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Normalization of a wave function

1 vote

In order to normalize the wave function I am aware of the following condition:

$$\int_{-\infty}^{\infty} | \psi (x,t) |^2 dx = 1$$

If I am not mistaken what I have to do first is obtaining A so as to ensure that this value meets the stated condition. But what I got does not make sense:

There has to be a mistake here but I do not see where.

My book has the following information about normalization:

Thank you

asked Aug 22, 2018 in Physics Problems by Jorge Daniel (606 points)

1 Answer

1 vote
Best answer

Your mistake is that you have not taken enough care about the definition of the function $|x|$.

For $-\infty \lt x \le 0$ the function $y=|x|$ is a reflection of the region $0 \le x \lt +\infty$. Likewise the function $y=e^{-2\lambda |x|}$ in the domain $-\infty \lt x \le 0$ is a reflection of the same function in the domain $0 \le x \lt +\infty$.

Therefore the integral over the domain $-\infty \lt x \lt +\infty$ is twice that over the domain $0 \le x \lt \infty$ : $$\int_{-\infty}^{+\infty} e^{-2\lambda |x|} dx=2\int_0^{+\infty} e^{-2\lambda x} dx=-\frac{1}{\lambda}[e^{-2\lambda x}]_0^{+\infty} =\frac{1}{\lambda}$$

answered Aug 23, 2018 by sammy gerbil (28,448 points)
edited Aug 23, 2018 by sammy gerbil
Alright I missed it, now got it. Did not you lack a factor of $\frac{1}{2}$? In order to cancel out the derivative of the exponential. I got A = $\sqrt{\lambda}$ but you got    A = $\sqrt{2\lambda}$
Yes you are correct, I missed that factor. I have updated my answer.