Chapter 5

Free Particle

5.1 One-Dimensional Quantum Mechanics

A 1D system can have both bound states (discrete spectrum) and unbound states (continuous spectrum).

Bound states (discrete spectrum)

Occur when the particle cannot escape to infinity and is confined to a limited region.
Condition (figure): \(V_{\min}<E<V_1\).
In 1D, bound-state energies are discrete and non-degenerate.
Node theorem: the \(n\)th state has \(n\) nodes if the ground state is labelled \(n=0\) (\(n-1\) nodes if it is labelled \(n=1\)).

Unbound states (continuous spectrum)

Motion not confined:
- \(V_1<E<V_2\): continuous, non-degenerate spectrum (infinite motion in one direction only).
- \(E>V_2\): continuous, doubly degenerate spectrum (infinite motion in both directions).

5.2 Free Particle (Constant Momentum)

Classical approach

Free particle: constant potential, taken as zero; \(H=\dfrac{p_x^2}{2m}\).
\(\dfrac{\partial H}{\partial x}=-\dot p_x=0\Rightarrow p_x=c\) (constant momentum).
\(\dfrac{\partial H}{\partial p_x}=\dot x\Rightarrow x=\dfrac cmt+\alpha_0\); at \(t_0\), \(x_0=\dfrac cmt_0+\alpha_0\).
Perfect localization: distribution \(f(x)=\delta(x-x_0)\) gives \(\left\langle x\right\rangle=x_0\), \(\left\langle x^2\right\rangle=x_0^2\), \(\Delta x=0\).
Classical motion is precise and deterministic: knowing \(x(t_0)\) fixes \(x(t_0+dt)\).

Quantum approach

Solve \(\hat H\phi=E\phi\) with \(\hat H=\dfrac{\hat P_x^2}{2m}\) (\(E>0\)): \[-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2}=E\phi\;\Rightarrow\;\frac{d^2\phi}{dx^2}+k^2\phi=0,\qquad k=\sqrt{\frac{2mE}{\hbar^2}}.\]
Solutions: \(\phi_+=Ae^{ikx}\), \(\phi_-=Ae^{-ikx}\) (plane waves).
\([\hat H,\hat P_x]=0\) \(\Rightarrow\) common eigenstates: \[\hat P_x\phi_\pm=\pm\hbar k\,\phi_\pm,\qquad E=\frac{p_x^2}{2m}=\frac{\hbar^2k^2}{2m}\text{ for both}.\]
Two eigenfunctions, one energy \(\Rightarrow\) doubly degenerate.

Example 5.1 Energy vs. momentum eigenstate

\(\psi(x)=Ae^{ikx}+Be^{-ikx}\). (a) Show \(\psi\) is an eigenfunction of \(\hat H\) but not of \(\hat P_x\). (b) What does an energy measurement give? (c) What does a momentum measurement give?

Solution

(a) \(\hat H\psi=\dfrac{\hbar^2k^2}{2m}\psi\); but \(\hat P_x\psi=\hbar k\big(Ae^{ikx}-Be^{-ikx}\big)\ne\lambda\psi\).
(b) Always \(\dfrac{\hbar^2k^2}{2m}\).
(c) \(+\hbar k\) or \(-\hbar k\) (probabilities \(\propto|A|^2\), \(|B|^2\)).

Probability density of a standing wave

\(\psi(0)=A(e^{ikx}+e^{-ikx})\) evolves as \[\psi(t)=2A\cos kx\;e^{-i\hbar k^2t/2m}\;\Rightarrow\;|\psi|^2=|c|^2\cos^2kx\quad(c=2A).\]
The state extends over \(-\infty<x<\infty\) \(\Rightarrow\Delta x\to\infty\); a plane wave’s Fourier transform is a delta function \(\Rightarrow\Delta p_x\to0\), consistent with \(\Delta x\Delta p_x\gtrsim\hbar\).

