Chapter 2

Simple Model Systems: Solving the Schrödinger Equation

2.1Introduction

Chapter 1 set out the postulates. This chapter applies them to the smallest set of problems that can be solved exactly, in one dimension, and it is worth being explicit about why these particular problems and no others.

Each one is a permanent tool rather than an exercise:

the free particle gives the continuum states that every ionization process ends in;
the infinite well gives quantization in its simplest form and the node-counting rule used to check every later solution;
the finite well gives the transcendental matching conditions that recur whenever a potential has more than one region;
the barrier gives tunnelling, and the exponential sensitivity that makes strong-field ionization possible;
the harmonic oscillator gives the ladder-operator method, which Chapter 3 reuses for angular momentum, and describes every molecular vibration in Chapter 7.

All five of these reappear later. The three-dimensional atoms of Chapters 4 and 5 reduce, after separating the angles, to a one-dimensional radial equation of exactly the form solved here.

2.2Stationary States

The time-dependent Schrödinger equation (TDSE) of Chapter 1 is

\begin{matrix} i ℏ \frac{\partial}{\partial t} | Ψ (t) ⟩ = \hat{H} | Ψ (t) ⟩ . \end{matrix}

Theorem 2.2.1Separation of the TDSE

If $\hat{H}$ does not depend on time, then Eq. (2.1) admits solutions of the form

\begin{matrix} | Ψ (t) ⟩ = e^{- i E t / ℏ} | ψ ⟩, \end{matrix}

where $| ψ ⟩$ satisfies the time-independent Schrödinger equation (TISE)

\begin{matrix} \hat{H} | ψ ⟩ = E | ψ ⟩ . \end{matrix}

Try $| Ψ (t) ⟩ = f (t) | ψ ⟩$ with $| ψ ⟩$ time-independent. Substituting into Eq. (2.1),

i ℏ \dot{f} (t) | ψ ⟩ = f (t) \hat{H} | ψ ⟩ .

Taking the inner product with $⟨ ψ |$ and dividing by $f ⟨ ψ | ψ ⟩$ ,

i ℏ \frac{\dot{f}}{f} = \frac{⟨ ψ | \hat{H} | ψ ⟩}{⟨ ψ | ψ ⟩} \equiv E,

a constant, since the left side depends only on $t$ and the right side not at all. Integrating gives $f = e^{- i E t / ℏ}$ , and substituting back gives Eq. (2.3).

Remark 2.2.1Why “stationary”

The state (2.2) does change in time — its phase rotates — but the phase cancels in every probability:

| Ψ (x, t) |^{2} = {| e^{- i E t / ℏ} |}^{2} | ψ (x) |^{2} = | ψ (x) |^{2},

and likewise for the expectation value of any time-independent operator. Nothing observable evolves. Motion requires a superposition of stationary states with different energies, whose relative phases advance at different rates — the mechanism behind wave packets (Sec. 2.3.1) and quantum beats (Chapter 11).

In one dimension, with $\hat{H} = {\hat{p}}^{2} / 2 m + V (x)$ and $\hat{p} = - i ℏ d / d x$ , Eq. (2.3) reads

\begin{matrix} - \frac{ℏ^{2}}{2 m} \frac{d^{2} ψ}{d x^{2}} + V (x) ψ (x) = E ψ (x) \end{matrix}

This is the equation solved five times below. Two general facts about it save work every time.

Theorem 2.2.2Continuity conditions

Wherever $V (x)$ is finite, $ψ$ and $d ψ / d x$ are both continuous. Where $V$ has an infinite jump, $ψ$ remains continuous but $d ψ / d x$ need not.

Rearranging Eq. (2.4), $ψ^{″} = \frac{2 m}{ℏ^{2}} (V - E) ψ$ . If $V$ is finite then $ψ^{″}$ is finite, so $ψ^{'}$ is continuous, and hence $ψ$ is too. If $V$ is infinite the argument fails for $ψ^{'}$ , but $ψ$ must still be continuous for $ψ^{″}$ to be integrable.

Theorem 2.2.3Node counting

The bound states of Eq. (2.4) in one dimension are non-degenerate, can be chosen real, and may be ordered so that the $n$ -th state ( $n = 1, 2, 3, \dots$ ) has exactly $n - 1$ nodes.

