We now know how to write the Schrödinger equation for a particle, and we saw in the previous lesson that solving the dynamics amounts to finding the stationary states, that is, solving the eigenvalue equation

In this lesson, we shall solve it explicitly for piecewise constant potentials: wells, steps and barriers. These models are of course idealisations, since no real potential has discontinuities. But they have two major advantages. First, the potential is constant on each interval and we already know the solutions: oscillatory, exponential or affine depending on the sign of $E-V$ (see Lesson 5). All the physics then lies at the matching points, and the calculations can be carried out entirely by hand. Second, they suffice to reveal the two most striking phenomena of the quantum mechanics of a point particle: the quantisation of energy for bound states, and tunnelling, that is, the possibility of crossing a barrier that is classically impassable.

We shall first work in one dimension, with a particle of constant mass $m$ and a time-independent potential. We shall begin by setting out the rules: normalisation, matching conditions and boundary conditions. We shall then study the bound states of two wells, one infinite and one finite, followed by scattering states at a step and a barrier. We shall finish by extending the discussion to three dimensions.

1. The general method

1.1. Bound states and scattering states

Let us first recall what classical mechanics would say. A particle with energy $E$ can only be found in regions where $E\geq V(x)$, since its kinetic energy $E-V(x)$ must be positive. If these regions are bounded, the particle is trapped and oscillates between two turning points: its motion is bound. If they extend to infinity, the particle arrives from far away, interacts with the potential and returns to infinity: this is scattering motion.

The same distinction appears in quantum mechanics, with two kinds of stationary state of very different nature.

• Bound states are normalisable eigenfunctions of $\hat H$, genuine vectors in Hilbert space. Their probability density is concentrated in the region of the well and decays at infinity. We shall see that the normalisability condition is satisfied only for certain values of the energy: the bound-state spectrum is discrete.
• Scattering states are generalised eigenstates, like the plane waves of the free particle. They oscillate out to infinity and are not normalisable; they exist for a continuum of energies. They do not themselves represent physical states, but rather components of normalisable wave packets, and are interpreted using the probability current introduced in Lesson 5.

1.2. The normalisation condition

For a bound state, the wavefunction must be square-integrable over the whole real line. This condition, though seemingly innocuous, is in fact highly restrictive. Suppose, for example, that in a region $x>b$ where $E<V$, the general solution of the stationary equation is $Ae^{\kappa x}+Be^{-\kappa x}$ with $\kappa>0$. If $A\neq0$, the growing exponential dominates and

A bound state must therefore have $A=0$: on the right, only the decaying exponential is allowed. On the left, by contrast, it is $e^{-\kappa x}$ that diverges as $x\to-\infty$ and must be excluded. These exclusions, combined with the matching conditions, will select the allowed energies.

For scattering states, the situation is different. A plane wave $e^{ikx}$ is not normalisable, but it is bounded, and a superposition of plane waves with neighbouring energies forms a normalisable packet. We therefore impose no normalisation on such a state; we fix the amplitude of the incident wave arbitrarily, and calculate the ratios of the reflected and transmitted currents to the incident current. A solution that grows exponentially at infinity is still excluded, however, because no finite-probability packet could contain it.

1.3. The matching conditions

On each interval where the potential is constant, we know how to write the general solution, which depends on two constants. We still need to know how to join these pieces at the points where the potential jumps. The stationary equation itself provides the answer.

These rules apply to a finite jump in the potential. A point interaction in $\delta$ produces a jump in the derivative, while an infinite wall, which we shall now consider, requires a different condition again. Continuity of $\phi'$ must therefore not be applied mechanically to every potential.

1.4. Infinite walls and Dirichlet conditions

A well with infinite walls models a particle perfectly confined to an interval, for example $[0,L]$. It may be viewed as the limit of a finite well whose depth tends to infinity: we shall see below that the wavefunction penetrates into the classically forbidden regions over a length $1/\kappa$ which tends to zero in this limit. The particle therefore cannot be found outside, the wavefunction vanishes there, and continuity of $\phi$ requires it to vanish at the boundaries:

These are called homogeneous Dirichlet conditions. The derivative $\phi'$, however, is not continuous at the walls: it generally takes a non-zero value just inside and is zero outside. Imposing $\phi'(0)=\phi'(L)=0$ as well would be an error, leaving no non-zero solution.

