Chemistry 460
Dr. Jean M. Standard
Problem Set 8 Solutions
1.
Determine the expectation values
Sˆx ,
Sˆy , and
Sˆz
for an electron in the spin up state. Use the
spin raising and lowering operators to evaluate the first two of these expectation values.
Sˆx
€
For this expectation value, we can combine Sˆ+ and Sˆ− to give
Expectation value
€
€
Sˆx =
€
€
Then, the expectation value
Sˆx
1
2
(Sˆ+ + Sˆ− ) .
€
for an electron in the spin up state can be written
€
Sˆ
x
€
=
α Sˆx α
= 12 α Sˆ+ + Sˆ− α
Sˆx
= 12 α Sˆ+ α
+ 12 α Sˆ− α .
Using the relations,
€
Sˆ+ α
Sˆ− α
= 0
=
β ,
the expectation value can be rewritten
€ as
Sˆx
= 12 α Sˆ+ α
+ 12 α Sˆ− α
= 0 + 12 α β
Sˆx
= 0.
The last relation comes about because the spin states are orthogonal.
€
2
1.
Continued
Expectation value
Sˆy
In this case, we can combine Sˆ+ and Sˆ− to yield
Sˆy =
€
€
i
2
(Sˆ− − Sˆ+ ) .
€
Substituting, the expectation value Sˆy for an electron in the spin up state can be written
€
Sˆy = α Sˆy α
€
= 2i α Sˆ− − Sˆ+ α
Sˆy
= 2i α Sˆ− α
− 2i α Sˆ+ α .
Applying the spin raising and lowering operators,
€
Sˆy
= 2i α Sˆ− α
= 2i α β
Sˆy
Expectation value
− 2i α Sˆ+ α
− 0
= 0.
€
Sˆz
For this expectation value, we can use
Sˆz α
€
Substituting, the expectation value Sˆz
€
!
2
α .
for an electron in the spin up state can be written
Sˆz
€
=
=
α Sˆz α
= α !2 α
= !2 α α
Sˆz
€
= !2 .
3
2.
Consider the two-electron spin wavefunction, ψ spin (1, 2 ) = α (1) β ( 2 ) − α ( 2 ) β (1) . Show that the
appropriate normalization constant is
1
.
2
To begin, we include the normalization constant N in the form of the wavefunction,
ψ spin (1, 2) = N "# α (1) β ( 2) − α ( 2) β (1) $% .
The normalization condition is
ψ spin ψ spin = 1 .
Substituting the form of the two-electron spin wavefunction given above yields,
N 2 α (1) β ( 2 ) − α ( 2 ) β (1) α (1) β ( 2 ) − α ( 2 ) β (1) = 1 .
Here we have assumed that the normalization constant can be chosen to be real. Expanding the terms and
separating into four integrals, we obtain,
N 2 { α (1) β ( 2 ) α (1) β ( 2 ) − α (1) β ( 2 ) α ( 2 ) β (1) − α ( 2 ) β (1) α (1) β ( 2 ) + α ( 2 ) β (1) α ( 2 ) β (1) } = 1 .
Note that the integrals involve double integrals over the spin coordinates of electrons 1 and 2. Breaking each
term up into a product of the integrals for electrons 1 and 2 yields,
N 2 { α (1) α (1) β ( 2 ) β ( 2 ) − α (1) β (1) β ( 2 ) α ( 2 )
− β (1) α (1) α ( 2 ) β ( 2 ) + β (1) β (1) α ( 2 ) α ( 2 ) } = 1 .
Next, we can use the orthonormal properties of the spin up and spin down functions,
α α = β β = 1, and α β = β α = 0 .
Substituting, the normalization expression for the two-electron spin wavefunction becomes,
N 2 { 1⋅1 − 0 ⋅ 0 − 0 ⋅ 0 + 1⋅1 } = 1 2N 2 = 1, or N = 1
.
2
Thus, the normalization constant has been demonstrated to be
ψ spin (1, 2 ) = 1
, and the normalized spin wavefunction is
2
1 "
# α (1) β ( 2 ) − α ( 2 ) β (1) $% .
2
4
3.
The permutation operator Pˆ12 interchanges the coordinates of electrons 1 and 2. That is,
Pˆ12 f (1, 2) = f (2, 1) .
€
Consider the spin functions
€
f (1,2) = α (1) + α (1) α (2) + α (2)
g(1,2) = α (1) β (2) + α (2) β (1)
u(1,2) = α (1) − α (2)
υ (1,2) = α (1) β (2) − β (1) β (2) .
Determine which of the functions are eigenfunctions of the permutation operator Pˆ12 . For the
eigenfunctions, indicate whether
€ they are symmetric or antisymmetric with respect to interchange of the
coordinates.
For the function f (1, 2) , the operation of the permutation operator is
€
Pˆ12 f (1, 2) = α (2) + α (2) α (1) + α (1)
= f (1, 2) .
