Physical Chemistry , 1st ed.



2 m

 2





x

2

    2 (Acos kxBsin kx)



2 m

 2





x

(kAcos kxkBsin kx)





2 m

^2

(k^2 Acos kxk^2 Bsin kx)

Factoring k^2 out of the terms in parentheses, we can get our original wave-

function back:



(

2 m

k^2 )^2

(Acos kxBsin kx)

The two negative signs cancel, and the collection of terms multiplying the

wavefunction are all constants. We have thus shown that operation of the

Hamiltonian operator on this wavefunction yields an eigenvalue equation;

the eigenvalue is the energy of a particle having that wavefunction:

E

k

2

2

m

^2

Note that at this point, we have no idea what the constant kis.

In the above example, the wavefunction determined is deficient in a few re-

spects, specifically the identities of several constants. Up to this point, nothing

has constrained those constants to any particular value. Classically, the con-

stants can have any value, indicating that the energy can have any value.

However, quantum mechanics imposes certain restrictions on allowed wave-

functions.

The first requirement of the wavefunction is that it must be continuous.

Since we recognize that the wavefunction in the regions x0 and xamust

be zero, then the wavefunction’s value at x0 and xamust be zero. This

is certainly true when approaching these limits ofxfrom outsidethe box, but

the continuity of the wavefunction requires that this must also hold when ap-

proaching these limits from insidethe box. That is,(0) must equal (a)

which must equal zero. This requirement, that the wavefunction must be a cer-

tain value at the boundaries of the system, is called a boundary condition.*

The boundary condition (0) is applied first: since x0, the wavefunction

becomes

(0)  0 Acos 0 Bsin 0

Since sin 0 0, the second term places no restriction on the possible value(s)

ofB.However,cos 0 1, and this is a problem unless A0. So, in order to

satisfy this first boundary condition,Amust be zero, meaning that the only ac-

ceptable wavefunctions are

(x) Bsin kx

Now we apply the other boundary condition:(a) 0. Using the wavefunc-

tion from above:

(a)  0 Bsin ka

where ahas been substituted for x. We cannot require that Bequal zero. If it

290 CHAPTER 10 Introduction to Quantum Mechanics

*Boundary conditions are also apparent for some classical waves. For example, a vibrat-

ing guitar string has a wave motion whose amplitude is zero at the ends, where the string is

tied down.