Part IV · Quantum Mechanics — Chapter 4.7

Wells, Barriers, and Tunnelling

Four potentials, solved exactly, with numbers at the end of each. The device that solves all four is a condition joining a wavefunction across a step, and the whole point of putting this chapter after Chapter 4.4 is that the condition is not a convenience. It is the domain of the Hamiltonian, and choosing it is choosing which operator the wall implements.

Tools you'll need  — Chapter 4.6 above all: §6 for the equation in the position representation, §8 for the probability current and the continuity equation, §8.3 for the current read off the phase, and §9 for stationary states and the time-independent equation. Chapter 4.4 §3 for what a domain is, §5 for one operator worked on three intervals, §6 for the count of self-adjoint extensions, and §7 for the four-parameter family a particle in a box has. Chapter 4.5 §2 for the spectrum of an operator with no eigenvectors, §7 for what $\ket x$ and $\ket p$ mean, §8.2 for when an expansion in eigenstates is legitimate, and its Worked example 1 for the free Hamiltonian. Chapter 4.2 §4.3 for compatible observables and quantum numbers, and §10.2 for the ammonia molecule. Chapter 0.9 §5 for the delta and its rules, and §6 for the bandwidth theorem. Chapter 0.8 §3 for a linear differential equation read as an eigenvalue problem. Chapter 0.5 §8 for commuting operators and simultaneous diagonalisation. Chapter 0.2 §3.2 for integration by parts with the boundary term kept.

1 · Bound states, and the boundary condition as a domain choice

Here is where this section is going. Every problem in this chapter is solved by writing down the general solution of the equation on each interval where the potential is constant and then joining the pieces at the places where it changes. The joining is the whole of the work, so the thing to know is what authorises it. The answer is that the joining conditions are the domain of $\hat H$ written out in position space, and that a function violating them is not a state the Hamiltonian is allowed to act on. By the end of this section you will have that derivation, you will know the one place it stops working, and you will have watched the infinite well being selected out of the four-parameter family Chapter 4.4 counted rather than assumed into existence.

1.1 · What is left to do after Chapter 4.6

Chapter 4.6 §9.1 reduced the equation of motion to an eigenvalue problem. In one dimension it reads

and its §9.3 gave the general solution of the time-dependent problem as a superposition of these with a phase $\ee^{-\ii E_nt/\hbar}$ attached to each. So the only thing standing between you and the complete solution of a quantum problem is the list of $E$ for which (4.7.1) has an acceptable solution, together with the solutions themselves. Chapter 0.5's closing list said the spectral theorem would come back in this part of the book as the statement that solving a system means diagonalising its Hamiltonian, and naming this chapter and the next as the places it happens. This is that.

Two words in the paragraph above are doing hidden work and both need unpacking before anything can be solved. The first is acceptable. Not every solution of a second-order differential equation is a state, and the ones that are not are what make the spectrum discrete. A second-order equation has a two-dimensional solution space at every energy. If every solution counted there would be a state at every $E$, and no quantisation at all. Section 3.2 is where acceptable is finally cashed, on the first problem the chapter solves. The second is $\hat H$ itself. Chapter 4.4 §3.1 established that an operator is a formula together with a domain, and (4.7.1) is a formula. Which functions $u$ are permitted is a separate question with a separate answer, and §1.3 below is where that answer comes from.

1.2 · Piecewise-constant potentials, and why they come first

Every potential in this chapter is constant on each of two or three intervals and jumps between them. That choice is not laziness, and here is why it is the right first case rather than a simplified one.

On an interval where $V$ takes the constant value $V_{0}$, equation (4.7.1) rearranges into a linear equation with constant coefficients,

whose solution space Chapter 0.8 §3.1 showed to have dimension two and whose solutions its §3.3 wrote down once and for all. The sign of the coefficient decides which of two shapes you get, and the two shapes are the entire vocabulary of this chapter. If $E\gt V_{0}$ the coefficient is negative, the solutions are $\ee^{\pm\ii kx}$ with $k=\sqrt{2m(E-V_{0})}/\hbar$, and the wavefunction oscillates with wavelength $2\pi/k$. If $E\lt V_{0}$ the coefficient is positive, the solutions are $\ee^{\pm\kappa x}$ with $\kappa=\sqrt{2m(V_{0}-E)}/\hbar$, and the wavefunction grows or decays over the length $1/\kappa$. A classical particle is forbidden from the second kind of region entirely, because its kinetic energy there would be negative. A quantum one is not forbidden, and everything interesting in this chapter comes from that single difference.

Piecewise-constant potentials are therefore the problems in which the solutions are known in advance on every piece and the only unknowns are the coefficients. That reduces a differential equation to simultaneous linear equations in a handful of constants, which is what "exactly solvable" means here. Chapter 4.10 handles a potential that varies smoothly, by an approximation that treats it as locally constant, and the exponential this chapter computes exactly is the thing that approximation has to reproduce.

1.3 · Where the matching conditions come from

Now for the joining, which is the step that decides whether this chapter is a set of recipes or a piece of physics. Suppose $V$ has a finite jump at a point $x_{0}$. The standard instruction is that $u$ and $u'$ are both continuous there. We want a reason rather than an instruction, so ask what could go wrong if either failed.

The requirement to hold on to is the one Chapter 4.4 §3.1 made central. For $u$ to be in the domain of $\hat H$, the function $\hat Hu$ has to land back in $L^{2}$, and near a finite jump in $V$ that is a condition on $u''$ alone, since $Vu$ is a bounded function there and cannot rescue anything. So suppose $u''$ is some $f$ lying in $L^{2}$ near $x_{0}$, and ask what that forces on $u'$. Integrating the supposition from a point on the left,

and the integral on the right tends to zero as $x$ approaches $x_{0}$ from either side. That is Cauchy–Schwarz doing the work, in the form Chapter 4.4 §5.2 stated: the integral of an $L^{2}$ function over a shrinking interval is bounded by $\norm f$ times the square root of the interval's length. So $u'$ is continuous at $x_{0}$, and it had no choice in the matter. A function whose slope jumps is thrown out, not because it looks wrong but because $\hat H$ carries it out of the space.

There is a second way of saying the same thing which is worth having, because the delta well of §4.6 below needs it and because it makes the mechanism visible. A jump of size $\sigma$ in $u'$ is $\sigma$ times a step function plus something smooth, and Chapter 0.9 §5.4 established that the derivative of a step is a delta. So differentiating once more gives $u''=\{\text{the ordinary second derivative}\}+\sigma\, \delta(x-x_{0})$, and $\hat Hu$ then contains $-\hbar^{2}\sigma\,\delta(x-x_{0})/2m$. Chapter 4.5 §7.1 established that a delta has no norm at all, so it is not a member of the space, which is the previous paragraph's conclusion arriving in the language this chapter uses later.

The condition on $u$ itself comes free with that. Once $u''$ is a function of the space, the display above makes $u'$ continuous, and a function with a continuous derivative is continuous. In the delta language the same point is made more emphatically: a jump in $u$ puts a delta into $u'$ and the derivative of a delta into $u''$, which is worse again. Both halves of the joining rule therefore come from one requirement, and the conclusion is best stated in the form used in every problem below.

Read (4.7.4) as a description of $\operatorname{dom}(\hat H)$ rather than as a technique. It is the same statement as Chapter 4.4's, in a setting where it happens to have a familiar name. Matching $u$ and $u'$ is what makes the Hamiltonian self-adjoint, and the smoothness of the wavefunction is a consequence of that rather than an aesthetic preference about how functions ought to look.

1.4 · The infinite wall, where that argument stops

Every step in §1.3 used the finiteness of the jump, and the place it was used has to be isolated, because the most-taught problem in quantum mechanics is the one where it fails. The delta produced by a jump in $u'$ survived because $Vu$ was an ordinary function that could not cancel it. If $V$ is infinite on one side, $Vu$ is not an ordinary function, and the cancellation is no longer impossible. Nothing in §1.3 then decides what happens to the slope.

The honest description of an infinitely high wall is not a potential at all. It is a restriction of the problem to the region where the particle is allowed to be, with the operator $-\hbar^{2}\dvn{2}{}{x}/2m$ acting on functions defined only there. That is precisely the situation Chapter 4.4 §7 analysed, and its answer was uncomfortable and correct. The formula $-\hbar^{2}\dvn{2}{}{x}/2m$ on an interval does not determine an operator. It determines a family of self-adjoint operators in one-to-one correspondence with $U(2)$, which is four real parameters, and its §7.4 showed by solving one slice of that family that different members have genuinely different energy levels. Chapter 4.4 closed by saying that this chapter would take its $U(2)$ and spend a paragraph rather than a clause on which member the infinite well is. Here is the paragraph.

1.5 · Which member of the family an impenetrable wall is

The condition everyone writes down for a box is $u(0)=u(L)=0$, which Chapter 4.4 §7.2 listed as the Dirichlet point of the family. What selects it is not that it is the simplest, since Neumann is equally simple and gives different numbers, and it is not that $\psi$ ought to vanish where the particle cannot go, since that is the conclusion rather than an argument. What selects it is a limit, and the limit can be taken.

