How to normalize Schrodinger equation

[warnexus]

Homework Statement

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?

Homework Equations

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

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)?

 

[SammyS]

Homework Statement

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?

Homework Equations

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

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)?

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.

 

[warnexus]

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.

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

 

[SammyS]

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

\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} \ . \

 

[warnexus]

\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} \ . \

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.

 

[SammyS]

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.

I just used a definition of what it means for a function to be odd.

cos(2θ)=1−2sin²(θ) is a double angle identity from trig. It's pretty basic.

http://en.wikipedia.org/wiki/Trig_Identity#Double-angle.2C_triple-angle.2C_and_half-angle_formulae