Quantum Harmonic Oscillator

One Dimensional Case

\(m\) is moving in a 1D potential from a force \(F=-\kappa x\), \(V(x)=\frac{1}{2}\kappa x^2 = \frac{1}{2}m\omega^2x^2\), \(\omega=\sqrt{\kappa/m}\).

Then, \(\hat{H}\Psi=-\frac{\hbar^2}{2m}\Psi'' + \frac{1}{2}m\omega^2 X^2\Psi = E\Psi\). \(-\frac{1}{2}\alpha\frac{d^2}{dq^2}+\frac{1}{2}\beta q^2 = E\).

Dimensionless Formulation

\(\alpha = \sqrt{\frac{m\omega}{\hbar}}, \xi=\alpha x, \lambda = \frac{2E}{\hbar\omega}\). \(-\frac{d^2}{d\xi^2}\Psi + (\lambda - \xi^2)\Psi = 0\).

For asymptotic behavior, \(\xi\gg\lambda\), \(\left(\frac{d}{d\xi^2}-\xi^2\right)\Psi=0\), \(\Psi = \xi^P\exp(\pm\xi^2/2)\). To have our expected behavior, \(\Psi = \xi^P\exp(-\xi^2/2)\).

So as an Ansatz, \(\Psi(\xi) = \exp(-\xi^2/2)H(\xi)\). \(\frac{d^2}{d\xi^2}H(\xi) - 2\xi\frac{d}{d\xi}H + (\lambda-1)H=0\).

Since the exponential is always even, we can construct even and odd \(H(\xi)\).

Series solution of this differential equation. \(H_e(\xi) = \sum_k^\infty c_k\xi^{2k}\) and \(H_o(\xi)=\sum_k^\infty d_k\xi^{2k+1}\).

Even Solution

For \(H_e\), \(\sum_k c_k((2k)(2k-1)\xi^{2k-2}+(\lambda - 1 - 4k)\xi^{2k})\) Shifting by 1, \(\sum_k (c_{k+1}2(k+1)(2k+1)+c_k(\lambda - 1 - 4k))\xi^{2k}=0\), \(c_{k+1} = \frac{4k+1-\lambda}{2(k+1)(2k+1)}c_k\).

\(\frac{c_{k+1}}{c_{k}} = \frac{4k+1-\lambda}{2(k+1)(2k+1)}\). For large \(k\), \(\sim \frac{1}{k}\).

Examining, \(\exp^{\xi^2}\), \(\sum_n \xi^{2n}{n!}\), \(\frac{c_{n+1}}{c_n} =\frac{n!}{(n+1)!}\sim \frac{1}{n}\). Thus, we need to introduce a cutoff. So, we need to make a \(N\) such that \(4N+1-\lambda=0\). Then \(c_N\neq 0\) but all future terms are zero.

Then \(\lambda=4N+1\) and \(H(\xi) = \sum_k^N c_k\xi^{2k}\).

Odd Solution

\(\lambda = 4N+3\).


\(\lambda = 2n+1, n\in\mathbb{N}\). So, \(E_n=\hbar\omega(2n+1)/2\). \(\Psi_n(\xi) = N_n\exp(-\xi^2/2)H_n(\xi)\).

  • Hermite Polynomials

    \(H_n''(\xi) -2\xi H_n'(\xi) + 2n H_n(\xi) = 0\) - derived from the ansatz in the Hamiltonian and removing the terms with the exponential. \(H_n(\xi) = (-1)^n\exp(\xi^2)\frac{d^n}{d\xi^n}\exp(-\xi^2)\)

    \(G(\xi,s)=\exp(-s^2+2s\xi) = \sum_n H_n(\xi) s^n/n!\) gives the power series of the polynomial. So, \(H_n(\xi)=\frac{\partial^n}{\partial s^n}G(\xi, s)\).