5.3 Probability Current Density

Electromagnetism: continuity equation \(\nabla\cdot\vec J+\dfrac{\partial\rho}{\partial t}=0\), with \(\vec J=\rho\vec v\) (charge conservation).
Quantum continuity equation: \(\nabla\cdot\vec J+\dfrac{\partial|\psi|^2}{\partial t}=0\), where the probability current density is \[J_x=\frac{\hbar}{2im}\Big(\psi^*\frac{\partial\psi}{\partial x}-\psi\frac{\partial\psi^*}{\partial x}\Big),\qquad \vec J=\frac{\hbar}{2im}\big(\psi^*\nabla\psi-\psi\nabla\psi^*\big).\]
Plane waves:
- \(\psi=Ae^{ikx}\): \(J_x=|A|^2\dfrac{\hbar k}{m}=|\psi|^2v\), \(v=\dfrac{\hbar k}{m}\).
- \(\psi=Ae^{-ikx}\): \(J_x=-|A|^2\dfrac{\hbar k}{m}\).
Real wave function, e.g. \(\psi=Ae^{-\gamma x}\): \(J_x=0\) though \(|\psi|^2=|A|^2e^{-2\gamma x}\ne0\).
- Formally \(\hat P_x\psi=-i\hbar(-\gamma)\psi\): imaginary momentum \(\Rightarrow\) negative kinetic energy, i.e. a classically forbidden region.
Summary:
- \(|\psi|^2\ne0\), \(J=0\): classically forbidden region — no de Broglie wavelength.
- \(|\psi|^2\ne0\), \(J\ne0\): classically allowed region — de Broglie wavelength \(\lambda=h/p\).

Example 5.2 Integrated current

Prove that for any normalized 1D wave function, \(\displaystyle\int_{-\infty}^\infty J(x)\,dx=\frac{\left\langle P_x\right\rangle}{m}\).

Solution

By parts, \(\displaystyle\int\psi\,\partial_x\psi^*dx=-\int\psi^*\partial_x\psi\,dx\) (boundary terms vanish).
\(\displaystyle\int J\,dx=\frac{\hbar}{2im}\cdot2\int\psi^*\partial_x\psi\,dx=\frac1m\int\psi^*\Big(-i\hbar\frac{\partial}{\partial x}\Big)\psi\,dx=\frac{\left\langle P_x\right\rangle}{m}\).

Example 5.3 Spherical wave

Find the current density for \(\psi(r)=\dfrac Are^{ikr}\).

Solution

Only the radial part of \(\nabla=\hat r\,\partial_r+\cdots\) acts: \(\partial_r\psi=\Big(ik-\dfrac1r\Big)\psi\).
\(\vec J=\dfrac{\hbar}{2im}\big(\psi^*\partial_r\psi-\psi\partial_r\psi^*\big)\hat r=\dfrac{|A|^2}{r^2}\dfrac{\hbar k}{m}\hat r\) — outgoing flux falling as \(1/r^2\).

5.4 Step Potential

\(V(x)=0\) for \(x<0\); \(V_0\) for \(x\ge0\).
Classically, from the left: \(E>V_0\) always crosses; \(E<V_0\) always reflects.
For electrons/protons the outcome is only probabilistic.

Case 1: \(E>V_0\)

