Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 3

Use additive notation for this finitely generated abelian group. The change of generators $A=a+3d$, $e=c-b$, together with $b,d$, is invertible over the integers: $a=A-3d$ and $c=b+e$. The second relation gives $2e=0$. Subtracting it from the first gives $4A=0$, while the third becomes $4b+2e=0$, hence $4b=0$. The last relation becomes $8A+4b+4e=0$ and is consequently redundant. Conversely $4A=4b=2e=0$ imply every original relation, so no further relation has been introduced or lost. Thus the cyclic group decomposition isThe infinite cyclic group factor contributes only zero to an element killed by two. Each finite factor has two such elements, so the kernel of multiplication by two has $2^3=8$ elements. Removing the identity leavesThe invertible generator change also supplies a direct presentation proof of the decomposition, without relying solely on a Smith normal form computation.

Set $D=1+u^2+v^2$. The two tangent vectors to the Enneper surface areTheir cross product is $D(-2u,2v,1-u^2-v^2)$. Since $4u^2+4v^2+(1-u^2-v^2{)}^2=D^2$, its length is $D^2>0$, proving regularity and giving the unit normalFor the first fundamental form, direct dot products give $E=G=D^2$ and $F=0$. The second derivatives are ${\sigma }_{uu}=(-2u,2v,2)$, ${\sigma }_{uv}=(2v,2u,0)$, and ${\sigma }_{vv}=(2u,-2v,-2)$. Taking their dot products with $\mathbf{n}$ gives the second fundamental form coefficients $e=2$, $f=0$, $g=-2$. Therefore the Gaussian curvature isHere curvature means Gaussian curvature; the opposite signs of the second-form diagonal entries also give zero mean curvature, as expected for this minimal surface.

Differentiability means existence of a linear map $A:{\mathbb{R}}^2\to \mathbb{R}$ such thatThis is the Fréchet derivative. Substituting increments on the coordinate axes shows $A(h,k)=f_x(x_0,y_0)h+f_y(x_0,y_0)k$. For the unit direction $v=(\cos \alpha ,\sin \alpha )$, the remainder is $r(tv)=o(|t|)$. Dividing the increment by $t$ proves differentiability of the restricted function and gives its directional derivative:For the supplied rational function, both coordinate-axis restrictions are zero, so $f_x(0,0)=f_y(0,0)=0$. Any Fréchet derivative would therefore be the zero map. But $f(t,t)=t$, and hence $|f(t,t)|/\sqrt{t^2+t^2}=1/\sqrt{2}$ for nonzero $t$. This contradicts the required vanishing remainder. ThusIn fact every directional derivative exists: along $v$ it equals ${\cos }^2\alpha \sin \alpha +\cos \alpha {\sin }^2\alpha$. These values do not depend linearly on the direction, illustrating why existence of directional derivatives is weaker than differentiability.

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently its only subsets that are both open and closed are the empty set and the whole space. For a subspace topology, open means relatively open within that subspace. A path-connected space is connected, since a separation would pull back along a path to a separation of an interval.

Use the continuous slope map $(x,y)\mapsto x/y$ on the given set, where $y>0$. The attained ratios are positive rational numbers. Choosing the irrational cut $\sqrt{2}$, the two subsetsare disjoint, relatively open, and cover the entire set. Both are nonempty: ratios one and two are attained inside the square. This irrational-cut separation of rational slopes is a separation, soThe individual rational-slope rays are connected, but their common origin is excluded; their separate connectivity does not imply connectivity of their union.

The set is a rational comb with a connecting spine: it contains the whole vertical segment $x=0$, and every horizontal segment at rational height. Any point with $x>0$ has rational height and can move horizontally to the vertical segment without leaving the set. It can then move vertically to $(0,0)$. Points already on the vertical segment use only the second step. Concatenating these polygonal paths connects any two points. ConsequentlyNo claim about a horizontal segment at irrational height is needed: at such a height, the only point in the set lies on the connecting spine.

