Unformatted text preview: 2
6 =
+ 24 − = ℏ
2 + + 2 + 24 = + = + = 1+
=0
39
=
4 ∴ = + 6 + + 6
5ℏ
13
=
+
2
32 6 + 6 + 6 24
24 24 =ℏ 24 2+ 1
39
+ 0+
2
6
24 4 To answer the last part of Problem 716,
is not included in the trial wavefunction because it
is odd; the book solution is incorrect in stating that
= 0. However, because it is odd
(while
and
are even) ,
=
=
=
= 0, and
is
and/or
in any meaningful way. In more
mathematically incapable of combining with
chemistryesque terms,
can not "hybridize" with
and/or , making its inclusion as a basis
function essentially pointless.
The 2 × 2 determinant can be solved symbolically before plugging the specific elements:
−
→
−
−
−
=0
0=
−
0=
−
+
−
Solving this expression, with the values determined above for the
integrals, in Mathcad gives
= 384ℏ + 56 ± 9 ℏ + 21 ℏ 8 57
+ 256 / 256
2
All of the constants in the first term are positive; the argu...
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 Spring '14
 SteveMiller
 Physical chemistry, Atom, pH, Conservation Of Energy, Energy, Characteristic polynomial, trial wavefunction

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