Unformatted text preview: t
should be 2 × 2:
−
−
0=
−
−
Knowing that the HO solutions are an orthogonal set allows us to simplify the determinant,
because
=
→
=
= 1 and
=
=0
−
∴0=
−
However, the
integrals must still be evaluated. The first,
, can be evaluated as follows:
= = = − ℏ
2 + − + = +
1+ 6 + 24 6 + 6
+ 6 24 24 + 24
In this expression,
and
are the expected values of and for the product
×
, respectively. Using Mathcad to evaluate the two integrals gives
= 0 (this can also be seen by inspection, since the integrand is odd and the integral is
taken over all space)
3
=
4
1
3
ℏ
∴
=
+
+
=ℏ 0+ + 0 +
=
+
6
24
2
6
24 4
2
32
The remaining
integrals can be evaluated in similar fashion:
= 6 + 2 24 = = + + 2 + ℏ
2 = − ℏ
2 + 2 + 6 + 24 − = ℏ
2 + +
+ = +
0+
=0
= ∴ = 3 + 6 6 = 6 0+ =
= − ℏ
2 −
+ + = +
0+
=0
=
= 6 3 + 6 3 ℏ
2 = 2
+ 8 2 + 2 + 6 + 24 6 + 6 + 6 = 24
= 24
24 24 2
+ 24 24 = = 24 2 + = + 6 24 24 24 = ∴ + 6 2
+ 6 24 = = + 2 6 0+
− 24 ℏ
2 3 = 2 + 2 8
+...
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 Spring '14
 SteveMiller
 Physical chemistry, Atom, pH, Conservation Of Energy, Energy, Characteristic polynomial, trial wavefunction

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