For $t>0$, use the Bromwich inversion formula with a vertical contour $\mathrm{Re}s=c>0$:Close the contour to the left, where the exponential decays. The large-arc contribution tends to zero by the usual exponential arc estimate, and the two simple poles are at $s=ia$ and $s=-ia$. The residues of the complete integrand are $e^{iat}/2$ and $e^{-iat}/2$. The residue theorem therefore givesThe right limit at zero is one. The sign of nonzero real $a$ has no effect on the answer.

Take the branch of the complex logarithm with $-\pi <\arg s<\pi$ and write $s^{-1/2}=\exp [-\log s/2]$. Deform the Bromwich contour into a Hankel contour about the negative real axis. The small circle at zero contributes $O(r^{1/2})$ and vanishes; the large closing arcs vanish by the exponential estimates for $t>0$. On the upper lip $s=-r+i0$, $s^{-1/2}=-i/\sqrt{r}$, while on the lower lip it is $i/\sqrt{r}$. Keeping the lip orientations in the Bromwich inversion with a square-root branch cut givesWith $r=u^2$, the supplied Gaussian integral evaluates this as $2{\pi }^{-1}{\int }_0^{\infty }{e}^{-t{u}^2}du$. HenceThe inverse is singular but locally integrable at zero, which is compatible with existence of its Laplace transform.

Assume real $k\ne 0$, and put $a=|k|$. Away from the source point the Green function solves $G^{\prime \prime }-a^{2}G=0$. The two decay conditions select $G=A{e}^{a(x-\xi )}$ on $x<\xi$ and $G=B{e}^{-a(x-\xi )}$ on $x>\xi$. Continuity at $\xi$ requires $A=B$: a jump in $G$ would create a derivative of the Dirac delta that is absent from the equation. Integrating the equation across the source gives $G^{\prime }(\xi +)-G^{\prime }(\xi -)=1$, so $-2aA=1$. Thus the negative of the one-dimensional modified Helmholtz Green function isIts distributional derivative verifies the delta equation, including its sign.

For a source for which the integral converges and the resulting function has the required decay, defineApplying the differential operator under the integral, or in the sense of distributions, gives $\mathcal{L}y=\int \delta (x-\xi )S(\xi )d\xi =S(x)$. A continuous integrable source, for example, makes this convolution vanish at both infinities and gives a classical solution. If there were two solutions, their difference would be $A{e}^{ax}+B{e}^{-ax}$; decay at positive infinity forces $A=0$, and at negative infinity forces $B=0$. Thus the decaying solution is unique whenever it exists.

The assumptions needed for existence should not be dropped: an arbitrary source need not admit a decaying solution, as $S=1$ demonstrates. Also the displayed formula presupposes $k\ne 0$; for $k=0$ no Green function for a point source can decay at both infinities, since its derivative must jump by one while both exterior linear pieces would have to be zero.

In the position representation, the orbital angular momentum operator isOn a common smooth domain use $[x_a,p_b]=i\hslash {\delta }_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator givesContracting with the two alternating tensors gives the orbital angular momentum commutation relations:For $L_{\pm }=L_1\pm i{L}_2$, these imply $[L_3,L_{\pm }]=\pm \hslash L_{\pm }$. Therefore $L_3\psi =0$ yieldsThusThe nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.

A 95 percent confidence set is a measurable data-dependent set $C(X_1,\dots ,X_n)$ whose repeated-sampling coverage satisfies ${\mathbb{P}}_{\theta }\{\theta \in C(X_1,\dots ,X_n)\}\ge 0.95$ for every parameter value, with equality for an exact 95 percent procedure. The parameter is fixed; the set is random. This is not a posterior probability statement about the parameter after the data have been observed.