Put the particle in a well of finite depth: $V=0$ inside an interval and $V=V_{0}$ outside, with $V_{0}$ large but finite. Section 1.3 applies throughout, because every jump is finite, so $u$ and $u'$ are continuous at both walls with no choice involved. Outside the well the energy is below the potential, so §1.2's second case applies and the wavefunction decays as $\ee^{-\kappa x}$ with $\kappa=\sqrt{2m(V_{0}-E)}/\hbar$. Now let $V_{0}$ grow with $E$ held fixed. The decay length $1/\kappa$ shrinks to zero, so the exterior wavefunction is crushed against the wall, and continuity of $u$ forces the interior value at the wall down with it. The slope does not go to zero at the same rate. In the exterior the slope is $-\kappa$ times the value, so as the value falls like $1/\kappa$ the slope stays finite. That is Dirichlet and it is nothing else: the value vanishes, the slope survives.

Worked example 2(d) does this limit with the numbers in hand, and Problem 3(b) checks the result against the exact finite-well levels. The point to carry now is the shape of the argument. The Dirichlet condition is a statement about what the wall is made of, namely that it is the limit of a very high finite barrier. Chapter 4.4 §7.4 made the same point from the other side. It exhibited members of the family whose walls attract rather than repel, with states bound to the boundary at negative energy. An experiment that measures the levels of a box is measuring which member of the family that box implements. What this chapter solves is the Dirichlet member, and it says so rather than assuming there was only one.

1.6 · Self-adjointness, checked rather than assumed

Section 1 has now delivered what it promised, and this last subsection is a debt paid rather than a step forward. Everything below solves an eigenvalue problem for an operator, and Chapter 4.5's spectral theorem applies only to operators that are self-adjoint, so somebody has to check that the three Hamiltonians in this chapter are. That is all this subsection does. A reader who is content to take it on trust can go to §2 and lose nothing. Nothing later uses anything from here except the fact itself, apart from Problem 1, which runs the same check on two other walls and gets two other spectra.

Chapter 4.6 §4.4 set a standard for this book and it applies to every operator below: no general theorem is quoted for the self-adjointness of $\hat p^{2}/2m+V$, and each case is checked where it arises. There are three cases in this chapter and all three are short.

The first is the infinite well. Take the formula $-\hbar^{2}\dvn{2}{}{x}/2m$ on $[0,L]$ with the domain of functions having $u''\in L^{2}$ and $u(0)=u(L)=0$. Chapter 4.4 §7.1 computed the boundary form for a second derivative, and symmetry is the demand that $\big[\overline{u}\,v'-\overline{u'}\,v\big]_{0}^{L}$ vanish for every pair in the domain. Every term there carries a factor $u$ or $v$ evaluated at an endpoint, and all four of those are zero, so the operator is symmetric. For self-adjointness, suppose $u$ satisfies the defining relation of Chapter 4.4 §3.3 against every $v$ in the domain. The bracket collapses to $\overline{u(L)}\,v'(L)-\overline{u(0)}\,v'(0)$, since $v$ itself vanishes at both ends, and the two slopes $v'(0)$ and $v'(L)$ can be given any pair of values by a cubic polynomial vanishing at both endpoints. So both coefficients vanish, $u(0)=u(L)=0$, and the adjoint's domain is the domain. This is Chapter 4.4 §7.3's argument for the Robin family with the arithmetic made shorter by the conditions being simpler.

The second is a bounded potential on the whole line, which covers the finite well, the step and the barrier. Chapter 4.6 §4.4 stated the reason in one clause and the clause deserves expanding once. The free operator $\hat p^{2}/2m$ is self-adjoint on the line, by Chapter 4.5 §4.3's verification through the Fourier transform. Adding multiplication by a bounded real $V$ adds an operator that is symmetric and defined on the whole space, and an operator like that changes no domain. The set of $u$ for which $v\mapsto\avg{u,(\hat H_{0}+V)v}$ extends continuously is the same as the set for which $v\mapsto\avg{u,\hat H_{0}v}$ does. The reason is that the difference between the two is $\avg{Vu,v}$, which is continuous in $v$ for every $u$ in the space. So the domain of the adjoint is unchanged and the sum is self-adjoint on the domain of $\hat H_{0}$. The finite well, the step and the barrier are all bounded step functions, so all three are covered by that sentence.

The third is the delta well of §4.6, which is neither bounded nor covered by the second case, and which is the one place in this chapter where the domain condition and the self-adjointness check are the same sentence. Take the free formula on the line with the origin removed, and the domain of functions with $u''\in L^{2}$ on each side, $u$ continuous at the origin, and the slope jumping by $u'(0^{+})-u'(0^{-})=-2m\lambda u(0)/\hbar^{2}$. Chapter 4.4 §7.1's boundary form now has one join instead of two ends, and what has to vanish is $\overline{u(0)}\,\big[v'\big]-\big[\overline{u'}\big]\,v(0)$, where the brackets are the jumps. Put both jump conditions in and the two terms are the same real multiple of $\overline{u(0)}v(0)$, so they cancel and the operator is symmetric. For self-adjointness, do not assume $u$ is continuous. Writing the form in terms of the jump in $u$ and the slope on one side leaves $\big[\overline{u}\big]\,v'(0^{-})+\big(-2m\lambda\overline{u(0^{+})}/\hbar^{2}-\big[\overline{u'}\big]\big)v(0)$, and $v'(0^{-})$ and $v(0)$ may be given any pair of values inside the domain. Both coefficients therefore vanish: $u$ is continuous, and $u$ has the same jump. The adjoint's domain is the domain. The check is three lines because the jump condition was built to make it three lines, which is the whole point of §1.3.

a natural place to stop  ·  the joining rule is earned; what follows is one new observable, which exists to halve the work in every problem after it

2 · Parity, and half the work

This section introduces the only new observable in the chapter. It is cheap to define, it costs one line to check, and it halves the algebra in §4 and again in the worked examples. It is here for a longer reason than that. Parity is the first instance in this book of a move that recurs constantly in Part IV and beyond. Find a symmetry of the Hamiltonian, turn it into an operator that commutes with $\hat H$, and use its eigenvalues as labels. The labels sort the eigenstates into families before any of them is computed. By the end of the section you will have the operator, the proof that its two eigenvalues exhaust the possibilities, the condition on $V$ under which it commutes with $\hat H$, and a proof that every bound state of a symmetric potential is either even or odd with nothing in between.

2.1 · The operator, and why it has no domain trouble

Define the parity operator by what it does to a function, which is to reflect its argument through the origin:

Chapter 4.4 taught you to ask about the domain before anything else, so ask. Reflection changes no modulus and merely relabels the points, so $\int\abs{\psi(-x)}^{2}\dd x=\int\abs{\psi(x)}^{2}\dd x$ by the substitution $x\to-x$, and $\hat\Pi$ maps $L^{2}(\R)$ onto itself with $\norm{\hat\Pi\psi}=\norm\psi$. It is a bounded operator defined on the whole space, so the difficulty of Chapter 4.4 §2, that an unbounded operator cannot be defined everywhere, does not arise. This is the first observable in Part IV with no domain question attached, and the reason is that it is bounded.

It is also self-adjoint, by the same substitution. In $\avg{\hat\Pi u,v}=\int\overline{u(-x)}\,v(x)\dd x$ change variables to $y=-x$, which leaves the measure alone and swaps the limits back, giving $\int\overline{u(y)}\,v(-y)\dd y=\avg{u,\hat\Pi v}$. Since $\hat\Pi$ is bounded and everywhere defined, symmetric and self-adjoint coincide here, which is Chapter 4.4 §4.1's statement that the two conditions come apart only when a domain has been restricted.

2.2 · Two eigenvalues, and there is no third

Reflecting twice returns every function to itself, so

That single equation fixes the spectrum. If $\hat\Pi\psi=\lambda\psi$ then applying $\hat\Pi$ again gives $\lambda^{2}\psi=\psi$, so $\lambda^{2}=1$ and $\lambda=\pm1$. Both values occur, and the eigenvectors are the functions you would guess: $\psi(-x)=+\psi(x)$ is an even function and $\psi(-x)=-\psi(x)$ is an odd one. Neither eigenspace is small, since any $\psi$ splits into an even and an odd piece by $\psi=\tfrac12(\psi+\hat\Pi\psi)+\tfrac12(\psi-\hat\Pi\psi)$, and the two pieces are eigenvectors with $\lambda=+1$ and $\lambda=-1$ respectively. So $L^{2}(\R)$ is the direct sum of the two eigenspaces and there is nothing outside them.

Chapter 4.2 §4.3 called a label of this kind a quantum number: a value of an observable that commutes with the Hamiltonian and therefore stays attached to a state as it evolves. Parity is the cheapest quantum number in the subject, taking two values and requiring no calculation to read off, and the next subsection says when it is attached to anything.

