Chapter 6: Quantum mechanics IIthe Schrödinger equation and its first solutions

6.1The Schrödinger wave equation

How do we calculate the wave function \(\Psi\)? Which wave equation does it obey? A sinusoidal wave moving in the \(+x\) direction, \(\Psi = A\sin(kx-\omega t+\phi)\), has a more general complex form

where \(A\) may also be complex. Light and sound obey \(\partial^2\Psi/\partial x^2 = v^{-2}\partial^2\Psi/\partial t^2\). For matter waves we require: (1) consistency with conservation of energy, \(E = K + V = p^2/2m + V(x,t)\); (2) consistency with the de Broglie and Einstein relations, \(p = h/\lambda = \hbar k\) and \(E = hf = \hbar\omega\); (3) linearity, so that any sum of two solutions \(\Psi_1\) and \(\Psi_2\) is also a solution, which lets wave functions interfere and form wave packets, and well-behaved (continuous, single-valued) solutions; and (4) we begin with a constant potential \(V(x,t) = V_0\), which exerts no net force. Nonrelativistically, \(E = \hbar^2k^2/2m + V\), so \(k^2 = 2m(E-V)/\hbar^2\). For \(\Psi = Ae^{i(kx-\omega t)}\), \(\partial\Psi/\partial x = ik\Psi\), so

the time-independent Schrödinger equation in one dimension, which we postulate also holds for potentials that vary in space (as experiment confirms). For the time dependence, \(\partial\Psi/\partial t = -i\omega\Psi = -(iE/\hbar)\Psi\), and with \(E\Psi = \frac{\hbar^2k^2}{2m}\Psi + V\Psi = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi\),

the time-dependent Schrödinger equation for a particle moving in a potential \(V\) (which can be \(V(x,t)\)) in one dimension. In three dimensions the second derivative becomes the Laplacian, \(\nabla^2\Psi = \partial^2\Psi/\partial x^2 + \partial^2\Psi/\partial y^2 + \partial^2\Psi/\partial z^2\). Unlike the classical wave equation, it has a first-order time derivative, \(i\hbar\,\partial/\partial t\): a somewhat different phenomenon. Example 6.1: the equation is linear, so superpositions \(a\Psi_1 + b\Psi_2\) of solutions are solutions. Example 6.2: \(Ae^{i(kx-\omega t)}\) satisfies it, given \(E = \hbar\omega = p^2/2m + V\). Example 6.3: \(A\sin(kx-\omega t)\) does not: it solves the classical wave equation, but the first time derivative turns the sine into a cosine.

Normalization and probability. The probability of finding the particle between \(x\) and \(x+dx\) is \(P(x)\,dx = \Psi^*(x,t)\Psi(x,t)\,dx\), so the probability between \(x_1\) and \(x_2\) is \(P = \int_{x_1}^{x_2}\Psi^*\Psi\,dx\). The particle must be somewhere:

Example 6.4: for \(\Psi = Ae^{-\alpha|x|}\), normalization gives \(A = \sqrt\alpha\). The probability of finding the particle between 0 and \(1/\alpha\) is \(\tfrac12(1-e^{-2})\approx0.432\), and between \(1/\alpha\) and \(2/\alpha\) only \(\tfrac12(e^{-2}-e^{-4})\approx0.059\).

A free-particle wave function \(e^{i(kx-\omega t)}\) cannot be normalized: the integral is infinite, because the particle has a finite probability of being anywhere along the \(x\) axis. It has precise \(k\) and \(\omega\), a definite energy and momentum, so by the uncertainty principle \(\Delta x = \infty\) and \(\Delta t = \infty\). Such wave functions are still useful if we restrict the particle to certain positions in space, as in a box or an atom, or combine them into localized wave packets.

Properties of valid wave functions (boundary conditions): (1) \(\Psi\) must be finite everywhere, to avoid infinite probabilities; (2) \(\Psi\) must be single valued; (3) for finite potentials, \(\Psi\) and \(\partial\Psi/\partial x\) must be continuous, because of the second-order derivative (there are exceptions where \(V\) is infinite); (4) to be normalizable, \(\Psi\) must approach zero as \(x\to\pm\infty\). Solutions that fail these do not generally correspond to physically realizable situations.