More formally, the model takes the state space to be $L^2(0,L)$ and the Hamiltonian to be the operator $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ acting on functions satisfying (3). It is precisely these conditions that make the boundary terms vanish in integrations by parts and render the Hamiltonian self-adjoint: on a bounded interval, the boundary conditions are part of the definition of the observable.

2. The infinite square well

2.1. The spectrum

Consider a particle free to move between two infinite walls placed at $x=0$ and $x=L$, with zero potential inside (Figure 1). This is the simplest model of a confined particle.

2.2. Comments

The origin of quantisation.

The energy can take only the discrete values $E_1,4E_1,9E_1,...$ It is important to see where this quantisation comes from: the boundary conditions leave only the standing waves for which an integer number of half-wavelengths fits inside the well, as for a vibrating string fixed at both ends. The free particle on the whole real line, which is subject to no such condition, instead had a continuous spectrum. As we noted in Lesson 3, quantisation is therefore not a general property of quantum mechanics, but a consequence of confinement.

Zero-point energy.

The ground level has energy $E_1>0$: the particle cannot be at rest at the bottom of the well. This can be understood from Heisenberg's uncertainty relation. Confined to the interval $[0,L]$, the particle has a position spread at most of order $L/2$, and hence a momentum spread at least of order $\hbar/L$. Since $\langle\hat P\rangle=0$ in a real state, the mean energy satisfies

This rough estimate gives the correct order of magnitude and the correct dependence on $m$ and $L$; the exact calculation simply replaces the factor $1$ by $\pi^2$. Confinement has a kinetic-energy cost, which increases as $1/L^2$ when the well is narrowed. For an electron in a $1$ nm well, one finds $E_1\approx0.38$ eV, which is the order of magnitude of the energies involved in semiconductor nanostructures.

Nodes.

The function $\phi_n$ vanishes $n-1$ times inside the well. The higher the energy, the more the wavefunction oscillates, consistent with the fact that kinetic energy is related to the curvature of the wavefunction: indeed, $E=\frac{\hbar^2}{2m}\int|\phi'|^2dx$ from the calculation in the proof. We shall encounter this node rule again for the finite well, the harmonic oscillator and the hydrogen atom.

The dynamics.

An arbitrary state can be expanded in the basis of the $\phi_n$, and evolves according to

The probabilities $|c_n|^2$ of the different energies are constant, but the position density generally depends on time because of the cross terms between levels.

2.3. A delta function at the centre of the well

To see the jump condition (2) at work, add a repulsive point interaction $g\,\delta(x-L/2)$, with $g>0$, at the centre of the well (Figure 3). This term models a very thin, very high barrier with integral $g$; the coefficient $g$ has the dimensions of energy times length.

On either side of the centre, the potential is zero and the solution is sinusoidal. Choosing from the outset solutions that vanish at the walls, we write

with $E=\hbar^2k^2/(2m)$. It remains to impose continuity of $\phi$ and the jump in its derivative at $L/2$. Since the potential is symmetric about the centre, we can seek separately solutions that are odd and even about this point (we shall justify this procedure below).

The solutions that are odd about the centre vanish there. The right-hand side of (2) is then zero: the delta function does not see them. These are the functions $\sin(2j\pi x/L)$ of the ordinary well, that is, the $\phi_n$ with even index $n$, with the same energies.