2.3 · When parity commutes with the Hamiltonian

Two operators commute or they do not, and the question here has an answer that depends on $V$ alone. Apply $\hat H$ then $\hat\Pi$ to a function and compare with the other order. Reflection turns $\dvn{2}{u}{x}$ evaluated at $-x$ into $\dvn{2}{}{x}$ of the reflected function, because two minus signs from the chain rule cancel, so the kinetic term commutes with $\hat\Pi$ whatever the potential is. The potential term does not commute automatically. Working both orders out at the point $x$,

The two agree for every $u$ exactly when $V(-x)=V(x)$ at every point. So the commutator $[\hat\Pi,\hat H]$ vanishes precisely for an even potential, and that is the only hypothesis parity ever needs. One detail belongs with it, because Chapter 4.4 spent a chapter insisting on it. A commutator of an unbounded operator with a bounded one is a statement about domains as well as formulae, and here the domain behaves: if $V$ is even then $\hat\Pi$ maps $\operatorname{dom}(\hat H)$ onto itself, since reflecting a function with $\hat Hu\in L^{2}$ produces a function with the same property. So the equality in (4.7.7) holds on the whole domain rather than on some smaller set where both sides happen to make sense.

2.4 · Every bound state of an even potential is even or odd

Commuting operators can be diagonalised together, which is Chapter 0.5 §8's theorem in finite dimensions and Chapter 4.2 §4.3's postulate in this setting. What we want here is sharper and it can be had directly, without appealing to either: for the potentials of this chapter, an even $V$ forces every bound state individually to be even or odd, with no choice and no mixing.

Start with the observation that makes the argument possible. If $u$ solves (4.7.1) at energy $E$ and $V$ is even, then $\hat\Pi u$ solves it at the same $E$, by (4.7.7). So the reflected state is a state, with the same energy. If the energy level has only one state in it up to scale, the reflected state must be a multiple of the original, and that is the whole argument. So the question becomes whether a bound state can share its energy with anything else.

It cannot, in one dimension, and the reason is an identity about second-order equations rather than anything quantum. Let $u_{1}$ and $u_{2}$ both solve (4.7.1) at the same $E$ and form their Wronskian $W=u_{1}u_{2}'-u_{1}'u_{2}$. Differentiate it. The terms with first derivatives cancel in pairs, and each second derivative can be replaced using the equation itself, which gives

So $W$ takes one value everywhere. Now use the fact that both functions are bound states. Every potential in this chapter is constant outside a bounded region, and a bound state has $E$ below that constant, so §1.2's second case applies far out and each $u_{i}$ is a combination of $\ee^{\pm\kappa x}$ there. Square-integrability kills the growing exponential and leaves the decaying one, so both $u_{i}$ and both $u_{i}'$ go to zero as $x\to\infty$. Hence $W\to0$, and a constant with limit zero is zero. Setting $W=0$ says that $\dv{}{x}(u_{2}/u_{1})=-W/u_{1}^{2}$ vanishes, so $u_{2}$ is a constant multiple of $u_{1}$ on every interval where $u_{1}$ has no zero. The zeros need one more line, because every excited state has them. At a zero $x_{0}$ of $u_{1}$, the vanishing of $W$ gives $u_{1}'(x_{0})u_{2}(x_{0})=0$; and $u_{1}'(x_{0})\neq0$, since a solution of a second-order equation whose value and slope both vanish at a point is identically zero. So $u_{2}(x_{0})=0$ as well, and the ratio $u_{2}/u_{1}$ approaches $u_{2}'(x_{0})/u_{1}'(x_{0})$ from both sides. The constant is the same on either side of every node, so it is the same constant throughout. Bound states in one dimension are non-degenerate.

Put the two halves together. For even $V$ the reflected state $\hat\Pi u$ has the same energy as $u$ and is therefore a multiple of it, $\hat\Pi u=cu$. Reflecting twice gives $c^{2}=1$, so $c=\pm1$, and every bound state is an eigenvector of parity. The practical consequence is the one that saves the labour in §4: search the even functions and the odd functions separately, and you have not missed anything. Each search is a one-parameter problem instead of a two-parameter one, because evenness or oddness fixes the solution on the left once it is known on the right.

Two cautions travel with the result. The non-degeneracy argument used the decay at infinity, so it says nothing about states at energies above the potential far away, and §5 will meet exactly that case: scattering states at a given energy come in pairs, one arriving from each side, and neither has definite parity on its own. And the argument is one-dimensional. Chapter 4.13 finds levels of hydrogen with several states in them, which is why an extra label is needed there and why Chapter 4.11 builds one.

2.5 · The move, named once, because it happens four more times

What was done above has three steps and none of them mentioned parity. A transformation leaves the Hamiltonian alone. That transformation is represented by an operator commuting with $\hat H$. The eigenvalues of that operator are labels sorting the eigenstates into families that can be searched separately. This is the standard route from a symmetry to a quantum number, and it will run again, because the next five chapters run it four more times with a different symmetry each time.

• Chapter 4.11 runs it on rotations. The transformation is turning the system in space, the commuting operators are the components of angular momentum, and the labels are $\ell$ and $m$. The algebra is longer because rotations about different axes do not commute with each other, which is the one feature parity has not got.
• Chapter 4.16 uses parity to kill matrix elements in the fine structure of hydrogen without computing them, since an integral of an odd function over a symmetric interval is zero and knowing the parities decides which entries can be skipped.
• Chapter 4.17 shows that parity is the electric-dipole selection rule. A transition driven by a term odd under reflection connects states of opposite parity and nothing else, which is why most of the transitions one could imagine do not happen.
• Chapter 4.18 runs the identical algebra on the exchange of two identical particles. The operator squares to the identity, so its eigenvalues are $\pm1$, and the two values are bosons and fermions. That chapter will name this section when it does it, because it is the same three lines.

The reason to name the pattern now, on the cheapest possible example, is that when it appears with rotations the algebra is heavy enough to hide the shape. Here there is no algebra at all and the shape is all there is.

3 · The infinite well

This is the first problem solved end to end, and it is the cheapest one in the subject: the answer comes out in four lines and there is no arithmetic beyond a square root. What earns it the space is everything around the four lines. The spectrum is discrete, and this section says which step of the calculation makes it so. The lowest energy is not zero, and this section says why without invoking a principle. And the operator being solved is one particular member of Chapter 4.4's family, so the answer is a measurement of the walls as much as of the particle.

3.1 · The operator, and the problem it poses

Take a particle confined to $0\le x\le L$ with nothing inside to push it around, which is $\hat H=-\hbar^{2}\dvn{2}{}{x}/2m$ on the domain of §1.6: functions with $u''\in L^{2}[0,L]$ and $u(0)=u(L)=0$. That is the Dirichlet member of the $U(2)$, selected in §1.5 by the limit of a high finite wall, and §1.6 checked that it is self-adjoint. Inside the well the potential is zero, so (4.7.2) with $V_{0}=0$ gives the oscillating case for every positive $E$, and the general solution is

Only positive energies are being written down here, and the reason to say so is that the other possibilities have not been ruled out yet. At $E=0$ the two solutions are $1$ and $x$ rather than a sine and a cosine, and at $E\lt0$ they are real exponentials. Both are settled at the end of §3.2, in three lines each, and §3.6 then shows that nothing has been left out of the list.

Two constants, two conditions. The condition at $x=0$ reads $B=0$ at once, since $\sin$ vanishes there and $\cos$ does not, and that leaves $u=A\sin kx$ with the condition at $x=L$ still to impose. This is the moment the spectrum becomes discrete, so watch it closely rather than passing through.

3.2 · Where quantisation actually happens

The remaining condition is $A\sin kL=0$, and it has two ways of being satisfied. Either $A=0$, which makes $u$ the zero function everywhere, or $\sin kL=0$, which restricts $k$. Discard the first, and it is important to be exact about why. The zero function is not a state. Chapter 4.2 §3 required a state to be a unit vector up to phase, and the zero vector has no unit multiple, so "the particle is in the state $u\equiv0$" describes nothing at all rather than describing a particle at rest. That is the entire reason the ground state cannot have $E=0$, and that form of the reason is the one to keep rather than an appeal to a principle, because the appeal would be circular here: §3.5 recovers the uncertainty bound from this answer, so it cannot also be what produced it.

With $A\ne0$ the condition is $\sin kL=0$, so $kL$ is an integer multiple of $\pi$. Writing $k_{n}=n\pi/L$ and putting that back into the relation between $k$ and $E$ gives the levels:

The normalising constant comes from $\int_{0}^{L}\sin^{2}(n\pi x/L)\dd x=L/2$, which holds for every $n$ because the average of $\sin^{2}$ over a whole number of half-periods is one half. The index starts at $n=1$ rather than $n=0$: the value $n=0$ gives $u\equiv0$, which the previous paragraph disposed of, and negative $n$ gives the same function with a sign, which Chapter 4.2 §3 says is the same state. So the levels are the squares of the positive integers, in units of $\pi^{2}\hbar^{2}/2mL^{2}$, and they get further apart as you go up.

That leaves the two energies §3.1 set aside, and the same argument disposes of both. At $E=0$ the solutions are $1$ and $x$; the condition at the origin kills the constant, and the condition at $L$ then kills what is left. At $E\lt0$ they are $\sinh\kappa x$ and $\cosh\kappa x$; the condition at the origin kills the cosine, and $\sinh\kappa L$ is never zero for $\kappa\gt0$, so the sine goes with it. In both cases the only survivor is the zero function, which is not a state. The list above is therefore the whole of the spectrum.