For independent normal random variables of known variance, $\overline{X}\sim N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The supplied quantile gives the exact confidence intervalIts deterministic length is $3.920\sigma /\sqrt{n}$. Requiring this to be at most $\epsilon$ givesAmong fixed-length intervals based equivariantly on the sample mean, the centred interval has greatest coverage: an interval of a given length captures most mass of a symmetric unimodal normal distribution when centred on its mean. Thus the width calculation is sharp for the usual fixed-length normal-mean construction.

A state of a Markov chain is a recurrent state if, starting there, the first strictly positive return time is finite with probability one. A recurrent Markov chain has every state recurrent; in an irreducible Markov chain recurrence is a class property. Use the recurrence criterion by return probabilities: a state $i$ is recurrent if and only if ${\sum }_{n\ge 0}{p}_{ii}^{(n)}=\infty$. To see the criterion, this sum is the expected number of visits including the initial one. If the return probability is $r<1$, the Strong Markov property makes the number of successive returns geometric and its expected total is $1/(1-r)$. If $r=1$, every fixed number of returns occurs with probability one, so the expectation is infinite.

For the symmetric simple random walk on the integers, odd-time return probabilities vanish, whileBy Stirling's formula, $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\sim 4^n/\sqrt{\pi n}$. Hence $p_{00}^{(2n)}\sim 1/\sqrt{\pi n}$, whose sum diverges. The return-probability criterion proves recurrence of zero. Translation invariance gives the same result at every integer, or one can use irreducibility. ThusThis proof establishes almost-sure return, not a finite expected return time; every state of the walk is in fact a null recurrent state.

A quadratic form over $\mathbb{R}$ or $\mathbb{C}$ is a function $q(x)=B(x,x)$ for a symmetric bilinear form $B$, equivalently a homogeneous polynomial of degree two. Its polarization is $B(x,y)=[q(x+y)-q(x)-q(y)]/2$. If $q$ is not identically zero, choose $v$ with $q(v)\ne 0$. Every $x$ decomposes aswhere the second term lies in $v^{\perp }$ and $kv\cap v^{\perp }=\{0\}$. The quadratic form is consequently a direct sum of its one-dimensional value on $v$ and its restriction to $v^{\perp }$. Induction on dimension diagonalizes it. If $q=0$, any basis suffices. Rescale every nonzero diagonal coefficient: over the reals it becomes $1$ or $-1$, while over the complexes a square root makes it $1$. HenceOver $\mathbb{C}$ the negative coefficients can all be avoided.

For a real quadratic form, let $p$ and $m$ be its positive and negative indices in a diagonalization. Its rank of a quadratic form is $p+m$ and its signature of a quadratic form is $p-m$, or the ordered pair $(p,m)$ in the alternative convention. These indices are intrinsic: the largest dimension of a subspace where the form is positive definite is $p$, since any larger subspace intersects the nonpositive coordinate subspace nontrivially; similarly for $m$. For the form here, completing squares givesThe two displayed linear coordinates, together with $x_1$, form an invertible coordinate change. ThereforeThere is one zero coordinate direction.

The PDF's final dimension condition is $n\ge 2^d$, not the fraction produced by the TeX conversion. We prove the requested common zero of complex quadratic forms by induction on $d$. A complex quadratic form of rank $r$ has canonical expression $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$ and add the $n-r$ radical coordinate vectors. Their span is a totally isotropic subspace, because the form is zero on every linear combination. This proves that complex quadratic forms have a large isotropic subspace: its dimension isFor $d=0$ there is a nonzero vector whenever $n\ge 1$. For $d\ge 1$, take such a subspace $E$ for $q_1$. If $n\ge 2^d$, then $\dim E\ge 2^{d-1}$. Apply the induction hypothesis to the restrictions of $q_2,\dots ,q_d$ to $E$. It supplies a nonzero $x\in E$ annihilated by those forms, and $q_1(x)=0$ by construction. ThusThe argument makes no nondegeneracy assumption on any of the forms.

