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

$\lambda = 4N+3$.

Combining

$\lambda = 2n+1, n\in\mathbb{N}$. 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, $X_H(t) = X_H(0) + [iHt/ℏ,X_H(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.