Chapter 0.8 §3 said that a linear differential equation is an eigenvalue problem and named this chapter among the places the observation would be spent. Equation (4.7.10) is what spending it looks like: a differential operator, a boundary condition, and a discrete list of numbers at which a non-zero solution exists. The discreteness did not come from the differential equation, which has solutions at every $E$. It came from the boundary condition, which is to say from the domain, which is to say from the wall.

3.3 · The same problem with parity switched on

Section 2 is unused so far because the box was placed on $[0,L]$, where the origin is at one wall and $V$ is not even. Slide it to $[-L/2,L/2]$ and the potential becomes symmetric, so §2.3 applies and every level must be even or odd. Substituting $x\to x+L/2$ in (4.7.10) turns each $\sin(n\pi x/L)$ into a cosine for odd $n$ and a sine for even $n$, so

The parities alternate up the ladder, starting even at the ground state. That pattern is not special to this potential. It holds for the finite well of §4 as well, where §4.3 derives it rather than reading it off: the roots of the matching condition arrive on alternating branches, the even ones and the odd ones in turn. Chapter 4.8 finds the same alternation for the oscillator by a third route. Nothing here needed the energies. Parity sorted the states before any of them was computed, which is what §2.5 said the move is for.

3.4 · Numbers

A formula is only as good as what it predicts, so put an electron in the box. With $\hbar=1.054571817\times10^{-34}\ \mathrm{J\,s}$ and $m_{e}=9.1093837\times10^{-31}\ \mathrm{kg}$, the unit $\pi^{2}\hbar^{2}/2m_{e}L^{2}$ is $0.376030\ \mathrm{eV}$ at $L=1\ \mathrm{nm}$, and the levels scale as $1/L^{2}$:

$L$	$E_{1}$	$E_{2}$	$E_{2}-E_{1}$	wavelength of that gap
$1\ \mathrm{nm}$	$0.3760\ \mathrm{eV}$	$1.5041\ \mathrm{eV}$	$1.1281\ \mathrm{eV}$	$1099\ \mathrm{nm}$
$0.5\ \mathrm{nm}$	$1.5041\ \mathrm{eV}$	$6.0165\ \mathrm{eV}$	$4.5124\ \mathrm{eV}$	$274.8\ \mathrm{nm}$
$0.2\ \mathrm{nm}$	$9.4008\ \mathrm{eV}$	$37.603\ \mathrm{eV}$	$28.202\ \mathrm{eV}$	$43.96\ \mathrm{nm}$

Read the last column, which is where the model stops being an exercise. Chapter 4.6 §9.4 showed that a superposition of two levels beats at $(E_{2}-E_{1})/\hbar$ and that this is what a spectral line is, so the last column is the light a box of that size would emit. A box of a nanometre lands in the near infrared and a box of a couple of Ångströms lands in the far ultraviolet, which is the right order of magnitude for an atom and the reason atomic transitions are where they are. Nothing in the model is a fit. The only inputs are $\hbar$, the electron mass and a length.

The $1/L^{2}$ scaling is the part to remember. Halving the box quadruples every level, so confinement energy is very sensitive to size, and the sensitivity is the reason a particle can be treated as free in a box of a micron and cannot in a box of a nanometre.

3.5 · The ground-state energy, read as an uncertainty

Chapter 0.9 §6.4 proved the bandwidth theorem, that a function and its Fourier transform cannot both be narrow, and said that quantum mechanics would add to it a single substitution. That substitution is $p=\hbar k$, and Chapter 4.9 will make the one-line argument in general. Here it can be checked against an exact answer instead, which is better than making it, because the exact answer came first.

Compute the two spreads in the ground state. The mean momentum is zero, and the reason for that comes from the physics rather than from inspection. Since $u_{1}$ is real, Chapter 4.6 §8.3 says it carries no current anywhere. That same subsection showed the integral of the current over all space to be $\avg{\hat p}/m$, so $\avg{\hat p}=0$ with no integral done. Then $\avg{\hat p^{2}}$ needs no integration at all, since $u_{1}$ is an eigenfunction of $\dvn{2}{}{x}$ with eigenvalue $-(\pi/L)^{2}$, giving $\avg{\hat p^{2}}=\hbar^{2}\pi^{2}/L^{2}$ and $\Delta p=\pi\hbar/L$. The position spread is one integral, $\avg{x}=L/2$ and $\avg{x^{2}}=L^{2}(\tfrac13-\tfrac{1}{2\pi^{2}})$, so

Against the bound $\hbar/2$ that is a ratio of $1.1357$, so the ground state of a box comes within fourteen per cent of the tightest state allowed. Now run the argument the other way, as an estimate rather than a check. Since $\avg{\hat p}=0$ the energy is $(\Delta p)^{2}/2m$, and the bound gives $\Delta p\ge\hbar/2\Delta x$, so $E_{1}\ge\hbar^{2}/8m(\Delta x)^{2}=3.826\,\hbar^{2}/mL^{2}$. The true value is $4.935\,\hbar^{2}/mL^{2}$. The estimate delivers $78\%$ of the answer from a general theorem about Fourier transforms and one substitution, which is a fair account of what that theorem is worth: the order of magnitude and the scaling, never the coefficient.

Chapter 4.6 §6.3 gave the same result a third reading and it is the one to keep. Written as an average the kinetic energy is $\avg T=(\hbar^{2}/2m)\int\abs{u'}^{2}$, so energy is total steepness. A function pinned to zero at two walls a distance $L$ apart has to rise and come back down within that distance, and the gentlest way to do it costs $(\pi/L)^{2}$ in mean square slope. The zero-point energy of a box is a fact about how gently a function can vanish at two places, and there is no more to it than that.

3.6 · Every state in the box, and the check that nothing is missing

Chapter 4.6 §9.3 gave the general solution as a sum over eigenstates with a phase on each, and attached a condition from Chapter 4.5 §8.2: the expansion is legitimate when the eigenvectors are complete. Here they are, and it costs one sentence to see it. Extend any $\psi$ on $[0,L]$ to $[-L,L]$ as an odd function. Chapter 4.3 §8 proved that the trigonometric system is an orthonormal basis of $L^{2}$ on an interval, and the odd members of that system are exactly the functions $\sin(n\pi x/L)$, so an odd function is the sum of its sine series and the restriction back to $[0,L]$ is the expansion wanted. Hence every state in the box is

The box is therefore completely solved: its spectrum is pure point, its eigenfunctions are complete, and the evolution of any initial condition is (4.7.13). That is as good as a quantum problem ever gets, and §5 will meet the first potential for which it is not available.

3.7 · A warning inherited from Chapter 4.4, and worth carrying

One temptation should be closed off before the next section, because it is the place this chapter is easiest to misread. It is natural to say that a particle in a box is a free particle with a restriction imposed on it, so that $\hat H=\hat p^{2}/2m$ with $\hat p$ the usual momentum operator cut down to the interval. Chapter 4.4's Worked example 2 showed that this is false and named this chapter as the place to carry the point.

Its argument fits in one sentence, since the conclusion matters more than the details. Momentum observables on an interval are not scarce, and Chapter 4.4 §5.4 found that every one of them is $\hat p_{\theta}$ for some phase $\theta$, with the condition $u(L)=\ee^{\ii\theta}u(0)$. Squaring one of them drags a condition on the slopes along with it, and the Dirichlet domain imposes nothing on the slopes, so the two operators have different domains and are different operators. Chapter 4.4 made that concrete by exhibiting two Dirichlet eigenfunctions demanding incompatible values of the same $\theta$. The infinite well is not a free particle with walls attached. It is a different operator, chosen from a family, and the choice is what the walls are. Everything in (4.7.10) is a property of that choice, and Chapter 4.4 §7.4's table of levels at other choices is what the alternative would have looked like.

4 · The finite well, and counting its bound states

Now make the walls finite. The equation is the same, the matching conditions are the honest ones of (4.7.4) rather than a choice from a family, and the answer stops being a formula. What comes out is a transcendental equation with no closed-form solution, which has to be read off a graph and then solved numerically. That is not a defect of the method. It is what almost every quantum problem looks like, and the value of this one is that the graph tells you the answer before the arithmetic does: how many bound states there are, how they appear one at a time as the well deepens, and why there is always at least one. By the end of the section you will also have the delta well, which is this problem in a limit and is the only bound-state problem in this book whose answer fits on one line.

4.1 · The potential, and one decision about where to put the origin

Take

and look for bound states, meaning solutions with $-V_{0}\lt E\lt0$. The reason to look in that window rather than anywhere else is §1.2. Below $-V_{0}$ the energy is under the potential everywhere, so the solution is a combination of rising and falling exponentials in all three regions and cannot be made to decay at both ends without vanishing identically. Above zero the energy is above the potential outside, so the solution oscillates all the way out, never falls off, and is not in $L^{2}$. Only the window between gives a function that oscillates in the middle and dies at both ends, which is what a bound state is. Section 5 takes up what happens above zero, where those non-decaying solutions turn out to be the whole story rather than a failure.

