How does the time-energy uncertainty principle give particles a non-well-defined mass?

It is example 3.7 from Griffith Quantum Mechanics 3ed:

The $$\Delta$$ particle last about $$10^{-23}\ \mathrm{s}$$, before spontaneously disintegrating. If you make a histogram of all measurements of its mass, you get a kind of bell-shaped curve centered at $$1232\ \mathrm{MeV}/c^{2}$$, with a width of about $$120\ \mathrm{MeV}/c^{2}$$ (figure: 3.2). Why does the rest energy ($$mc^{2}$$) sometimes come out higher than $$1232$$, and sometimes lower? Is this experimental error?

Ans: No, for if we take $$\Delta t$$ to be the lifetime of the particle (certainly one measure of "how long it takes the system to change appreciably"), $$\Delta E\Delta t = \left(\frac{120}{2}\ \mathrm{MeV}\right) \left(10^{-23}\right) = 6 \times 10^{-22}\ \text{MeV s}.$$ whereas $$\frac{\hbar}{2}$$ = $$3 \times10^{-22}\ \text{MeV s}$$. So the spread in $$m$$ is about as small as the uncertainty principle allows - a particle with so short a lifetime just doesn't have a very well-defined mass.

Now my question is what does Griffiths mean by: "So the spread in $$m$$ is about as small as the uncertainty principle allows"; and also why does he say: "a particle with so short a lifetime just doesn't have a very well-defined mass." What is the reason or logic behind this?

I would like an explanation from a physical point of view rather than a mathematical one.

• Isn't it to be taken that the spreading is merely dictated by the undeterminacy principle? Mar 31, 2021 at 11:12
• E and t aren't operators, but you can still derive $\Delta E \Delta t \geq \hbar / 2$. So if a particle has a very short lifetime then $\Delta E$ must be big enough to satisfy the uncertainty principle and since $E \propto m$ the mass won't be well defined. Mar 31, 2021 at 11:13
• @Wihtedeka so you mean as $\Delta E$ is constrained by the uncertainty principle and $E \propto m$ that why spread in m is wide !! Mar 31, 2021 at 14:06