Normalization of a Wave Function

[r16]

Homework Statement

I'm starting to (trying) teach myself some quantum mechanics out of the Griffiths book, and since there are no answers in the back I have no idea if I'm on the right track or not. Could you guys look over the answer to this equation to see if it looks right?

Consider the wave function $$\Psi(x,t) = A e^{-\lambda |x|}e^{-i \omega t}$$

a Normalize $$\Psi$$

Homework Equations

$$1 = \int^\infty_{-\infty} | \Psi |^2 dx$$

The Attempt at a Solution

$$|\Psi(x,t)|^2 = (\Psi*)\Psi$$
$$\Psi* = A e^{-\lambda |x|}e^{i \omega t}$$
$$|\Psi(x,t)|^2 = A^2e^{-2 \lambda |x|}$$

$$1 = \int^\infty_{-\infty} A^2e^{-2 \lambda |x|} dx = \frac{A^2}{\lambda}$$
$$A = \sqrt{\lambda}$$

Look correct?

 

[Cyosis]

That is correct. Don't forget you can check your answer by integrating the function with your normalization constant plugged in.

 

[flatmaster]

Here, the absolute value becomes a problem. The absolute value makes the integrand an even function. That means that the final function is symetric about the y axis. What you can do is drop the absolute value, Integrate from 0 to infinity, and multiply the result by two.

 

[flatmaster]

Oh wait, you did the integral already. Well, that's what you would do if you were actually doing the math rather than using a table or computer program.

 

[r16]

Here, the absolute value becomes a problem. The absolute value makes the integrand an even function. That means that the final function is symetric about the y axis. What you can do is drop the absolute value, Integrate from 0 to infinity, and multiply the result by two.

i did the integral mathematically exactly the way you said! I wasn't completely sure that this was the correct way to evaluate the absolute value, but I had a feeling it was. I'm a math major in addition to being a physics major =)

 

[flatmaster]

This lies within the tips and tricks that are learned slowly and never taught explicitly. When they are taught explicitly, they are often not understood at the time.