Theorem 2.2.3 is quoted rather than proved, but it is used constantly: it is how one confirms that a numerical solution has not skipped a level, and it reappears as the rule $n - l - 1$ for hydrogen in Chapter 4.

2.3The Free Particle

Take $V (x) = 0$ everywhere. Equation (2.4) becomes

\begin{matrix} \frac{d^{2} ψ}{d x^{2}} = - k^{2} ψ, k = \frac{\sqrt{2 m E}}{ℏ}, \end{matrix}

with general solution

\begin{matrix} ψ_{k} (x) = A e^{i k x} + B e^{- i k x} . \end{matrix}

Taking the two terms separately, $e^{i k x}$ moves to the right and $e^{- i k x}$ to the left, each with

\begin{matrix} E = \frac{ℏ^{2} k^{2}}{2 m}, p = ℏ k . \end{matrix}

There is no boundary condition and hence no quantization: every $E > 0$ is allowed. This is the continuum, and it is the first appearance of the object that Chapters 9 to 11 depend on.

Remark 2.3.1Normalization of continuum states

A plane wave is not square-integrable: $\int_{- \infty}^{\infty} | e^{i k x} |^{2} d x = \int d x$ diverges. Two standard repairs exist. One is to confine the particle to a large box of length $L$ and normalize to $1 / \sqrt{L}$ , taking $L \to \infty$ at the end. The other, used throughout the research literature and adopted in Chapter 10, is to normalize to a Dirac delta,

\begin{matrix} ⟨ k | k^{'} ⟩ = δ (k - k^{'}) or ⟨ ε | ε^{'} ⟩ = δ (ε - ε^{'}), \end{matrix}

the second being energy normalization. Which convention is in use changes factors of $2 π$ and densities of states in every formula, so it must always be stated.

2.3.1Wave Packets and Group Velocity

A single plane wave is spread over all space and describes nothing localized. A particle is built as a superposition:

\begin{matrix} Ψ (x, t) = \frac{1}{\sqrt{2 π}} \int_{- \infty}^{\infty} A (k) e^{i (k x - ω (k) t)} d k, ω (k) = \frac{E (k)}{ℏ} = \frac{ℏ k^{2}}{2 m} . \end{matrix}

Expanding $ω (k)$ about the central wavenumber $k_{0}$ , $ω ≃ ω_{0} + (k - k_{0}) d ω / d k$ , the packet moves without changing shape at the group velocity

\begin{matrix} v_{g} = {\frac{d ω}{d k} |}_{k_{0}} = \frac{ℏ k_{0}}{m} = \frac{p_{0}}{m}, \end{matrix}

which is the classical velocity — a first instance of the correspondence principle. Keeping the next term in the expansion makes the packet spread, because different $k$ components travel at different speeds.

2.4The Infinite Square Well

Let

\begin{matrix} V (x) = {\begin{cases} 0, & 0 < x < a, \\ \infty, & otherwise. \end{cases} \end{matrix}

Since $V = \infty$ outside, $ψ$ must vanish there, and by Theorem 2.2.2 $ψ$ is continuous, so

\begin{matrix} ψ (0) = ψ (a) = 0 . \end{matrix}

Inside, Eq. (2.4) is Eq. (2.5) again, with general solution most conveniently written

\begin{matrix} ψ (x) = A \sin k x + B \cos k x . \end{matrix}

Apply the boundary conditions in turn. At $x = 0$ , $ψ (0) = B = 0$ , leaving $ψ = A \sin k x$ . At $x = a$ ,

\begin{matrix} A \sin k a = 0 ⟹ k a = n π, n = 1, 2, 3, \dots \end{matrix}

( $n = 0$ gives $ψ \equiv 0$ , which is not a state; negative $n$ merely changes the sign.) Hence $k_{n} = n π / a$ and, from Eq. (2.5),

\begin{matrix} E_{n} = \frac{ℏ^{2} k_{n}^{2}}{2 m} = \frac{n^{2} π^{2} ℏ^{2}}{2 m a^{2}}, n = 1, 2, 3, \dots \end{matrix}