For the even solutions, take $B=A$. The derivative is $Ak\cos(kL/2)$ immediately to the left of the centre and $-Ak\cos(kL/2)$ immediately to its right. The jump condition reads

This is a transcendental equation, to be solved graphically or numerically. Its two limiting cases are instructive. For $g=0$, we recover $\cos(kL/2)=0$, that is, the $\phi_n$ with odd index in the ordinary well. For $g>0$, the right-hand side is negative and each solution is shifted towards a larger value of $k$: the energies of the even states increase, as expected since the barrier is repulsive and these states have non-zero amplitude at the centre. As $g\to+\infty$, we must have $\sin(kL/2)\to0$, that is, $k\to2j\pi/L$: each even state approaches the energy of its neighbouring odd state. We then obtain two independent wells of width $L/2$, whose levels are doubly degenerate. For large but finite $g$, the levels remain grouped in closely spaced pairs: this is our first encounter with the two-site model coupled by tunnelling, which we shall revisit in Lesson 11.

3. The finite square well

3.1. Solutions in the three regions

Now consider a well of finite depth $V_0>0$ and width $2a$:

This model is more realistic than the infinite well: the particle can escape if its energy is high enough. Here we seek bound states with energy $-V_0<E<0$ (Figure 4).

Why this range of energies? On the one hand, for $E\geq0$, the outer regions are classically allowed and the solutions there are oscillatory (or affine if $E=0$): they are not square-integrable and describe scattering states. On the other hand, the energy of a normalised state cannot fall below the minimum of the potential, since $E=\frac{\hbar^2}{2m}\int|\phi'|^2dx+\int V|\phi|^2dx$ and $V\geq-V_0$. We therefore set

The wavenumber $k$ describes the oscillation inside the well, where the classical kinetic energy is $E+V_0$, and $\kappa$ the decay outside. They are not independent, since

In region II, the equation is $\phi"=-k^2\phi$; in regions I and III, it is $\phi"=\kappa^2\phi$. The normalisation condition excludes the exponential $e^{-\kappa x}$ on the left and $e^{\kappa x}$ on the right. This leaves

The shifts $x+a$ and $x-a$ in the exponentials do not change the form of the solutions: they merely simplify the values at the boundaries, which are $A$ and $F$.

3.2. Matching conditions and parity

The potential has only finite jumps: the function and its derivative are continuous at $x=-a$ and $x=a$. This gives four linear equations for the four constants $A,B,C,F$:

We could solve this system directly by requiring its determinant to vanish for a non-zero solution to exist. But the calculation can be simplified considerably by using the symmetry of the problem. The potential is even, $V(-x)=V(x)$: if $\phi(x)$ is a solution with energy $E$, then so is $\phi(-x)$, and hence so are the combinations $\phi(x)\pm\phi(-x)$, which are respectively even and odd. We may therefore seek the eigenstates among even functions ($C=0$, $A=F$) and odd functions ($B=0$, $A=-F$), without losing any eigenvalue.

The corresponding wavefunctions are, up to a normalisation constant,

The normalisation must take all three regions into account. For the even state, for example, the condition $\int|\phi|^2dx=1$ reads

and similarly, for the odd state, we obtain

3.3. Graphical solution

The equations in Proposition 1 are transcendental: their solutions cannot be expressed in terms of elementary functions. The problem is nevertheless solved exactly, in the sense that the energies are the roots of explicit equations, which can be calculated numerically to any desired accuracy. To visualise these roots, introduce the dimensionless variables

The parameter $z_0$ measures both the depth and the width of the well. From (5), $\kappa a=\sqrt{z_0^2-z^2}$, and the equations become

The solutions are the intersections of the quarter-circle $\sqrt{z_0^2-z^2}$ with the positive branches of the curves $z\tan z$ and $-z\cot z$, shown in Figure 6.

Several conclusions can be read directly from this figure.

