Skip to main content

Physics Library

Free Physics Study Material for CSIR NET, IIT JAM, GATE, JEST & TIFR

by Pravegaa Education
Chapter 7

Harmonic Oscillator

Classical Harmonic Oscillator

  • Potential \(V(x)=\tfrac12kx^2=\tfrac12m\omega^2x^2\), \(\omega=\sqrt{k/m}\) — e.g. a spring–mass system.

  • Equation of motion: \(\ddot x+\omega^2x=0\Rightarrow x=x_0\sin\omega t\).

  • Energy \(E=\dfrac{p^2}{2m}+\dfrac12m\omega^2x^2\) is constant; turning points (\(p=0\)): \(x_0=\pm\sqrt{\dfrac{2E}{m\omega^2}}\).

Figure

Classical observations

  • SHM between turning points \(\pm x_0\); zero probability in the classically forbidden region \(|x|>x_0\).

  • Energy is continuous: \(0<E<\infty\).

  • Speed is maximum at the centre and zero at the turning points \(\Rightarrow\) probability is minimum at the centre and maximum near the edges.

7.1  Schrödinger (Analytic) Method

  • \(\hat H=\dfrac{\hat p^2}{2m}+\dfrac12m\omega^2\hat x^2\);   \(-\dfrac{\hbar^2}{2m}\psi''+\dfrac12m\omega^2x^2\psi=E\psi\).

  • Dimensionless variables: \(\xi=\sqrt{\dfrac{m\omega}{\hbar}}\,x\),   \(k=\dfrac{2E}{\hbar\omega}\) \(\Rightarrow\) \(\dfrac{d^2\psi}{d\xi^2}=(\xi^2-k)\psi\).

  • Large \(\xi\): \(\psi\approx Ae^{-\xi^2/2}\). Try \(\psi=H(\xi)e^{-\xi^2/2}\): \[H''-2\xi H'+(k-1)H=0.\]

  • Series \(H=\sum_ja_j\xi^j\) gives the recursion \[a_{j+2}=\frac{2j+1-k}{(j+1)(j+2)}a_j.\]

  • For large \(j\), \(a_{j+2}\approx\frac2ja_j\) \(\Rightarrow H\sim e^{\xi^2}\) and \(\psi\sim e^{\xi^2/2}\to\infty\): unacceptable.

  • The series must terminate at \(j=n\): \(2n+1-k=0\) \(\Rightarrow\) energy quantization: \[E_n=\Big(n+\frac12\Big)\hbar\omega,\qquad n=0,1,2,\ldots\]

  • With \(k=2n+1\): \(H_n''-2\xi H_n'+2nH_n=0\) — Hermite’s equation; \(H_0=1\), \(H_1=2\xi\), \(H_2=4\xi^2-2\).

  • Normalized eigenfunctions: \[\psi_n(\xi)=\Big(\frac{m\omega}{\pi\hbar}\Big)^{1/4}\frac{1}{\sqrt{2^nn!}}H_n(\xi)e^{-\xi^2/2}.\]

  • Zero-point energy: \(E_0=\dfrac{\hbar\omega}{2}\).

  • First three states:

    • \(\psi_0=\Big(\dfrac{m\omega}{\pi\hbar}\Big)^{1/4}e^{-\xi^2/2}\),   \(E_0=\dfrac{\hbar\omega}{2}\)

    • \(\psi_1=\Big(\dfrac{m\omega}{\pi\hbar}\Big)^{1/4}\dfrac{1}{\sqrt2}2\xi e^{-\xi^2/2}\),   \(E_1=\dfrac{3\hbar\omega}{2}\)

    • \(\psi_2=\Big(\dfrac{m\omega}{\pi\hbar}\Big)^{1/4}\dfrac{1}{\sqrt8}(4\xi^2-2)e^{-\xi^2/2}\),   \(E_2=\dfrac{5\hbar\omega}{2}\)

Figure

7.2  Operator (Algebraic) Method

  • Dimensionless operators: \(X=\sqrt{\dfrac{\hbar}{m\omega}}\hat X\),   \(P=\sqrt{m\omega\hbar}\,\hat P\),   \(H=\hbar\omega\hat H\) \(\Rightarrow\hat H=\dfrac{\hat P^2+\hat X^2}{2}\), with \([\hat X,\hat P]=i\).

