BioPHYSICAL chemistry

We can check the solutions for all values of xby substituting the dif-
ferent solutions into Schrödinger’s equation. For example, let us put the
ground-state solution into the equation and show that it holds. First,
we calculate the second derivative of the wavefunction. In taking these
derivatives note that the two terms are obtained and can be rewritten in
terms of the wavefunction.

ψ 0 (x) =N 0 e−x

(^2) /2α 2
ψ 0 (x) =N 0 e−x
(^2) /2α 2
(11.16)

(11.17)

(11.18)

Substitution of the second derivative into Schrödinger’s equation yields:

(11.19)

ψ (11.20)

(^0) αα
2
2
4
2
()xx 222 m 2 0
k
m

−+ E

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟+−

⎛

⎝

ZZ

⎜⎜⎜

⎞

⎠

⎟⎟

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

= 0

−−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

⎡

⎣

⎢

⎢

⎤

⎦

⎥

⎥

+

Z^2

0

2

42

2

2

1

m 2

x

xkx

ψ

αα

() xxxEx^2 ψψ 000 ()= ()

x

x

=− 0

2

4

1

ψ()

α αα^2

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

d

d

d

d

2

2 002

2

0

22

x

xN

x

x

ψ eNx

α

()=−⎛ /α

⎝

⎜⎜

⎞

⎠

− ⎟⎟= ee

−x x

−

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

222 2

42

/α^1

αα

d

d

d

x d

xN

x

eNe

xxx

ψ

α

αα

00

2

0

2

2

()==−^22 //^22 ⎛

⎝

−−⎜⎜

⎜

⎞

⎠

⎟⎟

CHAPTER 11 VIBRATIONAL MOTION 227

0

(a) U

x x

v  1

v  2

ψν

v  3

v  0

0

(b)

x x

v  1

v  2

ψ^2 ν

v  3

v  0

0

(c)

x x

Figure 11.3Wavefunctions and their energies of the simple harmonic oscillator.