I am given the following function:

$$\Psi (x,t) = Ae^{-\lambda|x|}e^{-iwt}$$

Where $A, \lambda$ and $\omega \in + \mathbb{R}$

I have to determine the expected value of x. Then:

$$< x > = \int_{-\infty}^{\infty} x |\Psi (x,t)|^2 dx$$

I previously normalized the function, thus I know:

$$A = +\sqrt{\lambda}$$

I arrived to a point where I have:

$$2\lambda \int_{0}^{\infty} xe^{-2\lambda|x|} dx$$

I got the following result:

$$< x > = \frac{1}{2\lambda}$$

I have checked other answers to this problem and some people argue < x > should be 0 as, and I quote, 'an odd function times an even one becomes an odd function' which is in a symmetric domain and because of that < x > is 0. I considered since I was in high school that $e^x$ was neither an odd nor even function. So I think this comment is misleading.

I have not studied much of function analysis so I am pretty confused here. I have checked the method I used and I think I am not mistaken but if the function analysis explanation is right please shed some light on it.

• 'I arrived at the point where I have ....': something has gone wrong here. The answer is $0$. – Kavi Rama Murthy Aug 24 '18 at 12:02

I can see how you would move from

$$< x > = \int_{-\infty}^{\infty} x |\Psi (x,t)|^2 dx$$

to

$$< x > = \lambda \int_{-\infty}^{\infty} xe^{-2\lambda|x|} dx.$$

If this is indeed what you did, then the real mystery is how you proceeded from

$$< x > = \lambda \int_{-\infty}^{\infty} xe^{-2\lambda|x|} dx$$

to

$$< x > = 2 \lambda \int_{0}^{\infty} xe^{-2\lambda|x|} dx.$$

It seems that you were thinking that

$$\int_{-\infty}^{0} xe^{-2\lambda|x|} dx = \int_{0}^{\infty} xe^{-2\lambda|x|} dx,$$

this isn't true. Instead, we have that

$$\int_{-\infty}^{0} xe^{-2\lambda|x|} dx = \color{red}{-} \int_{0}^{\infty} xe^{-2\lambda|x|} dx.$$

• @Vincent because of the definition of the function |x| – JD_PM Aug 24 '18 at 12:25
• And what is more, when you get the result of this integral by parts method you get a factor which is not multiplied by x and is an exponential. How can you get 0 then? (When you plug $\infty$ you get 0 though) – JD_PM Aug 24 '18 at 12:30
• yes the point is that one of the two $x$'s gets an absolute value sign around it, but the other one does not. It is the $x$ which is not $|x|$ but just itself that causes the sign flip. – Vincent Aug 24 '18 at 12:49
• If we agree on $< x > = \lambda \int_{-\infty}^{\infty} xe^{-2\lambda|x|} dx$ then it is easy to see that $< x > = \lambda \int_{-\infty}^{0} xe^{-2\lambda|x|} dx + \lambda \int_{0}^{\infty} xe^{-2\lambda|x|} dx$ – Vincent Aug 24 '18 at 12:50
• My argument is that we do not even need to compute the two terms because the last equality reads <x> = 'some number' + 'minus that same number'; this of course equals zero. – Vincent Aug 24 '18 at 12:51