# Clarification on inertial mass

I was talking to my friend the other day about the origin of inertia and it pondered both of our brains. I would like to know if the concept of inertia exists at the quantum level? I realize that quantized particles are probabilistic and cannot be understood at a classic level. So, does this mean inertia is applicable to only complicated systems like an atom and molecules, etc? There is such a huge gap in connection between classical and quantum systems it destroys my brain..

• "There is such a huge gap in connection between classical and quantum systems it destroys my brain.." made me laugh :D – Danu May 19 '15 at 20:41

## 1 Answer

The concept of inertia can be applied to quantum particles, although a localized quantum particle will never have a sharply defined momentum.

The core concept of inertia (in the sense, that bodies at rest stay at rest and bodies in motion remain in motion) is cleanly expressed by the Galilei invariance of a theory. And it is easy to show, that the Schrödinger equation is Galilei invariant, thus showing, that quantum particles have inertia in the usual sense.

Along another line, one can argue via the Ehrenfest theorem, that: $$m \partial_t^2 \langle x \rangle = -\langle \nabla V(x) \rangle$$ Therfore, the Newton axioms hold for the center of mass, if $V(x)$ scales on a scale much smaller than the wave packet (or $V$ is the harmonic oscillator potential, then $-\langle \nabla V(x) \rangle = - \langle kx \rangle = -k \langle x \rangle$). ($F = ma$ implies that the law on inertia holds).

• So does this mean that a particle could theoretically be isolated in a vacuum state and, without any forces acting on it, remain still? – obliv May 20 '15 at 20:34
• No. Expectation value of the position would stay put. The state would, however, spread (as any located state has non-zero momentum components). – Sebastian Riese May 20 '15 at 20:46
• Is the latter applicable only to quantum mechanics? I had no idea that any located state would have non-zero momentum components.. could you point me in a direction to learn more about this? – obliv May 20 '15 at 20:49
• This only holds for quantum mechanics and is due to the Heisenberg uncertainty relation. (Any introductory book on quantum mechanics will do nicely). – Sebastian Riese May 20 '15 at 20:51
• Some part of my layman's intuition says "this looks like the right answer" but by god I want to understand it and I want the answer to the same question. Is there any way you can make this comprehensible to me? Pretend I am a child with the ability to Google bra-ket and wotnot. – Sentinel Feb 4 '18 at 21:21