Normalization fixes $A$ :

\begin{matrix} 1 = \int_{0}^{a} | A |^{2} \sin^{2} (\frac{n π x}{a}) d x = | A |^{2} \frac{a}{2} ⟹ A = \sqrt{\frac{2}{a}}, \end{matrix}

using $\int_{0}^{a} \sin^{2} (n π x / a) d x = a / 2$ . Therefore

\begin{matrix} ψ_{n} (x) = \sqrt{\frac{2}{a}} \sin (\frac{n π x}{a}), 0 < x < a \end{matrix}

These are orthonormal, $\int_{0}^{a} ψ_{m} ψ_{n} d x = δ_{m n}$ , as Chapter 1's general theorem for Hermitian operators requires.

Remark 2.4.1Three lessons

First, confinement quantizes: the discrete spectrum came entirely from the two boundary conditions, not from any postulate. Second, the ground-state energy is not zero, $E_{1} = π^{2} ℏ^{2} / 2 m a^{2}$ ; a confined particle cannot be at rest, which is the uncertainty principle in energy form. Third, $ψ_{n}$ has $n - 1$ nodes inside the well, confirming Theorem 2.2.3.

Example 2.4.1Numbers for an electron and for a molecule

Evaluate $E_{1}$ for an electron confined to $a = 0.1$ nm (an atom) and for a nitrogen molecule confined to $a = 1$ cm (a laboratory box).

Solution.

Use $E_{1} = π^{2} ℏ^{2} / 2 m a^{2}$ . For the electron, with $ℏ = 1.055 \times 10^{- 34}$ J s and $m = 9.11 \times 10^{- 31}$ kg,

E_{1} = \frac{π^{2} (1.055 \times 10^{- 34})^{2}}{2 (9.11 \times 10^{- 31}) (1.0 \times 10^{- 10})^{2}} = 6.0 \times 10^{- 18} J = 38 eV .

Comparable to atomic binding energies — so quantization matters at this scale, as it must.

For N $_{2}$ , $m = 4.65 \times 10^{- 26}$ kg and $a = 10^{- 2}$ m:

E_{1} = \frac{π^{2} (1.055 \times 10^{- 34})^{2}}{2 (4.65 \times 10^{- 26}) (10^{- 2})^{2}} = 1.2 \times 10^{- 38} J = 7.5 \times 10^{- 20} eV .

Compare $k_{B} T \approx 0.025$ eV at room temperature: the level spacing is smaller by eighteen orders of magnitude, so the levels are utterly unresolvable and the motion is classical. The same formula covers both cases; only $m a^{2}$ differs.

2.5The Finite Square Well

Let

\begin{matrix} V (x) = {\begin{cases} - V_{0}, & | x | < a, \\ 0, & | x | > a, \end{cases} V_{0} > 0, \end{matrix}

and seek bound states, $- V_{0} < E < 0$ . Define

\begin{matrix} k = \frac{\sqrt{2 m (E + V_{0})}}{ℏ} (inside), κ = \frac{\sqrt{- 2 m E}}{ℏ} (outside), \end{matrix}

both real and positive.

The solutions are oscillatory inside and exponential outside. Because $V (- x) = V (x)$ , the Hamiltonian commutes with parity, so the eigenstates can be chosen even or odd — which halves the work.

Even solutions

Take

\begin{matrix} ψ (x) = {\begin{cases} C \cos k x, & | x | < a, \\ D e^{- κ | x |}, & | x | > a . \end{cases} \end{matrix}

Matching $ψ$ and $ψ^{'}$ at $x = a$ (Theorem 2.2.2):

\begin{matrix} C \cos k a = D e^{- κ a}, - C k \sin k a = - D κ e^{- κ a} . \end{matrix}

