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Looking at the proof on Wikipedia, the part that confuses me is as as follows. Firstly, $\sigma_x^2\sigma_p^2\ge |\langle f|g\rangle|^2$. Then, because $$|z|^2=(\text{Re}(z))^2+(\text{Im}(z))^2\ge(\text{Im}(z))^2=\left(\frac{z-z^*}{2i}\right)^2,$$ we rewrite $$|\langle f|g\rangle|^2\ge\text{Im}(|\langle f|g\rangle|^2)= \bigg(\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\bigg)^2.$$

And so ultimately, we have two inequalities where we solve for what's on the right hand side:

$$ \sigma_x^2\sigma_p^2\ge |\langle f|g\rangle|^2 \ge\text{Im}(|\langle f|g\rangle|^2) $$

where $$\text{Im}(|\langle f|g\rangle|^2)= \bigg(\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\bigg)^2.$$

This doesn't seem to be a tight lower bound on $\sigma_x^2\sigma_p^2$; is that true (does a better bound exist?) or am I missing something?

Tobias Fünke
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Marc Bacvanski
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    A better bound does not exist since there are states that satisfy $\sigma_x \sigma_p = \hbar /2 $. – jd27 Jun 15 '23 at 06:44

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This is the best bound for 1D position/momentum operators in quantum mechanical systems since there are states that satisfy $\sigma_x \sigma_p=\frac{\hbar}{2}$ - an example might be the coherent state solutions to a 1D quantum harmonic oscillator. $$|{\alpha}\rangle=e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$ See more in https://physics.stackexchange.com/a/457772/353463 or https://physics.stackexchange.com/a/353893/353463.

There is also an alternative (albeit a bit more involved) way you may prove that this is the tight lower bound for product of uncertainties in any system - using calculus of variations.

In 1D it goes as follows:

First: We construct the probability density measure which in nonrelativistic quantum mechanics is just the amplitude squared of the wavefunction in momentum space: $\rho(p)=\left|\tilde{\psi}(p)\right|^2$ satisfying the constraint $$\int\tilde{\rho}(p)dp=\int\left|\tilde{\psi}(p)\right|^2dp=N^2$$

Note that in 1D the momentum space wavefunction can be interpreted as the independent degree of freedom (i.d.o.f) of the system. $f(p) \leftrightarrow \tilde{\psi}(p)$

Second: We write the uncertainty of momentum $\Delta p^2$ and position $\Delta x^2$ in terms of the i.d.o.f: (recall the representation of position operator in momentum space: $\hat{x}=i\hbar \partial_p$) \begin{align} \Delta \hat{p}^2 =\frac{1}{N^2}\int dp p^2 |f(p)|^2\\ \Delta \hat{x}^2 =\frac{\hbar}{N^2}\int dp \frac{\partial f(p)}{\partial p}\frac{\partial f^*(p)}{\partial p} \\ N^2 = \int dp \left|f(p)\right|^2 \end{align}

Finally: This allows us to construct the uncertainty product and minimize it using variation with respect to $f^*(p)$ as follows: $$ \gamma^2=\frac{\Delta\hat{p}^2\Delta\hat{x}^2}{\hbar^2}, \quad \hbar^2\frac{\delta \gamma^2}{\delta f^*(p)}=\frac{\delta\Delta \hat{x}^2}{\delta f^*(p)}\Delta \hat{p}^2+\frac{\delta\Delta \hat{p}^2}{\delta f^*(p)}\Delta \hat{x}^2$$

from which you get (By Euler-Lagrange eqs): $$ \left[-\Delta \hat{p}^2\partial_p^2+\frac{\Delta\hat{x}^2}{\hbar^2}p^2-2\gamma^2\right]f(p) = 0$$ which can be transformed by a change of coordinates to: $$\left[-\partial_q^2+q^2\right]f(q)=2\gamma f(q)$$ for which you explicitly find the minimal solution $\gamma=\frac{1}{2}$ for $f(q)=e^{-q^2/2}$.

This reasoning of course is overkill for a 1D nonrelativistic system but can be easily extended to higher-dimensional systems as well as relativistic particles, where you find that the standard $\gamma\ge\frac{3}{2}$ in 3D is not a tight bound for relativistic systems and in fact you obtain $\gamma\ge 1+\frac{\sqrt{5}}{2}$ in the ultrarelativistic limit as the tight bound.

See for reference https://iopscience.iop.org/article/10.1088/1367-2630/ab3076 or https://journals.aps.org/pra/abstract/10.1103/PhysRevA.103.052211

Adam
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