# Finding minimum kinetic energy with Heisenberg uncertainty principle

A problem in my textbook states:

Using the uncertainty principle, find the minimum value (in MeV) of the kinetic energy of a nucleon confined within a nucleus of radius $$R = 5 \times 10^{-15}$$m.

Given the Heisenberg uncertainty principle: $$\Delta x \Delta p \geqslant \frac{\hbar}{2}$$

I understand one can, using either classical or relativistic formulae for kinetic energy, and plugging in the value for $$R$$, solve for the energy and find a solution of the form: $$\Delta pc \geqslant \frac{\hbar c}{2\Delta x}$$ or $$\frac{\Delta p^2}{2m} \geqslant \frac{\hbar^2}{2m\Delta x^2}$$

Now I fail to understand how these solutions (given to be correct by the textbook) actually imply a minimum energy. To me it seems that these solutions rather imply a minimum uncertainty in energy. Does this minimum energy-uncertainty imply an equal minimum energy, and if so, why?

I read elsewhere that since it is implied that the uncertainty of a quantity (in this case energy) is smaller than its magnitude, the minimum uncertainty also serves as a lower bound for the magnitude. I should therefore rephrase my question as follows:

Is it usually the case that one can assume such a quantity to have a larger magnitude than uncertainty, and if not, can this be assumed for a general case such as this one?

This question is pretty old now, so I would answer the following. Consider a particle inside some box. Its hamiltonian is $$\tag{1} H = \frac{p^2}{2m}.$$ The state of lowest energy is $$|E_{0}\rangle$$ and you have $$\tag{2} \Delta p \ge \frac{\hbar}{2 \Delta x},$$ where $$\Delta x$$ is about the size of the box. Then you have, since $$\Delta p^2 \equiv \langle p^2 \rangle - \langle p \rangle^2$$ : \begin{align} \langle E \rangle_0 \equiv \langle E_0 | H | E_0 \rangle = \frac{\langle p^2 \rangle}{2m} &\equiv \frac{\Delta p^2}{2 m} + \frac{\langle p \rangle^2}{2m} \\[12pt] &\ge \frac{\hbar^2}{8 m \Delta x^2} + \frac{\langle p \rangle^2}{2m} \ge \frac{\hbar^2}{8 m \Delta x^2}. \tag{3} \end{align} Then obviously there's a minimum value to the energy (the state vector $$| E_0 \rangle$$ is an eigen-ket of the hamiltonian (1) with minimal energy $$E_0$$).

For an ultra-relativistic particle, I would use the hamiltonian $$H = p c$$. Then $$H^2 = p^2 c^2$$ and \begin{align} E_0^2 \equiv \langle H^2 \rangle_0 = \langle p^2 \rangle c^2 &\equiv \Delta p^2 c^2 + \langle p \rangle^2 c^2 \\[12pt] &\ge \frac{\hbar^2 c^2}{4 \Delta x^2} + \langle p \rangle^2 c^2 \ge \frac{\hbar^2 c^2}{4 \Delta x^2}, \tag{4} \end{align} thus $$\tag{5} E_0 \ge \frac{\hbar c}{2 \Delta x}.$$

• You beat me to it, I was just about to post a very similar answer.
– user137661
Oct 28, 2019 at 20:39
• Although it does not address the relativistic case in the OP's question.
– user137661
Oct 28, 2019 at 20:40
• @SV, I've added something for a relativistic particle.
– Cham
Oct 28, 2019 at 21:02