Dividing the second by the first eliminates both constants:

\begin{matrix} k \tan k a = κ \end{matrix}

Odd solutions

Take $ψ = C \sin k x$ inside and $ψ = \pm D e^{- κ | x |}$ outside. The same division gives

\begin{matrix} - k \cot k a = κ \end{matrix}

Solving the conditions

Equations (2.22) and (2.23) are transcendental: they have no closed-form solution. The standard route is graphical. Introduce the dimensionless variables

\begin{matrix} z = k a, z_{0} = \frac{a \sqrt{2 m V_{0}}}{ℏ}, \end{matrix}

and note from Eq. (2.19) that $k^{2} + κ^{2} = 2 m V_{0} / ℏ^{2}$ , so $κ a = \sqrt{z_{0}^{2} - z^{2}}$ . The conditions become

\begin{matrix} \tan z = \frac{\sqrt{z_{0}^{2} - z^{2}}}{z} (even), - \cot z = \frac{\sqrt{z_{0}^{2} - z^{2}}}{z} (odd), \end{matrix}

to be solved for $0 < z < z_{0}$ . Plotting both sides against $z$ and reading off the intersections gives the bound-state energies.

Theorem 2.5.1Number of bound states

The finite well of Eq. (2.18) has

\begin{matrix} N = ⌈ \frac{2 z_{0}}{π} ⌉ \end{matrix}

bound states, alternating even, odd, even, . In particular $N \geq 1$ : a one-dimensional well always binds at least one state, however shallow.

The last statement is worth remembering, and it is special to one dimension — in three dimensions a shallow well may bind nothing at all.

Remark 2.5.1Penetration into the forbidden region

Outside the well $E < V$ , which is classically forbidden, yet Eq. (2.20) gives $ψ \propto e^{- κ | x |}$ , nonzero for all finite $x$ . The particle has a nonzero probability of being found where a classical particle could never be, over a characteristic depth $1 / κ$ . This is the same exponential tail that becomes tunnelling in the next section, and — in Chapter 4 — the reason a low- $l$ electron can reach a nucleus that a centrifugal barrier appears to exclude it from.

2.6The Potential Barrier and Tunnelling

Now invert the well. Let

\begin{matrix} V (x) = {\begin{cases} V_{0}, & 0 < x < L, \\ 0, & otherwise, \end{cases} \end{matrix}

and send a particle in from the left with $0 < E < V_{0}$ — classically unable to pass.

Write the solution in three regions:

\begin{matrix} ψ (x) = {\begin{cases} A e^{i k x} + B e^{- i k x}, & x < 0, \\ F e^{κ x} + G e^{- κ x}, & 0 < x < L, \\ C e^{i k x}, & x > L, \end{cases} \end{matrix}

with $k = \sqrt{2 m E} / ℏ$ and $κ = \sqrt{2 m (V_{0} - E)} / ℏ$ . The absence of a left-moving wave for $x > L$ encodes the physical setup: nothing comes back from the far side.

Matching $ψ$ and $ψ^{'}$ at $x = 0$ and at $x = L$ gives four linear equations in $A, B, F, G, C$ ; eliminating $B$ , $F$ and $G$ leaves the ratio $C / A$ , and the transmission coefficient $T = | C / A |^{2}$ is

\begin{matrix} T = {[1 + \frac{V_{0}^{2} \sinh^{2} (κ L)}{4 E (V_{0} - E)}]}^{- 1} \end{matrix}

For a thick or high barrier, $κ L ≫ 1$ , we may use $\sinh x ≃ \frac{1}{2} e^{x}$ , giving the form that is actually used:

\begin{matrix} T ≃ \frac{16 E (V_{0} - E)}{V_{0}^{2}} e^{- 2 κ L} \end{matrix}

Remark 2.6.1Exponential sensitivity

The dominant factor is $e^{- 2 κ L}$ : the transmission falls exponentially with barrier width and with the square root of the barrier height. That extreme sensitivity is what makes tunnelling both a precise tool and a violent process. It underlies the scanning tunnelling microscope, radioactive $α$ decay, and — the case relevant to this course — strong-field ionization, in which a laser field bends the atomic potential into a barrier of finite width so that the bound electron can escape without absorbing a photon.

Example 2.6.1Tunnelling through a thin barrier

An electron of energy $E = 1$ eV meets a barrier of height $V_{0} = 2$ eV and width $L = 0.5$ nm. Find $T$ , and then find it again for $L = 1.0$ nm.

Solution.