Separation of variables. Try \(\Psi(x,t) = \psi(x)f(t)\) with a time-independent potential \(V(x)\). Dividing the Schrödinger equation by \(\psi f\),

The left side depends only on \(t\), the right only on \(x\), so both equal a constant \(B\). Then \(i\hbar\,df/f = B\,dt\) gives \(f(t) = e^{-iBt/\hbar}\), and comparing with the free-particle \(e^{-i\omega t}\), \(B = \hbar\omega = E\):

the time-independent Schrödinger equation: \(\psi(x)\) is an eigenfunction of the operator \(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\) with eigenvalue \(E\). Since \(\Psi^*\Psi = |\psi(x)|^2e^{i\omega t}e^{-i\omega t} = |\psi(x)|^2\) is constant in time, such a solution is a stationary state, like a standing wave. Example 6.5: for free electrons in a metal (\(V = 0\)), \(d^2\psi/dx^2 = -k^2\psi\) with \(k^2 = 2mE/\hbar^2\): for \(E>0\), \(\psi = A\sin kx + B\cos kx\); for \(E\lt 0\), \(k\) is imaginary and \(\psi = Ce^{\pm|k|x}\).

Classical vs quantum mechanics. Newton's second law and Schrödinger's equation are both differential equations postulated to explain observed behavior, and experiments show both are successful; Newton's law can be derived from Schrödinger's equation, so the latter is more fundamental. Classical mechanics only appears more precise because it deals with macroscopic phenomena whose underlying uncertainties are too small to be significant, much as ray optics is a good approximation when wavelengths are small compared to apertures but wave optics is needed for interference. Classical mechanics is a good macroscopic approximation; as far as we know, there is only one correct theory, and that is quantum mechanics.

6.2Expectation values

The wave equation must determine measurable quantities, including position, momentum, and energy. In quantum mechanics we compute the expectation value: the expected average of many measurements of a given quantity, such as \(\langle x\rangle\). A physical observable must be real. For \(N_i\) particles at positions \(x_i\), the average is \(\bar x = \sum_iN_ix_i/\sum_iN_i\), and in the continuous version \(\bar x = \int xP(x)\,dx/\int P(x)\,dx\). In quantum mechanics \(P(x)\,dx = \Psi^*\Psi\,dx\), so for a normalized \(\Psi\)

(Without normalization, divide by \(\int\Psi^*\Psi\,dx\).) To find \(\langle p\rangle\) we must represent \(p\) in terms of \(x\) and \(t\). For the free particle, \(\partial\Psi/\partial x = ik\Psi = i(p/\hbar)\Psi\), so \(p\,\Psi = -i\hbar\,\partial\Psi/\partial x\). An operator transforms one function into another, \(\hat Qf(x) = g(x)\); \(-i\hbar\,\partial/\partial x\) acting on \(\Psi\) gives \(p\Psi\), so we call it the momentum operator, and similarly \(\partial\Psi/\partial t = -i\omega\Psi = -(iE/\hbar)\Psi\) gives the energy operator:

The operator must sit between \(\Psi^*\) and \(\Psi\) so that it acts on \(\Psi\), in the order shown; thus \(\langle p\rangle = -i\hbar\int\Psi^*\,\partial\Psi/\partial x\,dx\) and \(\langle E\rangle = i\hbar\int\Psi^*\,\partial\Psi/\partial t\,dx\). Although found for free particles, these operators are general results. Example 6.6: writing the energy \(E = p^2/2m + V\) with operators, \(\hat E\Psi = i\hbar\,\partial\Psi/\partial t\) and \(\big[\frac{1}{2m}\hat p^2 + V\big]\Psi = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} + V\Psi\) (where \(\hat p^2\) means two successive applications of \(\hat p\)), reproducing the Schrödinger equation: a check of the consistency of the definitions.

