One-Dimensional Wave Mechanics

One-dimensional wave mechanics¶

Learning objectives¶

After this chapter, you should be able to:

• solve the time-independent Schrödinger equation in regions of constant potential;

• impose boundary and matching conditions;

• distinguish bound states from scattering states;

• calculate spectra and wavefunctions for the infinite square well;

• use probability current to interpret reflection and transmission;

• explain tunneling through a classically forbidden region; and

• describe the harmonic oscillator with ladder operators.

Sections 7.1–7.6 form a core progression from boundary-value problems to probability flow and tunneling. The above-barrier resonance subsection is an optional extension. Section 7.7 introduces a second solution strategy—the oscillator’s algebraic ladder method—and can be taught after the well and current sections without requiring the resonance extension.

7.1 Schrödinger’s equation in position space¶

Chapter 6 developed the position representation in the abstract: operators become differential expressions, and a ket becomes a wavefunction. This chapter puts that machinery to work on the simplest possible setting, a single spatial dimension, where boundary conditions and matching conditions turn into concrete, solvable calculations. For one nonrelativistic particle of mass $m$ in a potential $V(x,t)$,

If $V$ is time independent, a separated solution $\psi(x,t)=u(x)e^{-iEt/\hbar}$ obeys

This is an eigenvalue equation for the Hamiltonian. A normalizable energy eigenfunction changes only by an overall phase, so its probability density is stationary.

The local behavior of a solution depends entirely on the sign of $E-V$:

• if $E>V$, solutions oscillate with $k=\sqrt{2m(E-V)}/\hbar$;

• if $E<V$, solutions grow or decay exponentially with $\kappa=\sqrt{2m(V-E)}/\hbar$; and

• at abrupt finite changes in $V$, both $u$ and $du/dx$ are continuous.

The last condition follows by integrating the differential equation across an arbitrarily narrow interval: a finite jump in $V$ cannot produce a $\delta$-function in $u''$, so $u'$ cannot jump either. Infinite potentials are the one exception, and it is worth being precise about why. An infinite wall forces $u=0$ at the wall, but $u'$ is allowed to jump there, because the argument just given assumed a finite jump in $V$. An infinitely deep well is really the limit of an ever-taller finite well, not a literal potential one can integrate across.

Physically, equation (2) is nothing more than the same eigenvalue problem introduced abstractly in Chapter 3, specialized to the position representation of Chapter 6. $\hat H=\hat p^2/2m+V(\hat x)$ acting on $|\psi\rangle$ becomes a second-order differential operator acting on $u(x)=\langle x|\psi\rangle$. Nothing new is being postulated here—only the concrete form that an old postulate takes once a continuous basis is chosen.

7.2 The infinite square well¶

This is the simplest bound-state problem in quantum mechanics, and it is worth solving completely: every later matching calculation in this chapter repeats the same basic steps on a harder potential. Let $V=0$ for $0<x<L$ and $V=\infty$ elsewhere. The wavefunction vanishes at $x=0$ and $x=L$. Inside,

The first boundary gives $B=0$; the second requires $\sin kL=0$. Thus

and

Boundary conditions have quantized the energy. There is no $n=0$ state: it would be the zero function and could not be normalized. The nonzero ground-state energy is required by confinement and the uncertainty principle—a state squeezed into a region of size $L$ cannot also have $p=0$, so it cannot have $E=0$.

The infinite well is the simplest model of confinement, and its qualitative lessons reappear whenever a particle is trapped: quantum dots, particles in optical lattices, and nucleons in a schematic nuclear well are all discussed using the same vocabulary of quantized levels and a nonzero ground-state energy. One more pattern is worth noting now, because it will recur for every bound-state problem in this chapter: the integer $n-1$ counts the number of interior nodes of $u_n$. More nodes means more curvature, and more curvature means higher energy, since curvature is exactly what converts potential energy into kinetic energy in equation (2).

Example 7.1: a superposition in the well¶

Suppose

Then

The energy probabilities stay $1/2$, but the position density contains an interference term proportional to $\cos[(E_2-E_1)t/\hbar]$. A superposition of stationary states need not itself have a stationary density.

Concept check 7.1¶

Why is $\psi(x,t)=u_n(x)e^{-iE_nt/\hbar}$ called stationary if it depends on time?

