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Asinka+Bcoska=0
n
−
which reduce to
Asinka=0
.
Bcoska=0
n
Since these sine and cosine expressions cannot be simultaneously zero, we
get the solutions
A=0 and coska=0 ka=nπ/2 with n odd
⇒ .
(cid:26)B =0 and sinka=0 ka=nπ/2 with n even
⇒
In fact it suffices to restrict the solutions to non-negative integers. The
reasonis thatthe solutionsfornegativeintegersarenotlinearlyindependent of
the solutions with positive integers. This is apparent from the explicit form of
the solutions
Bcos(nπx/2a), n odd
u (x)= . (4.14)
n (cid:26)Asin(nπx/2a), n even
Changing n to n in these formulas have no effect for the cosine solutions,and
−
for the sine solutions, the ensuing change of sign for sine can be absorbed into
the constant A.
TheconstantsAandB arenormalizationconstantstobedeterminedbythe
normalization condition
a
u n(x)∗u n(x)dx=1 (4.15)
Z
a
−
where ’*’ denotes complex conjugation. Clearly, some normalization of the
solutions is needed, and this particular one is related to the interpretation of
quantummechanicswherethewavefunctionsψ(x)areinterpretedasprobability
amplitudes. To say that ψ is a probability amplitude is to say that the integral
d
ψ(x) ψ(x)dx
∗
Z
c
is the probability of detecting the particle in the interval (c,d).
In fact, an even stronger property of our solutions can be inferred:
a 1, if n=m
u (x) u (x)dx=δ = . (4.16)
Z n ∗ m nm (cid:26)0, if n=m
a 6
−
Thisequationexpressestheorthonormality ofthesolutions,i.ethesolutions
are normalized and solutions with different index are orthogonal. This is a
general property of solutions to eigenvalue problems. The general theory will
be spelt out in chapter 5.
73
The solutionsu (x)cannowbe consideredtoformabasisofa linearvector
n
space. Any solution to the wave equation can be expressed as a linear combi-
nation (or superposition) of the basis functions
∞
ψ(x)= α u (x), (4.17)
n n
nX=0
where α are complex numbers. These numbers are arbitrary apart from a
n
globalnormalization. Sincethe totalprobabilityofdetecting the particleinthe
box must be 1, we get
a
∞
ψ(x)∗ψ(x)dx= α∗nα =1 (4.18)
n
Z
−a X0
This follows from a nice calculation involving several of the formulas given
in this section and some trigonometry. Performing this calculation gives quite
a lot of insight into the mathematics of quantum mechanics.
Since we now know k, we can get a formula for the separation constant E,
which is to be interpreted as the energy of the system
¯h2k2 ¯h2 nπ 2 π2¯h2
E = = = n2. (4.19)
n 2m 2m(cid:18)2a(cid:19) 8ma2
In this way we get quantization of the energy. The energy can only take
values determined by the integer n. This is symbolized by indexing the E with
n.
ButwhyisE energy? Thiscanbeunderstoodbygoingbacktotheclassical
n
equationforenergyinterms ofkinetic andpotentialenergyE =K+V. Inour
case, the potential energy is zero inside the box, and we have simply E = K.
Then using the quantization rules we got
1 ∂
E H = ( i¯h )2.
−→ 2m − ∂x
If this Hamiltonian operator is applied to any of the solutions, the result is
1 ∂ nπx
Hu (x)= ( i¯h )2Bcos( )=
n
2m − ∂x 2a
¯h2 nπ nπx ¯h2π2 nπx
( )2Bcos( )=( n2)Bcos( )=E u (x)
n n
2m 2a 2a 8ma 2a
In the next to last expression, we recognize the separation constant (eigen-