6.3Infinite square-well potential

The simplest system is the particle in a box of Section 5.8, now with the full quantum-mechanical solution. The potential is \(V = \infty\) for \(x\le0\) and \(x\ge L\), and \(V = 0\) for \(0\lt x\lt L\): an infinite square well. The particle moves freely (no force) between \(x = 0\) and \(L\), and the boundary conditions lead to energy quantization. Outside, \(\psi = 0\) is the only possible solution (zero probability of being there). Inside, \(d^2\psi/dx^2 = -k^2\psi\) with \(k = \sqrt{2mE}/\hbar\), solved by \(\psi = A\sin kx + B\cos kx\). Continuity at \(x = 0\) requires \(B = 0\), and at \(x = L\), \(A\sin kL = 0\), so \(k_nL = n\pi\). (\(n = 0\) gives \(\psi = 0\), physically uninteresting, and negative \(n\) give no new solutions.) Hence

the possible energies are quantized, the integer \(n\) is a quantum number, and the ground state is \(E_1 = \pi^2\hbar^2/2mL^2\), as in Section 5.8. (\(d\psi/dx\) is not continuous at the walls, allowed because \(V = \infty\) there.) Normalizing, \(A^2\int_0^L\sin^2(n\pi x/L)\,dx = A^2L/2 = 1\), so

identical to the modes of a vibrating string with fixed ends. The lowest energy cannot be zero because \(n = 0\) is ruled out; classically the particle could have zero or any positive energy. For a macroscopic object (a tennis ball in a court) \(E_1\) is tiny and adjacent levels are too close to distinguish: macroscopic objects have very large \(n\). Classically the probability density is uniform, \(P = 1/L\). The quantum \(P_n = (2/L)\sin^2(k_nx)\) oscillates, but for large \(n\) its average over the many oscillations is \((2/L)\times\tfrac12 = 1/L\), as the correspondence principle requires. Example 6.7: \(\Psi_n = \sqrt{2/L}\sin(k_nx)e^{-i\omega_nt} = \sqrt{2/L}\,\frac{e^{i(k_nx-\omega_nt)} - e^{-i(k_nx+\omega_nt)}}{2i}\), a standing wave made of waves traveling right and left.

Example 6.8 (first excited state, \(n = 2\)): \(\langle x\rangle = L/2\) (the middle of the box, even though \(|\psi_2|^2 = 0\) there); \(\langle x^2\rangle = 0.32L^2\) (so \(\sqrt{\langle x^2\rangle} = 0.57L\)); \(\langle p\rangle = 0\), because the particle moves left as often as right; and \(\langle p^2\rangle = 4\pi^2\hbar^2/L^2\), so \(\langle p^2\rangle/2m = E_2\), consistent with \(E = p^2/2m\) for \(V = 0\). Example 6.9: a proton in a nucleus of diameter \(10^{-14}\) m, treated as an infinite well, has \(E_1 = \pi^2(\hbar c)^2/2mc^2L^2 = 2.0\) MeV and \(E_2 = 4E_1 = 8\) MeV, a transition energy of 6 MeV, reasonable for nuclear protons. For an electron the energies would be \(\sim10^4\) MeV (relativistic corrections reduce this but leave it far above the energies of electrons emitted in beta decay), another sign that electrons do not exist inside the nucleus.

6.4Finite square-well potential

Now let \(V = V_0\) for \(x\le0\) (region I) and \(x\ge L\) (region III), and \(V = 0\) for \(0\lt x\lt L\) (region II), and consider a particle with \(E\lt V_0\), classically bound inside the well. Quantum mechanics allows the particle to be outside. In regions I and III, \(-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V_0\psi = E\psi\), i.e.

with exponential solutions \(e^{\alpha x}\) and \(e^{-\alpha x}\). The growing exponential is rejected in each region because it diverges, leaving \(\psi_{\rm I} = Ae^{\alpha x}\) (\(x\lt 0\)) and \(\psi_{\rm III} = Be^{-\alpha x}\) (\(x>L\)), both nonzero. Inside, \(d^2\psi/dx^2 = -k^2\psi\) with \(k = \sqrt{2mE}/\hbar\), so \(\psi_{\rm II} = Ce^{ikx} + De^{-ikx}\) (equivalently, sines and cosines). Matching \(\psi\) and \(d\psi/dx\) at \(x = 0\) and \(x = L\) is tedious but gives quantized energies, illustrated graphically below. There is a finite probability of being outside the well, falling to zero exponentially; and since the wave function spreads beyond the walls, the de Broglie wavelength is larger than for the infinite well, so the momentum and energy levels are lower. The probability of penetrating a distance greater than \(\delta x\approx1/\alpha\) begins to decrease markedly; this penetration depth is

important for the tunneling of electrons in semiconductors and for nuclear α decay. That the classically forbidden penetration is proportional to \(\hbar\) is consistent with the uncertainty principle: to be in the barrier region the particle needs a large energy uncertainty \(\Delta E\), which can last only a short time \(\Delta t\ge\hbar/2\Delta E\).