The origin is at the centre of the well, and that decision has to be named, because it is §2 being spent. Centred, $V$ is even, so (4.7.7) makes parity commute with $\hat H$, and §2.4 then says every bound state is even or odd. Two searches on half the line replace one search on the whole of it, and each search has one unknown ratio rather than several. That is the practical value of parity and this is where it is collected.

4.2 · The three regions, and what has to be matched

Inside the well the energy is above the potential, since $E\gt-V_{0}$, so §1.2's first case gives oscillation with

Both are real and positive in the window, and $\kappa$ is real precisely because $E$ is negative, which is the arithmetic form of "the particle is trapped". Outside the well the general solution is $C\ee^{-\kappa x}+D\ee^{+\kappa x}$, and on the right-hand side square-integrability forces $D=0$, since $\ee^{+\kappa x}$ blows up. On the left the surviving one is $\ee^{+\kappa x}$ by the same argument reflected. So each exterior region contributes one constant rather than two, and that is the second place in this chapter where an acceptability requirement rather than the differential equation does the restricting.

Take the even sector first. An even solution is $u=A\cos kx$ inside, and by evenness it is enough to match at the right-hand wall $x=a$, where (4.7.4) demands

Two equations, two unknown constants, and one of the constants is only an overall scale that normalisation will fix later. So the useful content is what is left when the scale is removed, which is obtained by dividing the second equation by the first. The constants cancel and a condition on the energy alone survives:

The odd sector runs identically with a sine in place of a cosine. Putting $u=A\sin kx$ inside and matching at $x=a$ gives $A\sin ka=C\ee^{-\kappa a}$ and $Ak\cos ka=-\kappa C\ee^{-\kappa a}$, and dividing the second by the first gives

Neither condition can be solved in closed form, because each mixes a trigonometric function of $E$ with a square root of $E$. What can be done is to see the whole solution set at once, and for that the two conditions want rewriting in variables with no units in them.

4.3 · One dimensionless picture, containing everything

Divide the problem by its own scales. Put $z=ka$ and $y=\kappa a$, both dimensionless, and notice from (4.7.15) that the two are not independent: squaring and adding removes $E$ entirely, since $k^{2}+\kappa^{2}=2mV_{0}/\hbar^{2}$. So

and every property of the well is carried by the single number $z_{0}$, which combines the depth and the width into the one quantity the answer depends on. Dividing (4.7.17) and (4.7.18) by $k$ and writing $y=\sqrt{z_{0}^{2}-z^{2}}$ puts both conditions in a form with one unknown:

The right-hand side is the same curve in both, and it is the useful one to picture. It starts at $+\infty$ as $z\to0$, falls steadily, and reaches zero at $z=z_{0}$, beyond which there is nothing: $z\gt z_{0}$ would make $\kappa$ imaginary, which is the energy rising above zero and out of the bound window. On the left sit $\tan z$ and $-\cot z$, each running from zero up to $+\infty$ on successive intervals of length $\pi/2$, with $\tan$ starting at $z=0$ and $-\cot$ starting at $z=\pi/2$. Every crossing of the falling curve with one of the rising branches is a bound state, and the parities alternate, exactly as (4.7.11) found for the infinite well.

Counting them is now a matter of asking how many rising branches begin before $z_{0}$. The branches begin at $z=0,\ \pi/2,\ \pi,\ 3\pi/2,\dots$, and each one that begins below $z_{0}$ climbs to infinity while the falling curve stays finite, so each contributes exactly one crossing. The last branch is the exception worth naming, because it is cut off at $z_{0}$ before it has climbed anywhere: there the falling curve reaches zero while the rising one is still positive, so the two have again swapped which is larger and there is again exactly one crossing. Hence

which says that states appear one at a time as the well is deepened or widened, each new one arriving exactly when $z_{0}$ passes a multiple of $\pi/2$. At that moment the new state has $\kappa=0$, so it is infinitely spread out and barely bound at all, and it tightens as the well deepens further. There is a figure below on which that can be watched happening.

4.4 · At least one, always, in one dimension

Look at the first branch, near $z=0$. The falling curve goes to $+\infty$ there, because it has $z$ in the denominator, while $\tan z$ starts at zero. At the other end of the branch, whichever of $z_{0}$ and $\pi/2$ comes first, the falling curve has dropped to zero or the tangent has risen to infinity. Either way the two have swapped which is larger, so they cross somewhere in between. A one-dimensional square well of any depth and any width has at least one bound state. No matter how shallow, no matter how narrow, something is trapped.

That is a stronger statement than it looks, and it fails in three dimensions, because the failure is what makes several later chapters non-trivial. Chapter 4.13 shows that a spherically symmetric potential reduces to a radial equation for the function $u(r)=r\,R(r)$, with one extra requirement: $u(0)=0$, because $R$ has to stay finite at the origin. For the lowest angular momentum that equation is exactly (4.7.1) with the same $V$; for higher angular momentum it carries an extra repulsive term, which can only make binding harder, so the lowest case is the one that decides whether anything binds at all. Compare that with the two sectors above. Requiring $u(0)=0$ is precisely the odd sector, and the odd sector has no crossing until $z_{0}$ passes $\pi/2$. So a three-dimensional well binds nothing at all unless

and a weak enough attraction in three dimensions binds nothing. One dimension is the exceptional case, not the general one, and the exception is entirely due to the even sector being available there and absent here.

4.5 · Numbers, and a check that does not use any of this

Fix a well of half-width $a=0.5\ \mathrm{nm}$ for an electron and take three depths. The count (4.7.21) and the roots of (4.7.20), found by bisection, give:

$V_{0}$	$z_{0}$	$N$	bound-state energies, in eV
$1\ \mathrm{eV}$	$2.561584$	$2$	$-0.809239,\ -0.297230$
$5\ \mathrm{eV}$	$5.727875$	$4$	$-4.728196,\ -3.922604,\ -2.620794,\ -0.940658$
$20\ \mathrm{eV}$	$11.455750$	$8$	$-19.68206,\ -18.72992,\ -17.14897,\ -14.94958,\ -12.15015,\ -8.784903,\ -4.929399,\ -0.865509$

Read two features of the table before the check. The lowest level never reaches the bottom of the well, which is §3.5's confinement energy again: the state has to curve, and curvature costs. And the levels crowd towards the top rather than being evenly spaced, because a state near zero energy leaks far outside the well and behaves as if the well were wider than it is.

Now the check, and the point of it is that it shares no algebra with what produced the table. Put the same potential on a fine grid, replace the second derivative by its three-point difference, and diagonalise the resulting tridiagonal matrix. That procedure knows nothing about parity, about (4.7.20), or about matching conditions. On a grid of $48\,000$ points across $12\ \mathrm{nm}$, extrapolated from two resolutions, the negative eigenvalues reproduce every entry in the table with a worst relative departure of $4.6\times10^{-10}$ over all fourteen states at all three depths. The count agrees too, which is the part (4.7.21) was for: the matrix returns two, four and eight negative eigenvalues and no others.

The figure below is where (4.7.21) is worth watching rather than reading. It loads on a well shallow enough to hold exactly one state. Drag $z_{0}$ upwards slowly and a new rising branch comes into range every time it passes a multiple of $\pi/2$, each new state arriving at the top of the well barely bound and tightening as you go on. The two counts printed underneath, the formula's and the root finder's, have to agree at every setting, including at the thresholds themselves, which is the only place they could disagree.

well strength z₀

z0 = 1.200

one state at a threshold six states

z0 = 1.200000 ceil(2 z0 / pi) = 1 roots actually found = 1 -- these agree, and nothing in the root finder knows the formula

E / V0 : -0.533865 (even) largest scaled residual of the matching condition = 8.24e-17

4.6 · The delta well, which is the limit with exactly one state

One special case earns its own name because it is the only bound-state problem in this book with a one-line answer. Shrink the well and deepen it together, holding the product of depth and width fixed:

so that the well tends to Chapter 0.9 §5's delta, $V(x)=-\lambda\,\delta(x)$, with $\lambda$ carrying the units of energy times length. What happens to the count is immediate: $z_{0}^{2}=2ma^{2}V_{0}/\hbar^{2}=ma\lambda/\hbar^{2}$ goes to zero with $a$, so (4.7.21) gives $N=1$. A delta well has exactly one bound state however strong it is, and by §4.4 it never has none.

Finding its energy takes the even condition in the same limit. With $z_{0}$ small, $z$ is close to it and small too and $\tan z\approx z$, so (4.7.20) becomes $z^{2}=\sqrt{z_{0}^{2}-z^{2}}$, whose solution is $\sqrt{z_{0}^{2}-z^{2}}=z^{2}\to z_{0}^{2}$. But $\sqrt{z_{0}^{2}-z^{2}}$ is $\kappa a$, so $\kappa a=z_{0}^{2}=ma\lambda/\hbar^{2}$, and the $a$ cancels off both sides. That leaves

The normalisation is $\int\kappa\,\ee^{-2\kappa\abs x}\dd x=1$, done in one line. Notice that the energy goes as the square of the strength, so a delta well twice as strong binds four times as deeply and half as far out, and there is no second state at any strength.

