Page 275 - Essentials of physical chemistry

P. 275

The Schrödinger Wave Equation 237

ð ð
d dc dc dc
¼ adx and then ¼ a dx and so ln(c) ¼ ax þ C. Now
dx dx c c
¼ ac )
a c ¼ 0 )
ax
take the antiln of the equation to ﬁnd c ¼ Ae where C ¼ ln A. Before we go further, please note
that ( a) in the equation became (þa) in the solution so when we encounter this type of differential
equation, we can simply write the solution as the constant with opposite sign as a power of base-e!
Using that idea we can write down the two solutions of our PIB problem.

p ﬃﬃﬃﬃﬃ p ﬃﬃﬃﬃﬃ
i 2mE x þi 2mE x
c(x) ¼ C 1 e h 2 þ C 2 e h 2 and that is the solution except we do not know C 1 or C 2
and we need to use boundary conditions to ﬁnd these constants.
iu
Now recall Euler’s rule that e ¼ cos(u) þ i sin(u) and apply it to both terms. We do not know
C 1 or C 2 but there would be a cos( ) part from both terms and a sin( ) part from each term. Let
the coefﬁcients of these two parts be new unknown constants so we have
" ! # " ! #
r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2mE 2mE
c(x) ¼ A cos 2 x þ iB sin 2 x . However, at x ¼ 0 the potential energy
h h
goes up to þ1 so physically no particle can be there and the wave function must be zero there;
so c(0) ¼ 0. Since cos(0) ¼ 1, the only way we can have c(0) ¼ 0isif A ¼ 0. The other constant B is
" ! #
r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2mE
still unknown but it can be nonzero. Now we have c(x) ¼ iB sin 2 x . The fact that
h
i appears as a factor may seem strange but we will ﬁnd a way around that soon.
Next we need to apply the boundary condition to the right side of the box at x ¼ L and again we
see that the wave function must go to zero at that wall. This is easily seen by considering

2
d c
2
dx 2 d c
. As long as is ﬁnite, an inﬁnite denominator makes c ¼ 0.
c ¼ 2
2m 2 [V E] dx
h
Up to this point the value of the energy E could be any value but we are about to see it become
" ! #
r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2mE
quantized! At x ¼ L we have c(L) ¼ iB sin L ¼ 0. We could set B to zero but then
h 2
there is no wave function at all! That is called the trivial solution and corresponds to no particle in
the box! Is there any other way this can be zero? Yes, the sin(u) is zero every time u ¼ p,2p,
!
r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ
2mE
3p, ..., np. Thus we ﬁnd the quantization condition for the energy as 2 L ¼ np, which
h
!
r ﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2 2 2 2 2 2 2 2
2mE np n p n p h n h
h
means that 2 ¼ and E n ¼ 2 ¼ ¼ 2 ¼ E n . Now the energy can
2 2
h L 2mL 2m(2p) L 8mL
have only certain values!
We still need to ﬁnd the value of B. There was a lot of philosophical groping in the 1920s as
scientists tried to interpret the meaning of the De Broglie wave but the physicists were well educated
in electromagnetic theory. It was known that while the oscillating electric ﬁeld of a light wave could
p
ﬃﬃﬃﬃﬃﬃﬃ
be written as E ¼ Ae 2pint , with the complex number i ¼ 1 embedded in the formula, only an
2
intensity I ¼ (Ae þ2pint ) * (Ae þ2pint ) ¼ A could be measured in the laboratory as a real number.
Further, the intensity is the square of the amplitude of the wave. We have already seen that the
p ﬃﬃﬃﬃﬃﬃﬃ
complex number i ¼ 1 has occurred in our c(x) wave function. Thus it was realized that c*c /
probability. We can only measure real numbers in the laboratory but c(x) can be complex so we need
to measure the probability as the product of the wave function and its complex conjugate where if any
ﬃﬃﬃﬃﬃﬃﬃ
p
1 it must be changed to i in the complex conjugate. Note if
part of the formula has i ¼
2
2
c ¼ a þ ib, then c* ¼ a ib and c*c ¼ (a ib)(a þ ib) ¼ a þ b which is a totally real number!
The next question is how to assign probability based on c*c? ‘‘Certainty’’ might have been
assigned 100% but instead c*c ¼ 1 was chosen as certainty with lesser values for other lower

270 271 272 273 274 275 276 277 278 279 280