ˆ
−k−ρ
ˆ
ˆ
#
ˆ
ˆ
−ikx eixz
e
+
dz = −2π i
+
2
2
2k
+k k − z
−∞
−k+ρ
k+ρ
(36)
The integrals around the two little semicircles are the same as before, so
we get, for x < 0:
k−ρ
∞
eixz
dz +
k2 − z 2
eixz
dz +
2
2
−k k − z
GREEN’S FUNCTION FOR ONE DIMENSIONAL SCHRÖDINGER EQUATION
ˆ
6
eixs
πi eikx − e−ikx
ds
=
(37)
2
2
k
2
−∞ k − s
π
= − sin (kx)
(38)
k
π
= sin (k |x|)
(39)
k
where the last line follows because x < 0.
The above method gives the same answer for any choice of the little semicircles around the singularities, even if we choose the contour so that there
are no singularities inside the contour.
Thus the integral form 32 becomes in general
ˆ
m
ψ (x) = ψ0 (x) + 2
sin (k |x − x0 |) V (x0 ) ψ (x0 ) dx0
(40)
2h̄ k
∞
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