While reading Shankar's book on Quantum Mechanics, I encountered an interesting problem:

Compute $\Delta T\cdot\Delta X$, where $T = P^2/2m$.

I found several solutions online which arrive at the result $\Delta T\cdot\Delta X \ge 0$.

My question is: does there exist a state $|{\psi}\rangle$ which saturates this inequality, i.e. for which $\Delta T\cdot\Delta X = 0$? We know $\Delta X\ne 0$ (from the uncertainty relation between $X$ and $P$), so then we must surely have that $\Delta T = 0$. But I'm struggling to imagine a physical state with a well-defined kinetic energy! If it is indeed possible, please provide an example of such a state. Thanks!


1 Answer 1


The minimum value of $\Delta x \Delta T$ is dictated by by the magnitude of the commutator $|[x,T]|=\hbar p/m$. Only if this vanishes can the inequality be saturated. This requires a momentum eigenstate with $p=0$, which means a constant wave function $\psi=c$.

Whether such a wave function can exist is a trickier question. Obviously, the domain of the wave function must be finite, or else $\psi$ will not be normalizable. If the particle is confined to a finite region by a potential, then $p=0$ will not be an energy eigenstate. This leaves us with the case of a finite region with periodic boundary conditions.

For a real particle, if $x$ is the Cartesian coordinate, this is not a realistic arrangement. (Periodic boundary conditions are often used in statistical mechanics, but there they are idealizations that are useful when the thermodynamic limit is taken.) However, if $x$ represents something like the position of a bead on a hoop, then the zero-momentum state is the real physical ground state. More generally, such a state will exist when the coordinate variable represents an angular position. For instance, in three dimensions, an S state wave function in a central potential has zero angular momentum and zero angular momentum uncertainty, although there is still position-energy uncertainty, because of the radial dependence of the kinetic energy.

  • $\begingroup$ I see. So for something like $\psi = c$, the momentum wave function would be a delta function at $x=0$, and so I guess it would follow that both momentum and kinetic energy are identically zero, hence $\Delta T = 0$. Right? The only shady thing though is that $\Delta X = \infty$ in this case... $\endgroup$
    – J-J
    May 21, 2019 at 18:00
  • $\begingroup$ It's seeming more likely that no such "physical" state exists (constant wave functions aren't normalisable, after all). I'm fine if that's the case. $\endgroup$
    – J-J
    May 21, 2019 at 18:02
  • $\begingroup$ +1: This is interesting tho. For the theory of a particle on a ring, a physical state can exist which saturates this bound. Since $\Delta x$ can only be finite as the eigenspectrum is compact, I can simply choose a momentum eigenstate with $p=0$ as you suggest and it would saturate this uncertainty bound, right? $\endgroup$
    – youpilat13
    May 21, 2019 at 18:27
  • 1
    $\begingroup$ It doesn't require $p=0$, it's sufficient to have $p=\mathrm{const}$. I.e. $\psi=\exp(ikx)$. $\endgroup$
    – Ruslan
    May 21, 2019 at 20:20
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    $\begingroup$ @J-J Yes, the eigenstates of the momentum operator are not physically realizable as they are not normalizable. But that is not the trick here as far as I understand. The point is that $\Delta T \Delta X$ is, in fact, $0\cdot \infty$ but, since $T\sim p^2$, if we regularize the product properly and evaluate this $0\cdot \infty$, it will turn out to be $0$. Intuitively, you can see this without explicitly performing the regularization via recalling that the $0\cdot \infty$ of $\Delta p \Delta x$ turns out to be finite after such a regularization. $\endgroup$
    – youpilat13
    May 28, 2019 at 12:28

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