7.3 General bound states¶

A bound state is normalizable and localized. For potentials approaching constants at infinity, bound-state energies lie below the asymptotic potential, so the wavefunction decays exponentially outside the classically allowed region.

Several qualitative facts are useful before solving anything further:

• one-dimensional bound-state energies are nondegenerate;

• the ground state has no interior nodes;

• higher states have successively more nodes; and

• for an even potential, eigenstates can be chosen with even or odd parity.

The nondegeneracy claim deserves a moment’s justification, and it is special to one dimension specifically. Two independent solutions of a second-order ordinary differential equation at the same $E$ would be related by their Wronskian. Normalizability at both infinities forces that Wronskian to vanish, which in turn forces the two solutions to be proportional to one another—so there is really only one independent bound state at that energy. Chapter 8 will show that this argument fails in three dimensions, where rotational symmetry routinely produces several independent states at one energy.

The finite square well¶

For a finite square well, oscillatory interior solutions must be matched to decaying exterior solutions. The matching conditions produce transcendental equations rather than the simple integer rule of the infinite well. The finite well has only finitely many bound states, and each leaks into the classically forbidden region.

Let $V(x)=-V_0$ for $|x|<L/2$ and $V(x)=0$ for $|x|>L/2$, with $-V_0<E<0$. Inside, the wavefunction oscillates with $k=\sqrt{2m(E+V_0)}/\hbar$; outside, it decays with $\kappa=\sqrt{-2mE}/\hbar$. Because the well is symmetric about $x=0$, every bound state can be chosen even or odd, and it is enough to match at $x=L/2$ and use the symmetry to fix the other boundary automatically. Matching $u$ and $u'$ gives

Neither equation can be solved in closed form, but both are constrained by energy conservation, $k^2+\kappa^2=2mV_0/\hbar^2$, which is independent of $E$. In terms of the dimensionless variables $\xi=kL/2$ and $\eta=\kappa L/2$, this constraint is the quarter circle

and the bound states are exactly the intersections of this circle with the curves $\eta=\xi\tan\xi$ (even) and $\eta=-\xi\cot\xi$ (odd) in the first quadrant. The dimensionless number $z_0$ measures the well’s “strength”: how many oscillations the interior wavefunction can complete before decay must take over.

This picture, without any further algebra, already answers the two questions that matter most:

• A one-dimensional well always binds at least one state. As $\xi\to0$, $\xi\tan\xi\to0$, so the even branch always starts at the origin and rises to meet the circle of any radius $z_0>0$. However shallow or narrow the well, there is always exactly one even, nodeless bound state. This is a genuinely one-dimensional fact; shallow three-dimensional wells can fail to bind anything at all.

• New states appear at regular intervals of $z_0$. The tangent branch diverges at $\xi=\pi/2,3\pi/2,\ldots$ and the cotangent branch changes sign there, so a new branch of the even curve enters the first quadrant every time $z_0$ passes a multiple of $\pi$, and a new branch of the odd curve enters every time $z_0$ passes an odd multiple of $\pi/2$. The total bound-state count, for $z_0>0$, is therefore

  $$N=\left\lceil\frac{2z_0}{\pi}\right\rceil.$$

  (11)

  At an exact threshold the prospective new solution has $E=0$ and is not normalizable, so it does not yet count as a bound state; this is why the ceiling formula, rather than “floor plus one,” handles threshold values correctly. Away from exact thresholds the two forms agree. The states alternate even, odd, even, odd, $\ldots$ as $z_0$ grows. Only once $z_0\gg1$ do the lowest levels approach the $n^2$-spaced infinite-well energies of equation (5); the topmost bound state, barely below $E=0$, always looks qualitatively different—loosely bound with a long decaying tail.

Example 7.2: counting bound states¶

Take a well with $z_0=4$. Since $\pi/2\approx1.57$, $\pi\approx3.14$, and $3\pi/2\approx4.71$, the value $z_0=4$ has passed the first two thresholds but not the third: one even branch exists from $z_0>0$, a second even branch appears once $z_0>\pi$, and one odd branch appears once $z_0>\pi/2$, while the second odd branch (threshold $3\pi/2$) has not yet appeared. Equation (11) confirms this directly: $N=\lceil 8/\pi\rceil=\lceil2.55\rceil=3$. This well supports exactly three bound states: two even and one odd.