Region I: \(\psi_1=Ae^{ik_1x}+Be^{-ik_1x}\), \(k_1=\sqrt{2mE/\hbar^2}\).
Region II: \(\psi_2=Ce^{ik_2x}\), \(k_2=\sqrt{2m(E-V_0)/\hbar^2}\) (no wave incoming from \(+\infty\), so \(D=0\)).
Currents: \(J_{\rm inc}=|A|^2\dfrac{\hbar k_1}{m}\), \(J_{\rm ref}=-|B|^2\dfrac{\hbar k_1}{m}\), \(J_{\rm tr}=|C|^2\dfrac{\hbar k_2}{m}\).
Reflection and transmission coefficients: \[R=\Big|\frac{J_{\rm ref}}{J_{\rm inc}}\Big|=\frac{|B|^2}{|A|^2},\qquad T=\Big|\frac{J_{\rm tr}}{J_{\rm inc}}\Big|=\frac{|C|^2}{|A|^2}\frac{k_2}{k_1}.\]
Continuity of \(\psi\) and \(\psi'\) at \(x=0\): \(A+B=C\), \(A-B=\dfrac{k_2}{k_1}C\) \[\frac CA=\frac{2k_1}{k_1+k_2},\qquad \frac BA=\frac{k_1-k_2}{k_1+k_2}.\]
Result: \[R=\Big(\frac{k_1-k_2}{k_1+k_2}\Big)^2=\Big(\frac{\sqrt E-\sqrt{E-V_0}}{\sqrt E+\sqrt{E-V_0}}\Big)^2,\qquad T=\frac{4k_1k_2}{(k_1+k_2)^2},\qquad R+T=1.\]
Analysis:
- \(E\gg V_0\): \(R\to0\).
- \(E=2V_0\): \(R=\Big(\dfrac{\sqrt2-1}{\sqrt2+1}\Big)^2\approx3\%\), \(T\approx97\%\).
- \(E=1.01V_0\): \(R=\Big(\dfrac{\sqrt{1.01}-\sqrt{0.01}}{\sqrt{1.01}+\sqrt{0.01}}\Big)^2\approx67\%\), \(T\approx33\%\).
- Even with \(E>V_0\) the particle may reflect; reflection grows as \(E\to V_0\).

Case 2: \(E<V_0\)

Region I: \(\psi_1=Ae^{ik_1x}+Be^{-ik_1x}\).
Region II: \(\psi_2''=\gamma^2\psi_2\), \(\gamma=\sqrt{2m(V_0-E)/\hbar^2}\) \(\Rightarrow\psi_2=Ce^{-\gamma x}\) (must vanish as \(x\to\infty\)).
Boundary conditions: \(A+B=C\), \(A-B=\dfrac{i\gamma}{k_1}C\) \(\Rightarrow\dfrac BA=\dfrac{k_1-i\gamma}{k_1+i\gamma}\).
\(|B/A|=1\Rightarrow R=1\); \(\psi_2\) real \(\Rightarrow J_{\rm tr}=0\), \(T=0\).
Yet \(|\psi_2|^2=|C|^2e^{-2\gamma x}\ne0\): the particle penetrates the classically forbidden region — tunnelling (consistent with the uncertainty principle).
Region I is a standing wave: \(|\psi_1|^2\) oscillates as \(\cos(2k_1x+\text{phase})\).
\(V_0\to\infty\): \(\gamma\to\infty\), \(|\psi_2|^2\to0\) — no penetration; \(\psi=0\) where \(V=\infty\).

Example 5.4 Reflection at a step

A particle of energy 9 eV meets a step of 8 eV. Find (a) the reflection coefficient; (b) the percentage transmitted.

Solution

(a) \(R=\Big(\dfrac{\sqrt9-\sqrt1}{\sqrt9+\sqrt1}\Big)^2=\Big(\dfrac{3-1}{3+1}\Big)^2=\dfrac14\).
(b) \(T=1-R=\dfrac34=75\%\).

Example 5.5 Tunnelling through a rectangular barrier

\(V(x)=V_0\) for \(0<x<a\), zero otherwise; \(E<V_0\). With \(k_1^2=\dfrac{2mE}{\hbar^2}\), \(k_2^2=\dfrac{2m(V_0-E)}{\hbar^2}\), show that for a thick barrier (\(k_2a\gg1\)) \[T\approx\frac{16k_1^2k_2^2}{(k_1^2+k_2^2)^2}\,e^{-2k_2a}.\]

