# Is the uncertainty principle a tight bound?

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?

• A better bound does not exist since there are states that satisfy $\sigma_x \sigma_p = \hbar /2$.
– jd27
Commented Jun 15, 2023 at 6:44

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.