€
The function f (1, 2) is an eigenfunction of the permutation operator and it is symmetric with respect to
interchange.
€
For the function g(1, 2) , the operation of the permutation operator is
€
Pˆ12 g(1, 2) = α (2) β (1) + α (1) β (2)
= g(1, 2) .
€
The function g(1, 2) is an eigenfunction of the permutation operator and it is symmetric with respect to
interchange.
€
For the function u(1, 2) , the operation of the permutation operator is
€
Pˆ12 u(1, 2) = α (2) − α (1)
= − u(1, 2) .
€
The function u(1, 2) is an eigenfunction of the permutation operator and it is antisymmetric with respect to
interchange.
€
For the function υ (1, 2) , the operation of the permutation operator is
€
Pˆ12 υ (1, 2) = α (2) β (1) − β (2) β (1)
≠ constant ⋅ υ (1, 2) .
€
The function υ (1, 2) is NOT an eigenfunction of the permutation operator.
€
€
5
4.
Consider a Slater determinant for a two-electron system given by:
χ1 (1)
1
2
Ψ SD = χ1 ( 2 )
χ 2 (1) χ 2 ( 2 )
.
Here, the functions χ1 and χ 2 correspond to spin orbitals. When the coordinates of the two electrons
are interchanged, the resulting Slater determinant has the form:
Ψ SD, interchange = 1
2
χ1 ( 2 )
χ1 (1)
χ 2 ( 2 ) χ 2 (1)
.
Show by expanding the original and interchanged Slater determinants that the interchanged function
equals –1 times the original function, as required by the Pauli Principle.
Expanding out the first Slater determinant gives
Ψ SD = = χ1 (1)
1
2
χ1 ( 2 )
χ 2 (1) χ 2 ( 2 )
1
{ χ1 (1) χ 2 (2) − χ 2 (1) χ1 (2) } .
2
Expanding out the interchanged Slater determinant gives
Ψ SD, interchange = 1
2
χ1 ( 2 )
χ1 (1)
χ 2 ( 2 ) χ 2 (1)
= 1
{ χ1 (2) χ 2 (1) − χ 2 (2) χ1 (1) }
2
= 1
{ − χ1 (1) χ 2 (2) + χ 2 (1) χ1 (2) }
2
= −
1
{ χ1 (1) χ 2 (2) − χ 2 (1) χ1 (2) }
2
Ψ SD, interchange = − Ψ SD .
Thus, interchanging the coordinates of the two electrons leads to a wavefunction that is –1 times the orginal;
that is, the wavefunction is properly antisymmetric as required by the Pauli Principle.
6
5.
The ground state configuration of the four-electron Be atom is 1s22s2. Construct the Slater determinant
corresponding to this configuration.
The spin orbitals for this system are 1sα , 1sβ , 2sα , 2sβ . This leads to a single 4×4 Slater determinant for the
4-electron system of the form
€
ΨSD =
1s(1)α (1)
1s(2)α (2)
1s(3)α (3) 1s(4)α (4)
1 1s(1) β (1) 1s(2) β (2) 1s(3) β (3)
4! 2s(1)α (1) 2s(2)α (2) 2s(3)α (3)
2s(1) β (1) 2s(2) β (2) 2s(3) β (3)
1s(4) β (4)
.
2s(4)α (4)
2s(4) β (4)
€
6.
An excited state configuration of Be is 1s22s3s. Determine all the possible Slater determinants for this
configuration.
The spin orbitals for this system are 1sα , 1sβ , 2sα , 2sβ , 3sα , 3sβ . This leads to a possibility of four
different 4×4 Slater determinants for the 4-electron system. The possible sets of spin orbitals are
1) 1sα , 1sβ ,
2) 1sα , 1sβ ,
3) 1sα , 1sβ ,
4) 1sα , 1sβ ,
€
2sα ,
2sα ,
2sβ ,
2sβ ,
3sα
3sβ
3sα
3sβ .
