One-Dimensional Problems

5.1 The Infinite Square Well

A particle of mass $m$ in a potential $V(x) = 0$ for $0 \lt x \lt L$ and $V(x) = \infty$ otherwise.

Derivation. Inside the well, the time-independent Schrodinger equation is:

$-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi \implies \frac{d^2\phi}{dx^2} + k^2\phi = 0$

Where $k = \sqrt{2mE}/\hbar$. The general solution is:

$\phi(x) = A\sin(kx) + B\cos(kx)$

Boundary conditions: $\phi(0) = \phi(L) = 0$.

From $\phi(0) = 0$: $B = 0$So $\phi(x) = A\sin(kx)$.

From $\phi(L) = 0$: $\sin(kL) = 0$Which requires $kL = n\pi$ for $n = 1, 2, 3, \ldots$

Therefore $k_n = n\pi/L$ and:

$E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2}$

Normalisation. $\int_0^L |A|^2\sin^2(n\pi x/L)\,dx = |A|^2 L/2 = 1$Giving $A = \sqrt{2/L}$.

Solutions:

$\phi_n(x) = \sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right), \quad E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}, \quad n = 1, 2, 3, \ldots$

Properties:

• The ground state ($n = 1$) has the lowest energy $E_1 > 0$ (zero-point energy).
• Energy levels are not equally spaced; $E_n \propto n^2$.
• There are $(n - 1)$ nodes in the $n$-th eigenstate.

:::caution Common Pitfall The ground state has $n = 1$Not $n = 0$. The solution $n = 0$ gives $\phi(x) = 0$ everywhere, Which is not normalisable. Furthermore, $E_1 > 0$ (zero-point energy) is a direct consequence of The uncertainty principle: confining the particle to a finite region requires kinetic energy. :::

5.2 The Quantum Harmonic Oscillator

$V(x) = \frac{1}{2}m\omega^2 x^2$.

5.2.1 Algebraic Method: Ladder Operators

Define the ladder operators (creation and annihilation operators):

$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right), \quad \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)$

Commutation relation. Using $[\hat{x}, \hat{p}] = i\hbar$:

$[\hat{a}, \hat{a}^\dagger] = \frac{m\omega}{2\hbar}\!\left[\hat{x} + \frac{i\hat{p}}{m\omega},\, \hat{x} - \frac{i\hat{p}}{m\omega}\right] = \frac{1}{2\hbar}(-i)(i\hbar) + \frac{1}{2\hbar}(i)(-i\hbar) = 1$

Inversion. From the definitions:

$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(\hat{a} + \hat{a}^\dagger), \quad \hat{p} = -i\sqrt{\frac{m\omega\hbar}{2}}(\hat{a} - \hat{a}^\dagger)$

Hamiltonian in terms of ladder operators. Substituting into $\hat{H} = \hat{p}^2/(2m) + m\omega^2\hat{x}^2/2$:

$\hat{H} = \hbar\omega\!\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)$

Where we used $\hat{a}\hat{a}^\dagger = [\hat{a}, \hat{a}^\dagger] + \hat{a}^\dagger\hat{a} = 1 + \hat{a}^\dagger\hat{a}$.

Number operator. $\hat{N} = \hat{a}^\dagger\hat{a}$So $\hat{H} = \hbar\omega(\hat{N} + 1/2)$.

Proof that $\hat{a}$ and $\hat{a}^\dagger$ lower and raise the energy. Compute $[\hat{H}, \hat{a}]$:

$[\hat{H}, \hat{a}] = \hbar\omega[\hat{a}^\dagger\hat{a}, \hat{a}] = \hbar\omega(\hat{a}^\dagger[\hat{a}, \hat{a}] + [\hat{a}^\dagger, \hat{a}]\hat{a}) = -\hbar\omega\,\hat{a}$

So $\hat{H}\hat{a}|n\rangle = (E_n - \hbar\omega)\hat{a}|n\rangle$: $\hat{a}$ lowers energy by $\hbar\omega$. Similarly, $[\hat{H}, \hat{a}^\dagger] = +\hbar\omega\,\hat{a}^\dagger$.