6.5Three-dimensional infinite-potential well

In three dimensions, \(\psi = \psi(x,y,z)\), \(E = K + V = p^2/2m + V\) with \(p^2 = p_x^2+p_y^2+p_z^2\), and the momentum operators \(\hat p_x = -i\hbar\,\partial/\partial x\), and so on, give

Example 6.10: a free particle in a box with sides \(L_1, L_2, L_3\). Trying \(\psi = A\sin(k_1x)\sin(k_2y)\sin(k_3z)\), the boundary conditions give \(k_i = n_i\pi/L_i\), and substitution gives \(E = \frac{\hbar^2}{2m}(k_1^2+k_2^2+k_3^2)\):

Not surprisingly, three dimensions need three quantum numbers. For a cube of side \(L\), \(E = \frac{\pi^2\hbar^2}{2mL^2}(n_1^2+n_2^2+n_3^2)\); the ground state (1,1,1) has \(E = 3\pi^2\hbar^2/2mL^2\), and the first excited state, (1,1,2), (1,2,1), or (2,1,1), has \(E = 6\pi^2\hbar^2/2mL^2\) for all three. A state is degenerate when more than one wave function has the same energy, here a result of the cube's symmetry. If the sides have three different lengths, the degeneracy is removed. Degeneracy also occurs classically (planetary orbits of different eccentricities with the same energy), and a perturbation of the potential, such as an external magnetic field (Zeeman effect) or electric field (Stark effect), can split degenerate levels.

6.6Simple harmonic oscillator

A spring exerts the restoring force \(F = -\kappa(x-x_0)\), with potential energy \(V = \tfrac12\kappa(x-x_0)^2 = -\int_{x_0}^xF\,dx\). Why simple harmonic motion? Because it is so common in nature: any analytical potential with a minimum, such as that of a diatomic molecule, is approximately harmonic near the minimum. Expanding about a stable equilibrium point \(x_0\),

and setting \(V_0 = 0\) leaves \(V = \tfrac12V_2(x-x_0)^2\), the origin of \(V = \tfrac12\kappa x^2\) that occurs so often. Inserting \(V = \kappa x^2/2\) (with \(x_0 = 0\)) into the time-independent equation,

with \(\alpha^2 = m\kappa/\hbar^2\) and \(\beta = 2mE/\hbar^2\). The particle is confined to the potential well, so \(\psi\to0\) as \(x\to\pm\infty\). What is the lowest possible energy? \(E = 0\) would require \(x = 0\) and \(p = 0\) simultaneously, violating the uncertainty principle, so all levels are positive. The classical turning points \(\pm a\), where \(E_0 = \kappa a^2/2\), define the classical limits of motion, but the wave function extends a little beyond them. The solutions are

where \(H_n\) are the Hermite polynomials (\(H_0 = 1\), \(H_1 = 2\xi\), \(H_2 = 4\xi^2-2\), …, \(H_n(\xi) = (-1)^ne^{\xi^2}\frac{d^n}{d\xi^n}e^{-\xi^2}\)) of order \(n\), up to normalization, and \(\omega = \sqrt{\kappa/m}\) is the classical angular frequency. The energy levels are equally spaced. In contrast with the particle in a box, the oscillations come from the polynomial, which dominates at small \(|x|\), and the tails from the Gaussian, which dominates at large \(|x|\). For \(n = 0\), \(E_0 = \tfrac12\hbar\omega\), the zero-point energy or ground-state energy, and \(\psi_0 = (\alpha/\pi)^{1/4}e^{-\alpha x^2/2}\) is the Gaussian wave packet: the Heisenberg limit. Example 6.11 estimates the minimum energy from the uncertainty principle: with \(\langle K\rangle = \langle V\rangle = E/2\), \(\Delta x\Delta p\ge\hbar/2\) gives exactly \(E_0 = \hbar\omega/2\). The zero-point energy is no mere oddity: it keeps \(^4\)He from freezing at atmospheric pressure, however cold.