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that, for any $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=bq+r$ and either $r=0$ or $\nu (r)<\nu (b)$. To prove that every ideal is principal, take a nonzero ideal $J$ and choose a nonzero $b\in J$ of minimal Euclidean value. Divide any $a\in J$ by $b$. The remainder belongs to $J$, so minimality forces it to be zero. Thus $J=(b)$; the zero ideal is also principal. Hence every Euclidean domain is a principal ideal domain.

In $R=\mathbb{Z}[\sqrt{-7}]$, consider $J=(2,1+\sqrt{-7})$. It is proper: the map $R\to {\mathbb{F}}_2$ sending $\sqrt{-7}$ to one is a well-defined surjective ring map, and annihilates both generators. Suppose $J=(\eta )$. The positive integer field norm $N(a+b\sqrt{-7})=a^2+7b^2$ is multiplicative, and $\eta$ divides two. Therefore $N(\eta )$ divides four. A proper ideal cannot have a unit generator, so $N(\eta )\ne 1$. There is no element of norm two. The elements of norm four are only $\pm 2$, but neither divides $1+\sqrt{-7}$ in this ring. This is a contradiction. ThusThe argument does not merely show failure of one proposed norm.

Now put $\omega =(1+\sqrt{-7})/2$ and $R^{\prime }=\mathbb{Z}[\omega ]$. Its algebraic field norm issince $\omega \overline{\omega }=2$ and $\omega +\overline{\omega }=1$. To construct Euclidean division, let $z\in \mathbb{C}$. Choose an integer $b$ so that $|\mathrm{Im}z-b\sqrt{7}/2|\le \sqrt{7}/4$, then choose an integer $a$ so that $|\mathrm{Re}z-b/2-a|\le 1/2$. The lattice point $q=a+b\omega$ satisfiesFor $A,B\in R^{\prime }$ with $B\ne 0$, apply this construction to $z=A/B$, and set $r=A-Bq\in R^{\prime }$. Multiplicativity gives $N(r)=N(B)|z-q{|}^2<N(B)$ unless $r=0$. ThereforeThis is the geometric division argument for the norm-Euclidean integers of discriminant minus seven.

For $|\lambda |<1$, let $F_{\lambda }(\zeta )=(\zeta -\lambda )/(1-\overline{\lambda }\zeta )$. Direct calculation givesIts denominator does not vanish in the disc, and this identity shows that it maps the disc into itself. Its inverse $(w+\lambda )/(1+\overline{\lambda }w)$ has the same property, so $F_{\lambda }$ is a Poincare disc automorphism. Multiplication by a complex number of modulus one is another such automorphism.

Conversely let $T$ be a Möbius transformation mapping the disc onto itself, and let $\lambda =T^{-1}(0)$. Then $T\circ F_{\lambda }^{-1}$ is a disc automorphism fixing zero, say $S(\zeta )=a\zeta /(c\zeta +d)$ with $d\ne 0$. Since it maps the boundary circle to itself, $|a{|}^2=|c\zeta +d{|}^2=|c{|}^2+|d{|}^2+2\mathrm{Re}(c\overline{d}\zeta )$ for every $|\zeta |=1$. This forces $c\overline{d}=0$, hence $c=0$ and $|a/d|=1$. Thus $S$ is a rotation $\eta \zeta$, and $T=\eta F_{\lambda }$. Absorb the denominator's minus sign into $\omega =-\eta$ to obtain precisely the paper's convention:Here mapping the disc to itself means bijectively onto itself; there are nonautomorphic Möbius transformations whose image is a proper subdisc.

For the Poincare disc model, radial distance from zero to Euclidean radius $r$ isIt is minimal: any curve has length at least the integral of $2|dr|/(1-r^2)$, which in turn is at least the net radial integral. Consequently a hyperbolic circle of radius $a$ centred at zero is $|\zeta |=\tanh (a/2)$, a Euclidean circle.