Let $|n\rangle$ be an eigenstate with $\hat{H}|n\rangle = E_n|n\rangle$. Then:

$\hat{a}|n\rangle = c_n|n-1\rangle, \quad \hat{a}^\dagger|n\rangle = c_{n+1}|n+1\rangle$

The constants follow from normalisation. Since $\hat{a}^\dagger\hat{a}|n\rangle = n|n\rangle$:

$\|c_n|n-1\rangle\|^2 = \langle n|\hat{a}^\dagger\hat{a}|n\rangle = n$

Therefore:

$\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle, \quad \hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$

Ground state. The lowering process must terminate: $\hat{a}|0\rangle = 0$. This gives the Differential equation:

$\left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\phi_0(x) = 0 \implies \phi_0(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\!\left(-\frac{m\omega x^2}{2\hbar}\right)$

Energy spectrum. $E_n = \hbar\omega(n + 1/2)$ for $n = 0, 1, 2, \ldots$ The zero-point energy $E_0 = \hbar\omega/2 \gt 0$ is a direct consequence of $[\hat{x}, \hat{p}] = i\hbar$.

5.2.2 Analytic Solution

The eigenfunctions involve Hermite polynomials $H_n$:

$\phi_n(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\!\left(\sqrt{\frac{m\omega}{\hbar}}\,x\right) e^{-m\omega x^2/(2\hbar)}$

The first few Hermite polynomials are $H_0(\xi) = 1$, $H_1(\xi) = 2\xi$, $H_2(\xi) = 4\xi^2 - 2$.

Example 5.1. Using the ladder operators, find $\phi_1(x)$ from $\phi_0(x)$.

5.3 The Free Particle

$V(x) = 0$ everywhere. The Schrodinger equation:

$-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi$

Solutions: $\phi_k(x) = \frac{1}{\sqrt{2\pi}} e^{ikx}$ with $E = \frac{\hbar^2 k^2}{2m}$.

The energy spectrum is continuous (all $E \geq 0$). The eigenfunctions are not normalisable (plane Waves); physical states are wave packets constructed by superposition.

5.3.1 Parity

The parity operator $\hat{\Pi}$ reflects the coordinate: $\hat{\Pi}\psi(x) = \psi(-x)$.

Properties:

• $\hat{\Pi}^2 = \hat{I}$So eigenvalues are $\pm 1$.
• Even functions ($\psi(-x) = \psi(x)$) have parity $+1$.
• Odd functions ($\psi(-x) = -\psi(x)$) have parity $-1$.
• If $V(x) = V(-x)$ (symmetric potential), then $[\hat{H}, \hat{\Pi}] = 0$So energy eigenstates can be chosen to have definite parity.

Theorem 5.1. For a symmetric potential $V(x) = V(-x)$The energy eigenstates are either even Or odd.

Proof. Since $[\hat{H}, \hat{\Pi}] = 0$There exists a simultaneous eigenbasis. Let $\hat{H}\phi = E\phi$ and $\hat{\Pi}\phi = \pi\phi$ where $\pi = \pm 1$. Then $\phi(-x) = \pi\phi(x)$ So $\phi$ is either even ($\pi = +1$) or odd ($\pi = -1$). $\blacksquare$

This theorem explains why the infinite square well, harmonic oscillator, and finite square well Eigenstates all have definite parity: their potentials are all symmetric about the origin.

5.3.2 The Virial Theorem

Theorem 5.2 (Virial Theorem). For a stationary state of a Hamiltonian $\hat{H} = \hat{p}^2/(2m) + V(\hat{x})$:

$2\langle T \rangle = \langle x\,V"(x) \rangle$

Where $T$ is the kinetic energy.

