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 / S18 (Advanced quantum mechanics)

Assume a real, time-independent potential with the same zero asymptote on both sides, so $k=\sqrt{2mE}/\hslash$ is the same incident and transmitted wave number. For unit incoming amplitudes,Multiplying the two Schrödinger equations crosswise and subtracting shows that the Wronskian $W={\psi }_L{\psi }_R^{\prime }-{\psi }_L^{\prime }{\psi }_R$ is constant. Its two asymptotic values are $-2ik{t}_R$ and $-2ik{t}_L$, hence $t_L=t_R$.

For a real potential the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ is conserved, giving $|r_L{|}^2+|t_L{|}^2=|r_R{|}^2+|t_R{|}^2=1$. ThereforeThe reflection amplitudes can have different phases. Reality or self-adjointness is essential for the reflection-probability conclusion: a complex absorbing potential need not have equal left and right reflectances. Equal asymptotic speeds are also used in identifying transmission with the squared amplitude.

Away from $x=x^{\prime }$, the free equation is $G^{\prime \prime }+k^{2}G=0$. The outgoing boundary condition selects $G=A{e}^{ik|x-x^{\prime }|}$, continuous at its source. Integrating the differential equation across the source gives the derivative jumpThe jump for the proposed form is $2ikA$, so $A=im/(k{\hslash }^2)$. ThusFor $x$ far to either side, the $x^{\prime }$-dependent phase is absorbed into the outgoing amplitude. The outgoing prescription is equivalently the boundary value of the resolvent with $E\to E+i0$; changing that prescription would give a different Green function.

Write $L_0\psi =-U\psi$ and use the outgoing inverse of $L_0$. Adding the homogeneous incoming wave yieldsFor the attractive delta potential, let $\eta =m\alpha /(k{\hslash }^2)$. The equation reduces to $\psi (x)=e^{ikx}+i\eta e^{ik|x|}\psi (0)$ and $\psi (0)=1/(1-i\eta )$. Separating the two half-lines givesBoth sides have the same group velocity, soThe same amplitudes satisfy continuity and the exact delta-potential derivative jump ${\psi }^{\prime }(0^{+})-{\psi }^{\prime }(0^{-})=-2m\alpha \psi (0)/{\hslash }^2$, providing an independent local check on the sign.

Let $\lambda =m{U}_0{a}^2/{\hslash }^2\ll 1$ and $ka\ll 1$. Inside a region of size $a$, the Schrödinger equation estimates the curvature as $a^2{\psi }^{\prime \prime }=O(\lambda +(ka{)}^2)\psi$. Matching to the incoming and reflected waves constrains the slope as well. To leading order the interior wave has nearly constant value $C$, while its small accumulated derivative change encodes the potential's integrated strength.

This is not a Born assumption that $C$ is close to one. At very low $k$ the wave can be strongly reflected even by a weak potential, and $C$ can be small. The matching solution in the next part gives $C\sim k/(k+im\gamma /{\hslash }^2)$. It implies a fractional variation across the range of order $ka+m|\gamma |a/{\hslash }^2$, which is small under the two specified conditions. ThusThe approximation requires a finite short-range potential with a controlled width and integral; it does not describe arbitrary long-range tails.

Replace the slowly varying interior wave by $C$ and $e^{ik|x-x^{\prime }|}$ by one when both arguments lie within the range. The integral equation givesHenceThe integrated strength has dimensions of energy times length. For $U=-\alpha \delta (x)$, $\gamma =-\alpha$ and this is exactly the result in part c. For a regular weak potential the approximation sums the important ratio $m\gamma /(k{\hslash }^2)$ without requiring it to be small, which is why it remains useful beyond the naive first Born expansion at small momentum.

Write $\nu =m\gamma /{\hslash }^2$, so $t(k)\simeq k/(k+i\nu )$. Its nontrivial pole isFor negative $\gamma$ this is on the positive imaginary $k$ axis, the physical bound-state sheet. For positive $\gamma$ it lies on the negative imaginary axis and is a virtual-state pole reached by analytic continuation through the continuum cut. These different sheets can have the same negative numerical energy; specifying only $E_{\ast }$ would conceal the distinction. As a function of $E$, the momentum square root also introduces the continuum threshold branch point at zero. If $\gamma =0$, numerator and denominator cancel at this order and no pole is predicted by this leading approximation.

For a nonzero everywhere attractive weak potential, $\gamma <0$, the pole has $k=i\kappa$ with $\kappa =m|\gamma |/{\hslash }^2>0$. The exterior waves decay as $e^{-\kappa |x|}$, giving one shallow bound state to leading weak-coupling order:Its size ${\kappa }^{-1}$ is much larger than the potential range because $\kappa a=O(\lambda )\ll 1$. For an everywhere repulsive potential there is no negative-energy normalizable bound state: the kinetic and potential quadratic forms are both nonnegative. Its lower-half-plane pole is a virtual state, not a stable bound state. More complicated sign-changing potentials with a vanishing integral require higher-order analysis and are not covered by this simple leading result.

