How Do You Calculate Particle Probability in a Quantum State?

[Ayham]

Homework Statement

What is the probability that a particle in the ground state will be found between L/2 and 2L/3?
im new guys so go easy :)

Homework Equations

∫ψ(x)^2 dx = ∫2/L (sin(πx/L))^2 dx
in attachment

The Attempt at a Solution

The answer should be 30.44%
i got 66.66% and sometimes a negative number
please show me the steps too :/

 

[dydxforsn]

So you have the integral:\int{\psi \psi^{*} dx} = \int^{\frac{L}{2}}_{\frac{2L}{3}}{\sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} \sqrt{\frac{2}{L}} \sin{\frac{n \pi x}{L}} dx} = \int^{\frac{L}{2}}_{\frac{2L}{3}}{\frac{2}{L} \sin^{2}{\frac{n \pi x}{L}} dx} You can get the integral of sine squared from an integral table:\int{\sin^{2}{ax} dx} = \frac{x}{2} - \frac{\sin{2ax}}{4a} Keep in mind that both \frac{x}{2} and \frac{\sin{2ax}}{4a} are evaluated at the limits of integration.

Doing all of this I obtained 30% for the answer when I plugged in n = 1 (for the ground state).

You're probably just messing up the minus sign on one of the 4 terms that come about when you evaluate the \frac{x}{2} - \frac{\sin{2ax}}{4a} term at the limits of integration. You 4 terms should be \int{\psi \psi^{*} dx} = \frac{2}{L} (\frac{L}{3} - \frac{L}{4} - \frac{\sin{(2\frac{n \pi}{L}\frac{2L}{3}})}{4 \frac{n \pi}{L}} + \frac{\sin{(2 \frac{n \pi}{L} \frac{L}{2}})}{4 \frac{n \pi}{L}})

Edit*** After a 3rd check, when n = 1 the answer is indeed 30%. Looks like we are both susceptible to math errors on this one >_< (I had edited my post thinking the answer was 60% when I double checked my original answer...)

 

[Ayham]

I love you :')