Atomic and Molecular Physics
Rafael Carrasco Carmona
Obtain the Schrödinger equation for a two electon atom (Eq.
(6.3))
In this exercise we are considering an atom composed by a nucleus of
mass “M ” and charge “Ze” (“Z” is the atomic number of the atom and “e”
is the charge of the electron), and two electrons of mass “m” and charge “e”.
We will note as:
R0 = X01 e1 + X02 e2 + X03 e3 = X0i ei
R1 = X11 e1 + X12 e2 + X13 e3 = X1i ei
R2 = X21 e1 + X22 e2 + X23 e3 = X2i ei
(1)
(2)
(3)
(where e1 = x, e2 = y, e3 = z) to the position vectors of the nucleus,
the first and the second electron respectively (note that in the previous expressions we are employing the Einstein summation convention and we will
continue with this criteria unless otherwise indicated, that is why we note
the basis vectors by this way instead the common one) and:
∂
∂
∂
∂
e1 +
e2 +
e3 = −ih̄
ei (4)
= −ih̄
1
2
3
∂(X0 )
∂(X0 )
∂(X0 )
∂X0i
∂
∂
∂
∂
e1 +
e2 +
e3 = −ih̄
ei (5)
= −ih̄
1
2
3
∂(X1 )
∂(X1 )
∂(X1 )
∂X1i
∂
∂
∂
∂
= −ih̄
e1 +
e2 +
e3 = −ih̄
ei (6)
1
2
3
∂(X2 )
∂(X2 )
∂(X2 )
∂X2i
P0 = −ih̄∇R0
P1 = −ih̄∇R1
P2 = −ih̄∇R2
to their momenta. If we only take into account the Coulomb interaction
between all the particles, we are allowed to establish that the Hamiltonian of
the system is the sum of the kinetic energies (T) and these interactions (V):
T =
P0 2 P1 2 P2 2
h̄2 2
h̄2 2
h̄2 2
+
+
=−
∇R0 −
∇R1 −
∇
2M
2m
2m
2M
2m
2m R2
(7)
Ze2
Ze2
e2
−
+
4π0 |R1 − R0 | 4π0 |R2 − R0 | 4π0 |R2 − R1 |
(8)
V =−
H=
P0 2 P1 2 P2 2
Ze2
Ze2
e2
+
+
−
−
+
=
2M 2m 2m 4π0 |R1 − R0 | 4π0 |R2 − R0 | 4π0 |R2 − R1 |
1
Atomic and Molecular Physics
−
Rafael Carrasco Carmona
h̄2 2
h̄2 2
h̄2 2
Ze2
Ze2
∇R0 −
∇R1 −
∇R2 −
−
2M
2m
2m
4π0 |R1 − R0 | 4π0 |R2 − R0 |
+
e2
4π0 |R2 − R1 |
(9)
It is important to mention that the potential energy (V) concerns the
interaction between the nucleus and the two electrons and the interaction
between these two particles.
In order to simplify the previous expression we change to the center of
mass coordinates defined as:
R =
1
1
(M R0 + mR1 + mR2 ) =
(M · X01 + m · X11 + m · X21 )e1
M + 2m
M + 2m
1
1
+
(M · X02 + m · X12 + m · X22 )e2 +
(M · X03 + m · X13 + m · X23 )e3
M + 2m
M + 2m
3
X
1
=
(M · X0j + m · X1j + m · X2j )ej
(10)
M + 2m j=1
ri = Ri − R0 = (Xi1 − X01 )e1 + (Xi2 − X02 )e2 + (Xi3 − X03 )e3 =
3
X
i = 1, 2
(Xij − X0j )ej
(11)
j=1
We will denote the components of R as X j and the ones of ri as xji .
