# Schrodinger equation, normalizing

## Homework Statement

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

$$V(x) = \begin{cases} 0, & 0 < x < a,\\ \infty, & \text{otherwise,} \end{cases},$$

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.

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

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

Did I do this correctly?

Last edited:

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ehild
Homework Helper
It looks correct.

ehild

Thanks.

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...