Let $x_{<}=\min (x,x^{\prime })$ and $x_{>}=\max (x,x^{\prime })$. A homogeneous solution vanishing at the left wall is $u(x)=\sin [k(x+a/2)]$; one vanishing at the right wall is $v(x)=\sin [k(a/2-x)]$. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-k\sin ka$. Continuity and the source derivative jump $G_{+}^{\prime }-G_{-}^{\prime }=-2m/{\hslash }^2$ therefore giveThis is exactly the printed form because $x_{<}=(x+x^{\prime }-|x-x^{\prime }|)/2$ and $x_{>}=(x+x^{\prime }+|x-x^{\prime }|)/2$, with a minus sign from reversing the argument of the second sine.

The true poles occur at $k_n=n\pi /a$, $n=1,2,\dots$, corresponding toThe apparent zero at $k=0$ is removable: the numerator and $k\sin ka$ both vanish quadratically. Negative values of $k_n$ duplicate the same energies, rather than describing extra levels. The boundary conditions and source jump uniquely determine the inverse whenever $E$ is not an eigenvalue.

Expand $G(x,x^{\prime })$ in a complete orthonormal eigenbasis in its first argument. Applying the operator yields ${\sum }_n(E_n-E)c_n{\phi }_n(x)=\delta (x-x^{\prime })$. Completeness expands the Dirac delta distribution as ${\sum }_n{\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$, soConsequently ${\lim }_{E\to E_n}(E_n-E)G={\phi }_n(x){\phi }_n^{\ast }(x^{\prime })$.

For the explicit formula, use $E_n-E\simeq -({\hslash }^2{k}_n/m)(k-k_n)$ and $\sin ka\simeq (-1{)}^{n}a(k-k_n)$. Also
$\sin [k_n(a/2-x)]=(-1{)}^{n+1}\sin [k_n(x+a/2)]$. The resulting limit isThe expression is symmetric in $x,x^{\prime }$ and holds on either side of the source. The usual complex-analysis residue with respect to $E-E_n$ has the opposite sign, since the source requests the limit with $E_n-E$.

Factorizing the projector found in part b gives the normalized eigenfunctionswith zero continuation outside. Their squared integrals are one and different positive integers give orthogonal functions. These are the familiar infinite-square-well states shifted so that the well centre is at zero. Each eigenfunction has an arbitrary overall constant phase, which cancels from the projector.

For $E<0$, inversion of the free operator gives $G_0(x,y)=m{e}^{-\kappa |x-y|}/(\kappa {\hslash }^2)$, with $\kappa =\sqrt{-2mE}/\hslash$. The bound-state equation is $\psi =-G_{0}U\psi$, henceFor the separable kernel $V(y,z)=-\alpha F(y)F^{\ast }(z)$, defineThenAn overall normalization still has to be chosen after finding a permitted energy. A nontrivial bound state must have $C\ne 0$: if $C=0$, the potential acts as zero and a square-integrable negative-energy free solution on the whole line cannot exist. Decay or square-integrability assumptions on $F$ are needed to make these integrals well defined; smoothness alone would not ensure convergence.

Multiply the expression for $\psi$ by $F^{\ast }(x)$ and integrate. Cancelling the nonzero $C$ givesThe kernel is real and symmetric, so relabelling the integration variables puts the expression in the printed order:This is the scalar consistency condition for the rank-one potential, rather than an independent normalization equation. The double integral is real and nonnegative because the Fourier transform of $e^{-\kappa |x|}$ is $2\kappa /(q^2+{\kappa }^2)>0$. Thus attractive $\alpha >0$ is needed for a nontrivial solution. Under the usual integrability assumptions the corresponding free resolvent expectation decreases with $\kappa$, giving at most one bound state for this rank-one attraction.

For a free particle the Dirac matrices are constant and the momenta commute. Squaring the Hamiltonian cancels all mixed terms by their anticommutators:Applying another time derivative to the Dirac equation therefore givesfor every component. This is the Klein-Gordon equation. Its componentwise satisfaction is a necessary consequence of the first-order Dirac equation, not a converse selecting all spinor solutions of the squared equation.

If $Av=\lambda v$ for $A={\alpha }_i$ or $\beta$, then $A^2=I$ implies ${\lambda }^2=1$. The polynomial $x^2-1$ has distinct roots, so such a matrix is diagonalizable andAnticommutation with another invertible Dirac matrix maps a $+1$ eigenspace to a $-1$ eigenspace, so both signs occur with equal multiplicities in a nontrivial representation. A Hermitian operator is appropriate for the physical Hamiltonian and additionally guarantees an orthonormal eigenbasis.

