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, $XH(t) = XH(0) + [iHt/ℏ,XH(0)] + ⋯ =$ 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.

Created: 2024-05-30 Thu 21:20

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