There is a second route to the same answer that says something §1 was building towards, so take it. Integrate (4.7.1) across a vanishing interval around the origin. The $Eu$ term contributes nothing, since $u$ is bounded and the interval shrinks. The potential term contributes $-\lambda u(0)$, by the defining property of the delta. And the kinetic term contributes the jump in the slope, by the fundamental theorem. So

and substituting $u=\sqrt\kappa\,\ee^{-\kappa\abs x}$, whose slope jumps by $-2\kappa u(0)$, returns $\kappa=m\lambda/\hbar^{2}$ again. Read (4.7.25) the way §1.3 asked you to read (4.7.4). It is a joining condition with a jump in it, and the jump is allowed because the delta in $Vu$ is exactly what cancels the delta that $u''$ acquires. The size of the allowed jump is fixed by $\lambda$, so (4.7.25) is one more domain condition, and turning $\lambda$ turns it into a different one. The delta well is not a strange potential. It is a one-parameter family of boundary conditions at a point, sitting inside the same classification Chapter 4.4 §6 counted, and the parameter is the strength. Which member is self-adjoint is not settled by the counting, so §1.6 checked this one directly, and it is the third of that subsection's three cases.

a natural place to stop  ·  the bound states are finished; what follows is the same equation at energies where nothing is trapped, and where the right question is not what the levels are but what fraction gets through

5 · The step, the barrier, and what transmission means

Everything so far has been about trapped particles, and the whole apparatus of §§3 and 4 rested on being able to demand that the wavefunction decay at both ends. This section removes that demand. Above the top of any potential that is constant far away, no solution decays and there are no bound states at all, and the question worth asking changes shape: not what the levels are, but what fraction of an incoming particle gets through. Making that question precise is the work of this section, and the answer turns out not to be the ratio of amplitudes anyone would write down first. By the end you will have transmission and reflection defined as a ratio of fluxes, a proof that they add to one taken from Chapter 4.6 §8's continuity equation, and a number for something a classical particle cannot do.

5.1 · What kind of solution exists above the top

Take a potential that is zero for $x\lt0$ and equal to $V_{0}$ for $x\gt0$, with the particle's energy $E$ above $V_{0}$. By §1.2 the energy is above the potential on both sides, so the solution oscillates on both sides and there is no growing exponential to discard. Nothing forces any coefficient to vanish, and in particular nothing makes the solution square-integrable: $\int\abs{\ee^{\ii kx}}^{2}\dd x$ diverges, as Chapter 0.9 §5.3 pointed out and Chapter 4.5 §2 turned into a theorem.

So a two-dimensional space of solutions survives at every $E$ above $V_{0}$, and none of them is a state. That is not a failure of the method and Chapter 4.5 said in advance that it would happen. Its §8.2 attached a condition to the expansion in eigenstates, that the spectrum be pure point and the eigenvectors complete, and warned that the condition fails for a Hamiltonian with a continuous part, where an expansion in bound states alone misses the scattering states entirely, naming this chapter as where that case is met. This is it. A finite well has a handful of bound states and, above zero, an entire continuum with no eigenvectors in it, and any honest description of a particle in that well needs both.

5.2 · What is being computed, given that these are not states

Chapter 4.5's Worked example 1 did the free Hamiltonian $\hat p^{2}/2m$ in exactly this situation and its answer is the one to import. The spectrum is the half-line $[0,\infty)$, and there are no eigenvectors, because the equation $\hbar^{2}k^{2}/2m=E$ picks out two points and a set of two points has measure zero. The familiar statement that each energy above zero is "doubly degenerate" survives as a statement about the preimage. The set of wavenumbers giving energies in a narrow window has two components, one moving right and one moving left. The scattering solutions below are those two components. The formal way to hold them is as labels on pieces of a spectral measure rather than as vectors, and that sentence is here for the record: nothing in §§5 and 6 turns on it, and the working statement is the one before it, that the preimage has a right-moving piece and a left-moving piece.

What makes the calculation legitimate anyway is that the quantity being computed is a ratio, and ratios survive the failure of normalisation. Chapter 4.5 §7.4 set out the procedure once: work with the non-normalisable object, keep only quantities that have a limit, and check that the answer does not depend on the regularisation. Here the quantity is a flux ratio and the check is immediate, since multiplying every coefficient by the same constant changes nothing in it. That is the whole of the legitimacy argument, and it is enough for everything this chapter computes.

What the ratio is for is a separate statement, and it is worth being exact about its status. Build a genuine normalised packet out of these solutions, narrowly peaked around one $k$, and send it in. Long after it has hit the step it separates into a piece moving left and a piece moving right, and the probability of finding the particle on the far side tends to the flux ratio computed below. That is a theorem of scattering theory and this book does not prove it. Chapter 4.6 §10 builds a packet and watches it spread, which is the picture, but it does that for a free particle and never sends one at a step. So take the last sentence as an interpretation of the number rather than as a result derived here, and note that nothing in §§5 and 6 rests on it: every quantity computed below is the ratio itself. Chapter 4.3's Problem 3 drew that picture, of a travelling bump escaping to infinity, and said this chapter would take the unbound states of a genuine potential. These are them.

5.3 · The step, matched

Write the solution with one incoming wave, arriving from the left. On the left there is the incoming wave and whatever comes back; on the right there is only an outgoing wave, since nothing is sent in from that side:

The choice to put no left-moving wave on the right is a statement about the experiment rather than about the equation, and the difference has to be said out loud: the equation has a second solution with a wave arriving from the right, and §2.4's remark about parity applies here, since neither of the two is even or odd on its own. The jump in $V$ at the origin is finite, so (4.7.4) holds with no choices to make, and it gives two equations:

Solve them the way §4.2's were solved, by removing the overall scale. Divide through by $A$ and write $r=B/A$ and $t=C/A$ for the two ratios that survive. Adding and subtracting the pair gives $2=t(1+k'/k)$ and $2r=t(1-k'/k)$, so

Both are real here and both are pieces of arithmetic in the two wavenumbers alone. The whole physics of the step is that the wavelength changes across it, and $r$ measures how big the change is: if $k'=k$ there is no step and $r=0$, and if $k'\to0$, meaning the energy sits right at the top of the step, then $r\to1$ and everything comes back. This is the same algebra as any wave meeting a region where it travels at a different speed, and the matching conditions are the reason.

5.4 · The current these solutions carry

Now the question that the rest of the section turns on. Given $r$ and $t$, what fraction of the particle gets through? The tempting answer is $\abs t^{2}$, and it is wrong. To see what the right answer is we need the probability current for these solutions, so compute it rather than quoting a formula. Chapter 4.6 §8 built the current out of the wavefunction and, in one dimension, it reads $J=(\hbar/m)\operatorname{Im}\!\big(u^{*}u'\big)$.

Take the left-hand region first, where two waves are superposed. Differentiating (4.7.26) and multiplying by the conjugate gives four terms, two of them direct and two of them cross terms:

The bracket in the second group is $\bar z-z$ for $z=A^{*}B\,\ee^{-2\ii kx}$, which is $-2\ii$ times the imaginary part of $z$ and is therefore purely imaginary. Multiplied by the $\ii k$ in front it becomes real, so it contributes nothing at all to the imaginary part of $u^{*}u'$. That cancellation is the whole reason a current is the right bookkeeping device for scattering, and Chapter 4.6's Problem 2 set it up for exactly this moment: the density of a superposition has interference fringes in it and the current has none. What survives is

the second following from the same calculation with a single wave in it. Each term has the form "density times velocity", since a wave $\ee^{\ii kx}$ has $\abs\psi^{2}=$ its squared amplitude and a particle of momentum $\hbar k$ moves at $\hbar k/m$. So the left-hand current is the incident flux minus the reflected flux, and the right-hand current is the transmitted flux. Nothing has been assumed about the step; only the form of the solution far from it was used.

5.5 · Why the two are equal, and what that forces

Chapter 4.6 §8 proved $\partial_{t}\rho+\partial_{x}J=0$ for every solution of the equation of motion, and that statement is local, so it does not care whether the solution is normalisable. Attach the time factor $\ee^{-\ii Et/\hbar}$ of Chapter 4.6 §9.1 to $u$ and the density is $\abs u^{2}$ with no time in it, exactly as its §9.2 found for a bound state. So $\partial_{t}\rho=0$ and the continuity equation reduces to $\dv{J}{x}=0$: the current is the same number at every point.

Check it directly as well, because it takes two lines and it shows precisely which hypotheses are doing the work. Differentiate $\operatorname{Im}(u^{*}u')$ and the product rule gives $\operatorname{Im}(\abs{u'}^{2})+\operatorname{Im}(u^{*}u'')$. The first term is the imaginary part of a non-negative real number and vanishes. In the second, the equation replaces $u''$ by $2m(V-E)u/\hbar^{2}$, so $u^{*}u''$ is a real multiple of $\abs u^{2}$ and its imaginary part vanishes too. The two hypotheses used are that $V$ is real and that $E$ is real, which are the same two facts Chapter 4.6 §8.5 identified as the ones a complex potential breaks.