Given a centre $p$, use a Poincare disc automorphism carrying $p$ to zero. Since it is an isometry, it takes the given hyperbolic circle to the preceding Euclidean circle. A Möbius transformation takes generalized circles to generalized circles. The preimage here is compact and contained in the disc, so it cannot be a straight line. ThusThis argument does not say that their Euclidean and hyperbolic centres coincide.

Rotate the disc so that $P=e^{i\alpha }$ and $Q=e^{-i\alpha }$, with $0<\alpha \le \pi /2$ representing half the smaller angle. For $\alpha <\pi /2$, the hyperbolic line joining them is the arc of a circle orthogonal to the unit circle. Its Euclidean centre is real, say $c$, and orthogonality gives squared radius $c^2-1$. Substituting $P$ into $|P-c{|}^2=c^2-1$ gives $c=\sec \alpha$, with radius $\tan \alpha$.

The arc's point nearest zero is on the positive real axis, atDistance from zero in the Poincare disc model increases with Euclidean modulus, by the radial-length calculation. ThereforeFor $\alpha =\pi /2$, the hyperbolic line is a diameter through zero and its distance is zero, agreeing with the continuous limit of the formula. Choosing the smaller endpoint angle avoids a spurious negative value caused by an inappropriate oriented-angle convention.

Suppose an open hyperbolic disc centred at $X$ is contained in a hyperbolic triangle. Move $X$ to zero by an isometry. Extend the radial geodesics from zero through the three vertices to ideal points $P_1,P_2,P_3$. Because zero lies inside the original triangle, the three vertex directions are not contained in a semicircle. The ideal hyperbolic triangle with those endpoints therefore contains zero: otherwise a separating geodesic through zero would put all three endpoints in one semicircle. It also contains each finite vertex, which lies on the segment from zero to the corresponding ideal endpoint. Geodesic convexity then makes it contain the original triangle.

The three angular gaps between successive ideal endpoints sum to $2\pi$, and each is at most $\pi$. One has size at least $2\pi /3$, so its half-angle $\alpha$ is at least $\pi /3$. The distance in part (iii) decreases as $\alpha$ increases. Thus the corresponding side of the ideal triangle is at distance at mostfrom zero. If the disc radius exceeds this distance, it contains a point on that side and points beyond it, and cannot be contained in the ideal triangle, hence cannot be contained in the original one. This proves the hyperbolic triangle inradius bound:The symmetric ideal triangle has equality, explaining the constant. Finite triangles have strictly smaller inradius.

Let $p=(x_0,y_0)$, $a=f_x(p)$ and $b=f_y(p)$. For sufficiently small $h,k$, decompose the increment into an $x$-segment at height $y_0+k$ and a $y$-segment at $x_0$. The one-variable mean value theorem applies on each segment because existence of its partial derivative gives continuity along the segment. Thus, when both increments are nonzero,for some $0<\theta ,\eta <1$. The same formula with the absent segment omitted covers zero increments. Continuity of both partial derivatives at $p$ makes the remainder after subtracting $ah+bk$ at most $\epsilon (|h|+|k|)$ for small increments, and hence at most $\sqrt{2}\epsilon \sqrt{h^2+k^2}$. Therefore the Fréchet derivative is $(h,k)\mapsto ah+bk$. Continuity throughout the disc is not required; continuity at the point suffices.

For the determinant, expand by multilinearity of the determinant in the columns of $I+H$. Taking no perturbed column gives one; taking exactly one gives $\mathrm{tr}H$; all other terms have degree at least two in the entries of $H$. Since matrix-entry functionals are bounded in any finite-dimensional norm, their sum is $O(\Vert H{\Vert }^2)$. HenceFor invertible $A$, multiplicativity of the determinant gives $\det (A+H)=\det A\det (I+A^{-1}H)$. The bounded linear map $H\mapsto A^{-1}H$ transfers the preceding remainder estimate, giving the derivative of the determinantTo justify an open invertible neighborhood, let $\Vert K\Vert <1$. Completeness and submultiplicativity make the Neumann series $S={\sum }_{j=0}^{\infty }(-K{)}^j$ converge in matrix norm. Multiplying finite partial sums by $I+K$ gives $I-(-K{)}^{N+1}$, which tends to $I$. Continuity of multiplication proves $(I+K)S=S(I+K)=I$. Thus $I+K$ is invertible, and the series gives $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$ near zero.

