Newton's law states that for every action there is an equal and opposite reaction. This law explains how rockets fly in space and it also accounts for the majority of the lift action generated by an airplane's wings.

Is there a fundamental quantum explanation of the third law of motion?


3 Answers 3


Newton's third law states that if object A acts on object B with force $\mathbf{F}_{AB}$, then object B must act on object A with force:


When expressed in terms of A and B's momentum, the same equation can be written as:

$$\frac{\mathrm{d} \mathbf{p}_A}{\mathrm{d}t} = -\frac{\mathrm{d} \mathbf{p}_B}{\mathrm{d} t}$$


$$\frac{\mathrm{d}}{\mathrm{d} t} \left [ \mathbf{p}_A + \mathbf{p}_B \right ] = 0$$


$$\mathbf{p}_A + \mathbf{p}_B = constant$$

This says that A and B must act on each other such that the sum of their momenta is constant over time. So Newton's third law is just an expression of the conservation of momentum. Momentum conservation itself can be seen as a consequence of spacial translation symmetry, as shown by Noether's Theorem. This symmetry remains in quantum mechanics, so momentum conservation still applies in QM.


Fundamentally Newton's 3$^{\text{rd}}$ law is a statement of the conservation of four-momentum, which can be implied through fundamental symmetries in nature by Noether's Theorem. There are no quantum mechanical effects at play here as space/time symmetries (associated with the conservation of four-momentum) are assumed in calculations at the quantum scale.


A Patronizing Comment

Every classical law has a quantum mechanical origin--in principle. So, it is always a little bit better to think as to what is the quantum mechanical origin of a certain classical law rather than thinking if there is a quantum mechanical origin or not. Of course, it is not necessary that an explicit explanation of such kind is always known to us.

As many answers have pointed out, it is the Noether theorems that make sure that the momentum is conserved and as a direct consequence, (as long as the degrees of freedom that can carry the momentum are only the particle degrees of freedom) the third law of Newton follows. This has been beautifully demonstrated in the first half of the answer by @Jold.

But the reason I am writing this answer is that this has been a fully classical story so far. It is not really pleasant to say that the Noether theorems remain valid in quantum mechanics as well and thus, the third law of Newton has been traced back to its quantum origins. I am not saying this is not correct but I am just saying that this all sounds too hand-wavy if one is looking for details--the OP has not explicitly asked for the details but I would like to provide some details anyway for the sake of completeness to the extent I am competent.

Thus, I plan to show in a little more detail as to how the conservation of classical momentum arises from the application of symmetry arguments in quantum mechanics. Once this is done, the rest of the game is straightforward as summarized above and in the answer by @Jold.

First, we show that the momentum operator is the generator of the spatial translations in quantum mechanics. Second, we (almost trivially) show that the commutator of the momentum operator vanishes if the system (i.e. the Hamiltonian) respects translational invariance. Finally, using the Ehrenfest theorem we show that the expectation value of the momentum operator is thus constant in time--and this means that the classical momentum is conserved.

Momentum Operator is the Generator of the Spatial Translation in Quantum Mechanics

Let's say there is a class of operators $T_\epsilon$ (where the infinitesimal real parameter $\epsilon$ parametrizes the class) specified (completely and consistently) via its action on the complete orthonormal position eigenbasis as the following $$T_\epsilon|x\rangle=|x+\epsilon\rangle$$ Thus, such an operator $T_\epsilon$ is defined as the translation operator that translates a position eigenket by $\epsilon$.


That the action of the operator $T_\epsilon$ on a generic physical state $|\psi\rangle$ physically corresponds to the classical translation of the corresponding classical particle by $\epsilon$ can be verified through proving that $$\langle\psi|T_\epsilon^\dagger X T_\epsilon|\psi\rangle=\langle\psi|X|\psi\rangle+\epsilon$$ and $$\langle \psi|T_\epsilon^\dagger P T_\epsilon|\psi\rangle=\langle\psi|P|\psi\rangle$$ where $X$ and $P$ are the position and momentum operators respectively.

Now, in order to explicitly construct the operator $T_\epsilon$, we expand it in $\epsilon$ to the first order as $$T_\epsilon=I-\dfrac{\iota\epsilon}\hbar G$$where $G$ is an operator defined by the above equation. It is called the generator of translation. Now, it can be seen that $$\langle x|T_\epsilon|\psi\rangle=\psi(x+\epsilon)$$Thus, we can now expand both the sides to the first order in $\epsilon$ to get $$\langle x|I|\psi\rangle-\dfrac{\iota\epsilon}{\hbar}\langle x |G|\psi\rangle=\psi(x)+\epsilon\dfrac{\partial}{\partial x}\psi(x)$$ Or, $$\langle x |G|\psi\rangle=\dfrac{\hbar}{\iota}\dfrac{\partial}{\partial x}\psi(x)$$ Thus, we conclude that the generator $G$ of the spatial translation is the momentum operator $P$. Thus, $$T_\epsilon=I-\dfrac{\iota \epsilon}{\hbar}P$$ Momentum Operator Commutes with the Hamiltonian Provided the Translational Symmetry

The translational symmetry of the system is represented in the form of the condition $$\langle\psi|T_\epsilon^\dagger H T_\epsilon|\psi\rangle=\langle\psi|H|\psi\rangle$$ Therefore, we require $$\bigg\langle\psi\bigg|\bigg(1+\dfrac{\iota \epsilon}{\hbar}P\bigg) H \bigg(1-\dfrac{\iota \epsilon}{\hbar}P\bigg)\bigg|\psi\bigg\rangle=\langle\psi|H|\psi\rangle$$ Or, equivalently, $$\langle\psi|H|\psi\rangle+\dfrac{\iota\epsilon}{\hbar}\langle\psi |[P,H]|\psi\rangle+\mathcal{O}(\epsilon^2)=\langle\psi|H|\psi\rangle$$ Thus, we require$$\langle\psi|[P,H]|\psi\rangle=0$$ Now, it can be seen that thus $[P,H]=0$ using the fact that $\iota[P,H]$ is Hermitian and then employing the standard argument that if an operator $O$ is Hermitian and it satsfies $\langle\psi|O|\psi\rangle=0$ for all $\psi$ then it can be shown that the operator is the zero operator. Anyway, this last bit is not important, all we need is $\langle\psi|[P,H]|\psi\rangle=0$.

Finally, Going Classical from Quantum Using the Ehrenfest Theorem

Since we have $$\langle\psi|[P,H]|\psi\rangle=0$$ according to the Ehrenfest theorem, $$\dfrac{d}{dt}\langle\psi|[P,H]|\psi\rangle=0$$ Thus, we have shown that the expectation value of the momentum operator is constant in time--which translates to saying that the classical momentum is conserved (as the expectation values of the quantum Hermitian operators are supposed to capture the behavior of the corresponding classical observables). Fin.

My answer is based on the discussion of symmetries and conservation laws from Principles of Quantum Mechanics, R. Shankar, $3^{rd}$ Edition, Chapter $11$.


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