• There is always at least one bound state. On the interval $0<z<\min(z_0,\pi/2)$, the function $z\tan z$ starts at zero and increases, while $\sqrt{z_0^2-z^2}$ starts at $z_0>0$ and decreases: the two curves must intersect. Thus, in one dimension, an attractive well has at least one even bound state, however shallow or narrow it may be¹.
  Note 1: This result is specific to one dimension. In three dimensions, a spherical well that is too weak has no bound state, as may be checked later.
• The number of bound states is finite, and is $1+\lfloor 2z_0/\pi\rfloor$, where $\lfloor\cdot\rfloor$ denotes the integer part. Indeed, the positive branches of $z\tan z$ and $-z\cot z$ occupy in turn the intervals $[0,\pi/2[$, $[\pi/2,\pi[$, $[\pi,3\pi/2[$, and so on. On each of them, the trigonometric function starts at zero and increases, while the quarter-circle decreases until it vanishes at $z_0$: each interval that begins before $z_0$ therefore contains exactly one intersection. For $z_0=4$, there are indeed three states. A wider or deeper well contains more.
• Even and odd states alternate as the energy increases, and the $n$th state has $n-1$ nodes, as in the infinite well.
• The infinite well is recovered when $V_0\to+\infty$. In this limit $z_0\to+\infty$, the quarter-circle intersects the branches near their asymptotes $z=\pi/2,\pi,3\pi/2,...$, that is, $k\simeq n\pi/(2a)$. The energies measured from the bottom of the well therefore tend to $E+V_0=\hbar^2\pi^2n^2/(2m(2a)^2)$, those of the infinite well of width $2a$.

3.4. Penetration into the forbidden region

Unlike in the infinite well, the wavefunction does not vanish at the boundary of the well: it extends into the classically forbidden regions as a decaying exponential, called an evanescent wave. This name describes the spatial dependence of the amplitude, not a decrease in probability over time: the state is stationary. The amplitude decays over the characteristic length $1/\kappa$, and the probability density as $e^{-2\kappa d}$ at a distance $d$ from the boundary. As the energy approaches zero from below, $\kappa\to0$ and the tail extends farther and farther.

The probability of finding the particle outside the well is therefore non-zero. For the even state, it is

This is a purely quantum effect: a classical particle with energy $E<0$ could never reach the region $|x|>a$. This does not mean that the particle escapes: the density is stationary and the probability current is zero everywhere, since the wavefunction is real. The particle remains bound, but its wavefunction “spills out” of the well. As we shall see, it is this spillover that makes it possible to cross a barrier of finite width.

Finally, note that the bound states do not form a basis of $L^2(\R)$: to describe an arbitrary state, the scattering states with energy $E>0$ must be added. Here the spectrum of the Hamiltonian has a finite discrete part and a continuous part.

4. Parity and non-degeneracy

We used the parity of the potential to seek even or odd solutions. This procedure has a general justification: the parity operator $(\hat\Pi\psi)(x)=\psi(-x)$ commutes with the Hamiltonian whenever $V$ is even, and $\hat H$ and $\hat\Pi$, whose eigenvalues are $\pm1$, can therefore be diagonalised simultaneously. But we obtained a stronger result: every bound state is automatically even or odd. This follows from the next result.

If $V$ is even and $\phi$ is a bound state, $\phi(-x)$ is a bound state with the same energy, so $\phi(-x)=c\,\phi(x)$. Applying this relation twice gives $c^2=1$: the state is even or odd.

The assumptions matter. A particle on a ring, whose wavefunction must be periodic and does not tend to zero, has states of the same energy with opposite momenta, $e^{\pm ikx}$. And we shall see at the end of this lesson that degeneracies are common in three dimensions.

5. The potential step

We now turn to scattering states, beginning with the potential step: $V=0$ for $x<0$ and $V=V_0>0$ for $x>0$ (Figure 7). A particle approaches from the left with energy $E$. Classically, the situation is simple: if $E>V_0$, the particle crosses the step and slows down; if $E<V_0$, it bounces back. Let us see what quantum mechanics says.

5.1. Above the step

For $E>V_0$, the solutions are oscillatory on both sides, with wavenumbers

On the left, there is the incident wave $e^{ikx}$, whose amplitude we set to one, and a possible reflected wave $e^{-ikx}$. On the right, there is only a transmitted wave $e^{iqx}$ travelling to the right: a term in $e^{-iqx}$ would describe a wave arriving from $+\infty$, which is not the situation under study. We therefore write

Continuity of $\phi$ and $\phi'$ at $x=0$ gives

Substituting the first equation into the second gives $k(1-r)=q(1+r)$, hence

These coefficients are amplitudes. To obtain reflection and transmission probabilities, the probability fluxes must be compared, and this is where the current from Lesson 5 becomes essential.