Apply the first-derivative formula at $I+H$ to a test matrix $K$:The last estimate is uniform for $\Vert K\Vert \le 1$, so it proves differentiability of the derivative in the operator norm, rather than merely a directional second derivative. The second derivative of the determinant is the symmetric bilinear mapIts symmetry follows from $\mathrm{tr}(HK)=\mathrm{tr}(KH)$.

Rouché's theorem states that if $f,g$ are holomorphic on a neighborhood of a closed disc with boundary $C$ and $|g|<|f|$ on $C$, then $f$ and $f+g$ have the same number of zeros in the disc, counted with multiplicity. More generally the conclusion holds for a simple closed contour whose interior lies in their domain. To prove it, set $f_t=f+tg$ for $0\le t\le 1$. The strict boundary inequality ensures $f_t\ne 0$ on $C$. By the argument principle,is the integer number of zeros inside. The integrand is continuous in $t$ uniformly on $C$, since its denominator is bounded away from zero on the compact product $C\times [0,1]$. Thus $N(t)$ is continuous and integer-valued, so constant. This proves the theorem.

On $|z|=1$, the term $8z^6$ dominates the remaining terms because $3+1+2+1=7<8$. Thus the polynomial has six zeros inside the unit circle. On $|z|=2$, $3z^9$ dominates because its modulus is $1536$, whereas the sum of the other bounds is $512+32+16+1=561$. There are therefore nine zeros inside radius two. The strict inequalities exclude boundary zeros, so

For the fixed-degree polynomial limit, choose $d+1$ distinct complex nodes ${\zeta }_0,\dots ,{\zeta }_d$. Write $p_n(z)={\sum }_{j=0}^d{a}_{j,n}{z}^j$, inserting zero coefficients if necessary. The vector of values at the nodes equals the fixed Vandermonde matrix times the coefficient vector. Its determinant is ${\prod }_{i<j}({\zeta }_j-{\zeta }_i)\ne 0$. Convergence of the values therefore implies coefficient convergence, say $a_{j,n}\to a_j$. For $p(z)={\sum }_{j=0}^d{a}_j{z}^j$ and a compact set with $|z|\le R$,This proves that the given limit is a polynomial of degree at most $d$ without invoking any theorem about uniform limits of holomorphic functions.

When the limit has $d$ distinct roots, choose pairwise disjoint small closed discs around them, with no limit root on their boundaries. On each boundary $|p|$ has a positive minimum. Uniform convergence makes $|p_n-p|<|p|$ there for all sufficiently large $n$, so Rouché's theorem puts exactly one root $z_{i,n}$ of $p_n$ in each disc. For every smaller disc about $z_i$ the same argument applies eventually. The unique root in the original disc must then lie in the smaller one, provingThe literal demand for a root at every initial index needs this finite-index qualification: $p_1=1$, $p_n=z$ for $n\ge 2$ satisfies the hypotheses with limit $z$, but $p_1$ has no root. If every initial polynomial has a root, arbitrary choices at the finitely many exceptional indices extend the convergent sequences; otherwise the sequences necessarily start at an eventual index.

For the real Sturm-Liouville problem, multiply the equations for $y_n$ and $y_m$ by $y_m$ and $y_n$, respectively, and subtract. Integration by parts givesThe boundary term vanishes because both functions are zero at both endpoints. Distinct eigenvalues therefore imply weighted orthogonality. For equal indices the integral is the squared weighted norm, so

Let $c_n={\int }_0^{1}w{y}_n^{2}dx>0$. For $y=\sum a_n{y}_n$ in the operator domain, with convergence sufficient to evaluate its quadratic energy, weighted orthogonality givesFor finite sums these identities follow immediately; the corresponding convergent energy expansion follows by approximation in the operator or quadratic-form domain. For nonzero $y$, the Rayleigh quotient is a weighted average of the eigenvalues, soEquality occurs precisely when only a lowest-eigenvalue component is present.