  • Ladder operators: \[a=\frac{1}{\sqrt2}(\hat X+i\hat P),\qquad a^\dagger=\frac{1}{\sqrt2}(\hat X-i\hat P).\]

  • Number operator \(N=a^\dagger a\) (Hermitian, \(N^\dagger=N\)); \(N\left|n\right\rangle=n\left|n\right\rangle\).

  • \(N=\frac12\big(\hat X^2+\hat P^2+i[\hat X,\hat P]\big)=\hat H-\frac12\) \(\Rightarrow\) \[H=\Big(N+\frac12\Big)\hbar\omega,\qquad H\left|n\right\rangle=\Big(n+\frac12\Big)\hbar\omega\left|n\right\rangle.\]

Commutator algebra

  • \([a,a^\dagger]=\frac12\big(-i[\hat X,\hat P]+i[\hat P,\hat X]\big)=1\).

  • \([N,a]=[a^\dagger,a]a=-a\),   \([N,a^\dagger]=a^\dagger[a,a^\dagger]=a^\dagger\).

Lowering operator

  • \(Na\left|n\right\rangle=(aN-a)\left|n\right\rangle=(n-1)a\left|n\right\rangle\) \(\Rightarrow a\left|n\right\rangle=c_-\left|n-1\right\rangle\).

  • \(|c_-|^2=\left\langle n\middle|a^\dagger a\middle|n\right\rangle =n\Rightarrow a\left|n\right\rangle=\sqrt n\left|n-1\right\rangle\); in particular \(a\left|0\right\rangle=0\), so \(n\ge0\) (lowest state \(n=0\)).

Raising operator

  • \(Na^\dagger\left|n\right\rangle=(n+1)a^\dagger\left|n\right\rangle\Rightarrow a^\dagger\left|n\right\rangle=c_+\left|n+1\right\rangle\).

  • \(|c_+|^2=\left\langle n\middle|aa^\dagger\middle|n\right\rangle =\left\langle n\middle|N+1\middle|n\right\rangle =n+1\Rightarrow a^\dagger\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle\).

Summary

  • \(a\left|n\right\rangle=\sqrt n\left|n-1\right\rangle\),   \(a^\dagger\left|n\right\rangle=\sqrt{n+1}\left|n+1\right\rangle\),   \(N\left|n\right\rangle=n\left|n\right\rangle\).

  • \(E_n=\big(n+\tfrac12\big)\hbar\omega\), \(n=0,1,2,\ldots\); discrete, with zero-point energy \(\tfrac12\hbar\omega\).

  • \(\left|n\right\rangle=\dfrac{(a^\dagger)^n}{\sqrt{n!}}\left|0\right\rangle\).

  • \(\hat x=\sqrt{\dfrac{\hbar}{2m\omega}}(a+a^\dagger)\),   \(\hat p=i\sqrt{\dfrac{m\omega\hbar}{2}}(a^\dagger-a)\).

Ground-state wave function from \(a\left|0\right\rangle=0\)

  • \(\Big(\sqrt{\dfrac{m\omega}{\hbar}}x+\sqrt{\dfrac{\hbar}{m\omega}}\dfrac{d}{dx}\Big)\phi_0=0\Rightarrow\dfrac{d\phi_0}{\phi_0}=-\dfrac{m\omega}{\hbar}x\,dx\).

  • \(\phi_0=A\exp\Big(-\dfrac{m\omega x^2}{2\hbar}\Big)\); normalization gives \(A=\Big(\dfrac{m\omega}{\pi\hbar}\Big)^{1/4}\).