For any one of the four anticommuting matrices $A$, choose another $B$. Since $B^{-1}=B$ and $BAB=-A$, cyclic invariance of the trace givesThereforeEquivalently, the paired positive and negative unit eigenvalues cancel. This argument does not depend on a particular matrix representation.

The zero trace and unit eigenvalues imply equal numbers of $+1$ and $-1$ eigenvalues, so $N$ is even. A two-dimensional representation is impossible: all four matrices ${\alpha }_1,{\alpha }_2,{\alpha }_3,\beta$ would be traceless $2\times 2$ matrices. Anticommutation gives $\mathrm{Tr}(AB)=0$ for distinct matrices, whereas $\mathrm{Tr}(A^2)=2$. They would therefore be four linearly independent members of the three-dimensional complex space of traceless $2\times 2$ matrices, a contradiction.The usual irreducible $3+1$ dimensional Dirac representation has $N=4$. Larger representations can contain multiple copies; the requested lower bound is not itself a proof of irreducibility.

In two spatial dimensions there are only two kinetic matrices and one mass matrix. The three Pauli matrices square to identity and anticommute pairwise, sois a two-component Dirac Hamiltonian with $H_{2+1}^2=c^2(p_x^2+p_y^2)+m^2{c}^4$. It therefore reproduces the stated first-order equation and its Klein-Gordon square. Requiring a fourth anticommuting matrix in three spatial dimensions is precisely what prevented this two-component construction above.

Printed angular-momentum qualification. Define the dimensionless spin $s_i=-i{\epsilon }_{ijk}{\alpha }_j{\alpha }_k/4$ and the physical orbital angular momentum ${\ell }_i={\epsilon }_{ijk}{x}_j{p}_k$, with $[x_i,p_j]=i\hslash {\delta }_{ij}$. ThenThe source prints $L_i=i{\ell }_i$ instead. Taken literally, it is anti-Hermitian, has incompatible units with the dimensionless spin when they are added, and givesIt cannot yield the requested usual angular-momentum algebra or conservation law. For a concrete counterexample set $\hslash =1$ and take a wave with momentum along $z$. Sincethe printed $J_i=i{\ell }_i+s_i$ obeys $[H_D,J_x]=(1+i)c{\alpha }_y{p}_z\ne 0$.

The physically consistent operator is ${\mathbf{J}}_{\mathrm{phys}}=\mathbf{\ell }+\hslash \mathbf{s}$. It satisfiesThe cancellation follows directly from the two mixed commutators above. In a spherical basis the free spinor can be labelled by energy, $j=1/2,3/2,\dots$, and $m_j=-j,\dots ,j$, together with a parity or equivalent Dirac angular label distinguishing the two orbital couplings $\ell =j\pm 1/2$. Alternatively one can choose a momentum and helicity basis. One cannot simultaneously assign all components of momentum and all angular labels as independent commuting observables.

Use $\mathbf{p}=-i\hslash \nabla$ and regard $e$ as the signed charge in the printed minimal coupling. Multiplying the first Dirac equation by $c{\gamma }^0$ gives $H=c\mathbf{\alpha }\cdot \mathbf{\pi }+\beta m{c}^2+e\Phi$, with $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c$. Removing the rest-energy phase yieldsFor the alternative equation use its specified field conventions literally. They giveConsequently its Hamiltonian is $c\mathbf{\alpha }\cdot \mathbf{p}+\beta m{c}^2-\kappa \beta \mathbf{\Sigma }\cdot \mathbf{B}+(i\kappa /c)\beta \mathbf{\alpha }\cdot \mathbf{E}$. The two shifted equations areThese electric-field factors depend on the source's convention $E_i=c{F}_{0i}$; substituting a different Gaussian or SI definition without changing the equation would give incorrect powers of $c$.

For kinetic and field energies small compared with $m{c}^2$, the minimally coupled lower spinor is $\tilde{\chi }\simeq (\mathbf{\sigma }\cdot \mathbf{\pi })\tilde{\phi }/(2mc)$. Inserting it into the upper equation and using $[{\pi }_i,{\pi }_j]=i(e\hslash /c){\epsilon }_{ijk}{B}_k$ gives the Pauli equationThe spin moment is $e\hslash /(2mc)$ multiplying $\mathbf{\sigma }$, or gyromagnetic factor two relative to physical spin $\hslash \mathbf{\sigma }/2$.

For the alternative equation, elimination givesTo leading kinetic order and leading magnetic interaction this yieldsKeeping the source electric coupling during the same elimination writes the kinetic part more explicitly asThis product includes derivatives acting on the field and must not be treated as an ordinary scalar product of commuting variables. The parameter $\kappa$ is the intrinsic magnetic-moment coefficient: the moment operator is $\kappa \mathbf{\sigma }$. As printed, the alternative equation has no minimal charge coupling, so it can describe a neutral particle carrying a magnetic moment. If a Pauli term were added to the charged equation instead, its coefficient would represent an additional, anomalous contribution to the charge-generated Dirac moment.