# Expectation value of position of a particle

As you might be able to tell from my question that I'm just starting to learn about quantum mechanics. My question is the following:

With a wave function like this: $\displaystyle{\Psi (x, t) = Ae^{-\lambda|x|}e^{-i\omega t}}$, what happens to the variable $t$ when using $$\displaystyle{<x> = \int_{-\infty}^{\infty} x |\Psi |^2dx}~ ?$$

Can you assume $t=0$ like with the normalization to $1$, or do you absorb it in the constant because you integrate with respect to $x$? (I'm not sure how this works if this is the correct way). Basically I'm unsure about how to work this out even if my second thought would work.

Any help would be appreciated!

• $\Psi (x, t) = Ae^{-\lambda|x|}e^{-i\omega t}$ so $\Psi (x, t)^{*} = Ae^{-\lambda|x|}e^{+i\omega t}$. Hence $\psi^2$ is independent of time. – AMS Sep 17 '16 at 13:00
• A good resource, imo, is a book based totally on worked examples, but unfortunately with enough typo's to keep you on your toes ,: ). Quantum mechanics demystified, by David McMahon, look through it on Amazon, I keep going back to it for the basics, which happens with me quite a lot : ). – user108787 Sep 17 '16 at 13:08

In general if a state is time dependent, the corresponding expectation values are time dependent, so you should write something like $$\langle x(t)\rangle= \int x |\Psi(x,t)|^2 \mathrm{d}x$$
However note that in this case $$|\Psi(x,t)|^2 = |A|^2 \mathrm{e}^{-2\lambda |x|}$$ does not actually depend on $t$.
• Thank you! I failed to see that $e^{-i \omega t}$ would of course cancel with the complement to give $e^0 = 1$ – jopie Sep 17 '16 at 13:06