Proof. Using Ehrenfest’s theorem for the operator $\hat{G} = \hat{x}\hat{p}$:

$\frac{d}{dt}\langle\hat{x}\hat{p}\rangle = \frac{i}{\hbar}\langle[\hat{H}, \hat{x}\hat{p}]\rangle = 0$

For a stationary state. Computing the commutator:

$[\hat{H}, \hat{x}\hat{p}] = \left[\frac{\hat{p}^2}{2m} + V, \hat{x}\hat{p}\right] = \frac{1}{2m}[\hat{p}^2, \hat{x}]\hat{p} + [\hat{x}\hat{p}, V] + \hat{x}[V, \hat{p}]$

$= \frac{-i\hbar}{m}\hat{p}\hat{p} + \hat{x}[V, \hat{p}] + \hat{x}[V, \hat{p}] = \frac{-i\hbar\hat{p}^2}{m} + 2i\hbar\hat{x}\,V'(x)$

Setting $d\langle\hat{x}\hat{p}\rangle/dt = 0$ and dividing by $i\hbar$:

$-\frac{\langle\hat{p}^2\rangle}{m} + 2\langle\hat{x}\,V'(\hat{x})\rangle = 0$

$-2\langle T \rangle + \langle x\,V'(x) \rangle = 0 \implies 2\langle T \rangle = \langle x\,V'(x) \rangle \qquad \blacksquare$

Applications. For the harmonic oscillator ($V \propto x^2$): $2\langle T \rangle = 2\langle V \rangle$ So $\langle T \rangle = \langle V \rangle = E/2$. For the hydrogen atom ($V \propto -1/r$): $2\langle T \rangle = -\langle V \rangle$So $\langle T \rangle = -E$ and $\langle V \rangle = 2E$.

5.4 The Finite Square Well

Consider $V(x) = -V_0$ for $|x| \lt a$ and $V(x) = 0$ for $|x| \gt a$Where $V_0 \gt 0$.

5.4.1 Bound States ($E \lt 0$)

Define $k = \sqrt{2m(E + V_0)}/\hbar$ (inside) and $\kappa = \sqrt{-2mE}/\hbar$ (outside). Note that $k^2 + \kappa^2 = 2mV_0/\hbar^2$.

Even parity solutions. Inside: $\phi(x) = A\cos(kx)$. Outside: $\phi(x) = Be^{-\kappa|x|}$.

Matching $\phi$ and $\phi'$ at $x = a$ and dividing the two conditions:

$k\tan(ka) = \kappa$

Odd parity solutions. Inside: $\phi(x) = A\sin(kx)$. Outside: $\phi(x) = Be^{-\kappa|x|}$ (with sign For $x \lt 0$). Matching gives:

$-k\cot(ka) = \kappa$

These are transcendental equations solved graphically. Define $z = ka$ and $z_0 = a\sqrt{2mV_0/\hbar^2}$.

The even condition becomes $\tan z = \sqrt{z_0^2/z^2 - 1}$ and the odd condition becomes $-\cot z = \sqrt{z_0^2/z^2 - 1}$. The number of bound states is $N = \lfloor 2z_0/\pi \rfloor + 1$. There is always at least one bound state (the even ground state).

5.4.2 Scattering States ($E \gt 0$)

For $E \gt 0$The particle has enough energy to escape. Define $k_1 = \sqrt{2mE}/\hbar$ (outside) And $k_2 = \sqrt{2m(E + V_0)}/\hbar$ (inside). The solutions are oscillatory everywhere. The Transmission coefficient is:

$T = \frac{1}{1 + \dfrac{V_0^2}{4E(E + V_0)}\sin^2(2k_2 a)}$

Resonances occur when $2k_2 a = n\pi$ (integer multiples of $\pi$), giving $T = 1$: the well Becomes perfectly transparent.

Example 5.3. A finite square well has $V_0 = 5\,\mathrm{eV}$ and $2a = 1\,\mathrm{nm}$. Estimate the Number of bound states for an electron.

5.5 The Delta Function Potential

Consider $V(x) = -\alpha\delta(x)$ where $\alpha \gt 0$.

5.5.1 Bound State ($E \lt 0$)

The wave function is $\psi(x) = Ae^{\kappa x}$ for $x \lt 0$ and $\psi(x) = Be^{-\kappa x}$ for $x \gt 0$ Where $\kappa = \sqrt{-2mE}/\hbar$.

Matching conditions.

