Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 / 2

Choose the overall phase so that $A$ is positive. Normalization requiresThereforeAn arbitrary constant phase multiplying this answer has no effect on any expectation value.

The probability density is even, so its first moment vanishes. Its second moment follows by integrating the polynomial:Using the normalization from part a givesThe variance in position is consequently $a^2/7$.

The wavefunction is real and vanishes continuously at both endpoints. HenceThe momentum second moment is most safely evaluated from its quadratic form:ThusThe derivative jumps where the wavefunction joins zero. On the whole real line the strong second-derivative operator therefore produces endpoint delta distributions, so the state is not in the usual operator domain of $p^2$. The finite momentum second moment, equivalently the kinetic-energy quadratic form or the integral of $p^2$ against the momentum probability density, remains well defined and gives the value above.

Both first moments vanish, so the standard deviations follow directly from the second moments:The Heisenberg uncertainty relation is satisfied and is not saturated. A Gaussian wave packet can saturate the position-momentum bound; this compactly supported polynomial is not Gaussian.

For a single slit of width $d$, the far-field amplitude is proportional to $\mathrm{sinc}(\pi d\sin \theta /\lambda )$. The first zeros satisfy $d\sin \theta =\pm \lambda$. At small angles the angular width between those zeros is $2\lambda /d$, or $2L\lambda /d$ on a screen a distance $L$ away.The result assumes the wavelength, propagation distance and far-field conditions stay fixed. If the slit becomes comparable to the wavelength, the exact sine relation replaces the small-angle scaling.