Past exam of the physics A course of the University of Cambridge / 2011 / Paper 2 / Section A

For a normalizable state take $a>0$. inverse construction of a quantum potential gives $V=E+{\hslash }^2{\Psi }^{\prime \prime }/(2m\Psi )$. differentiation yields ${\Psi }^{\prime \prime }/\Psi =16a^2{x}^6-12a{x}^2$, so the chosen potential energy zero fixes $E=0$. ThusThis energy lies above the negative minima of the potential energy; its zero value does not mean zero kinetic energy. The nodeless eigenfunction is the ground state.

The Sturm oscillation theorem gives three nodes of a wavefunction for the fourth bound state. reflection symmetry and nondegeneracy of one-dimensional bound states make it odd: one node of a wavefunction is at the centre, and the other two lie at opposite positions. In each flat allowed region the wavefunction is a sinusoid, with $k=\sqrt{2m(E-V)}/\hslash$; its wavelength is shortest in the central, deepest region. Outside the well it decays exponentially, with opposite signs on the two sides. Both $\Psi$ and ${\Psi }^{\prime }$ remain continuous at every finite step, as required by wavefunction matching at a finite potential step; only the curvature jumps. In particular, the wavefunction need not vanish at a step.

This original wavefunction matching at a finite potential step illustration uses freely chosen well dimensions and energies. It shows the required parity, node count, wavelength ordering and decaying tails; the source drawing supplies no numerical scale from which to fix exact node positions.

The canonical commutation relation is $[p_x,x]=-i\hslash I$. Applying the commutator product rule on a common operator domain,Equivalently, applying both sides to a twice differentiable test function gives $-{\hslash }^2[(xf{)}^{\prime \prime }-x{f}^{\prime \prime }]=-2{\hslash }^2{f}^{\prime }$, the same result.

The central estimate is $z=3.2743$. propagation of uncertainty must include the occurrence of $x$ in both numerator and denominator. With angular derivatives measured per radian,Interpreting the two quoted errors as independent standard deviations, convert them to radians before calculatingIf the quoted errors are instead absolute bounds, the first-order worst-case estimate is $|\delta z|\lesssim |z_x|\delta x+|z_y|\delta y=0.60$. Correlated errors add $2z_x{z}_y\mathrm{Cov}(x,y)$ to the variance. Treating $x+y$ and $x$ as independent would lose their shared measurement uncertainty.

In the high-resolution quantization approximation, a smooth input distribution varies little across one conversion interval, so rounding to the nearest level gives an approximately uniform quantization error $ϵ\in [-\Delta V/2,\Delta V/2]$. ThereforeThe input must stay within the analogue-to-digital converter's range. A large standard deviation alone does not guarantee uniform quantization error: an input concentrated exactly on widely separated conversion levels has zero error. Truncation without a half-step correction also has a different mean and root mean square.