First $κ$ . With $V_{0} - E = 1$ eV $= 1.602 \times 10^{- 19}$ J,

κ = \frac{\sqrt{2 (9.11 \times 10^{- 31}) (1.602 \times 10^{- 19})}}{1.055 \times 10^{- 34}} = 5.12 \times 10^{9} m^{- 1} .

For $L = 0.5$ nm, $κ L = 2.56$ , so $e^{- 2 κ L} = e^{- 5.12} = 6.0 \times 10^{- 3}$ . The prefactor is $16 E (V_{0} - E) / V_{0}^{2} = 16 (1) (1) / 4 = 4$ , so

T ≃ 4 \times 6.0 \times 10^{- 3} = 2.4 \times 10^{- 2} .

About one electron in forty gets through.

Doubling the width to $1.0$ nm gives $κ L = 5.12$ and $e^{- 2 κ L} = e^{- 10.24} = 3.6 \times 10^{- 5}$ , so

T ≃ 1.4 \times 10^{- 4} .

Doubling the width cut the transmission by a factor of $170$ . That is Remark 2.6.1 in numbers, and it is why a tunnelling current measures distance to a fraction of an atomic diameter.

2.7The Harmonic Oscillator

Let

\begin{matrix} V (x) = \frac{1}{2} m ω^{2} x^{2}, so \hat{H} = \frac{{\hat{p}}^{2}}{2 m} + \frac{1}{2} m ω^{2} {\hat{x}}^{2} . \end{matrix}

This system earns its place three times over: it is exactly solvable; any smooth potential near a stable minimum is approximately harmonic, which is why it describes every molecular vibration in Chapter 7; and its solution introduces the ladder-operator method that Chapter 3 reuses for angular momentum.

2.7.1Ladder Operators

Definition 2.7.1Annihilation and creation operators

\begin{matrix} \hat{a} = \sqrt{\frac{m ω}{2 ℏ}} (\hat{x} + \frac{i}{m ω} \hat{p}), {\hat{a}}^{†} = \sqrt{\frac{m ω}{2 ℏ}} (\hat{x} - \frac{i}{m ω} \hat{p}) . \end{matrix}

Theorem 2.7.1Commutator and Hamiltonian

\begin{matrix} [\hat{a}, {\hat{a}}^{†}] = 1, \end{matrix}

and

\begin{matrix} \hat{H} = ℏ ω ({\hat{a}}^{†} \hat{a} + \frac{1}{2}) = ℏ ω (\hat{N} + \frac{1}{2}), \hat{N} \equiv {\hat{a}}^{†} \hat{a} . \end{matrix}

For the commutator, expand using $[\hat{x}, \hat{p}] = i ℏ$ :

[\hat{a}, {\hat{a}}^{†}] = \frac{m ω}{2 ℏ} [\hat{x} + \frac{i \hat{p}}{m ω}, \hat{x} - \frac{i \hat{p}}{m ω}] = \frac{m ω}{2 ℏ} (- \frac{i}{m ω} [\hat{x}, \hat{p}] + \frac{i}{m ω} [\hat{p}, \hat{x}]) .

Both terms equal $ℏ / m ω$ , so the bracket is $2 ℏ / m ω$ and the whole expression is $1$ .

For the Hamiltonian, multiply out:

{\hat{a}}^{†} \hat{a} = \frac{m ω}{2 ℏ} ({\hat{x}}^{2} + \frac{{\hat{p}}^{2}}{m^{2} ω^{2}} + \frac{i}{m ω} [\hat{x}, \hat{p}]) = \frac{m ω {\hat{x}}^{2}}{2 ℏ} + \frac{{\hat{p}}^{2}}{2 ℏ m ω} - \frac{1}{2} .

Multiplying by $ℏ ω$ ,

ℏ ω {\hat{a}}^{†} \hat{a} = \frac{m ω^{2} {\hat{x}}^{2}}{2} + \frac{{\hat{p}}^{2}}{2 m} - \frac{ℏ ω}{2} = \hat{H} - \frac{ℏ ω}{2},

which rearranges to Eq. (2.34).

