Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / Section D

The displacement is along $x$, transverse to propagation in the membrane's $yz$ plane. Write $A=|A|e^{i\varphi }$; then $d=|A|\cos (\omega t-k_{y}y-k_{z}z+\varphi )$. Here $|A|$ is the displacement amplitude, $\varphi$ the initial phase, $\omega$ the angular frequency and $\mathbf{k}$ the wavevector normal to constant-phase lines. Its magnitude is $2\pi /\lambda$, and its direction specifies phase propagation for the chosen sign convention.

For equal-phase two-wave membrane modes, two equal complex amplitudes with opposite $k_y$ sum toThe edge condition gives $\cos (k_{y}a/2)=0$, hence $k_y=(2j+1)\pi /a$, $j=0,1,\dots$, for this even-in-$y$ family. Opposite relative phase gives the sine family; together they yield all Dirichlet transverse modes $k_y=n\pi /a$, $n\ge 1$. For the equal-phase family,The lowest mode has $k_y=\pi /a$. ThereforeIt approaches $c$ for $\lambda \to 0$ and diverges as $\lambda \to 2a^{-}$; there is no real travelling-mode $k_z$ beyond the cutoff. This is an axial phase velocity, not the group velocity: $v_g=c\sqrt{1-(\lambda /2a{)}^2}<c$. Here $c$ denotes the membrane wave speed, not the speed of light.

For the evanescent membrane response below cutoff, below $f_c=c/(2a)$, put $k_z=-i\kappa$ for decay along positive $z$, with $\kappa =\sqrt{(\pi /a{)}^2-(\omega /c{)}^2}$. A localized central drive creates an evanescent, sinusoidally oscillating deformation decaying away in both axial directions. Far enough from the drive the leading profile is proportional to $\cos (\pi y/a)e^{-\kappa |z-z_0|}\cos (\omega t+{\varphi }_0)$. Higher evanescent modes contribute near the point source. The pattern has no steadily advancing axial phase front or far-field propagating wave; decreasing frequency increases its spatial localization.

Direct differentiation gives ${\psi }^{\prime \prime }/\psi =x^2/a^4-1/a^2$. Inserting this into the Schrödinger equation and separately matching the $x^2$ and constant terms yieldsFor $V_0=m{\omega }_0^2/2$, $a^2=\hslash /(m{\omega }_0)$ and $E=\hslash {\omega }_0/2$. The normalizable Gaussian has no nodes, as expected for the nondegenerate one-dimensional ground state of this confining potential.

The normalization integral is $|C{|}^{2}a\sqrt{\pi }=1$. Choose its irrelevant global phase so thatFor the oscillator ground-state density in a comparison wavelength, choose a free comparison particle with the same total energy. It has $p=\sqrt{2mE}=\hslash /a$, hence comparison wavelength $\lambda =h/p=2\pi a$. ConsequentlyThe bound density is localized on scale $a=\lambda /(2\pi )$, with standard deviation $\lambda /(2\pi \sqrt{2})$, rather than being spatially uniform like the density of a single free plane wave. The comparison does not assign constant free-particle momentum to the bound state; its total energy also includes potential energy.