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- Homework Statement:
- An electron in a carbon nanotube of length ##L## is described by the wavefunction ##\psi(x) = (2/L)^{1/2}\sin{(\pi x/L)}##. Compute the expectation value of the kinetic energy of the electron.

- Relevant Equations:
- ##\left \langle \Omega \right \rangle = \int \psi^* \hat{\Omega}\psi ~d\tau##

If ##\hat{T} = -\frac{\hbar}{2m}\frac{\mathrm{d^2} }{\mathrm{d} x^2}##, then the expectation value of the kinetic energy should be given as:

$$\begin{align*}

\left \langle T \right \rangle &= \int_{0}^{L} \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)} \hat{T}\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)} dx \\

&= \frac{-\hbar^2}{mL} \int_{0}^{L} \sin{\left(\frac{\pi x}{L}\right)} \frac{\mathrm{d^2} }{\mathrm{d} x^2} \sin{\left(\frac{\pi x}{L}\right)} dx \\

&= \frac{\pi^2 \hbar^2}{mL^3} \int_{0}^{L} \sin^2{\left(\frac{\pi x}{L}\right)} dx \\

&=\frac{h^2}{8mL^2}

\end{align*}

$$

Are my calculations correct?

To solve the integral ##\int \sin^2(f(x))) dx## I use the half angle identity for ##\sin(x)##.

$$\begin{align*}

\left \langle T \right \rangle &= \int_{0}^{L} \sqrt{\frac{2}{L}} \sin{\left(\frac{\pi x}{L}\right)} \hat{T}\sqrt{\frac{2}{L}}\sin{\left(\frac{\pi x}{L}\right)} dx \\

&= \frac{-\hbar^2}{mL} \int_{0}^{L} \sin{\left(\frac{\pi x}{L}\right)} \frac{\mathrm{d^2} }{\mathrm{d} x^2} \sin{\left(\frac{\pi x}{L}\right)} dx \\

&= \frac{\pi^2 \hbar^2}{mL^3} \int_{0}^{L} \sin^2{\left(\frac{\pi x}{L}\right)} dx \\

&=\frac{h^2}{8mL^2}

\end{align*}

$$

Are my calculations correct?

To solve the integral ##\int \sin^2(f(x))) dx## I use the half angle identity for ##\sin(x)##.