Heisenberg's Uncertainty Principle — Question 6

Question 6

A quadratic phase adds position-momentum correlation without changing position density

Consider $\psi=(2\pi a^2)^{-1/4}e^{-x^2/(4a^2)}e^{ibx^2}$, with real b. Its position variance is $a^2$, and both means vanish. Define $C_{xp}=\operatorname{Re}\langle xp\rangle$. The stronger bound is $\sigma_x^2\sigma_p^2-C_{xp}^2\ge\hbar^2/4$.

1. Find $\sigma_p^2$ from $\langle p^2\rangle=\hbar^2\int|\psi\prime|^2dx$.

2. Find $C_{xp}$, check the stronger bound and evaluate the ordinary product at $b=1/(4a^2)$.

Question 6 - Solution

A quadratic phase adds position-momentum correlation without changing position density

(a) Method and calculation

$$\psi\prime=\left(-\frac{x}{2a^2}+2ibx\right)\psi\quad\Rightarrow\quad\boxed{\sigma_p^2=\frac{\hbar^2}{4a^2}+4\hbar^2b^2a^2}.$$$$\boxed{\sigma_x\sigma_p=\frac\hbar2\sqrt{1+16b^2a^4}}.$$

(b) Interpretation and further calculation

$$C_{xp}=\operatorname{Re}\int\psi^*x(-i\hbar\psi\prime)dx=\boxed{2\hbar ba^2}.$$$$\boxed{\sigma_x^2\sigma_p^2-C_{xp}^2=\hbar^2/4},\quad b=\frac1{4a^2}:\ \boxed{\sigma_x\sigma_p=\hbar/\sqrt2}.$$

Physical check

The quadratic phase changes momentum statistics while leaving the position density fixed. Saturation of the covariance bound need not imply saturation of the simpler product bound.