    The recurrence relation is then, \(2nH_{n-1}(\xi)=H_{n}'(\xi)\)

    Finding normalization: \(\int_\mathbb{R}dx|\Psi_n(x)|^2 = \int d\xi\exp(-\xi^2)G(\xi,s)G(\xi,t)\).

Number Operator

  • \(\hat{a}=\sqrt{\frac{m\omega}{2\hbar}}(X+\frac{i}{m\omega}P)\), annihilation operator
  • \(\hat{a}^\dagger=\sqrt{\frac{m\omega}{2\hbar}}(X-\frac{i}{m\omega}P)\), creation operator
  • \(N=\hat{a}^\dagger\hat{a},[\hat{a},\hat{a}^\dagger]=\mathbb{I}\)
  • \(H=\hbar\omega(2N+1)/2\)
  • \(N|n\rangle = n|n\rangle \Rightarrow H|n\rangle = \hbar\omega(2n+1)/2|n\rangle, E_n=\hbar\omega(2n+1)/2\).
  • \(X=\sqrt{\frac{\hbar}{2m\omega}}(a+a^\dagger)\)
  • \(P=i\sqrt{\frac{m\omega\hbar}{2}}(a^\dagger-a)\)

This implies that the eigenstate energies are equally spaced by \(\hbar\omega\) starting at \(\hbar\omega/2\).

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

Deriving some properties

\(a^\dagger|n\rangle = |a^\dagger n\rangle= |?\rangle\). \(Na^\dagger|n\rangle = (a^\dagger N+a^\dagger)|n\rangle = (n+1)a^\dagger|n\rangle\) Thus, \(N|a^\dagger n\rangle = (n+1)|a^\dagger n\rangle\). Thus, \(|a^\dagger n\rangle = \tilde{c}|n+1\rangle\).

Similarly, \(a|n\rangle = \tilde{d}|n-1\rangle\).

\(\langle N\rangle = n = \langle n|a^\dagger a|n\rangle = \langle an|an\rangle = \langle n-1|\tilde{d}^*\tilde{d}|n-1\rangle\) Thus, \(\tilde{d}=\sqrt{n}\).

\(\langle n|a a^\dagger |n\rangle = \langle n|N+I|n\rangle=\langle a^\dagger n|a^\dagger n\rangle = \langle n+1||\tilde{c}|^2|n+1\rangle\). Thus, \(\tilde{c}=\sqrt{n+1}\).

General Forms

\(a^2|n\rangle=\sqrt{n(n-1)}|n-2\rangle\). \(a^m|n\rangle=\sqrt{\frac{n!}{(n-m)!}}|n-m\rangle\)

\(a^{\dagger m}|n\rangle = \sqrt{\frac{(n+m)!}{n!}}|n+m\rangle\).

\(a^\dagger |n\rangle = \sqrt{n+1}|n\rangle = \sqrt{(n+1)!}{n!}\).

\(|n\rangle = (a^\dagger)^n/\sqrt{n!}|0\rangle\).

\(a|0\rangle = \sqrt{0}|-1\rangle = 0\).

\(\langle n'|a|n\rangle = \delta^{n'+1}_n \sqrt{n}\)

\(\langle n'|a^\dagger|n\rangle = \delta^{n'-1}_n n\sqrt{n+1}\)

\(\langle n'|X|n\rangle\), \(n'=n\pm 1\).

\(\langle n'|P|n\rangle\), \(n'=n\pm 1\).

Thus, \(X\) and \(P\) are non-diagonal in this basis. \([H,X] = 2i\hbar P, [H,P]=-2i\hbar X\) (Note check this commutator, this was put on the fly)

Wavefunction Correspondance

\(n\rangle \Leftrightarrow \Psi_n(x)\).