2.7.2The Spectrum

Let $| n ⟩$ be an eigenket of $\hat{N}$ with $\hat{N} | n ⟩ = n | n ⟩$ . From Eq. (2.33),

\begin{matrix} [\hat{N}, \hat{a}] = - \hat{a}, [\hat{N}, {\hat{a}}^{†}] = + {\hat{a}}^{†}, \end{matrix}

so $\hat{a} | n ⟩$ has eigenvalue $n - 1$ and ${\hat{a}}^{†} | n ⟩$ has $n + 1$ : the operators step down and up the ladder.

The ladder must have a bottom, because

\begin{matrix} n = ⟨ n | {\hat{a}}^{†} \hat{a} | n ⟩ = ‖ \hat{a} | n ⟩ ‖^{2} \geq 0 . \end{matrix}

If $n$ were not a non-negative integer, repeated application of $\hat{a}$ would eventually produce a state with negative eigenvalue, contradicting Eq. (2.36). The only escape is that the chain terminates on a state with $\hat{a} | 0 ⟩ = 0$ , whence $n = 0, 1, 2, \dots$

Theorem 2.7.2Harmonic oscillator spectrum

\begin{matrix} E_{n} = ℏ ω (n + \frac{1}{2}), n = 0, 1, 2, \dots \end{matrix}

with

\begin{matrix} \hat{a} | n ⟩ = \sqrt{n} | n - 1 ⟩, {\hat{a}}^{†} | n ⟩ = \sqrt{n + 1} | n + 1 ⟩, | n ⟩ = \frac{({\hat{a}}^{†})^{n}}{\sqrt{n!}} | 0 ⟩ . \end{matrix}

The normalization constants follow from Eq. (2.36): $∥ \hat{a} | n ⟩ ∥^{2} = n$ gives $\sqrt{n}$ , and $∥ {\hat{a}}^{†} | n ⟩ ∥^{2} = ⟨ n | \hat{a} {\hat{a}}^{†} | n ⟩ = ⟨ n | (\hat{N} + 1) | n ⟩ = n + 1$ gives $\sqrt{n + 1}$ .

Remark 2.7.1Two features to carry forward

The levels are equally spaced by $ℏ ω$ — unique among the systems in this chapter, and the reason a vibrational spectrum shows one dominant line rather than a converging series (Chapter 8). And the ground-state energy is $\frac{1}{2} ℏ ω$ , not zero: the zero-point energy. A chemical bond is never still, even at absolute zero.

2.7.3Position Representation

To obtain the wave functions, write $\hat{a} | 0 ⟩ = 0$ in position space. With $\hat{p} = - i ℏ d / d x$ , Def. 2.7.1 gives

\begin{matrix} (x + \frac{ℏ}{m ω} \frac{d}{d x}) ψ_{0} (x) = 0 ⟹ \frac{d ψ_{0}}{ψ_{0}} = - \frac{m ω}{ℏ} x d x, \end{matrix}

which integrates to a Gaussian. Introducing the natural length scale and the dimensionless coordinate

\begin{matrix} ξ = x \sqrt{\frac{m ω}{ℏ}}, \end{matrix}

and normalizing with $\int e^{- ξ^{2}} d ξ = \sqrt{π}$ ,

\begin{matrix} ψ_{0} (x) = {(\frac{m ω}{π ℏ})}^{1 / 4} e^{- ξ^{2} / 2} . \end{matrix}

Applying ${\hat{a}}^{†}$ repeatedly, as in Eq. (2.38), generates the rest. The result is

\begin{matrix} ψ_{n} (x) = {(\frac{m ω}{π ℏ})}^{1 / 4} \frac{1}{\sqrt{2^{n} n!}} H_{n} (ξ) e^{- ξ^{2} / 2} \end{matrix}

where $H_{n}$ are the Hermite polynomials, given by

\begin{matrix} H_{n} (ξ) = (- 1)^{n} e^{ξ^{2}} {(\frac{d}{d ξ})}^{n} e^{- ξ^{2}}, \end{matrix}

the first few being

\begin{matrix} H_{0} = 1, H_{1} = 2 ξ, H_{2} = 4 ξ^{2} - 2, H_{3} = 8 ξ^{3} - 12 ξ, H_{4} = 16 ξ^{4} - 48 ξ^{2} + 12 . \end{matrix}

$H_{n}$ has $n$ real zeros, so $ψ_{n}$ has $n$ nodes — consistent with Theorem 2.2.3 once the counting is shifted for the label starting at $n = 0$ . The parity is $(- 1)^{n}$ , since $H_{n}$ contains only even or only odd powers.