Solution

I (\(x<0\)): \(\psi_{\rm I}=Ae^{ik_1x}+Be^{-ik_1x}\)
II (\(0<x<a\)): \(\psi_{\rm II}=Ce^{k_2x}+De^{-k_2x}\)
III (\(x>a\)): \(\psi_{\rm III}=Fe^{ik_1x}\) (no incoming wave from the right)

Match \(\psi,\psi'\) at \(x=0\): \(A+B=C+D\), \(ik_1(A-B)=k_2(C-D)\).
Match at \(x=a\): \(Ce^{k_2a}+De^{-k_2a}=Fe^{ik_1a}\), \(k_2\big(Ce^{k_2a}-De^{-k_2a}\big)=ik_1Fe^{ik_1a}\).
Eliminating \(B,C,D\): \[\frac AF=\Big[\frac12+\frac i4\Big(\frac{k_2}{k_1}-\frac{k_1}{k_2}\Big)\Big]e^{(ik_1+k_2)a}+\Big[\frac12-\frac i4\Big(\frac{k_2}{k_1}-\frac{k_1}{k_2}\Big)\Big]e^{(ik_1-k_2)a}.\]
For \(k_2a\gg1\) keep only \(e^{k_2a}\): \[\Big|\frac AF\Big|^2\approx\Big[\frac14+\frac1{16}\Big(\frac{k_2}{k_1}-\frac{k_1}{k_2}\Big)^2\Big]e^{2k_2a}=\frac{(k_1^2+k_2^2)^2}{16k_1^2k_2^2}e^{2k_2a}.\]
\(T=\dfrac{|F|^2}{|A|^2}\approx\dfrac{16k_1^2k_2^2}{(k_1^2+k_2^2)^2}e^{-2k_2a}\) — dominated by \(e^{-2k_2a}\).

5.5 Particle in a One-Dimensional Infinite Box

\(V(x)=0\) for \(0<x<a\); \(\infty\) otherwise.
Classically: constant speed, equal time in equal \(\Delta x\); any energy \(0<E<\infty\) (continuous).
Quantum: discrete energies, a non-zero minimum energy, and position-dependent probability.

Solution

\(\psi=0\) outside (\(V=\infty\)). Inside: \(\psi''+k^2\psi=0\), \(k=\sqrt{2mE/\hbar^2}\); \(\psi=Ae^{ikx}+Be^{-ikx}\).
\(\psi(0)=0\Rightarrow B=-A\Rightarrow\psi=C\sin kx\).
\(\psi(a)=0\Rightarrow\sin ka=0\); \(k=0\) gives \(\psi\equiv0\), so \(ka=n\pi\): \[k_n=\frac{n\pi}{a},\qquad E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}=n^2E_1,\quad n=1,2,3,\ldots\]
Zero-point energy: \(E_1=\dfrac{\pi^2\hbar^2}{2ma^2}\) — the minimum possible energy; no classical analogue.
Spectrum: \(E_1\) (ground), \(E_2=4E_1\) (1st excited), \(E_3=9E_1\) (2nd excited), \(E_4=16E_1\) (3rd excited); \(E_n\) is the \((n-1)\)th excited state.
Normalization: \(|C|^2\displaystyle\int_0^a\sin^2\frac{n\pi x}{a}dx=1\Rightarrow C=\sqrt{\frac2a}\), and \[\psi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}{a},\qquad \int_0^a\psi_n^*\psi_m\,dx=\delta_{nm}.\]
\(\left\langle x\right\rangle=\dfrac2a\displaystyle\int_0^ax\sin^2\frac{n\pi x}{a}dx=\frac a2\).
Dirac notation: \(\hat H\left|\phi_n\right\rangle=n^2E_1\left|\phi_n\right\rangle\).

