= Chirped Gaussian wave packet
{title2=$\psi_{\sigma,\kappa}(x)=(2\pi\sigma^2)^{-1/4}e^{-(1-i\kappa)x^2/(4\sigma^2)}$}
A <chirped Gaussian wave packet> adds a quadratic <quantum phase> to the zero-mean <Gaussian wave packet> already evaluated above. Here $\sigma>0$ and $\kappa$ is a real dimensionless parameter. Its <probability density function> $|\psi_{\sigma,\kappa}(x)|^2$ is unchanged for every $\kappa$, so $\langle x\rangle=0$ and $(\Delta x)^2=\sigma^2$. Nevertheless, applying the <momentum operator> gives
$$
p\psi_{\sigma,\kappa}=\frac{\hbar(\kappa+i)x}{2\sigma^2}\psi_{\sigma,\kappa},
\qquad \langle p\rangle=0,
\qquad (\Delta p)^2=\|p\psi_{\sigma,\kappa}\|^2=\frac{\hbar^2(1+\kappa^2)}{4\sigma^2}.
$$
The sign of the <quantum covariance> follows directly, without assigning a classical trajectory to the <wavefunction>:
$$
\langle xp\rangle=\frac{\hbar}{2}(\kappa+i),
\qquad C_{xp}=\operatorname{Re}\langle xp\rangle=\frac{\hbar\kappa}{2},
\qquad r=\frac{C_{xp}}{\Delta x\,\Delta p}=\frac{\kappa}{\sqrt{1+\kappa^2}}.
$$
Here $r$ is a normalized symmetrized <quantum covariance>, analogous to a <correlation coefficient>. Positive $\kappa$ gives positive $C_{xp}$, and reversing $\kappa$ reverses $C_{xp}$ while leaving both <variances> unchanged. In the negative-exponent <Fourier transform> convention used above, the <momentum> density is
$$
|\widetilde\psi_{\sigma,\kappa}(p)|^2
=\sqrt{\frac{2\sigma^2}{\pi\hbar^2(1+\kappa^2)}}
\exp\!\left(-\frac{2\sigma^2p^2}{\hbar^2(1+\kappa^2)}\right).
$$
Consequently the <quantum uncertainty> product and the <covariance matrix> <determinant> tell different stories:
$$
\Delta x\,\Delta p=\frac{\hbar}{2}\sqrt{1+\kappa^2},
\qquad
\det\Sigma=\frac{\hbar^2(1+\kappa^2)}4-\frac{\hbar^2\kappa^2}4=\frac{\hbar^2}4.
$$
Every member of this <pure state> family saturates the <Robertson-Schrödinger uncertainty relation>. Only $\kappa=0$ saturates the <Heisenberg uncertainty relation> for this fixed pair of <position> and <momentum> axes. Thus being a <Gaussian wave packet> alone does not imply a minimum $\Delta x\,\Delta p$.
For a concrete comparison, choose a reference <length> $L$, put $X=x/L$, $P=pL/\hbar$, and set $\sigma=L$. The choices $\kappa=0,+2,-2$ all have $(\Delta X)^2=1$, but give $(\Delta P)^2=1/4,5/4,5/4$ and $C_{XP}=0,+1,-1$, respectively. The two nonzero cases have $r=\pm2/\sqrt5$ and product $\sqrt5/2$; all three <covariance matrices> have <determinant> $1/4$.
\Image[/chirped-gaussian-covariance-ellipse.png]
{title=Quadratic phase preserves the position density but broadens the momentum density: opposite chirps tilt covariance ellipses in opposite directions while preserving their area}
The <phase space> curves in the figure satisfy $z^T\Sigma^{-1}z=1$ in the dimensionless coordinates, where $z=(X,P)^T$. They display second moments, not observed simultaneous <position> and <momentum> pairs. Their common area is $\pi\sqrt{\det\Sigma}=\pi/2$, although their coordinate projections differ.
This distinction is useful for predicting focusing. Under <free particle> evolution with <mass> $m$, $x(t)=x(0)+tp(0)/m$, so
$$
(\Delta x(t))^2=\sigma^2+\frac{2t}{m}C_{xp}+\frac{t^2}{m^2}(\Delta p)^2.
$$
A negative chirp initially narrows the <position> distribution; positive chirp initially broadens it. For $\kappa<0$, the minimum occurs at $t_*=-2m\sigma^2\kappa/[\hbar(1+\kappa^2)]>0$ and has <variance> $\sigma^2/(1+\kappa^2)$. At that time the <quantum covariance> vanishes, and the unchanged <momentum> spread gives $\Delta x(t_*)\,\Delta p=\hbar/2$. The extra initial product was a usable position-momentum correlation, not extra <measurement backaction>.
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