One-Dimensional Problems | Physics

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.

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

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 $[\hat{x}, \hat{p}] = i\hbar$.

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 $\phi_0(x)$.

$V(x) = 0$ 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.

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.

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 $\langle V \rangle = 2E$.

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

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).

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.

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

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.

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.

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.

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.

flowchart TD
    A[5_One Dimensional Problems] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

One-dimensional quantum problems are the laboratory where quantum weirdness becomes visible. The infinite square well is like a ball bouncing in a perfectly elastic box: the standing waves that fit inside determine the allowed energies, and the zero-point energy means the ball can never be perfectly still. The harmonic oscillator is the quantum version of a pendulum: the ladder operators let you climb up and down the energy ladder one rung at a time, and each rung costs exactly one quantum of energy. The delta function potential is an infinitely sharp spike that still manages to bind a particle, showing that even an infinitely narrow potential can trap a quantum state. Quantum tunneling is the most dramatic departure from classical physics: a particle can pass through a barrier it classically cannot climb over, like a ball rolling through a wall. The thinner and lower the barrier, the more likely the tunnel, which is how nuclear decay and scanning tunneling microscopes work.

• Wave Functions and the Schrodinger Equation: The Schrodinger equation provides the foundation for solving 1D quantum problems.

• Operators and Observables: Operators like the Hamiltonian act on wave functions to extract energy eigenvalues.

• Angular Momentum and the Hydrogen Atom: The hydrogen atom extends 1D techniques to three dimensions with central potentials.

• Calculus

• Linear Algebra

• Vector Calculus

• Quantum Computing

Mistake 1: Assuming energy quantisation always occurs Energy is quantised only for bound states (particles trapped in a potential well). For scattering states ($E > 0$ for a finite potential), the energy spectrum is continuous. Students often assume all quantum systems have discrete energy levels, which is only true for confined particles.

Mistake 2: Forgetting the zero-point energy The ground state energy of a quantum system is never zero for a confining potential. For the infinite square well, $E_1 = \pi^2\hbar^2/(2mL^2) > 0$. This is a consequence of the uncertainty principle: confining a particle to a region of size $L$ requires momentum $\sim \hbar/L$, giving kinetic energy $\sim \hbar^2/(2mL^2)$. Students sometimes set the lowest energy to zero as in classical mechanics.

Mistake 3: Misunderstanding quantum tunnelling Tunnelling does not require the particle to “have” energy greater than the barrier height inside the barrier. The wave function extends into the classically forbidden region with exponentially decreasing amplitude. The particle’s energy is $E < V_0$ throughout, and the tunnelling probability decreases exponentially with barrier width and height.