Momentum eigenstates inside the box

Inside the box the particle is free, so energy states can be expanded in momentum states \(\pm\hbar k=\pm\dfrac{n\pi\hbar}{a}\): \[\left|\phi_{n,\pm}\right\rangle=\frac1{\sqrt a}e^{\pm in\pi x/a},\qquad \left|\phi_n\right\rangle=\frac{1}{i\sqrt2}\Big[\left|\phi_{n,+}\right\rangle-\left|\phi_{n,-}\right\rangle\Big].\]
A momentum measurement on \(\left|\phi_n\right\rangle\) gives \(\pm n\pi\hbar/a\) with equal probability (strictly, since the box is finite the momentum distribution is spread around these values).

5.6 Symmetric Infinite Box

\(V(x)=0\) for \(-\frac a2<x<\frac a2\); \(\infty\) otherwise.
\(V\) is symmetric \(\Rightarrow[\hat\pi,\hat H]=0\) \(\Rightarrow\) solutions are even or odd.

Even solutions	Odd solutions
\(\psi(-x)=\psi(x)\Rightarrow A=B\): \(\psi=C\cos kx\)	\(\psi(-x)=-\psi(x)\Rightarrow B=-A\): \(\psi=D\sin kx\)
\(\cos\frac{ka}{2}=0\Rightarrow\frac{ka}{2}=\frac\pi2,\frac{3\pi}2,\ldots\)	\(\sin\frac{ka}{2}=0\Rightarrow\frac{ka}{2}=\pi,2\pi,\ldots\)
\(k_n=\frac{n\pi}{a}\), \(n=1,3,5,\ldots\)	\(k_n=\frac{n\pi}{a}\), \(n=2,4,6,\ldots\)
\(\psi_n=\sqrt{\frac2a}\cos\frac{n\pi x}{a}\)	\(\psi_n=\sqrt{\frac2a}\sin\frac{n\pi x}{a}\)
\(\left\langle x\right\rangle=0\)	\(\left\langle x\right\rangle=0\)

Same spectrum \(E_n=n^2E_1\), \(E_1=\dfrac{\pi^2\hbar^2}{2ma^2}\); \(\left\langle \phi_n\middle|\phi_m\right\rangle =\delta_{nm}\).
Shifting the box horizontally changes neither the energies nor the normalization; the eigenfunctions re-orient to be even (\(n\) odd) or odd (\(n\) even) about the centre, and \(|\psi|^2\) stays symmetric about the centre.
\(\left\langle x\right\rangle\) = centre of the box: \(a/2\) for \([0,a]\), \(0\) for \([-\frac a2,\frac a2]\).

Example 5.6 Identifying a state from its shape

A box state has one node inside the box (one full sine period) with energy 2 eV. Find the ground- and next-state energies.

Solution

One node \(\Rightarrow n=2\) (first excited): \(4E_1=2\) eV \(\Rightarrow E_1=0.5\) eV.
Second excited: \(E_3=9E_1=4.5\) eV.

Example 5.7 Uniform state in a box

In the box \([0,a]\), \(\psi(x)=A\) on \(0<x<a\). Find (i) the probability of momentum \(\pi\hbar/a\) (i.e. overlap with \(\frac1{\sqrt a}e^{i\pi x/a}\)); (ii) the probability of the ground state.

Solution

(i)\(\left\langle \phi_{1,+}\middle|\psi\right\rangle =\dfrac{A}{\sqrt a}\displaystyle\int_0^ae^{-i\pi x/a}dx=\frac{A}{\sqrt a}\cdot\frac{2a}{i\pi}\)

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Quantum Mechanics – Chapter 5: Free Particle

Chapter 5

Free Particle

5.1 One-Dimensional Quantum Mechanics

Bound states (discrete spectrum)

Unbound states (continuous spectrum)

5.2 Free Particle (Constant Momentum)

Classical approach

Quantum approach

Probability density of a standing wave

5.3 Probability Current Density

5.4 Step Potential

Case 1: \(E>V_0\)

Case 2: \(E<V_0\)

5.5 Particle in a One-Dimensional Infinite Box

Solution

Momentum eigenstates inside the box

5.6 Symmetric Infinite Box

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