Now set the two expressions in (4.7.30) equal, divide by the incident flux $\hbar k\abs A^{2}/m$, and read off what the equality says:

Conservation of probability is where $T+R=1$ comes from, and the point to be clear about is that it was not assumed anywhere. The definitions of $T$ and $R$ are flux ratios, the continuity equation makes the flux uniform, and the identity follows. Substituting (4.7.28) confirms it in algebra as well: $T=4kk'/(k+k')^{2}$ and $R=(k-k')^{2}/(k+k')^{2}$, and the two numerators add to $(k+k')^{2}$.

5.6 · A number a classical particle cannot produce

Put energies into (4.7.28) and the interesting feature is visible at once. Write $\epsilon=E/V_{0}$, so that $k'/k=\sqrt{1-1/\epsilon}$, and

$E/V_{0}$	$k'/k$	$R$	$T$	classical $T$
$1.01$	$0.0995$	$0.6708$	$0.3292$	$1$
$1.1$	$0.3015$	$0.2880$	$0.7120$	$1$
$2$	$0.7071$	$0.02944$	$0.97056$	$1$
$10$	$0.9487$	$0.000693$	$0.999307$	$1$

Read the last two columns against each other. A classical particle with energy above a step always gets over it, slows down, and carries on, so its transmission probability is one at every entry in the table. The quantum particle does not. At twice the step height nearly three per cent of the probability comes back, and as the energy is lowered towards the top of the step the reflection climbs towards certainty even though the particle has enough energy to cross. Nothing is blocking it. What reflects is a wave meeting a change in its own wavelength, and the matching conditions of §1.3 are what make that reflection compulsory.

This is the first of the chapter's two departures from classical mechanics and it is the milder one. Section 6 takes the other, where the energy is below the top and classical mechanics says the transmission is exactly zero rather than merely one.

6 · Tunnelling, with a number

Now put the energy below the top of the wall. Classically that ends the discussion: a particle with less energy than the barrier cannot be inside it, since its kinetic energy there would be negative, and it certainly cannot be on the far side. Quantum mechanically the equation is unchanged and §1.2 already said what happens, which is that the wavefunction decays inside the wall instead of oscillating. It decays, but it does not reach zero in a finite distance, so what emerges on the far side is small rather than absent. This section computes exactly how small, gets a number for an electron and a wall you could build, and hands the exponential forward to the chapter that turns it into a general method.

6.1 · The barrier, and the one change from §5

Take a rectangular barrier of height $V_{0}$ and width $w$, with the same potential on both sides:

Outside the barrier the energy is above the potential, so the solution oscillates with $k=\sqrt{2mE}/\hbar$ exactly as in §5. Inside, the energy is below the potential, so §1.2's second case applies and the solution is a combination of $\ee^{\pm\kappa x}$ with

Both exponentials have to be kept this time. In §4 the growing one was discarded because the region was infinite and it would have destroyed square-integrability, but the barrier has finite width, so a function that grows across it is bounded and perfectly acceptable. That is the whole technical difference between this problem and the bound-state problems, and it is why a barrier of finite width transmits and an infinite one does not.

The solution therefore has three pieces and five constants, and there are four matching conditions, two at each face. One overall scale is free, so four conditions on four ratios determine everything.

6.2 · Solving it

The algebra is longer than anything else in this chapter and it has nothing in it but elimination, so the main line states what goes in and what comes out and the grind box does the eliminating. Write

with the same convention as §5.3, one wave arriving from the left and nothing sent in from the right. Impose (4.7.4) at $x=w$ first, which expresses $F$ and $G$ in terms of $C$, and then at $x=0$, which expresses $A$ in terms of those. Eliminating $F$ and $G$ gives the transmission amplitude directly:

Grind box — eliminating the two interior constants

The two conditions at $x=w$ are $F\ee^{\kappa w}+G\ee^{-\kappa w}=C\ee^{\ii kw}$ and $\kappa\big(F\ee^{\kappa w}-G\ee^{-\kappa w}\big)=\ii kC\ee^{\ii kw}$. Adding and subtracting,

$$F\,\ee^{\kappa w} \;=\; \tfrac12\,C\,\ee^{\ii kw}\Big(1+\tfrac{\ii k}{\kappa}\Big), \qquad G\,\ee^{-\kappa w} \;=\; \tfrac12\,C\,\ee^{\ii kw}\Big(1-\tfrac{\ii k}{\kappa}\Big).$$

The two conditions at $x=0$ are $A+B=F+G$ and $\ii k(A-B)=\kappa(F-G)$. Adding them eliminates $B$ and gives $2A=F\big(1+\tfrac{\kappa}{\ii k}\big)+G\big(1-\tfrac{\kappa}{\ii k}\big)$, which is $2A=F\big(1-\tfrac{\ii\kappa}{k}\big)+G\big(1+\tfrac{\ii\kappa}{k}\big)$. Substituting the two lines above,

$$2A \;=\; \tfrac12 C\,\ee^{\ii kw}\left[\ee^{-\kappa w}\Big(1+\tfrac{\ii k}{\kappa}\Big)\Big(1-\tfrac{\ii\kappa}{k}\Big) \;+\; \ee^{\kappa w}\Big(1-\tfrac{\ii k}{\kappa}\Big)\Big(1+\tfrac{\ii\kappa}{k}\Big)\right].$$

Multiply out the two products. Each gives $2$ from the real terms, since $(\ii k/\kappa)(-\ii\kappa/k)=+1$, plus an imaginary part $\pm\ii\Sigma$ with $\Sigma=k/\kappa-\kappa/k$. So the bracket is $\ee^{-\kappa w}(2+\ii\Sigma)+\ee^{\kappa w}(2-\ii\Sigma)$, which collects into $4\cosh\kappa w-2\ii\Sigma\sinh\kappa w$. Hence $A=C\ee^{\ii kw}\big[\cosh\kappa w-\tfrac{\ii}{2}\Sigma\sinh\kappa w\big]$, which is (4.7.36) once $-\Sigma$ is written as $\kappa/k-k/\kappa$.

For the modulus, the two terms in the bracket are real and imaginary respectively, so

$$\frac{1}{T} \;=\; \left|\frac{1}{t}\right|^{2} \;=\; \cosh^{2}\kappa w+\tfrac14\Sigma^{2}\sinh^{2}\kappa w \;=\; 1+\Big(1+\tfrac{\Sigma^{2}}{4}\Big)\sinh^{2}\kappa w,$$

using $\cosh^{2}=1+\sinh^{2}$. Finally $\Sigma=(k^{2}-\kappa^{2})/k\kappa$, so $1+\Sigma^{2}/4=(k^{2}+\kappa^{2})^{2}/4k^{2}\kappa^{2}$, and putting $\hbar^{2}k^{2}/2m=E$ and $\hbar^{2}\kappa^{2}/2m=V_{0}-E$ turns the numerator into $V_{0}^{2}$ and the denominator into $4E(V_{0}-E)$. $\blacksquare$

Note that $T=\abs t^{2}$ here with no correction factor, because the potential is the same on both sides and $k'=k$, which is the situation §5.5's warning said would hide the distinction. Taking the modulus squared of (4.7.36) and simplifying, as the grind box does at the end, gives the result this chapter exists for.

Before taking the modulus, notice what the modulus throws away. The amplitude carries an explicit factor $\ee^{\ii kw}$ and a complex bracket, so the transmitted wave emerges with a phase shift as well as a reduced size, and the shift is not the trivial one a free particle would pick up crossing the same distance. That phase is used twice below. Worked example 3(c) shows it is the only thing distinguishing a delta well from a delta barrier of the same strength, and §6.5 finds the bound states of a well by asking where the bracket vanishes, which is a question about the amplitude and not about $T$. Keep the amplitude; the modulus is what the next subsection reads, but it is not all there is.

6.3 · Reading the answer

Three quantities can be changed, the height of the wall, its width and the energy you send in, and the formula says something different about each. So here is the result with the factors labelled:

The decay length in the second factor is §6.1's, $1/\kappa=\hbar/\sqrt{2m(V_{0}-E)}$, so the width is being counted in units the wall itself sets.

The transmission is not zero at any energy, any width or any height, and that is the headline. Three readings of (4.7.37) are worth taking before any numbers go in.

The first factor is bounded below by one and is of order one unless the energy is squeezed against the floor or the top of the barrier. Writing $\epsilon=E/V_{0}$ it is $1/4\epsilon(1-\epsilon)$, which is smallest at $\epsilon=\tfrac12$ where it equals one, and it grows without limit at either end. So it matters, but it never contributes an order of magnitude in the middle of the range.

The second factor is where everything happens. For a barrier more than a couple of decay lengths thick, $\sinh\kappa w$ is very close to $\tfrac12\ee^{\kappa w}$, so the whole expression is dominated by an exponential, and

The exponent is $2\kappa w$ rather than $\kappa w$ because $T$ is a probability and the wavefunction amplitude is what falls by $\ee^{-\kappa w}$ across the wall. Everything else is in the prefactor, which is $16\epsilon(1-\epsilon)$ and lies between zero and four. So the transmission through a thick barrier is an exponential in the width, an exponential in the square root of the height above the energy, and an exponential in the square root of the mass. That last dependence is easy to miss and it is why tunnelling is an electron's phenomenon rather than a proton's.

