1. The problem statement, all variables and given/known data Find the expectation value of position as a function of time. 2. Relevant equations This is in the latter half of a multi-part question, previously we were given that: Eqn 1: Ψ(x, t) = A(ψ_{1}(x)e^{−iE1t/h¯} + iψ_{2}(x)e^{−iE2t/h¯}) and in an even earlier part: Eqn 2: ψ_{n}(x) = sqrt(2/L)sin(n*pi*x/L) Note: h¯ = hbar 3. The attempt at a solution As you can tell, I'm not too awesome at formatting on here so I'm going to quickly explain my method. I said: <x> = Integral from 0 to L of Ψ*(x, t)Ψ(x, t) dx So I told wolframalpha (we're allowed to use it) to simplify Eqn 1, before subbing in Eqn 2. This gave me: Eqn 3: -2(ψ_{1}ψ_{2}sin(t(E_{1}-E_{2})/h¯) + ψ_{1}^{2} + ψ_{2}^{2} I then subbed in Eqn 2 into Eqn 3 and integrated. No matter how many times I do this I always end up with a bunch of sines. These sines are all something like sin(n*pi/L) so when I sub in the limits of integration they become sin(n*pi) or sin(0), both of which are 0! This means that I'm just left with an answer of 2, which is not time dependent. Can anyone see where I've gone wrong?
Okay so I just found out I've had the formula wrong from the start. FML. It's now: <x> = Integral from 0 to L of Ψ*(x, t)xΨ(x, t) dx However, this still didn't help me. I'm now left with: A^{2}(L - L/(2pi^{2}) + L/(4pi^{2}) + L/(16pi^{2})) And I am thoroughly sick of this.
What happened to the factor of [itex]\sin\left[\frac{(E_1-E_2)t}{\hbar}\right][/itex]? After you multiply the integrand out, you should have \begin{align*} \langle x \rangle &= A^2\int_0^L \left(x\psi_1^2(x) + x\psi_2^2(x) - 2x\psi_1(x)\psi_2(x) \sin [(E_1-E_2)t/\hbar]\right)\,dx \\ &= A^2\left(\int_0^L x\psi_1^2(x)\,dx + \int_0^L x\psi_2^2(x)\,dx - 2\sin [(E_1-E_2)t/\hbar]\int_0^L x\psi_1(x)\psi_2(x)\,dx\right) \end{align*} As long as that last integral isn't 0, the expectation value will vary with time.