Concept check 7.2¶

Does a shallow, narrow finite well always fail to bind a particle if $V_0$ is small enough?

7.4 Probability current¶

Scattering problems ask how probability flows, not merely how large a wave-amplitude coefficient is. Equation (1) implies the continuity equation

where

For $Ae^{ikx}$, $j=(\hbar k/m)|A|^2$. Reflection and transmission probabilities are therefore ratios of outgoing to incoming current. This velocity weighting is essential when two regions have different wave numbers, as the step problem shows next.

7.5 Scattering from a step¶

Bound states are only half the story: a particle with enough energy to escape to infinity produces a scattering state instead, and the relevant questions become “how much reflects?” and “how much transmits?” rather than “what are the allowed energies?” Consider a particle incident from the left on

For $E>V_0$, write

where $k_1=\sqrt{2mE}/\hbar$ and $k_2=\sqrt{2m(E-V_0)}/\hbar$. Matching $u$ and $u'$ at $x=0$ gives

Even though $E>V_0$, reflection occurs because the wavelength changes abruptly. The amplitude $|t|^2$ is not by itself the transmission probability when the speeds differ. Using the current from Section 7.4 gives

Probability conservation plays the role here that particle-number bookkeeping played in Stern–Gerlach chains.

Concept check 7.3¶

Can a particle reflect from an upward step when $E>V_0$?

Total reflection below the step¶

Now suppose $E<V_0$. Region II no longer oscillates. Boundedness as $x\to\infty$, together with the absence of an exponentially growing incoming term from the right, leaves the purely decaying solution $u_{II}=Ce^{-\kappa x}$, with $\kappa=\sqrt{2m(V_0-E)}/\hbar$. Matching $u$ and $u'$ at $x=0$ as before gives

Both numerator and denominator have the same magnitude $\sqrt{k_1^2+\kappa^2}$, so $|r|=1$ exactly: every particle reflects, exactly as a classical particle would. What is not classical is that $u_{II}$ is nonzero. The wavefunction penetrates into the classically forbidden region with an exponentially decaying tail, even though the probability current carried by a purely real exponential vanishes identically,

since $u_{II}$ and $u_{II}'$ differ only by the real factor $-\kappa$. The forbidden region has nonzero probability density but carries no net probability flow toward $x=+\infty$, consistent with $R=1$. This statement does not assign a trajectory in which an individual particle travels in and then returns.

Writing $r$ in polar form exposes a reflection phase shift with no classical analog. If $\kappa\to\infty$ (an infinitely high step, i.e. an infinite wall), $r\to-1$: the familiar node-forcing hard-wall reflection. For finite $\kappa$, the phase interpolates continuously between this hard-wall value and $r\to+1$ as $E\to V_0^-$—which matches the $E>V_0$ formula for $r$ evaluated at $k_2\to0$, so the two regimes join smoothly at threshold even though their functional forms look unrelated.

Example 7.3: how a soft wall differs from a hard one¶

For $\kappa=k_1$ (a step exactly twice the particle’s kinetic energy), equation (18) gives $r=(ik_1+k_1)/(ik_1-k_1)=(i+1)/(i-1)=-i$. The reflection probability is still $|r|^2=1$, but $r=-i=e^{-i\pi/2}$ differs by a quarter cycle from the $\kappa\to\infty$ hard-wall answer $r=-1=e^{i\pi}$. A detector counting reflected atoms cannot see this phase directly; an interference experiment that recombines the reflected wave with another coherent path can.

Concept check 7.4¶

For $E<V_0$ at a step, is the probability density zero in the classically forbidden region?

7.6 Tunneling through a barrier¶

Section 7.5 showed total reflection when $E<V_0$ for a step that stays high forever. What changes if the classically forbidden region has finite width instead? Let a rectangular barrier have height $V_0>E$ and width $a$. Inside the barrier, solutions are exponential rather than oscillatory, yet matching at both boundaries leaves a nonzero transmitted amplitude. The exact transmission probability is

For a thick or high barrier, $\kappa a\gg1$,

The exponential dependence explains why tunneling is highly sensitive to barrier width and particle mass. Tunneling does not mean the particle borrows energy and violates conservation. An energy measurement in the stationary scattering state still gives $E$; the spatial wavefunction simply extends through a region where a classical particle of that energy could not travel.

