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Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 24, November 10, 2010
Solved Homework
We are given a perturbing potential
V
=
k
( 1 –
x
) for a particle in an otherwise
"normal" box of length 1. The first order correction to the groundstate energy is simply
the expectation value of the perturbing operator for the exact eigenfunction that would
exist in the
absence
of the perturbation. That is
a
1
1
( )
=
1
0
(
)
x
( )
k
1
−
x
( )
Ψ
1
0
(
)
x
( )
where
Ψ
1
is the ground state particleinabox wave function
Ψ
1
0
(
)
x
( ) =
sin
π
x
( )
0
≤
x
≤
1
E
1
0
(
)
=
π
2
2
=
a
1
0
(
)
(note that although our formal derivations used subscript "0" to represent a ground state,
for the particle in a box the ground state quantum number is
n
= 1). The energy of the
ground state in the absence of the perturbation is the zeroorder eigenvalue plus the first
order correction.
We may expand the expectation value above as
a
1
1
( )
=
k
1
0
(
)
x
( )
Ψ
1
0
(
)
x
( )
−
1
0
(
)
x
( )
x
Ψ
1
0
(
)
x
( )
=
k
1
−
1
2
=
k
2
where in making the first set of simplifications we have made use of the normalization of
the groundstate wave function (so the first integral is 1) and our prior knowledge that
<
x
> for a particle in a box wave function is always
L
/2, which in this case is 1/2.
Note that the result is completely intuitive. It says that, to first order, the ground
state energy goes up by
k
/2. This is the value that the perturbing function has in the
middle of the box, and it is what we would obtain as an
exact
answer if the bottom of the
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box had potential
k
/2 instead of the usual convention of zero. Since the perturbing
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This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.
 Fall '08
 Staff
 Physical chemistry, pH

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