The third reading is about $\kappa$ itself. The length $1/\kappa$ is how far the state reaches into a region it is classically forbidden from, and it is set by how far the top of the wall is above the energy rather than by how high the wall is in absolute terms. Raise the barrier and the energy together, keeping the difference fixed, and the reach does not change.

6.4 · A number, and a check that shares no algebra with it

Take an electron of energy $0.5\ \mathrm{eV}$ meeting a barrier $1\ \mathrm{eV}$ high and $1\ \mathrm{nm}$ wide, which is a barrier one could build. Then $V_{0}-E=0.5\ \mathrm{eV}$ and

so the decay length is $1/\kappa=0.2760\ \mathrm{nm}$ and the barrier is $\kappa w=3.6226$ decay lengths thick. Putting that into (4.7.37) with $\epsilon=\tfrac12$, so that the first factor is exactly one, gives

which is a suppression by a little under three orders of magnitude rather than by everything. The thick barrier estimate (4.7.38) gives $4\ee^{-7.245253}=2.854216\times10^{-3}$, which is high by $0.14\%$, so at three and a half decay lengths the exponential form is already the whole answer to two figures.

That number was obtained by solving four linear equations, so it deserves a check that does not solve any. Integrate the differential equation itself. Start at $x=w$ with the outgoing wave $u=\ee^{\ii kw}$ and $u'=\ii k\ee^{\ii kw}$, run a numerical integrator backwards through the barrier and out to $x=0$ with a tolerance of $10^{-13}$, and read the coefficients $A$ and $B$ off the value and slope there. The integrator knows nothing about $\sinh$, about $\kappa$, or about matching conditions. It returns

agreeing with (4.7.40) to a relative $3\times10^{-16}$, which is the last digit a double-precision number holds. The vanishing of $T+R-1$ is the more interesting entry, since nothing in the integration enforced it: §5.5 derived it from the continuity equation, and here it comes back out of a computation that used only the differential equation and two starting values. The second figure in this chapter, at the end of §6.5, runs both calculations live and prints their difference and $T+R-1$ alongside; its opening preset is this barrier, this electron and this energy, so the number above is the one on screen when it loads.

The sensitivity is what makes this formula useful, so tabulate it rather than asserting it. Holding the same electron and the same barrier height and moving only the width:

$w$	$\kappa w$	$T$	factor per extra $0.5\ \mathrm{nm}$
$0.5\ \mathrm{nm}$	$1.811$	$1.014\times10^{-1}$	—
$1.0\ \mathrm{nm}$	$3.623$	$2.850\times10^{-3}$	$35.6$
$1.5\ \mathrm{nm}$	$5.434$	$7.624\times10^{-5}$	$37.4$
$2.0\ \mathrm{nm}$	$7.245$	$2.037\times10^{-6}$	$37.4$

The right-hand column settles at $\ee^{2\kappa\times0.5\,\mathrm{nm}}=37.44$ as soon as the thick barrier form takes over, which is (4.7.38) read as a rate rather than a value. Half a nanometre of extra wall costs a factor of thirty-seven, every time.

Chapter 4.2 §10.2 promised this. Working the ammonia molecule as a two-state system, it found the nitrogen atom moving from one side of the hydrogen plane to the other through a barrier of $0.2504\ \mathrm{eV}$ that it does not have the energy to cross. The measured splitting between the two lowest levels is $98.72\ \mu\mathrm{eV}$, a ratio of $2536$. It said that the coupling being suppressed by three orders of magnitude rather than being zero is the quantitative content of tunnelling, and that this chapter would compute the coupling for a barrier. Worked example 1 does the computation on a double well and finds the splitting falling off exponentially with the width of the barrier between the two sides, in exactly the way (4.7.38) makes $T$ fall off. Three orders of magnitude is what an exponential of a moderate number looks like.

6.5 · When a barrier disappears

Nothing in the derivation of (4.7.36) required $E$ to be below $V_{0}$. Everything was algebra in $k$ and $\kappa$, and $\kappa$ becomes imaginary when the energy rises above the top. Put $\kappa=-\ii q$ with $q=\sqrt{2m(E-V_{0})}/\hbar$ real, use $\sinh(-\ii qw)=-\ii\sin qw$, and the same expression turns into

The hyperbolic sine has become an ordinary sine, and an ordinary sine has zeros. So there are energies at which the second term vanishes entirely and $T=1$ exactly, with nothing reflected at all:

A wall that a particle has more than enough energy to cross is normally partly reflecting, by §5.6, and at these particular energies it is not reflecting at all. The mechanism is interference and nothing else. Two reflections occur, one at each face, and the condition $qw=n\pi$ makes the round trip inside the barrier a whole number of wavelengths, so the two reflected waves arrive back at the front face exactly out of step and cancel. Everything then has to go forward, because §5.5 leaves it nowhere else to go.

The condition should look familiar. A whole number of half-wavelengths fitting across a width $w$ is (4.7.10), the infinite well of that width, so the barrier is transparent exactly at the energies at which its interior, walled off, would have a level. The same statement holds for a well rather than a barrier, with $V_{0}$ negative in (4.7.42) and $q=\sqrt{2m(E+\abs{V_{0}})}/\hbar$, and it is easier to see there because the well also has genuine bound states below zero to compare against.

The connection between the two is exact, and it is the cleanest thing in the section. One warning about letters before it, because the two problems label the same two regions in opposite orders. In §6 so far, $k$ has been the wavenumber outside the barrier and $\kappa$ the decay constant inside it. Below zero it is the other way round: outside is where the state decays and inside is where it oscillates, which is §4's arrangement. The next two paragraphs use §4's letters, so $\kappa$ now means the decay outside and $q$ the oscillation inside. Nothing about the algebra changes, and the same two symbols keep meaning decay and oscillation, but they change regions at the moment the energy crosses zero.

Now continue (4.7.36) to negative energy, where $k=\ii\kappa$ and the particle would be bound, and ask where the transmission amplitude $t$ becomes infinite. Setting the bracket to zero and substituting $k=\ii\kappa$ turns the condition into

Now compare that with §4. The well of full width $w$ had the even condition $\cot(qw/2)=q/\kappa$ and the odd condition $\cot(qw/2)=-\kappa/q$, and the double-angle identity $\cot2\theta=(\cot^{2}\theta-1)/2\cot\theta$ turns either of them into (4.7.44). So the poles of the transmission amplitude at negative energy are exactly the bound states, both parities, with nothing left over. Scattering and binding are one problem looked at on two parts of the same energy axis, and the function that answers one answers the other.

A well $0.6\ \mathrm{nm}$ wide and $3\ \mathrm{eV}$ deep makes that concrete. Its two bound states, from (4.7.20), sit at $-2.457819\ \mathrm{eV}$ and $-0.985835\ \mathrm{eV}$. Solving (4.7.44) numerically for the poles of $t$ returns $-2.457819$ and $-0.985835$, agreeing in every digit printed. Above zero the same well is perfectly transparent at $1.178113\ \mathrm{eV}$ and $6.400754\ \mathrm{eV}$, where $qw$ is $2\pi$ and $3\pi$, and elsewhere it is not: at $0.3\ \mathrm{eV}$ it transmits $0.5151$ and reflects the rest.

barrier height V₀ / eV

1.00 eV

width w / nm

1.000 nm

energy E / eV

0.500 eV

the worked barrier on a resonance a well

T = 2.850146716e-3 R = 9.971498533e-1 T + R - 1 = 0.000e+0 -- each computed from its own amplitude, never imposed

T from the closed form = 2.850146716397e-3 T from integrating the equation = 2.850146716397e-3 relative difference = 1.14e-14 below the top, kappa w = 3.622626

6.6 · What goes forward

Two chapters take the exponential from here, and each takes something different.

Chapter 4.10 §4 generalises it. A real barrier is not rectangular, and the WKB approximation replaces $2\kappa w$ by the integral $\tfrac{2}{\hbar}\int\abs{p}\,\dd x$ across the classically forbidden region, with $\abs p=\sqrt{2m(V(x)-E)}$ the magnitude of the momentum the particle would need. For a rectangular barrier that integral is $2\kappa w$ exactly, so the two expressions have to agree there, and that comparison is worth making because it is the one place in this book where an approximation and an exact answer for the same quantity can be set beside each other. The prefactor is where they differ, and (4.7.38) is what the approximation has to be scored against.

Chapter 4.17 needs the resonance of §6.5 for a different purpose. A resonance in (4.7.42) is a state that is almost bound, in the sense that it would be bound if the leakage were switched off, and the pole of $t$ that sits at a bound state moves off the real axis when the well is opened up. That is the mathematical origin of a finite lifetime, which is the imaginary part of an energy, which is the model Chapter 4.6 §8.5 built with a complex potential and Worked example 3 computed the line shape of.

a natural place to stop  ·  all four potentials are solved; what follows spends the machinery on three more problems and then hands it to you

7 · Worked examples

8 · Your turn