  • Higher states: \(\phi_n=\Big(\dfrac{m\omega}{\pi\hbar}\Big)^{1/4}\dfrac{1}{\sqrt{2^nn!}}H_n(\alpha x)e^{-\alpha^2x^2/2}\),   \(\alpha=\sqrt{m\omega/\hbar}\).

Quantum vs. classical oscillator

  • Energies are discrete: \(E_n=(n+\frac12)\hbar\omega\).

  • Non-zero probability in the classically forbidden region (tunnelling).

  • Ground state: maximum probability at the centre (opposite to classical). For large \(n\) (e.g. \(n=10\)), \(|\psi_n|^2\) peaks near the turning points — approaching the classical distribution (correspondence principle).

Figure   Figure

7.3  Worked Examples

Example 7.1   Changed frequency

\(V(x)=2m\omega^2x^2\). Find the energies and wave functions.

Solution

  • \(V=\frac12m(2\omega)^2x^2\) \(\Rightarrow\) frequency \(2\omega\):   \(E_n=\big(n+\tfrac12\big)2\hbar\omega\).

  • \(\phi_n=\Big(\dfrac{2m\omega}{\pi\hbar}\Big)^{1/4}\dfrac{1}{\sqrt{2^nn!}}H_n(\alpha x)\exp\Big(-\dfrac{m\omega x^2}{\hbar}\Big)\),   \(\alpha=\sqrt{\dfrac{2m\omega}{\hbar}}\).

Example 7.2   Energy from the wave function (1)

\(\phi(x)\propto x\exp\Big(-\dfrac{m\omega x^2}{\hbar}\Big)\). Find its energy.

Solution

  • \(\exp\Big(-\dfrac{m(2\omega)x^2}{2\hbar}\Big)\) \(\Rightarrow\) oscillator frequency \(2\omega\); factor \(x\propto H_1\) \(\Rightarrow n=1\).

  • \(E=\big(1+\tfrac12\big)2\hbar\omega=3\hbar\omega\).

Example 7.3   Energy from the wave function (2)

\(\psi(x)\propto x^2\exp\Big(-\dfrac{m\omega x^2}{\hbar}\Big)\). What energies can be measured?

Solution

  • Frequency \(2\omega\); \(x^2=\dfrac{H_2+2H_0}{4}\) \(\Rightarrow\psi\propto c_2\left|2\right\rangle+c_0\left|0\right\rangle\) — not a single eigenstate.

  • Outcomes: \(E_2=\tfrac52\cdot2\hbar\omega=5\hbar\omega\) or \(E_0=\tfrac12\cdot2\hbar\omega=\hbar\omega\).

Example 7.4   Measurement on a superposition

\(\left|\psi\right\rangle=3\left|0\right\rangle+4\left|1\right\rangle\) in \(V=\frac12m\omega^2x^2\). Outcomes and probabilities of energy?

Solution

  • \(P\big(\tfrac12\hbar\omega\big)=\dfrac{9}{25}\),   \(P\big(\tfrac32\hbar\omega\big)=\dfrac{16}{25}\).

Example 7.5   Half oscillator

\(V=\infty\) for \(x\le0\); \(\frac12m\omega^2x^2\) for \(x>0\). Find the ground and first excited energies.

Solution

  • Allowed solutions must vanish at \(x=0\): only the odd oscillator states, \(n=1,3,5,\ldots\)

  • Ground: \(E=\tfrac32\hbar\omega\); first excited: \(E=\tfrac72\hbar\omega\).

Example 7.6   Average energy in a half oscillator

\(V=\infty\) (\(x\le0\)), \(2m\omega^2x^2\) (\(x>0\)); \(\left|\psi\right\rangle=3\left|g\right\rangle+4\left|f\right\rangle\) (ground, first excited). Find \(\left\langle E\right\rangle\).

