From my understanding, inertia is typically taken as an axiom rather than something that can be explained by some deeper phenomenon. However, it's also my understanding that quantum mechanics must reduce to classical, Newtonian mechanics in the macroscopic limit.

By inertia, I mean the resistance to changes in velocity -- the fact of more massive objects (or paticles, let's say) accelerating more slowly given the same force.

What is the quantum mechanical mechanism that, in its limit, leads to Newtonian inertia? Is there some concept of axiomatic inertia that applies to the quantum mechanical equations and explains Newtonian inertia, even if it remains a fundamental assumption of quantum theory?


4 Answers 4


I'd go with the Ehrenfest theorem, for the time derivative of the expectation value of an operator $A$:

$$\frac d{dt}\langle A \rangle = \frac 1 {i\hbar}\langle [A, H]\rangle +\big\langle \frac{\partial A}{\partial t} \big\rangle$$

which gives:

$$ m \frac d{dt}\langle x \rangle=\langle p \rangle$$


$$ \frac d{dt}\langle p \rangle=-\big\langle V'(x) \big\rangle$$

Also note that force is:

$$ F=-\big\langle V'(x) \big\rangle $$

Combining it all and you get:

$$ \frac{d^2}{dt^2} \langle x \rangle = \frac{F}{m} $$

or: mass suppresses changes in velocity at fixed force. Aka: inertia.

  • 2
    $\begingroup$ interia -> inertia ;) $\endgroup$ Aug 22, 2022 at 18:30

Following Susskind. Quantum Mechanics. Theoretical Minimum. Page 286.

Define velocity as the time derivative of the average position.

$$ v=\frac{d\left<X\right>}{dt}= \frac{d}{dt} \int \psi(x,t)^* x \psi(x,t)dx $$

An operator $L$ evolves according to the following formula where $H$ is the Hamiltonian.

Assuming $L$ has no explicit time dependence,

$$ \frac{d}{dt}\left<L\right>=\frac{i}{\hbar}\left<[H,L]\right> $$

For a free particle, $H=\frac{p^2 }{2m}$ therefore

$$ v=\frac{i}{2m\hbar}\left<[P^2,X]\right> $$

Note that $[P^2 , X]=P[P,X]+[P,X]P.$

Substitute $[P,X]=-i\hbar$.

Then $v=\frac{\left<P\right>}{m}$.

There is also a path integral approach.


I submit that in quantum mechanics, just as in newtonian mechanics, the phenomenon of inertia is granted in order to formulate the theory at all.

(An example of an attempt to formulate a deeper theory that describes inertia as emerging from deeper concepts is the attempt by Erik Verlinde.)

We have that the following property of Nature carries over from newtonian mechanics to quantum mechanics:

The sum of potential energy and kinetic energy is a constant value.

$$ E_k + E_p = C \tag{1} $$

In the following I will discuss the implications of that for the status of inertia in quantum mechanics.

As we know, in the newtonian formalism there are two statements about inertia:

  • The existence of inertia is asserted
  • A quantitative description of inertia is provided

Of course, those two statements are in the form of a single expression:

$$ F = ma \tag{2} $$

Next we do a transformation; transformation from expressing in terms of force and acceleration to expressing in terms of potential and kinetic energy. The operation is integration with respect to the position coordinate.

For both sides of $F=ma$: evaluate the integral with respect to position, from starting point $s_0$ to point $s$:

$$ \int_{s_0}^s F \ ds = \int_{s_0}^s ma \ ds \tag{3} $$

The right hand side can be developed, capitalizing on the fact that position and acceleration are not independent.

I omit the factor $m$ temporarily, it is a multiplicative factor that is just carried over each step.

$$ \int_{s_0}^s a \ ds \tag{4} $$

Substitution, using $ds = v \ dt$, with corresponding change of limits.

$$ \int_{t_0}^t a \ v \ dt \tag{5} $$

Change the order:

$$ \int_{t_0}^t v \ a \ dt \tag{6} $$

Using $a \ dt = dv$, with corresponding change of limits.

$$ \int_{v_0}^v v \ dv \tag{7} $$

So we have:

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{8} $$

We multiply both sides with $m$, and then the right hand side of (8) gives us the right hand side of (3). The result: the Work-Energy theorem:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{9} $$

We define the quantity on the left hand side, $\int_{s_0}^s F \ ds$, as work done, and we define change of potential energy $E_p$ as the negative of work done


$$ - \Delta E_p = \Delta E_k \tag{10} $$

Hence the sum of potential energy and kinetic energy is a constant value.

$$ E_k + E_p = C \tag{1} $$


Theories of physics can usually be formulated in various forms that are intertransformable. These transformations are invertable.

As we know: the transformation between Lagrangian mechanics and Hamiltonian mechanics is Legendre transformation. Legrendre transformation is its own inverse; applying Legendre transformation twice recovers the original function.

The transformation from newtonian mechanics to Energy mechanics is integration with respect to the position coordinate. To recover F=ma we differentiate with respect to the position coordinate.

I submit that when the tranformation between two forms A and B is invertable the two forms are on par, neither being a deeper formulation than the other.

Quantum Mechanics

Quantum mechanics is formulated in such a way that it will predict outcomes in accordance with (1)

In other words, in formulating quantum mechanics it is granted that (1) will hold good.

If you are granting (1) you are implictly granting the phenomenon of inertia.

So: in the above sense inertia does not emerge from quantum mechanics. Inertia is granted as is.


You seem to have it a little backward. Newtonian mechanics are a set of laws that interpret and define the idea of inertial motion, including quantum mechanics, rather than the other way around.

The work of Lagrange, Hamilton, Euler, and Cauchy gives insight and reveals a deep structural layer of physical systems. They are the connection between classical and quantum theory.

You should also look at Cartan's reconstruction where gravitation is a result of the curvature of space-time rather than a force causing acceleration. This is also present in Einstein's theory. Carton's ideas introduce both an intuitive yet natural way of reviewing Newtonian theory.

  • $\begingroup$ I think you got caught up in jargon. $\endgroup$
    – sato
    Aug 26, 2022 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.