Schrodinger equation, normalizing

[SoggyBottoms]

Homework Statement

Consider the infinite well, a particle with mass m in the potential

V(x) =<br /> \begin{cases}<br /> 0, &amp; 0 &lt; x &lt; a,\\<br /> \infty, &amp; \text{otherwise,}<br /> \end{cases},<br />

At t = 0 the particle is in the state:

\Psi(x,0) = B \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right]
with b a real number and l a whole number. Use normalization to show how B depends on b.

1 = B^2 \int_0^a \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right]^2 dx \\<br /> = B^2 \left(\frac{1}{2} \int_0^a (1 - \cos{\left(\frac{2 l \pi}{a}x\right)})dx + \frac{b^2}{2} \int_0^a (1 - \cos{\left(\frac{4 l \pi}{a}x\right)})dx + <br /> b\int_0^a \cos{\left(\frac{l \pi}{a}x\right)}dx + b\int_0^a \cos{\left(\frac{3 l \pi}{a}x\right)}dx\right)\\<br /> = B^2(\frac{a}{2} + \frac{b^2a}{2})

B = \sqrt{\frac{2}{a(1 + b^2)}}

Did I do this correctly?

 

[ehild]

It looks correct.

ehild

 

[SoggyBottoms]

Thanks.

 

[SoggyBottoms]

Now I have to calculate the expectation value of the energy. E = \frac{p^2}{2m}, so \langle E \rangle = \langle \frac{p^2}{2m} \rangle and with \langle p \rangle = \int_{-\infty}^{\infty} \Psi^{\ast} \frac{\hbar}{i} \frac{\partial}{\partial x} \Psi dx we get:

\langle E \rangle = \frac{-\hbar^2}{2m} \frac{2}{a(1 + b^2)} \int_0^a \Psi^{\ast} \frac{\partial^2}{\partial x^2} \Psi dx

Is the integrand simply equal to \frac{d^2}{dx^2} \left[\sin{\left(\frac{l \pi}{a}x\right)} + b\sin{\left(\frac{2l \pi}{a}x\right)}\right]^2? Because that becomes a really long calculation, so I was wondering if there is another way...