Assume completeness as permitted and take a sufficiently regular solution of the diffusion equation. It belongs to the weighted expansion space at each positive time; its modal coefficients satisfy ${\dot{a}}_n=-{\lambda }_n{a}_n$. Alternatively differentiate its energy directly and use the preceding inequality:Thus the squared weighted norm is at most its initial value times $e^{-2{\lambda }_{1}t}$. This describes decay when ${\lambda }_1>0$; positivity of $p,w$ alone does not require ${\lambda }_1>0$ if the potential $q$ can be negative. The displayed inequality remains valid regardless of that sign.

The one-dimensional probability current isAt a finite jump of the potential, both $\psi$ and ${\psi }^{\prime }$ are continuous. A jump in $\psi$ would create a delta derivative in the stationary Schrödinger equation, and integrating that equation across an increasingly small interval then shows continuity of ${\psi }^{\prime }$. Accordingly impose continuity of both quantities at zero and at $a$.

For an incident wave of positive energy, define $k=\sqrt{2mE}/\hslash$. With no wave incident from the right, the left region hasThe interference terms cancel from the imaginary part in the current formula. If $E>V_0$, let $k_0=\sqrt{2m(E-V_0)}/\hslash$. In the right region,For $0<E<V_0$, let ${\kappa }_0=\sqrt{2m(V_0-E)}/\hslash$. Boundedness at positive infinity excludes the growing exponential, soThe decaying solution has zero current even if its amplitude is nonzero. These exterior forms do not require $E$ to lie on a particular side of $V_1$: in the finite middle layer the solution is oscillatory for $E>V_1$, exponential for $E<V_1$, and linear at $E=V_1$; matching determines the amplitudes.

Define the reflection coefficient as reflected current magnitude divided by incident current, and the transmission coefficient as outward right current divided by incident current. ThenThe denominator is $j_{\mathrm{incident}}=\hslash k|A{|}^2/m$. For a real potential, differentiation givesby the stationary Schrödinger equation. Continuity at each interface preserves this conserved current, so $j_{\mathrm{left}}=j_{\mathrm{right}}$. Dividing by incident current yieldsThere can be penetration into the barrier without transmission to infinity, because the entire right half-line remains classically forbidden.

Take the curl of the vacuum Faraday law and use the vacuum Ampere-Maxwell law:Since the electric divergence vanishes, the double-curl identity makes the left side $-\Delta \mathbf{E}$. Therefore the electromagnetic wave equation isThe same calculation interchanging the fields gives the wave equation for $\mathbf{B}$.

Substitute the monochromatic plane waves into the two divergence equations of Maxwell's equations. They become $ik{E}_{0z}=0$ and $ik{B}_{0z}=0$. For nonzero wavenumber,Thus both fields are transverse to the direction of propagation. The wave equation also gives the vacuum dispersion relation ${\omega }^2=c^2{k}^2$.

For these plane waves, the vacuum Faraday law readsCancel the nonzero factor $i\omega$ to obtainFor positive-frequency propagation in the positive $z$ direction, $k/\omega =1/c$.

Vacuum Maxwell's equations are invariant under spatial rotations. Rotate the preceding coordinate system so that ${\mathbf{e}}_z$ becomes the propagation direction $\hat{\mathbf{k}}$, and rotate the transverse electric direction to a unit polarization vector $\hat{\mathbf{n}}$. Dot and cross products transform covariantly under this rotation, so transversality and the field relation are unchanged. With $\omega =c|\mathbf{k}|$,These complex fields encode real fields through their real parts. The displayed form describes linear electromagnetic polarization; a general transverse complex amplitude can describe elliptical or circular electromagnetic polarization as well.