The four Slater determinants constructed from these sets of spin orbitals are
€
€
€
€
€
ΨSD1 =
1s(1)α (1)
1 1s(1) β (1)
4! 2s(1)α (1)
3s(1)α (1)
1s(2)α (2)
1s(2) β (2)
2s(2)α (2)
3s(2)α (2)
1s(3)α (3) 1s(4)α (4)
1s(3) β (3) 1s(4) β (4)
2s(3)α (3) 2s(4)α (4)
3s(3)α (3) 3s(4)α (4)
ΨSD2 =
1s(1)α (1)
1 1s(1) β (1)
4! 2s(1)α (1)
3s(1) β (1)
1s(2)α (2)
1s(2) β (2)
2s(2)α (2)
3s(2) β (2)
1s(3)α (3) 1s(4)α (4)
1s(3) β (3) 1s(4) β (4)
2s(3)α (3) 2s(4)α (4)
3s(3) β (3) 3s(4) β (4)
ΨSD3 =
1s(1)α (1) 1s(2)α (2) 1s(3)α (3) 1s(4)α (4)
1 1s(1) β (1) 1s(2) β (2) 1s(3) β (3) 1s(4) β (4)
4! 2s(1) β (1) 2s(2) β (2) 2s(3) β (3) 2s(4) β (4)
3s(1)α (1) 3s(2)α (2) 3s(3)α (3) 3s(4)α (4)
ΨSD4 =
1s(1)α (1)
1s(1)
β (1)
1
4! 2s(1) β (1)
3s(1) β (1)
1s(2)α (2)
1s(2) β (2)
2s(2) β (2)
3s(2) β (2)
1s(3)α (3) 1s(4)α (4)
1s(3) β (3) 1s(4) β (4)
.
2s(3) β (3) 2s(4) β (4)
3s(3) β (3) 3s(4) β (4)
7
7.
Determine all the possible Slater determinants for the 2s12p1 configuration of a two-electron system. You
will need to consider all the possible p orbitals.
The spin orbitals for this system are 2sα , 2sβ , 2 pxα , 2 px β , 2 pyα , 2 py β , 2 pzα , 2 pz β . Selecting
appropriate combinations for the two-electron system would lead to a possibility of 12 different 2×2 Slater
determinants! The possible sets of spin orbitals are
€
1) 2sα , 2 pxα
5) 2sα , 2 pyα
9) 2sα , 2 pzα
2) 2sα , 2 px β
6) 2sα , 2 py β
10) 2sα , 2 pz β
3) 2sβ , 2 pxα
7) 2sβ , 2 pyα
11) 2sβ , 2 pzα
4) 2sβ , 2 px β
8) 2sβ , 2 py β
12) 2sβ , 2 pz β
The first four of the 12 Slater determinants constructed from these sets of spin orbitals are
€
€
€
€
ΨSD1 =
1
2
2s(1)α (1)
2s(2)α (2)
2 px (1)α (1) 2 px (2)α (2)
ΨSD2 =
1
2
2s(1)α (1)
2s(2)α (2)
2 px (1) β (1) 2 px (2) β (2)
ΨSD3 =
1
2
2s(1) β (1)
2s(2) β (2)
2 px (1)α (1) 2 px (2)α (2)
ΨSD4 =
1
2
2s(1) β (1)
2s(2) β (2)
2 px (1) β (1) 2 px (2) β (2)
You get the idea. . . There are eight more like this.
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8
8.
Consider an electron configuration 1s3. We know that this configuration cannot exist. Use two spin up
functions and one spin down function and construct a Slater determinant for this configuration. By
expanding the determinant, show that this wavefunction is identically zero, and thus the electron
configuration 1s3 cannot exist.
For this one, the possible spin orbitals are 1sα , 1sβ . Using two spin up functions and one spin down functions,
the 3×3 Slater determinant is
1s(1)α (1) 1s(2)α (2) 1s(3)α (3)
1
1s(1)α (1) 1s(2)α (2) 1s(3)α (3) .
3!
1s(1) β (1) 1s(2) β (2) 1s(3) β (3)
€
ΨSD =
In order to evaluate this, the method of cofactors must be used to expand the 3×3 determinant. Using rhe
method of cofactors
€ to expand about the first row,
1s(2)α (2) 1s(3)α (3)
1
(−1)1+1 1s(1)α (1) 1s(2) β (2) 1s(3) β (3)
3!
ΨSD =
+
1s(1)α (1) 1s(3)α (3)
1
(−1)1+2 1s(2)α (2) 1s(1) β (1) 1s(3) β (3)
3!
+
1s(1)α (1) 1s(2)α (2)
1
(−1)1+3 1s(3)α (3) 1s(1) β (1) 1s(2) β (2) .
3!
Expanding the 2×2 determinants yields
€
1
1s(1)α (1) [ 1s(2)α (2)1s(3) β (3) − 1s(2) β (2)1s(3)α (3)]
3!
ΨSD =
−
1
1s(2)α (2) [ 1s(1)α (1)1s(3) β (3) − 1s(1) β (1)1s(3)α (3)]
3!
+
1
1s(3)α (3) [ 1s(1)α (1)1s(2) β (2) − 1s(1) β (1)1s(2)α (2)] .
3!
Simplifying the expression gives
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ΨSD =
1
1s(1)1s(2)1s(3) {α (1)α (2) β (3) − α (1) β (2)α (3) − α (1)α (2) β (3)
6
+ β (1)α (2)α (3) + α (1) β (2)α (3) − β (1)α (2)α (3)}
ΨSD = 0.
Since the Slater determinant is zero, the probability density for this state is zero, and therefore the 1s3
configuration
cannot exist.
€