# Gaussian Wave Packets

From QM Translations, $$\langle x|p\rangle = N\exp\left[\frac{i}{\hbar}px\right]$$.

$$\langle x'|x''\rangle = \delta(x'-x'') = \int dp'\langle x'|p'\rangle p'|x''\rangle = |N^2|\int dp' \exp\left[\frac{i}{\hbar}p'(x'-x'')\right]$$

From the Dirac Delta Function, $$\delta(x-x_0)=\frac{1}{2\pi}\int_\mathbb{R}\exp(i\kappa (x-x_0)) dk$$.

So, $$\delta(x-x_0)=\frac{1}{2\pi\hbar}\int_\mathbb{R}\exp(i\hbar \kappa (x-x_0)) d(\hbar k)=\frac{1}{2\pi\hbar}\int_\mathbb{R}\exp(ip(x-x_0)) dp$$.

Thus, $$\delta(x'-x'')=|N|^2\cdot 2\pi\hbar \delta (x'-x'') = |N^2|\int dp' \exp\left[\frac{i}{\hbar}p'(x'-x'')\right]$$. Hence, $$N = \frac{1}{\sqrt{2\pi\hbar}}$$.

Therefore, $$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}}\exp(ipx/\hbar)$$.

### Relating Position Wavefunction to Momentum Wavefunction

$$\Psi_\alpha(x)=\langle x|\alpha\rangle = \int dp\langle x|p\rangle \langle p|\alpha\rangle = \int dp \frac{1}{\sqrt{2\pi\hbar}}\exp(ipx/\hbar)\Phi_\alpha(p)$$.

This is just a Fourier, transform.

$$\Phi_\alpha(p) = \langle p|\alpha\rangle = \int dx \langle p|x\rangle\langle x|\alpha\rangle = \int dx \frac{1}{\sqrt{2\pi\hbar}}\exp(-ipx/\hbar)\Psi_\alpha(x)$$.

## Definition

$$\langle x|\alpha\rangle = \Psi_\alpha(x)=\frac{1}{\sqrt{d\sqrt{\pi}}|}\exp(-x^2/(2d^2))$$

### Momentum Space

$$\Phi_\alpha(p)= \frac{1}{\sqrt{2\pi\hbar}}\int_\mathbb{R}\exp(-ipx/\hbar)\Psi_\alpha(x)dx = \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}}\int_\mathbb{R}dx\exp(-ix/\hbar(p-\hbar k)-x^2/(2d)^2)$$.

After completing the square we get,

\begin{align*} \Phi_\alpha(p) &= \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}}\int_\mathbb{R}dx\exp(-ix/\hbar(p-\hbar k)-x^2/(2d)^2) \\ &= \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}}\int_\mathbb{R}dx\exp -\left(\frac{x}{\sqrt{2}d}+\frac{i}{\hbar}(p-\hbar k)\frac{d}{\sqrt{2}}\right)^2 - \frac{-(p-\hbar k)^2 d^2}{2\hbar^2} \\ &= \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}}\int dx \exp(-a^2)\exp(-(p-\hbar k)^2d^2/(2\hbar^2)) \\ &= \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}} \int da \exp(-a^2)\exp(-(p-\hbar k)^2d^2/(2\hbar^2)) \\ &= \frac{1}{\sqrt{2\pi\hbar}\sqrt{d\sqrt{\pi}}} \sqrt{\pi}\exp(-(p-\hbar k)^2d^2/(2\hbar^2)) \\ &= \sqrt{\frac{d}{\hbar \sqrt{\pi}}}\exp\left[-\frac{(p-\hbar k)^2d^2}{2\hbar^2}\right] \\ &=\Phi_\alpha(p) \end{align*}

The Gaussian Integral is, $$\int dx \exp(-\alpha x)=\sqrt{\frac{\pi}{\alpha}}$$.

## Properties

### Probability

$$\mathcal{P}_{[x,x+dx]} = |\Psi_\alpha(x)|^2dx = \frac{1}{d\sqrt{\pi}}\exp(-x^2/d^2)dx$$

$$\mathcal{P}_{[p+p+dp]} = |\Phi_\alpha(p)|^2dp = \frac{d}{\hbar\sqrt{\pi}} \exp(-(p-\hbar k)^2d^2/(2\hbar^2))dp$$.

### Expectation Value

$$\langle X\rangle = 0 = \int_\mathbb{R}dx\psi^*_\alpha(x)x\psi_\alpha(x)$$

$$\langle P\rangle = \hbar k = \int_{\mathbb{R}^2} d^2x \Psi^*_\alpha(x'')(-i\hbar\frac{\partial}{\partial x})\delta(x'-x'')\Psi_\alpha(x)$$

### Functional Properties

• Width at $$1/e$$ times the height of the Gaussian is $$2d$$.

### Minimum Uncertainty

• $$\Psi_\alpha(x) = \frac{1}{\sqrt{d\sqrt{\pi}}}\exp(-x^2/(2d^2))$$
• $$\Phi_\alpha(x) = \sqrt{\frac{d}{\hbar\sqrt{\pi}}}\exp(-(p-\hbar k)^2d^2/(2\hbar^2))$$
• $$\langle X\rangle = 0$$
• $$\langle P\rangle = \hbar k$$
• $$\Delta a = \sqrt{\langle a^2\rangle - \langle a\rangle^2}$$
• $$\langle x^2\rangle = \frac{d^2}{2}$$, $$\langle p^2\rangle = \frac{\hbar^2}{2d^2}+\hbar^2k^2$$
• $$\Delta x = \frac{d}{\sqrt{2}}$$
• $$\Delta p = \frac{\hbar}{d\sqrt{2}}$$
• $$\Delta x \Delta p = \frac{\hbar}{2}$$

Created: 2023-06-25 Sun 02:34

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