Quantum Mechanics - Finding expectation value

[Kyle91]

Homework Statement

Find the expectation value of position as a function of time.

Homework Equations

This is in the latter half of a multi-part question, previously we were given that:

Eqn 1: Ψ(x, t) = A(ψ₁(x)e^{−iE₁t/h¯} + iψ₂(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

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(ψ₁ψ₂sin(t(E₁-E₂)/h¯) + ψ₁² + ψ₂²

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?

 

[Kyle91]

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:

A²(L - L/(2pi²) + L/(4pi²) + L/(16pi²))

And I am thoroughly sick of this.

 

[vela]

What happened to the factor of $\sin\left[\frac{(E_1-E_2)t}{\hbar}\right]$?

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.

 

[Kyle91]

I love you vela. Thank you so much.