The equality $R+T=1$ is no accident: it expresses conservation of probability. For a stationary state, the continuity equation reduces to $\partial_xj=0$, and the current is the same everywhere. On the left, it is $j_{\rm inc}+j_{\rm refl}$ (the cross terms between $e^{ikx}$ and $e^{-ikx}$ may be checked not to contribute to the current); on the right, it is $j_{\rm trans}$. Their equality gives precisely $R+T=1$.

Notice the factor $q/k$ in $T$: it reflects the fact that the transmitted wave propagates more slowly than the incident wave. This is the example announced in the previous lesson: the squared modulus of the transmitted amplitude $|t|^2$, which can exceed one, is not a transmission probability. Note also that, contrary to classical intuition, the reflection is non-zero even though the energy exceeds the height of the step. It vanishes only in the limit $E\gg V_0$, where $q\simeq k$. This is a typically wave-like phenomenon, analogous to the partial reflection of light at the interface between two media with different refractive indices.

5.2. Below the step

For $0<E<V_0$, the region on the right is classically forbidden. Set $\kappa=\sqrt{2m(V_0-E)}/\hbar$. The solution there is exponential, and we retain only the decaying exponential, since the growing exponential could not describe the tail of any finite-probability packet:

The matching conditions at zero now read $1+r=t$ and $ik(1-r)=-\kappa t$. They give

The numerator and denominator of $r$ are complex conjugates: $|r|=1$, and reflection is total, $R=1$. The same result follows from the current: in the region on the right, $\phi$ is a real exponential multiplied by a constant, so $\phi^*\phi'$ is real and the current is zero, $T=0$. As in the finite well, the wavefunction penetrates the forbidden region over a length $1/\kappa$, but this penetration is accompanied by no transmission to infinity. Reflection is accompanied only by a phase shift, $r=e^{-2i\theta}$ with $\tan\theta=\kappa/k$, corresponding to a slight delay of the reflected packet.

6. The potential barrier and tunnelling

6.1. Statement of the problem

What happens now if the forbidden region is no longer semi-infinite, but has finite width? Consider the barrier $V=V_0$ for $0<x<a$ and $V=0$ elsewhere, and a particle approaching from the left with energy $0<E<V_0$ (Figure 8). Classically, the particle always bounces off the barrier. We have just seen that the wavefunction penetrates the forbidden region: if the barrier is thin enough, the evanescent wave is not yet negligible on the other side and can give rise to a transmitted wave there. This is tunnelling.

Using the same notation as for the step,

the solutions in the three regions are written

This time both exponentials must be retained in the barrier: since the region has finite length, the growing exponential presents no normalisation problem there. Continuity of $\phi$ and $\phi'$ at $x=0$ and $x=a$ gives four equations,

which determine the four unknowns $r$, $A$, $B$ and $t$.

6.2. The transmission coefficient

One point deserves clarification. We saw that a single real exponential carries zero current, and transmission beyond a step was zero. How, then, can the barrier transmit a non-zero current? The answer is that, inside the barrier, the wavefunction is a combination of the two exponentials, with complex coefficients. A direct calculation gives, for $\phi=Ae^{\kappa x}+Be^{-\kappa x}$,

which is non-zero whenever $A$ and $B$ do not have the same phase. The matching conditions impose precisely the combination that carries the transmitted current: there is no contradiction with the step, where the growing exponential was excluded.

6.3. Thick barrier and orders of magnitude

When the barrier is thick compared with the penetration length, $\kappa a\gg1$, we may approximate $\sinh\kappa a\simeq e^{\kappa a}/2$, and the term $1$ becomes negligible in (6). This leaves

The prefactor is of order unity: everything is controlled by the exponential. This can be understood simply: the amplitude of the evanescent wave is reduced by a factor $e^{-\kappa a}$ in crossing the barrier, and the probability by a factor $e^{-2\kappa a}$.