Now we are going to obtain the Laplacian in the new coordinates, for
what we will need to calculate the following partial derivatives:
∂
∂X j ∂
∂xj1 ∂
∂xj2 ∂
M
∂
∂
∂
=
+
δij
−δij j −δij j =
j+
j =
i
i
i
i
j
j
∂X0
∂X0 ∂X ∂X0 ∂x1 ∂X0 ∂x2
M + 2m ∂X
∂x1
∂x2
=
∂
∂
M
∂
−
−
i
M + 2m ∂X i ∂x1 ∂xi2
(12)
And these second derivatives (taking into account Schwarz’s theorem):
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Atomic and Molecular Physics
∂2
∂
=
i 2
∂(X0 )
∂(X0i )
M
=
M + 2m
−
Rafael Carrasco Carmona
M
∂
∂
∂
− i − i
i
M + 2m ∂X
∂x1 ∂x2
M
∂
∂
∂
− i − i
i
M + 2m ∂X
∂x1 ∂x2
M
∂
∂
∂
− i − i
i
M + 2m ∂X
∂x1 ∂x2
∂
=
∂xi2
=
∂
∂
∂
∂
M
∂
−
− i − i
i
i
∂X
M + 2m ∂X
∂x1 ∂x2 ∂xi1
M
M + 2m
2
∂2
∂2
M
−
∂(X i )2 M + 2m ∂xi1 ∂X i
M
∂2
M
∂2
∂2
∂2
M
∂2
−
+
+
−
M + 2m ∂xi2 ∂X i M + 2m ∂X i ∂xi1 ∂(xi1 )2 ∂xi2 ∂xi1 M + 2m ∂X i ∂xi2
2
M
∂2
∂2
∂2
2M
∂2
2M
∂2
+ i i+
=
−
−
∂x1 ∂x2 ∂(xi2 )2
M + 2m ∂(X i )2 M + 2m ∂xi1 ∂X i M + 2m ∂xi2 ∂X i
−
+
∂2
∂2
∂2
+
2
+
∂(xi1 )2
∂xi2 ∂xi1 ∂(xi2 )2
(13)
Hence, the expresion of the Laplacian is:
∇2R0
=
3
X
i=1
∂2
=
∂(X0i )2
M
M + 2m
2 X
3
i=1
3
3
3
2M X ∂ 2
∂2
−
∂(X i )2 M + 2m i=1 ∂xi1 ∂X i
3
3
X ∂2
X ∂2
X ∂2
2M X ∂ 2
−
+
+
2
+
=
M + 2m i=1 ∂xi2 ∂X i i=1 ∂(xi1 )2
∂xi2 ∂xi1 i=1 ∂(xi2 )2
i=1
M
M + 2m
2
M
M + 2m
∇2R −
2
2M
2M
∇R ·∇r1 −
∇R ·∇r2 +∇2r1 +2∇r1 ·∇r2 +∇2r2 =
M + 2m
M + 2m
∇2R −
2M
2M
∇R ·∇r1 −
∇R ·∇r2 +(∇r1 +∇r2 )2 (14)
M + 2m
M + 2m
where
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Atomic and Molecular Physics
Rafael Carrasco Carmona
3
X
∂
∇R =
e
i i
∂X
i=1
∇r1
3
X
∂
=
e
i i
∂x
1
i=1
∇r2
3
X
∂
=
ei
∂xi2
i=1
On the other hand, we must calculate the next derivatives:
m
∂
∂X j ∂
∂xj1 ∂
∂
∂
=
+
δij
+ δij j =
j =
i
i
i
j
j
∂X1
∂X1 ∂X
∂X1 ∂x1
M + 2m ∂X
∂x1
m
∂
∂
+
M + 2m ∂X i ∂xi1
(15)
And the second derivatives:
∂
∂2
=
i 2
∂(X1 )
∂X1i
m
∂
∂
+
=
M + 2m ∂X i ∂xi1
m
∂
∂
∂
m
∂
∂
∂
m
+ i
+
+ i
=
i
i
i
M + 2m M + 2m ∂X
∂x1 ∂X
M + 2m ∂X
∂x1 ∂xi1
2
m
2m
∂2
∂2
∂2
+
+
(16)
M + 2m ∂(X i )2 M + 2m ∂X i ∂xi1 ∂(xi1 )2
So the Laplacian is:
∇2R1
=
3
X
i=1
∂2
=
∂(X1i )2
m
M + 2m
m
M + 2m
2
2 X
3
∇2R +
i=1
3
3
X
∂2
2m X ∂ 2
∂2
+
+
=
∂(X i )2 M + 2m i=1 ∂X i ∂xi1 i=1 ∂(xi1 )2
2m
∇R · ∇r1 + ∇2r1
M + 2m
The expresion for ∇2R2 is similar to the previous one:
2
m
2m
2
∇R2 =
∇2R +
∇R · ∇r2 + ∇2r2
M + 2m
M + 2m
4
(17)
(18)
Atomic and Molecular Physics
Rafael Carrasco Carmona
If we substitute Eqs. (10), (11), (14), (17), (18) in Eq. (9) we obtain:
h̄2
H=−
2M
M
M + 2m
2
∇2R +
2M
h̄2
2M
h̄2
∇R · ∇r1 +
∇R · ∇r2
2M M + 2m
2M M + 2m
h̄2
h̄2
−
(∇2r1 +∇2r2 +2∇r1 ·∇r2 )−
2M
2m
h̄2
−
2m
m
M + 2m
2
∇2R −
m
M + 2m
2
∇2R −
2m
h̄2 2
Ze2
h̄2
Ze2
∇R ·∇r2 −
∇r2 −
−
2m M + 2m
2m
4π0 |r1 | 4π0 |r2 |
h̄2 (M + 2m) 2 h̄2
e2
=−
∇ −
+
4π0 |r2 + X0 − r1 − X0 |
2(M + 2m)2 R 2
−
=−
h̄2
2m
h̄2 2
∇R ·∇r1 −
∇
2m M + 2m
2m r1
1
1
+
M
m
(∇2r1 +∇2r2 )
h̄2
Ze2
e2
Ze2
∇r1 · ∇r2 −
−
+
=
M
4π0 |r1 | 4π0 |r2 | 4π0 |r12 |
Ze2
h̄2
h̄2 2
h̄2
Ze2
−
∇2R −
(∇r1 + ∇2r2 ) − ∇r1 · ∇r2 −
2(M + 2m)
2µ
M
4π0 r1 4π0 r2
+
e2
4π0 r12
(19)
mM
is the reduced mass of the nucleus and one electron,
M +m
= r2 − r1 , r1 = |r1 |, r2 = |r2 | and r12 = |r12 | .