1. Continuity: $A = B$ at $x = 0$.

2. Discontinuity in derivative (integrating the Schrodinger equation across $x = 0$):

$\psi'(0^+) - \psi'(0^-) = -\frac{2m\alpha}{\hbar^2}\psi(0)$

This gives $-\kappa B - \kappa A = -2m\alpha A/\hbar^2$And since $A = B$:

$\kappa = \frac{m\alpha}{\hbar^2}$

The bound state energy is:

$E = -\frac{\hbar^2\kappa^2}{2m} = -\frac{m\alpha^2}{2\hbar^2}$

The normalised wave function is $\psi(x) = \sqrt{\kappa}\,e^{-\kappa|x|}$. There is exactly one bound state.

5.5.2 Scattering States ($E \gt 0$)

For a particle of energy $E = \hbar^2 k^2/(2m)$ incident from the left:

$\psi(x) = \begin{cases} e^{ikx} + Re^{-ikx} & x \lt 0 \\ Te^{ikx} & x \gt 0 \end{cases}$

Applying the matching conditions at $x = 0$:

$1 + R = T, \quad ik(T - 1 - R) = -\frac{2m\alpha}{\hbar^2}T$

Solving:

$T = \frac{ik}{ik - m\alpha/\hbar^2} = \frac{1}{1 + im\alpha/(\hbar^2 k)}$

$R = \frac{-m\alpha/\hbar^2}{ik - m\alpha/\hbar^2} = \frac{-im\alpha/\hbar^2}{ik + m\alpha/\hbar^2}$

The transmission and reflection coefficients:

$|T|^2 = \frac{1}{1 + (m\alpha)^2/(\hbar^4 k^2)} = \frac{1}{1 + m\alpha^2/(2\hbar^2 E)}, \quad |R|^2 = 1 - |T|^2$

Note that even for very high energies ($E \to \infty$), $|R|^2 \to (m\alpha)^2/(\hbar^4 k^2) \neq 0$: The delta function always reflects some probability, unlike a smooth potential which becomes Transparent at high energies. This is because the delta function has an infinitely sharp feature At $x = 0$ that scatters waves of all wavelengths.

5.6 Quantum Tunnelling

Consider a rectangular barrier $V(x) = V_0$ for $0 \lt x \lt a$ and $V(x) = 0$ otherwise, with $E \lt V_0$.

Inside the barrier, the Schrodinger equation gives exponentially decaying and growing solutions:

$\psi(x) = Ce^{\kappa x} + De^{-\kappa x}, \quad \kappa = \sqrt{\frac{2m(V_0 - E)}{\hbar^2}}$

For a thick barrier ($\kappa a \gg 1$), the growing solution $Ce^{\kappa x}$ is negligible at the Far edge, and the transmission coefficient simplifies to:

$T \approx \frac{16E(V_0 - E)}{V_0^2}\,e^{-2\kappa a}$

The exponential factor $e^{-2\kappa a}$ is the hallmark of quantum tunnelling: the probability of Penetration decreases exponentially with barrier width and height.

:::caution Common Pitfall Tunnelling does not violate energy conservation. The particle does not “have” energy $V_0$ inside The barrier; rather, the wave function extends into the classically forbidden region with Exponentially decreasing amplitude. The particle’s energy is $E \lt V_0$ throughout. :::

Example 5.2. An electron with $E = 5$ eV approaches a barrier of height $V_0 = 10$ eV and Width $a = 0.5$ nm. Calculate $T$.

Application: alpha decay. Alpha decay can be understood as quantum tunnelling through the Coulomb Barrier. The Geiger-Nuttall law, which relates the decay constant to the alpha particle energy, Follows directly from the exponential dependence of $T$ on the barrier width.

Application: scanning tunnelling microscope (STM). In an STM, a small voltage is applied between A sharp tip and a conducting surface. Electrons tunnel across the gap, producing a current that Depends exponentially on the tip-surface distance: $I \propto e^{-2\kappa d}$. This allows atomic- Resolution imaging of surfaces, as a change in distance of $0.1$ nm changes the current by a factor Of about 10.