Example 2.7.1Zero-point energy of a real bond

The H $_{2}$ molecule has a vibrational wavenumber $\tilde{ν} = 4401$ cm $^{- 1}$ . Find $ℏ ω$ in eV and the zero-point energy, and compare with the bond dissociation energy of $4.48$ eV.

Solution.

Convert using $1$ cm $^{- 1} = 1.2398 \times 10^{- 4}$ eV:

ℏ ω = 4401 \times 1.2398 \times 10^{- 4} = 0.546 eV,

so the zero-point energy is

E_{0} = \frac{1}{2} ℏ ω = 0.273 eV .

This is $6 %$ of the dissociation energy — small but not negligible, and it must be included whenever bond strengths are compared. It also explains why the measured dissociation energy of H $_{2}$ differs from that of D $_{2}$ : the heavier isotope has smaller $ω = \sqrt{k / μ}$ and hence a lower zero-point energy, so it sits deeper in the same potential well and is harder to break. The potential curve is identical; only the nuclear mass differs.

2.8Summary

A time-independent $\hat{H}$ gives stationary states $Ψ = e^{- i E t / ℏ} ψ$ (Thm. 2.2.1); nothing observable evolves in one, so motion requires a superposition.
$ψ$ and $ψ^{'}$ are continuous wherever $V$ is finite (Thm. 2.2.2); the $n$ -th bound state has $n - 1$ nodes (Thm. 2.2.3).
Free particle: $ψ = e^{\pm i k x}$ , $E = ℏ^{2} k^{2} / 2 m$ , no quantization — the continuum. It is normalized either in a box or to $δ (ε - ε^{'})$ ; the convention matters. A packet moves at $v_{g} = p_{0} / m$ .
Infinite well: $E_{n} = n^{2} π^{2} ℏ^{2} / 2 m a^{2}$ , $ψ_{n} = \sqrt{2 / a} \sin (n π x / a)$ . Confinement quantizes, and $E_{1} > 0$ .
Finite well: matching gives the transcendental conditions $k \tan k a = κ$ (even) and $- k \cot k a = κ$ (odd), solved graphically in terms of $z = k a$ and $z_{0} = a \sqrt{2 m V_{0}} / ℏ$ . There are $⌈ 2 z_{0} / π ⌉$ bound states, always at least one, and the wave function penetrates the classically forbidden region as $e^{- κ | x |}$ .
Barrier: $T = [1 + V_{0}^{2} \sinh^{2} (κ L) / 4 E (V_{0} - E)]^{- 1}$ , reducing to $T ≃ 16 E (V_{0} - E) V_{0}^{- 2} e^{- 2 κ L}$ for a thick barrier. The exponential sensitivity underlies the tunnelling microscope and strong-field ionization.
Harmonic oscillator: $[\hat{a}, {\hat{a}}^{†}] = 1$ , $\hat{H} = ℏ ω (\hat{N} + \frac{1}{2})$ , $E_{n} = ℏ ω (n + \frac{1}{2})$ , and $ψ_{n} \propto H_{n} (ξ) e^{- ξ^{2} / 2}$ . Equal spacing and a nonzero zero-point energy are its two signature features.

