Past exam of the physics A course of the University of Cambridge / 2014 / Paper 1 / Section B

Set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(U-E)}/\hslash$. For unit incident amplitude, use $\psi =e^{ikx}+r{e}^{-ikx}$ on the left and $\psi =t_s{e}^{-\kappa x}$ on the right; the growing exponential is excluded at infinity. Continuity at zero gives $1+r=t_s$ and $ik(1-r)=-\kappa t_s$. ThusThe phase is chosen in $(-\pi ,0)$ for $0<E<U$. The evanescent wave is $\psi (x)=2k{e}^{-\kappa x}/(k+i\kappa )$ for $x>0$. Its probability density is nonzero, but its probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is zero. Penetration into a forbidden region therefore does not require transmitted flux through a semi-infinite step.

For a finite rectangular barrier, matching both interfaces givesThe approximation requires $\kappa a\gg 1$, or a width much greater than the evanescent penetration length. The tunnelling exponent is $-2a\sqrt{2m(U-E)}/\hslash$; its factor of two converts an amplitude decay to a probability decay. For the electron parameters, $\kappa =1.984\times 10^9{\mathrm{m}}^{-1}$ and $\kappa a=19.84$. ThereforeKeeping only the exponential gives $5.83\times 10^{-18}$; the matched-interface prefactor accounts for the factor of three.

In spherical coordinates, the angular Laplacian has spherical harmonics as eigenfunctions, $L^2{Y}_{lm}={\hslash }^{2}l(l+1)Y_{lm}$ and $L_z{Y}_{lm}=\hslash m{Y}_{lm}$. Separation leavesRegularity and normalizability give principal quantum number $n=1,2,\dots$, orbital quantum number $l=0,\dots ,n-1$, and magnetic quantum number $m=-l,\dots ,l$. Here the $m$ in the kinetic coefficient is the electron mass, distinct from the magnetic label.

For the proposed angular factor, $\cos \theta \propto Y_{10}$, so $l=1$ and magnetic quantum number zero. Direct differentiation gives ${\nabla }^2(r{e}^{-\alpha r}\cos \theta )=({\alpha }^2-4\alpha /r)\psi$. The $1/r$ term in the Schrödinger equation vanishes when $2{\hslash }^2\alpha /m=C$. With Bohr radius $a_0={\hslash }^2/(mC)$,wavefunction normalization uses $\int d\Omega {\cos }^2\theta =4\pi /3$ and ${\int }_0^{\infty }{r}^4{e}^{-2\alpha r}dr=24/(2\alpha {)}^5$, givingAn arbitrary constant phase is allowed. This is the $2p_z$ state. Under spatial inversion, $r$ is unchanged and $\cos \theta$ changes sign, so its parity eigenvalue is $-1$, also given by $(-1{)}^l$.

Rotational invariance means rotated states have the same energy; it does not require each member of a degenerate energy eigenspace to be invariant. rotations mix the three $l=1$ magnetic substates. An orbital oriented along $z$ can therefore have an anisotropic probability density while the Hamiltonian remains spherically symmetric.

The angular momentum commutation relations are $[L_x,L_y]=i\hslash L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hslash (L_y{L}_x+L_x{L}_y)$ and $[L_z,L_y^2]=-i\hslash (L_x{L}_y+L_y{L}_x)$, so $[L_z,L^2]=0$. Simultaneous eigenstates have eigenvalues ${\hslash }^{2}l(l+1)$ and $\hslash m$; orbital angular momentum has integer $l\ge 0$ and $m=-l,\dots ,l$.

By cyclic symmetry, $L^2$ commutes with every component and hence every squared component. Therefore $[H,L^2]=0$: an energy eigenbasis can be chosen with definite $l$. For unequal transverse coefficients, $L_z$ need not commute with $H$, so definite $m$ is not generally available.

Expanding the ladder operators gives $L_{+}^2+L_{-}^2=2(L_x^2-L_y^2)$ and $L^2-L_z^2=L_x^2+L_y^2$. Comparing coefficients givesFor $l=1,m=0$, both squared ladder operators vanish, since two raising or lowering steps would leave the multiplet. Thus $E_0=2\beta {\hslash }^2={\hslash }^2(a_x+a_y)$. In the remaining basis $\{|1,1⟩,|1,-1⟩\}$ the Hamiltonian isIts symmetric and antisymmetric combinations have energies ${\hslash }^2(a_x+a_z)$ and ${\hslash }^2(a_y+a_z)$, respectively. ConsequentlyThe energies are related by permutation of the principal axes, not generally equal. Two coincide when the corresponding coefficients coincide; all three become $2a{\hslash }^2$ in the isotropic limit.

Use $[x,p]=i\hslash$. Acting on a smooth test wavefunction, $[f(x),p]\psi =f(-i\hslash {\psi }^{\prime })+i\hslash (f\psi {)}^{\prime }=i\hslash f^{\prime }\psi$. Take $f$ real and use the self-adjoint operator domain with vanishing integration-by-parts boundary terms. Writing $Q=p-if$ givesThe energy form is $⟨\psi ,H\psi ⟩=\Vert Q\psi {\Vert }^2/(2m)\ge 0$, so every normalizable energy eigenstate has $E\ge 0$. If a normalizable zero-energy state exists, its nonnegative norm must vanish, $Q{\psi }_0=0$. HenceZero energy is an additional existence assumption, not a consequence of factorization alone.

For ${\psi }_1=Q^{†}{\psi }_0$, the commutator $[Q,Q^{†}]=2\hslash f^{\prime }$ givesThus $\alpha (x)=-i{\hslash }^2{f}^{\prime \prime }(x)/m$ and $\beta (x)=\hslash f^{\prime }(x)/m$ in the requested relation. Generally these are multiplication functions, not constant coefficients, and ${\psi }_1$ is not automatically an energy eigenstate.

If $f(x)=c{x}^n+\cdots$, its integral has leading term $c{x}^{n+1}/(n+1)$. Decay at both infinities on the full real line requires odd $n$ and positive leading coefficient $c$. Odd degree alone is not sufficient. For $f=kx$ take $k=m\omega >0$. Then $H=H_{\mathrm{SHO}}-\hslash \omega /2$, $\alpha =0$ and $\beta =\hslash \omega$. After normalization and removal of an irrelevant phase,The corresponding factorized-Hamiltonian energies are zero and $\hslash \omega$; the constant shift must be restored for the stated oscillator Hamiltonian.