Solution

  • Frequency \(2\omega\); \(\left|g\right\rangle=\left|1\right\rangle\): \(E=\tfrac32\cdot2\hbar\omega=3\hbar\omega\);   \(\left|f\right\rangle=\left|3\right\rangle\): \(E=\tfrac72\cdot2\hbar\omega=7\hbar\omega\).

  • \(\left\langle E\right\rangle=\dfrac{9\cdot3+16\cdot7}{25}\hbar\omega=\dfrac{139}{25}\hbar\omega\).

Example 7.7   Time evolution

At \(t=0\), \(\left|\psi\right\rangle=3\left|1\right\rangle+4\left|2\right\rangle\). Find the state at \(t=\pi/\omega\).

Solution

  • \(\left|\psi(t)\right\rangle=3e^{-3i\omega t/2}\left|1\right\rangle+4e^{-5i\omega t/2}\left|2\right\rangle\).

  • At \(\omega t=\pi\): \(e^{-3i\pi/2}=i\), \(e^{-5i\pi/2}=-i\) \(\Rightarrow\left|\psi\right\rangle=3i\left|1\right\rangle-4i\left|2\right\rangle\).

Example 7.8   Charged oscillator in an electric field

A charge \(q\) on a spring (\(k\), \(m\)) is placed in a uniform field \(E_0\) (\(V=0\) at \(x=0\)). (a) Write \(H\). (b) Find the energies.

Solution

  • (a)  \(L=\tfrac12m\dot x^2-\tfrac12kx^2+qE_0x\) \(\Rightarrow H=\dfrac{p^2}{2m}+\dfrac12m\omega^2x^2-qE_0x\).

  • (b)  Complete the square: \[H=\frac{p^2}{2m}+\frac12m\omega^2\Big(x-\frac{qE_0}{m\omega^2}\Big)^2-\frac{q^2E_0^2}{2m\omega^2}.\]

  • Same frequency, shifted centre: \(E_n=\big(n+\tfrac12\big)\hbar\omega-\dfrac{q^2E_0^2}{2m\omega^2}\).

Example 7.9   \(\left\langle x\right\rangle\) and \(\left\langle x^2\right\rangle\) in \(\left|n\right\rangle\)

Solution

  • \(\left\langle x\right\rangle=\sqrt{\dfrac{\hbar}{2m\omega}}\big(\sqrt n\left\langle n\middle|n-1\right\rangle +\sqrt{n+1}\left\langle n\middle|n+1\right\rangle \big)=0\).

  • \(x^2=\dfrac{\hbar}{2m\omega}\big(a^2+a^{\dagger2}+aa^\dagger+a^\dagger a\big)=\dfrac{\hbar}{2m\omega}\big(a^2+a^{\dagger2}+2N+1\big)\).

  • \(\left\langle x^2\right\rangle=\dfrac{\hbar}{2m\omega}(2n+1)\) (only the \(2N+1\) term survives).

Example 7.10   \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\) in a superposition (1)

\(\left|\psi\right\rangle=\sqrt{\tfrac23}\left|1\right\rangle+\sqrt{\tfrac13}\left|2\right\rangle\). Find \(\left\langle x\right\rangle\), \(\left\langle x^2\right\rangle\).

Solution

  • \(\left\langle a\right\rangle=\left\langle a^\dagger\right\rangle=\sqrt{\tfrac23}\sqrt{\tfrac13}\sqrt2=\tfrac23\) \(\Rightarrow\left\langle x\right\rangle=\sqrt{\dfrac{\hbar}{2m\omega}}\cdot\dfrac43\).

  • \(\left\langle a^2\right\rangle=0\) (no states differing by 2);

Continue reading — it's free

Register with your mobile number to unlock the full chapter and all 100+ chapters of the Physics Library.

Register / Login with OTP

Already registered? Use the same button — just verify your number.