BioPHYSICAL chemistry

the solution to use is based upon experience in solving differential equa-
tions. For this situation, consider the general solution:

ψ(x) =Asinkx+Dcoskx (10.2)

where Aand Bare the amplitudes of the sine and cosine functions, respec-
tively, and kis related to the wavelength according to k= 2 π/λwhere λ
is the wavelength. These parameters will be determined by substitution
of eqn 10.2 into Schrödinger’s equation (eqn 10.1). Remember that the
derivatives of the trigonometric terms are given by:

(10.3)

Taking the derivative of eqn 10.2 yields:

(10.4)

and

(10.5)

Notice that the second derivative was written in terms of the wavefunction;
this is something that we will do for all of the problems. Substitution of
the second derivative into Schrödinger’s equation gives:

(10.6)

We now have a solution to the problem and have found that the para-
meter kis related to the energy. To determine the values of the other
parameters we make use of the boundary conditions. As discussed in
Chapter 9, the wavefunctions are required to be everywhere continuous.
Due to the infinite potential outside of the box, the wavefunction outside
of the box is everywhere zero. To be continuous, we must set the wave-
functions to also be zero at the walls of the box. There we can write:

ψ(x=0) = 0 =Asin 0 +Bcos 0 →B= 0 (10.7)

ψ(x=L) = 0 =AsinkL→kL=nπ where n=1, 2, 3... (10.8)

→=E

k

m

Z^22

2

−=+=

ZZ^22

2

2

2

22 mx

x

m

kx Ex

d

d

ψψψ() () ()

d

d

2

2

22

x

( sinA kx+=− −Bcoskx) ( Aksinkx Bkcoss)kx =−k^2 ψ()x

d

d

d

d

2

x^2 (sinAkxB kx+=cos ) x( cosAk kx−Bksinn)kx

d

dx

(sinA kx+= −Bcos ) (coskx k A kx Bsin )kx

d

d

and

d

x d

kx k kx

x

sin ==cos coskx −ksinkx

CHAPTER 10 PARTICLE IN A BOX AND TUNNELING 199