Calculating <ψ(t)|x|ψ(t)> in a Harmonic Oscillator Potential

[Rococo]

Homework Statement

A particle in a harmonic oscillator potential in the following state after a time t:

$| ψ(t) > = \frac{1}{\sqrt{2}} [e^{(-iE_0 t/\hbar)} |ψ_0> + e^{(-iE_1 t/\hbar)} |ψ_1> ]$

I want to write an expression for $<ψ(t)| \hat{x} | ψ(t) >$.

Homework Equations

The answer is meant to be:

$<ψ(t)| \hat{x} | ψ(t) > = \frac{1}{2} [ <ψ_0| \hat{x} | ψ_1> e^{-i(E_1 - E_0)t/\hbar)} + <ψ_1| \hat{x} | ψ_0> e^{-i(E_0 - E_1)t/\hbar)}]$

The Attempt at a Solution

[/B]
$<ψ(t)| \hat{x} | ψ(t) > = \int{ψ^{*}(t) \hat{x} ψ(t)}$
$= \int{\frac{1}{\sqrt{2}} [ e^{(iE_0 t/\hbar)} ψ^{*}_0 + e^{(iE_1 t/\hbar)} ψ^{*}_1}] \hat{x} \frac{1}{\sqrt{2}}[ e^{-(iE_0 t/\hbar)} ψ_0 + e^{-(iE_1 t/\hbar)} ψ_1]$

$= \int{\frac{1}{2} [ e^{(iE_0 t/\hbar)} ψ^{*}_0 \hat{x} e^{-(iE_0 t/\hbar)} ψ_0 + e^{(iE_0 t/\hbar)} ψ^{*}_0 \hat{x} e^{-(iE_1 t/\hbar)} ψ_1 + e^{(iE_1 t/\hbar)} ψ^{*}_1} \hat{x} e^{-(iE_0 t/\hbar)} ψ_0 + e^{(iE_1 t/\hbar)} ψ^{*}_1} \hat{x} e^{-(iE_1 t/\hbar)} ψ_1$

$= \frac{1}{2} [<ψ_0| \hat{x} | ψ_0 > + e^{-i(E_1 - E_0)t/\hbar} <ψ_0| \hat{x} | ψ_1 > + e^{i(E_1 - E_0)t/\hbar} <ψ_1| \hat{x} | ψ_0 > + <ψ_1| \hat{x} | ψ_1> ]$

This is a different answer than it should be, where am I going wrong?

 

[Orodruin]

No, it is correct. You are just missing one step of simplification.

 

[Rococo]

No, it is correct. You are just missing one step of simplification.

Do you say that $<ψ_0| \hat{x} | ψ_0 >$ and $<ψ_1| \hat{x} | ψ_1 >$ are expectation values of position, which for the simple harmonic oscillator, are zero?

 

[Orodruin]

I did not say it, I wanted you to do it.