Example 7.4: compare two barrier widths¶

In the thick-barrier regime, increasing $a$ by $\Delta a$ multiplies $T$ by $e^{-2\kappa\Delta a}$. If $\kappa\Delta a=1$, the transmission falls by $e^{-2}\approx0.135$. This sensitivity underlies scanning tunneling microscopy and many decay processes.

Optional extension: above-barrier resonant transmission¶

Equation (20) was derived for $E<V_0$, but the same matching calculation applies for $E>V_0$, where the interior solution oscillates instead of decaying, with $k_2=\sqrt{2m(E-V_0)}/\hbar$. Formally replacing $\kappa\to ik_2$—so that $\sinh(\kappa a)\to i\sin(k_2a)$—turns equation (20) into

Unlike the tunneling regime, this transmission is not monotonic in the barrier width. Whenever $k_2a=n\pi$ for an integer $n$—the barrier width equals a whole number of half wavelengths of the interior wave—$\sin(k_2a)=0$ and $T=1$: the barrier becomes completely transparent, exactly as an antireflection coating of the right thickness eliminates reflected light. Away from these resonances, some reflection persists even though $E>V_0$ classically guarantees transmission. The two formulas meet consistently at $E=V_0$, where both $\kappa$ and $k_2$ vanish and a careful limit of either expression gives the finite value $T=[1+mV_0a^2/(2\hbar^2)]^{-1}$ rather than an ambiguous $0/0$.

Concept check 7.5¶

If a barrier is thick enough that $\kappa a\gg1$, can it ever be perfectly transparent?

7.7 The harmonic oscillator¶

Every bound state so far was found by directly solving a differential equation and matching boundary conditions—a fresh calculation for each new energy level. The harmonic oscillator admits a much slicker, purely algebraic route that sidesteps this repetition entirely. For

define

They satisfy $[\hat a,\hat a^\dagger]=1$, and

The spectrum does not follow merely by naming these operators, so here is the essential argument. Define $\hat N=\hat a^\dagger\hat a$. For every normalized state,

so its eigenvalues cannot be negative. The commutator above gives

Therefore $\hat a^\dagger$ raises any energy eigenvalue by $\hbar\omega$, while $\hat a$ lowers it by the same amount. Starting from a number eigenstate with eigenvalue $\nu$, repeated lowering would produce $\nu-1,\nu-2,\ldots$. Because number eigenvalues cannot be negative, this sequence must terminate at a state annihilated by $\hat a$. For that state, $\|\hat a|0\rangle\|^2=\langle0|\hat N|0\rangle=0$, so its number eigenvalue is zero. The original $\nu$ must therefore have been a nonnegative integer. Repeated raising gives every $n=0,1,2,\ldots$, and the norms of the raised and lowered states fix the ladder factors:

with

The ground state is defined by $\hat a|0\rangle=0$. In position space this first-order equation gives a Gaussian, as Example 7.5 works out explicitly. The ladder method obtains the complete spectrum without solving a new differential equation for every level: every excited state is reached from $|0\rangle$ by repeated algebra with $\hat a^\dagger$, rather than by solving a new second-order boundary-value problem for each $n$ as the earlier well problems required.

Example 7.5: the ground state in position space¶

Writing $\hat p=-i\hbar\,d/dx$, the condition $\hat a\,\langle x|0\rangle=0$ becomes the first-order equation

Integrating and normalizing gives

This is a Gaussian of exactly the form studied in Chapter 6 as a minimum-uncertainty wave packet, with width $\sigma_x=\sqrt{\hbar/(2m\omega)}$ and zero mean momentum. It therefore saturates $\Delta x\,\Delta p=\hbar/2$, just like the minimum-uncertainty packets of that chapter. But this particular Gaussian does not spread with time the way a free-particle packet does: it is an exact energy eigenstate, so $|\psi(x,t)|^2=|u_0(x)|^2$ is stationary. Confining harmonic forces, unlike free propagation, can hold a minimum-uncertainty packet in a permanent balance between spreading and restoring force.

Concept check 7.6¶

Why does the oscillator not have $E=0$?

Summary¶

• Stationary wavefunctions solve a Hamiltonian boundary-value problem.

