# Uncertainity relation of Kinetic energy with position

In R. Shankar's Principle of quantum mechanics book in the problem

Now $$\Delta T = \frac{-\hbar^2}{2m} \Delta( p^2)$$ And I don't arrive anywhere using this, but I also know that $$\Delta A \Delta B = \left|\frac{1}{2i}\left<[A,B]\right>\right|$$. So, using this I find the commutation relation \begin{align*} \frac{1}{2m}[p^2,x] &= \frac{-i p\hbar}{m}\\ \end{align*} And, $$\Delta T \cdot \Delta X = \frac{\left\hbar}{2m}$$ Here the author asks why this relation is not so famous. Though this seems nothing special to me. Am I donig anything wrong here?

• is p an operator in the final expression? Or an expectation value? Perhaps it is not so famous because T is not a partner variable to x. These relations compare uncertainty between canonical variables and their "momentum" as derived from Largrange-Hamilton formalism and T is simply not related to x in this manner. One can use [p, x] to derive uncertainty relations between f(p) and g(x) so why treat these as special?
– user196418
May 28, 2019 at 11:19
• – ACat
May 28, 2019 at 11:54

The relation will give you $$[x, T] = \frac{i \hbar}{m} p$$

Now, $$\Delta T \Delta X \geq \dfrac{\langle p \rangle \hbar}{2m}$$. Notice that for a state with zero momentum, the product of uncertainties can have the minimum value zero, unlike the case for any canonically conjugate pair (like momentum, position).

• To be sure, there wouldn't be a physical (normalizable) state where this uncertainty vanishes, right? Just that you can make it as close to zero as you want with a state more and more cleverly.
– ACat
May 28, 2019 at 12:00
• In principle, an eigne state gives zero uncertainty, the standard deviation for the corresponding operator is zero. Not sure if it is possible to prepare in reality. Is that what you wanted to point out @FeynmansOutforGrumpyCat ? May 28, 2019 at 12:08
• @Galilean Yes, precisely. An eigenstate of the momentum operator is not normalizable (and thus, cannot be realized experimentally even in theory).
– ACat
May 28, 2019 at 12:13
• Ah, one more comment. The state that saturates the uncertainty bound of vanishing uncertainty would be the eigenstate of the momentum operator. This is interesting because of the fact that the uncertainty in the measurement of position for such a state would still be infinitely large. So, $\Delta T\Delta X$ is as such $0\cdot \infty$ (just like in the case of $\Delta p\Delta x$), however, since $T\sim p^2$, a systematic regularization of that product tells us that this $0\cdot \infty$ is actually $0$.
– ACat
May 28, 2019 at 12:20