How to normalize Schrodinger equation

1. Mar 13, 2013

warnexus

1. The problem statement, all variables and given/known data

The solution to the Schrodinger equation for a particular potential is psi = 0 for absolute x > a and psi = Asin(pi*x/a) for -a <= x <= a, where A and a are constants. In terms of a, what value of A is required to normalize psi?

2. Relevant equations

psi = 0 for absolute value of > a
psi = Asin(pi*x/a) for -a<= x <= a

3. The attempt at a solution

The textbook gave the normalization as:

[integral] psi^2 dx = 1. The integral has upper limit positive infinity and lower limit negative infinity.

Can I just start by taking the integral of both sides of psi = Asin(pi*x/a)?

which makes psi = A-cos(pi*x/a) * pi*x^2/(2a)?

Last edited: Mar 13, 2013
2. Mar 13, 2013

SammyS

Staff Emeritus
No.

The integral of ψ(x) from -∞ to +∞ is zero. After all, ψ(x) is an odd function.

(ψ(x))2 ≥ 0 for all x and it's greater than zero for some values of x, so its integral is definitely positive.

3. Mar 13, 2013

warnexus

How did you know the integral is zero. Why is psi an odd function?

4. Mar 13, 2013

SammyS

Staff Emeritus
$\displaystyle \Psi(x)=A\sin\left(\frac{\pi x}{a}\right)$

Therefore, $\displaystyle \ \Psi(-x)=-\Psi(x)\,,\$ so $\displaystyle \ \Psi(x)\$ is odd.

That's pretty much beside the point except that it definitely shows that in general, $\displaystyle \ \int \left(f(x)\right)^2\,dx\ne\left(\,\int f(x)\,dx\right)^2\ .$

Solving your integral may be made easier by noting that $\displaystyle \ \cos(2\theta)=1-2\sin^2(\theta) \ .\ \$ And solving for sin2(θ) gives $\displaystyle \ \sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \ . \$

Last edited: Mar 13, 2013
5. Mar 13, 2013

warnexus

I know I stated -a<= x <= a. But x can be positive or negative. How did you know to choose x to be negative?

How is cos(2θ)=1−2sin2(θ)? I don't remember this being a trig identity. I am weak in math.

6. Mar 13, 2013

SammyS

Staff Emeritus