• Confinement produces discrete energies and a nonzero ground-state energy.

• A finite well always binds at least one state, and gains new states at regular intervals of its dimensionless strength $z_0$.

• Scattering probabilities are ratios of probability currents; total reflection below a step can still carry a nonclassical phase shift and a nonzero forbidden-region density.

• Wavefunctions can penetrate forbidden regions and tunnel through finite barriers without violating energy conservation, while the same matching calculation above a barrier predicts resonant, perfectly transparent energies.

• Ladder operators reveal the equally spaced harmonic-oscillator spectrum and generate every excited state algebraically from one Gaussian ground state.

Exercises¶

Exercises 1–12 form the core set on wells, probability flow, tunneling, and the oscillator algebra. Exercises 13–19 deepen boundary matching and below-step reflection. Exercises 20–24 are synthesis or optional-extension problems.

1. Derive equations (4) and (5) from both boundary conditions.

2. Find $\langle x\rangle$ for every infinite-well energy eigenstate by symmetry.

3. For $(u_1+u_2)/\sqrt2$, calculate $|\psi(x,t)|^2$ explicitly and identify its oscillation period.

4. Show that no two distinct normalizable bound states in one dimension can have the same energy by analyzing their Wronskian.

5. Derive $r$ and $t$ for the potential step and verify equation (17).

6. Calculate the current for $Ae^{ikx}+Be^{-ikx}$.

7. Derive the continuity equation from the Schrödinger equation and its complex conjugate.

8. Estimate the change in thick-barrier transmission when the particle mass is multiplied by four.

9. Verify $[\hat a,\hat a^\dagger]=1$ from $[\hat x,\hat p]=i\hbar$.

10. Derive equation (25).

11. Show that $\hat a^\dagger|n\rangle$ has energy $E_n+\hbar\omega$.

12. Example 7.5 derives the ground state $u_0(x)$. Apply $\hat a^\dagger$ to $u_0(x)$ in position space to find $u_1(x)$ explicitly, and verify by direct substitution that it solves equation (2) with $E_1=\tfrac32\hbar\omega$.

Boundary-matching consolidation¶

13. Starting from continuity of $u$ and $u'$ at $x=L/2$, and using the symmetry of the well to relate $u(-L/2)$ to $u(L/2)$, derive the even- parity condition in equation (8).

14. Derive the odd-parity condition in equation (9) following the same steps as Exercise 13.

15. For a well with $z_0=7$, find the total number of bound states and how many are even and how many are odd.

16. Explain, using the $\xi\to0$ limit of $\xi\tan\xi$, why a one-dimensional finite well can never have zero bound states, no matter how small $V_0$ or $L$ is.

17. Derive equation (18) by matching $u$ and $u'$ at $x=0$ for $E<V_0$.

18. Show directly from equation (18) that $|r|=1$ for every $\kappa>0$, and confirm that $r\to-1$ as $\kappa\to\infty$.

19. Show that the probability current $j_{II}$ vanishes identically for $u_{II}=Ce^{-\kappa x}$, for any constant $C$, and explain why this is consistent with $R=1$.

Extensions and synthesis¶

20. Derive equation (22) from equation (20) by the substitution $\kappa\to ik_2$, and find the smallest barrier width $a$ giving $T=1$ for a given $k_2$.

21. Take the limit $E\to V_0$ of equation (20) carefully (expanding $\sinh(\kappa a)\approx\kappa a$) to show that $T(E{=}V_0)=[1+mV_0a^2/(2\hbar^2)]^{-1}$, and check that the same value follows from equation (22) in the limit $k_2\to0$.

22. Using the Gaussian ground state of equation (31), calculate $\Delta x$ and $\Delta p$ explicitly and confirm $\Delta x\,\Delta p=\hbar/2$.

23. Explain, in terms of Ehrenfest’s theorem from Chapter 6, why the mean position of an oscillator energy eigenstate is time independent even though a classical particle of the same energy oscillates back and forth.

24. Compare the ground-state energy $\hbar\omega/2$ of the harmonic oscillator with the ground-state energy of an infinite well of width $L=2\sqrt{\hbar/(m\omega)}$ (a rough width match). Which is larger, and why might that be expected from the shape of the two potentials near their minima?

Selected exercise guidance¶

Use these checks only after making a complete attempt.