Normalization factor in wave equation

1. Aug 18, 2011

hnicholls

(Note: although arising in QM, this is essentially a calculus question)

Ѱ (x) = A sin (n╥x/a)

1 = ∫ l Ѱ (x) l^2 dx with limits of integration a to 0

1 = ∫ A^2 sin^2 (n╥x/a) dx with limits of integration a to 0

Indefinite integral ∫ sin^2 x dx = x/2 - sin2x/4

I know this integral should reduce to 1 = A^2 a/2

But what I get is

I = A^2 [a/2 - sin (2n╥a/4a) - 0/2 - sin (2n╥0/4a)]

Third and fourth term within bracket are 0, but second term is 1

i.e., sin (n╥/2) = 1

I assume I'm missing something, but not sure what it is.

Thanks

2. Aug 19, 2011

G01

Your integral is not right. In the second term, the 4 is not inside the argument of the sine function:

$$\int A^2 \sin^2(x) dx = \frac{x}{2}-\frac{\sin(2x)}{4} + C$$

3. Aug 20, 2011

hnicholls

Thanks.

That was my problem.

Correct value now.