where µ =
r12
So the Schrödinger equation would be:
h̄2
h̄2 2
h̄2
Ze2
Ze2
2
2
−
∇R −
(∇r1 + ∇r2 ) − ∇r1 · ∇r2 −
−
2(M + 2m)
2µ
M
4π0 r1 4π0 r2
e2
+
ψtotal (R, r1 , r2 ) = Etotal ψtotal (R, r1 , r2 )
4π0 r12
5
(20)
Atomic and Molecular Physics
Rafael Carrasco Carmona
As we can see, it is possible to express the Hamiltonian as the sum of
two terms, the first concerning only the movement of the center of mass, and
the second one concerning only the relative coordinates, so it is interesting
to seek solutions of the form ψtotal (R, r1 , r2 ) = U (R)ψ(r1 , r2 ). Substituting
this expression and dividing by ψ(R, r1 , r2 ) we get:
2
h̄2
U (R)
h̄
ψ(r1 , r2 )
2
−
∇R U (R)+
− (∇2r1 + ∇2r2 )
U (R)ψ(r1 , r2 )
2(M + 2m)
U (R)ψ(r1 , r2 )
2µ
h̄2
Ze2
Ze2
e2
− ∇r1 · ∇r2 −
−
+
ψ(r1 , r2 ) = Etotal
M
4π0 r1 4π0 r2 4π0 r12
(21)
So:
2
1
h̄2
1
h̄
2
∇R U (R) +
−
− (∇2r1 + ∇2r2 )
U (R)
2(M + 2m)
ψ(r1 , r2 )
2µ
Ze2
Ze2
e2
h̄2
−
+
ψ(r1 , r2 ) = Etotal
− ∇r1 · ∇r2 −
M
4π0 r1 4π0 r2 4π0 r12
(22)
Consequently, as each expression depends on different variables and their
sum is equal to Etotal , they has to be independiently equal to a constant so
that the sum of them is Etotal .
h̄2
2
∇ U (R) = ECM U (R)
−
(23)
2(M + 2m) R
2
h̄2
h̄
Ze2
Ze2
− (∇2r1 + ∇2r2 ) − ∇r1 · ∇r2 −
−
2µ
M
4π0 r1 4π0 r2
e2
+
ψ(r1 , r2 ) = Eψ(r1 , r2 )
(24)
4π0 r12
where E + ECM = Etotal .
As we can see, the center of mass behaves like a free particle of mass
M + 2m and energy ECM . If we consider that the mass of the nucleus is
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Atomic and Molecular Physics
Rafael Carrasco Carmona
infinitely large (that is M = ∞), the reduced mass is equal to the mass of
2
the electron, and the term − h̄M ∇r1 · ∇r2 becomes negligible (this is known as
the “mass polarisation” term). As consequence, the equation we must solve
is:
2
Ze2
Ze2
h̄
−
− (∇2r1 + ∇2r2 ) −
2µ
4π0 r1 4π0 r2
e2
+
ψ(r1 , r2 ) = Eψ(r1 , r2 )
(25)
4π0 r12
Let us recall that in atomic units 4π0 = 1, h̄ = 1, e = 1 and m = µ = 1,
so the last equation reduces to:
1 2
1 2
Z
Z
1
− ∇r1 − ∇r2 − − +
ψ(r1 , r2 ) = Eψ(r1 , r2 )
(26)
2
2
r1 r2 r12
which is the equation we were searching for.
7