2.9Exercises

The infinite well, start to finish. Derive Eqs. (2.15) and (2.17) yourself, applying both boundary conditions and carrying out the normalization integral. Then verify $\int_{0}^{a} ψ_{1} ψ_{2} d x = 0$ explicitly.
Matching at a step. Derive Eq. (2.22) from Eq. (2.21), and Eq. (2.23) by the same route for the odd solutions. Explain why dividing the two matching equations is legitimate and what it accomplishes.
Counting bound states. A well has $V_{0} = 5$ eV and half-width $a = 0.2$ nm, for an electron.
1. Compute $z_{0}$ from Eq. (2.24).
2. Use Theorem 2.5.1 to predict the number of bound states, and say which are even and which odd.
3. Solve Eq. (2.25) graphically or numerically for the lowest $z$ , and convert it to an energy in eV.
Tunnelling. Repeat Example 2.6.1 for a proton instead of an electron, at the same energy, height and width $L = 0.5$ nm. By what factor does $T$ change, and which quantity in $κ$ is responsible? Comment on why tunnelling is an electronic phenomenon far more often than a nuclear one.
Oscillator algebra.
1. Reproduce the proof of Theorem 2.7.1.
2. Derive Eq. (2.35) from Eq. (2.33).
3. Using Eq. (2.38), compute $⟨ n | {\hat{x}}^{2} | n ⟩$ by writing $\hat{x}$ in terms of $\hat{a}$ and ${\hat{a}}^{†}$ , and hence verify $⟨ V ⟩ = \frac{1}{2} E_{n}$ — the virial theorem for a quadratic potential.
Hermite polynomials. Use Eq. (2.43) to generate $H_{1}$ and $H_{2}$ and check them against Eq. (2.44). Then verify that $ψ_{1}$ from Eq. (2.42) is normalized, using $\int ξ^{2} e^{- ξ^{2}} d ξ = \sqrt{π} / 2$ .

2.10Project: Solving One-Dimensional Problems Numerically

The problem. Build a general numerical solver for Eq. (2.4) and validate it against every exact result in this chapter. This solver is reused in Chapters 3, 4, 5 and 7, so build it to be trusted.

Work in atomic units ( $ℏ = m = 1$ ), where Eq. (2.4) is $- \frac{1}{2} ψ^{″} + V ψ = E ψ$ .

On paper. Discretize on a uniform grid $x_{i} = x_{0} + i h$ using $ψ^{″} (x_{i}) ≃ (ψ_{i + 1} - 2 ψ_{i} + ψ_{i - 1}) / h^{2}$ . Write down the matrix $𝐇$ explicitly, giving its diagonal and off-diagonal entries.
On the computer. Validate against the infinite well: take $V = 0$ on $(0, a)$ with $ψ = 0$ at both ends, and compare the lowest five eigenvalues with Eq. (2.15). Report the relative error and how it scales when you halve $h$ — you should see it fall by about four, since the finite difference is second-order accurate.
On the computer. Validate against the harmonic oscillator: take $V = \frac{1}{2} x^{2}$ on a domain wide enough that $ψ$ has decayed, and confirm $E_{n} = n + \frac{1}{2}$ for $n = 0 \dots 5$ . Plot your $ψ_{0}$ and $ψ_{2}$ against Eqs. (2.42) and (2.44).
On the computer. Now the finite well. Take $V = - V_{0}$ on $| x | < a$ and zero outside, choose $V_{0}$ and $a$ giving $z_{0} \approx 5$ , and find all bound states. Check the count against Theorem 2.5.1, check the parity alternation, and compare each energy with a numerical solution of Eq. (2.25).
On the computer. Plot one bound state on a logarithmic vertical axis and confirm that the tail outside the well is a straight line of slope $- κ$ , with $κ$ from Eq. (2.19). This is Remark 2.5.1 made quantitative.
On the computer. Tunnelling. Rather than eigenvalues, compute $T (E)$ from Eq. (2.29) for a barrier of your choice and plot it against $E$ from $0$ to $2 V_{0}$ on a logarithmic scale. Mark $E = V_{0}$ . Comment on what happens above the barrier, where classically $T$ would be exactly $1$ .
What has to move. One animation: a Gaussian wave packet built as in Eq. (2.9) approaching your barrier, with $| Ψ (x, t) |^{2}$ shown as it splits into a reflected and a transmitted part. Caption it with the fraction that got through, and compare that fraction with Eq. (2.29) evaluated at the packet's central energy.

The check. Part (b) is the one that certifies the solver. If the error does not fall by a factor of four when $h$ is halved, the discretization is wrong — most often a missing factor of $\frac{1}{2}$ or $h^{2}$ in the matrix. Fix it before going further; every later chapter depends on this code.

Be ready to answer. In part (f), transmission above the barrier is not $1$ , and oscillates. Explain the oscillation physically — what is interfering with what — and identify the energies at which $T$ returns exactly to $1$ .