\(0=\langle x'|a|0\rangle = \langle x'\sqrt{\frac{m\omega}{2\hbar}}(X+iP/m\omega)|0\rangle = \langle x'\langle x'|0\rangle + \frac{i}{m\omega}(-i\hbar)\frac{d}{dx'}\langle x'|0\rangle\). Thus, \(x'\Psi_0(x') + \frac{\hbar}{m\omega}\Psi_0'(x')\). This gives, \(\Psi_0(x')=\sqrt[4]{\frac{m\omega}{\pi\hbar}}e^{-\frac{\left(\frac{x'}{\sqrt{\frac{\hbar}{m\omega}}}\right)^2}{2}}\).

Uncertainty Relations

\(\Delta X^2=\langle X^2\rangle - \langle X\rangle^2 = \langle X^2\rangle, \Delta p^2 = \langle P^2\rangle\).

\(\langle X^2\rangle = \frac{\hbar}{2m\omega}\langle n|a^2+aa^\dagger+a^\dagger a+a^{\dagger 2}|n\rangle = \frac{\hbar}{2m\omega}\langle 2N+1\rangle = \frac{\hbar}{2m\omega}(2N+1)\)

\((\Delta X)^2(\Delta P)^2 \geq \frac{\hbar}{4}\).

What is Oscillating

Time Evolution of QM HO States

Heisenberg Picture

\(\frac{d A_H}{dt}=\frac{i}{\hbar}[H,A_H]\).

For the QMHO, \(P_H,X_H\).

\(\frac{d P_H}{dt} = \frac{i}{\hbar}\left[\frac{P_H^2}{2m},+\frac{m\omega^2}{2}X^2,P_H\right]=-m\omega^2 X_H\)

Similarly for \(X\): \(\frac{d X_H}{dt} = \frac{P_H}{m}\).

\(\frac{d^2 X_H}{dt^2} = \frac{1}{m}\frac{d P_H}{dt} = -\omega^2X_H\). Thus, \(X_H(t) = \frac{P_H(0)}{m\omega}\sin(\omega t) + X_H(0)\cos(\omega t)\). \(P_H(t) = -m\omega X_H(0)\sin(\omega t) + P_H(0)\cos(\omega t)\).

\(X_H(t) = U^\dagger(t,t_0)X_H(0)U(t,t_0) = \exp(i H t/\hbar)X_H(0)\exp(-i Ht/\hbar)\)

Recall: \(e^BAe^{-B} = A + [B,A] + \frac{1}{2!}[B,[B,A]] + \cdots\).

Then, $XH(t) = XH(0) + [iHt/ℏ,XH(0)] + ⋯ = $ same thing from a series expansion of a sine and cosine.

Expectation Value

\(\langle n|X(t)|n\rangle = \langle n|\frac{P_H(0)}{m\omega}\sin\omega t+X_H(0)\cos\omega t|n\rangle = \cos\omega t\langle n|X_H(0)|n\rangle + \frac{1}{m\omega}\langle n|P_H(0)|n\rangle = 0\).

\(\langle \alpha|X(t)|\alpha\rangle = \cos\omega t\sum_{nm}\langle n|X_H(0)|m\rangle c_n^*c_m + \frac{1}{m\omega}\sin\omega t\sum_{nm}\langle n|P_H(0)|m\rangle c_n^*c_m\).

\(\Delta X(0)\Delta X(t)\geq \frac{1}{2}|\langle[X(0),X(t)]\rangle| = \frac{1}{2}|\langle\frac{1}{m\omega}\sin\omega t \cdot i \hbar\rangle| = \frac{\hbar}{2m\omega}|\sin\omega t|\).

  • Nobel Prize in 2005

    Glauber, Hall, Hansch. \(a|\alpha\rangle = \alpha|\alpha\rangle\), eigenstate of annihilation operator, \(|c_n|^2=\frac{\overline{n}^n}{n!}\exp(-\overline{n})\). So if your coefficients are poisson-distributed then you have an eigenstate of annihilation operator and you have a coherent system. Sakurai 2.21.

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:34