Example 6.12: normalizing \(\psi_0 = Ae^{-\alpha x^2/2}\) gives \(A = (\alpha/\pi)^{1/4}\); \(\langle x\rangle = 0\) by symmetry, and \(\langle x^2\rangle = 1/2\alpha = \hbar/2m\omega\), consistent with Example 6.11, so \(\langle V\rangle = \tfrac12\kappa\langle x^2\rangle = \hbar\omega/4 = \tfrac12E_0\). Similarly \(\langle p^2\rangle = -\hbar^2\int\psi_0\,\frac{d^2\psi_0}{dx^2}\,dx = \hbar^2\alpha/2\), so \(\langle K\rangle = \langle p^2\rangle/2m = \hbar\omega/4 = \tfrac12E_0\), and \(\langle V\rangle + \langle K\rangle = E_0 = \tfrac12\hbar\omega\).

Classically, a mass on a spring moves fastest through equilibrium and slowest (momentarily at rest) at the ends, so it is most likely found near the ends. The quantum ground state is the opposite, most likely at the center. By the correspondence principle the two should agree for large \(n\), and by \(n = 10\) the average of the quantum probability already resembles the classical one.

6.7Barriers and tunneling

A particle of energy \(E\) approaches a barrier of height \(V_0\) and width \(L\): \(V = 0\) in region I (\(x\lt 0\)) and region III (\(x>L\)), \(V = V_0\) in region II. When \(E>V_0\), the wave numbers are \(k_{\rm I} = k_{\rm III} = \sqrt{2mE}/\hbar\) and \(k_{\rm II} = \sqrt{2m(E-V_0)}/\hbar\). As in optics, where light entering glass changes wavelength and is partly reflected and partly transmitted, the wave function consists of incident, reflected, and transmitted waves, fixed by the boundary conditions; the difference from classical wave theory is that the wave function lets us compute only probabilities. Classically there is no reflection if \(E>V_0\) and total reflection if \(E\lt V_0\); quantum mechanics predicts almost total transmission for \(E\gg V_0\), almost complete reflection for \(E\ll V_0\), and unusual nonclassical behavior when \(E\) is comparable to \(V_0\). With particles coming only from the left,

the incident (\(Ae^{ik_{\rm I}x}\)), reflected (\(Be^{-ik_{\rm I}x}\)), and transmitted (\(Fe^{ik_{\rm I}x}\)) waves. The reflection and transmission probabilities are

and the boundary conditions give

The transmission probability is 1 when \(k_{\rm II}L = n\pi\): the waves reflected at \(x = 0\) and \(x = L\) are then precisely out of phase and cancel. A similar effect occurs for a potential well, where the particle's wavelength is shorter inside; when the well width matches half-integral or integral numbers of wavelengths, reflections can cancel, giving almost pure transmission or reflection for certain wavelengths.

When \(E\lt V_0\), define \(\kappa = \sqrt{2m(V_0-E)}/\hbar\). Now \(\psi_{\rm II} = Ce^{\kappa x} + De^{-\kappa x}\), and quantum mechanics allows \(\psi_{\rm III}\ne0\): the particle can actually be on the other side of the barrier. This is tunneling, with

an exponential decay with characteristic length \(\sim\kappa^{-1}\), where \(\kappa\propto\sqrt{V_0-E}\) (the height) and \(L\) is the width. The uncertainty principle helps: inside the barrier \(|\psi_{\rm II}|^2\sim e^{-2\kappa x}\), so over \(\Delta x = \kappa^{-1}\) the probability drops markedly (\(e^{-2} = 0.14\)), and \(\Delta p\ge\hbar/\Delta x = \hbar\kappa\) gives \(K_{\min} = (\Delta p)^2/2m = \hbar^2\kappa^2/2m = V_0-E\): exactly the "negative" kinetic energy required. The particle is allowed by quantum mechanics and the uncertainty principle to penetrate the classically forbidden region. The optical analog is frustrated total internal reflection: light totally reflected inside a prism can pass into a second prism brought very close, with an intensity that falls exponentially with the gap.

