Page 279 - Essentials of physical chemistry

The Schrödinger Wave Equation 241

probable position ‘‘x’’ the particle will have in the lowest energy state. Although we can treat the
coordinate as an operator, x ! x op , the function is not an eigenfunction of this operator! We can see
r ffiffiffi

2 1px
1
that xc ¼ x sin is just another function of x, not an eigenfunction. However, there is
L L
an alternate way to obtain the average value of the coordinate x or the expectation of the average
ð
c*Ocdt
value of x as < O > ð where O is any operator, so < x >¼< njxjn >
c*cdt
r ffiffiffi r ffiffiffi
L
ð
2 npx 2 npx
sin (x) sin dx. When we use previously normalized functions we do not
L L L L
0
need the denominator since it will be 1, but we should recall the analogy to the average grade of a
class where we have to divide by the number of students in the class. Thus there is a way to do a
weighted average of a quantum mechanical operator with the slight difference of inserting the operator
between c*and c. In undergraduate slang, we sometimes call this the ‘‘sandwich integral’’ where the
operator is sandwiched between c*and c, but really it is the ‘‘expectation value of the operator’’.
r ffiffiffi r ffiffiffi
L L
ð ð
2 px 2 px 2 2 px
sin (x) sin x sin dx:
Now let us evaluate < x >¼ L L L L dx ¼ L L
0 0

ð L ð L 2px
2 2 px 2 1 cos L
x sin x
<x> ¼ dx ¼ dx and again do two integrals. <x> ¼
L 0 L L 0 2
( " L
# )
ð L ð L L 2px ð L 2px
1 2px 1 x 2 x sin L sin L dx
x cos ¼ þ where we
xdx
L 0 0 L L 2 0 2p 0 0 2p
L
L
have used integration by parts for the second integral. To be complete we show details.
( " L ( " #)
# )
2 2px
2
1 L 0 cos 1 L 1 1 L
L
L 2 2p L 2 2p 2
< x > ¼ (0 0) 2 ¼ 0 2 ¼ and
L 0 L
so we find by calculation that the average value of the x coordinate is <x> ¼ L=2. In fact this
is true for any level n. Now in case you think you can just use common sense to guess properties
2
all of the time let us ask what is the average value of <x >. Again we can set up the
sandwich integral and carry out the integration by parts (twice) and we find that
r ffiffiffi r ffiffiffi
L 2
ð
2 2 npx 2 2 npx L 3
<x > ¼ sin (x ) sin dx ¼ 1 2 2 , leaving the proof to the
0 L L L L 3 2n p
homework. Recall that in classical mechanics (sophomore physics) there is a coordinate for every
momentum so let us consider <P x > as the average value of the x-momentum.
ð L r ffiffiffi r ffiffiffi ð L
2 npx h d 2 npx hnp npx npx
sin sin sin cos dx
<P x > ¼ dx ¼
L L i dx L L iL L L
0 0
ð L " # L
2 npx
hnp npx npx hnp sin L
sin cos ¼ 0, so <P x > ¼ 0.
np
<P x > ¼ dx ¼
iL L L iL
L 0
0
That makes sense because momentum is a vector and the average direction of the momentum
2
averages out to be exactly zero as the particle moves back and forth, but, what is <P > ¼ 0?
x
ð Lr ffiffiffi r ffiffiffi 2 2 2
2

2 2 npx h d 2 npx nph n h 2 2
x
x
L L i dx L L L 4L
< P > ¼ sin sin dx ¼ ¼ 2 ,so ( <P x > ) 6¼<P >.
0
2 2
2 2
P 2 n h n h
x 2
x
2m 8mL 4L
This checks with the quantized energy as E n ¼ ) P ¼ 2mE n ¼ (2m) 2 ¼ 2 .