Expectation value of energy in infinite well

[dyn]

Homework Statement

Given the following normalised time-independent wave function the question asks for the expectation value of the energy of the particle. The well has V(x)=0 for 0<x<a

Homework Equations

ψ( x ) = √(1/a) ( 1+2cos(∏x/a) )sin(∏x/a)

The Attempt at a Solution

I disagree with the given answer but we both start off the same way. We rearrange the equation as
ψ ( x ) = √(1/a) ( sin(∏x/a) + sin(2∏x/a) )

The given solution then performs then calculates <E> = ∫ ψ H ψ to arrive at <E> = E(1) + E(2).

I wrote ψ down as the superposition of the wavefunction √(2/a) sin (n∏x/a) which means ψ(1) and ψ(2) both have a coefficient of 1/(√2) in front of them. I then squared this coefficient and multiplied by E(1) +E(2) to get <E> = 1/2 ( E(1) + E(2) )

Is my method ok ? and is my answer correct ?

Thanks

 

[TSny]

Yes your method and answer are correct.

<E> = ∫ ψ H ψ = 1/2 ( E(1) + E(2) )