Example 6.14: electrons accelerated to 5 eV meet a barrier 0.8 nm wide and 10 eV high. Then \(\kappa = \sqrt{2mc^2(V_0-E)}/\hbar c = 1.15\times10^{10}\ \text{m}^{-1}\) and \(\kappa L = 9.2\), so the approximation gives \(T = 16(0.5)(0.5)e^{-18.4} = 4.1\times10^{-8}\), and the exact formula also gives \(4.1\times10^{-8}\). (With \(\kappa L\) unrounded, 9.16, both give \(4.4\times10^{-8}\), the value the demo shows.) Example 6.15: writing \(T = Me^{-2\kappa L}\) with \(M = 16(E/V_0)(1-E/V_0)\), at most 4 (for \(E/V_0 = 0.5\)) and about 1.4 for \(E/V_0 = 0.1\), a typical \(M\approx2\) gives the order-of-magnitude estimate \(T\approx2e^{-2\kappa L}\) (\(2\times10^{-8}\) here). Example 6.16: for a step barrier, the probability of penetrating a distance \(\ell\) falls to \(1/e\) where \(2\kappa\ell = 1\), so \(\ell = 1/2\kappa = \hbar/2\sqrt{2m(V_0-E)}\): for a 5 eV electron at a 10 eV step, 0.044 nm. Electrons do not penetrate very far into the classically forbidden region.

Alpha decay and other applications

Many nuclei heavier than lead emit α particles, with decay rates spanning a factor of \(10^{13}\) while their energies range only from 4 to 8 MeV. Inside the nucleus the α particle feels the short-range attractive nuclear force and the repulsive Coulomb force; outside, only the Coulomb force. Its kinetic energy (\(\sim5\) MeV) is far below the Coulomb barrier, so classically it is trapped, but quantum mechanically it can tunnel through, and small changes in the barrier's height and width change the rate enormously because of the exponential \(e^{-2\kappa L}\) (Conceptual Example 6.13). Example 6.17: for \(^{238}\)U, a nuclear radius of 7 fm and a 4.2 MeV α particle give a Coulomb barrier of 38 MeV and an outer turning point at 63 fm. A square barrier of the full height and width gives \(\kappa L\approx140\) and \(T\approx4\times10^{-121}\); a more realistic barrier of half height (19 MeV) and width (28 fm) gives \(T\approx1.5\times10^{-41}\). The α particle moves at \(0.047c\), crossing the nucleus about \(10^{21}\) times per second, so tunneling takes about \(10^{20}\) s, compared with the measured half-life of \(4.5\times10^9\) y \(\approx10^{17}\) s: not bad for such a rough estimate.

In a tunnel diode, discovered by Leo Esaki (Nobel Prize 1973), electrons tunnel through a junction only about 10 nm wide much faster than they could diffuse, and the applied bias controls the barrier, making a very fast switch. The scanning tunneling microscope (STM) of Binnig and Rohrer (Nobel Prize 1986) holds a sharp tip an atom's diameter above a surface; electrons tunnel across the gap, and the current changes by a factor of about \(10^4\) when the gap changes by only 0.4 nm. A feedback system keeps the current constant by moving the tip up and down, so the tip traces the atomic contours of the surface. The atomic force microscope (AFM) works similarly with the tiny force between tip and atoms, and so also images insulators and biological samples.

This completes Chapter 6. Chapter 7 applies the Schrödinger equation to the hydrogen atom.

Further reading

• S. T. Thornton, A. Rex, and C. E. Hood, Modern Physics for Scientists and Engineers, 5th ed. (Cengage Learning), Chapter 6.
• E. Schrödinger, "Quantisierung als Eigenwertproblem," Ann. Phys. 79, 361 (1926).
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed. (Cambridge University Press, 2018), Chapter 2.
• G. Gamow, "Zur Quantentheorie des Atomkernes," Z. Phys. 51, 204 (1928).
• G. Binnig and H. Rohrer, "Scanning tunneling microscopy," Helv. Phys. Acta 55, 726 (1982).