This exponential dependence makes tunnelling extraordinarily sensitive to the width of the barrier and the mass of the particle. Consider an electron and a barrier exceeding its energy by $V_0-E=1$ eV. We find $1/\kappa\approx0.2$ nm. For a barrier of $1$ nm, $e^{-2\kappa a}\approx4\times10^{-5}$; for a barrier of $2$ nm, it falls to about $10^{-9}$: doubling the width reduces the transmission by several tens of thousands. For a proton, $\kappa$ is multiplied by $\sqrt{m_p/m_e}\approx43$, and the transmission through the $1$ nm barrier becomes of order $e^{-440}$, entirely negligible. Tunnelling is therefore a phenomenon of light particles and atomic distances. It underlies the scanning tunnelling microscope, whose current varies by an order of magnitude for a tip displacement of about one ångström, and it also explains $\alpha$ radioactivity, in which an $\alpha$ particle crosses the Coulomb barrier of the nucleus.

One point of interpretation bears emphasis. It is sometimes said that the particle temporarily “borrows” energy to cross the barrier. This picture is misleading: the Hamiltonian is time-independent, the state is stationary, and the energy is $E$ on both sides of the barrier. Tunnelling is a direct consequence of the wave nature of the wavefunction and the continuity imposed by the matching conditions.

6.4. Above the barrier

For $E>V_0$, the barrier region is classically allowed, with wavenumber $q=\sqrt{2m(E-V_0)}/\hbar$. The solution there is $Ae^{iqx}+Be^{-iqx}$, and the preceding calculation remains valid if $\kappa$ is replaced by $iq$. Since $\sinh(iqa)=i\sin(qa)$ and $\kappa^2=-q^2$, we obtain

The transmission is not equal to one, as it would be classically, but oscillates with energy. It becomes perfect, $T=1$, when $qa=n\pi$ with $n\geq1$, that is, when the width of the barrier contains an integer number of half-wavelengths: the waves reflected from the two edges of the barrier then interfere destructively. This is the same interference mechanism as in an anti-reflection coating, and a further illustration of the rules for adding amplitudes seen in Theme 1. At the threshold $E=V_0$, the solution inside the barrier is affine; taking the limit in either formula, with $\sinh\kappa a\simeq\kappa a$, gives

6.5. An application: flash memory

Tunnelling is not merely a laboratory curiosity: it lies at the heart of the flash memory in our USB drives, phones and solid-state drives. A flash-memory cell contains a region capable of storing electric charge, separated from the rest of the transistor by an insulating layer a few nanometres thick, which acts as a potential barrier for electrons. To write or erase, a control voltage is applied that deforms the barrier and makes it sufficiently probable for electrons to cross by tunnelling. Once the voltage is removed, the barrier returns to its original shape, transmission becomes negligible again, and the charge remains trapped for years, even without a power supply. The stored charge changes the voltage required to make the transistor conduct, allowing the information to be read.

Our square barrier explains the principle: a modest change in the barrier changes the transmission by several orders of magnitude, making both rapid writing and very long retention possible. In the real device, the electric field makes the barrier triangular rather than rectangular, and charge may be stored in an isolated gate or in traps in the material; formula (6) therefore does not directly give the current through a cell. But the quantum mechanism that allows electrons to cross the insulator is indeed the one we have just calculated².

7. Extension to three dimensions

7.1. Separation of variables

In three dimensions, the stationary equation becomes a partial differential equation:

It is generally much more difficult to solve than in one dimension. There is, however, one case that reduces directly to the preceding results: when the potential is a sum of three terms, each depending on only one coordinate, $V=V_x(x)+V_y(y)+V_z(z)$, and the boundary conditions are themselves separable. We then seek solutions in the form of a product

Substitution and division by $uvw$ wherever this product does not vanish gives

The first term depends only on $x$, the second only on $y$ and the third only on $z$. Since their sum is constant for all values of $x$, $y$ and $z$, each of them must be constant; denote these constants by $E_x$, $E_y$ and $E_z$, with $E=E_x+E_y+E_z$. This gives three one-dimensional stationary equations, which we know how to solve. To assert that these products provide a basis for all states, each of the three families of solutions must also be complete, as will be the case in the example below.