# What does Conservation of momentum mean in Quantum mechanics?

In quantum mechanics why do we say that momentum in conserved when different measurements on particle give different values of it ?

For example in ground state of Harmonic oscillator I know that expectation value of momentum is independent of time but that is "expectation" value and not the actual momentum of particle or the momentum that we actually measure .

In quantum mechanics why do we say that momentum in conserved when different measurements on particle give different values of it ?

Italics mine.

A measurement has to happen with an interaction. Conservation laws, momentum, energy, angular momentum, imposed axiomatically so as to have continuity between quantum state measurements and macroscopic classical measurements, are in the vectorial sum of the momenta of the individual particles involved in the interaction. Sum before should equal sum after the interaction, exactly as in classical mechanics.

This axiomatic assumption has been tested implicitly with innumerable experiments and no violations have been reported by experiments.

If momentum is conserved (which only happens when the Hamiltonian commutes with the translation operator, which only happens when the potential is a constant), then what momentum conservation guarantees is that if you start the system in a momentum eigenstate, it will always remain in a momentum eigenstate (although the phase will change).

• It can also be in a superposition of momentum eigenstate and in this case not only the phase will change over time. Commented Oct 18, 2021 at 6:17
• @NicolasSchmid Correct. Commented Oct 18, 2021 at 17:37

Conservation of momentum means nothing more and nothing less than that, left to its own devices, the total system momentum space probability distribution does not change with time:

$$\frac{df_p}{dt}(\mathbf{p}) = 0$$

where $$[f_p(t)](\mathbf{p}) = |[\psi_p(t)](\mathbf{p})|^2$$ is the momentum space probability distribution derived from the time series of momental wave functions $$\psi_p$$.

When a measurement is performed, the system is no longer isolated for the time it is happening. The agent doing the measurement is interacting therewith, so we cannot assume the system afterward is unaltered.

The subtlety with quantum measurement is not that it changes things - it's that if you try to make the disturbance smaller, there comes a point where you cannot make it any smaller still without starting to sacrifice information gain from the measurement, and this point is wholly independent of the measuring technique.

The conservation law for momentum states that if there is a translation symmetry in our system, the momentum will be conserved. Translation symmetry requires that the potential should be constant.

Now think about how do the solutions for the energy eigenstates with potential $$V(x)=0$$ look like ? If you solve the time independent Schrödinger equation in 1 dimension you will find $$\psi_p(x) =e^{\frac{i p x}{\hbar}}$$. But those are also eigenstates of the momentum operator, which means that when you measure the momentum of $$\psi_p(x) =e^{\frac{i p x}{\hbar}}$$, you should get $$p$$ with probability 1. (The momentum is related to the energy with the equation $$p=\sqrt{2mE}$$ since there is only kinetic energy)

So since the eigenstates of the energy are the same as the eigenstates of the momentum, the momentum is conserved in time !

You might know that it is actually impossible to know precisely the momentum of a particle. Indeed it is not possible to be in an eigenstate of the momentum, since it is not normalisable. But any wave function is a superposition of these $$\psi_p(x) = e^{\frac{i p x}{\hbar}}$$ (you know that from the Fourier transform). So if you had several particles prepared with the same initial wave function in a world where $$V(x)=0$$ everywhere, you would probably measure different momentum for different particles. So what is conserved ? The expectation value! In other word, conservation of momentum states that for any initial wave function in a world where $$V(x)=0$$ the expectation value $$\langle p \rangle$$ will not change over time.

Now to your example with the Harmonic oscillator. In QM, we say that the momentum is conserved when its expectation value is conserved over time. So for an eigenstate of the energy in the HO, the momentum IS indeed conserved over time, but it is not true for a superposition of these energy eigenstates, whereas if $$V(x) = constant$$ all possible wave function will have a constant expectation value for the momentum.

With this I answered the title "What does conservation of momentum mean in quantum mechanics?". Now the question "Why do we say that momentum in conserved when different measurements on particle give different values of it ?" We say the momentum is conserved since its expectation value doesn't change over time for a closed system (without interaction). When there is a measurement, there is an interaction which changes the wave function and probably also the expectation value of the momentum. Different interactions can lead to different expectation values of the momentum, in these cases, the momentum is not conserved. But as long as there was no measurement, it makes sense to say that the momentum of the particle is conserved.

• So if expectation value of p is constant for some potential (like HO) then that doesn't necessarily mean the momentum is conserved . But if momentum is conserved then expectation value of p must be constant ,(like free particle) . Right ?
– mum
Commented Oct 18, 2021 at 6:34
• Right, and you gave the example of the harmonic oscillator, where the expectation of the momentum for an eigenstate of energy is constant over time. But if you take a superposition of the energy eigenstates, the momentum expectation value might not be constant over time. But in the case of $V(x)=0$ since the energy eigenstates are the same a s the momentum eigenstates, you can conclude that for ANY wave function the expectation value of the momentum will be conserved. Commented Oct 18, 2021 at 6:40
• Oh . That makes sense. So the momentum expectation value is time dependent for superposition of energy eigenstates of HO . One last point of confusion is that in energy eigenstate of HO, expectation value of momentum is constant but momentum is not actually conserved . So is it Just because the probability distribution is symmetric or something else. Like how do I differentiate between a state in which momentum is actually conserved and vs a State in which momentum is not conserved but it's expectation value is constant (like HO energy eigenstates )?
– mum
Commented Oct 18, 2021 at 6:51
• I updated my answer since the first sentence was wrong (I said V(x) has to be constant for the momentum to be conserved, but it is the other way around. Momentum is conserved if V(x)= constant. And I updated the end too Commented Oct 18, 2021 at 7:19
• And I would add that in the HO the expectation value of the momentum of an energy eigenstate is constant 0 because the particle goes as much in one direction as in the other (superposition). So momentum IS conserved for this special case. Commented Oct 18, 2021 at 7:25

There are several ways to think of momentum and energy conservation in quantum mechanics:

Ehrenfest theorem
The equations of motion for average quantities obey the Ehrenfest theorem - in case of linear system they are simply identical with the classical equations of motion, and the conservation of average momentum works in the same way as in classical mechanics.

Note that momentum is not conserved in the case of a Harmonic oscillator, mentioned in the OP, although the time-averaged momentum of a classical oscillator is the same as the quantum mechanical average (that is zero).

Conservation in matrix elements
If matrix elements of operators are expanded in a basis that is translationally invariant (such as plane waves or Bloch waves), the conservation of momentum applies to elementary processes. E.g., a typical photon/phonon emission/absorption by electrons is described by Hamiltonian like: $$H_{e-ph}=\sum_\mathbf{k}\sum_\mathbf{q}M_{\mathbf{k}}(\mathbf{q})c_{\mathbf{k}+\mathbf{q}}^\dagger c_\mathbf{k}a_\mathbf{q} + h.c.$$ Two particles with momenta $$\hbar\mathbf{k}$$ and $$\hbar\mathbf{q}$$ are annihilated and instead a single particle with momentum $$\hbar(\mathbf{k}+\mathbf{q})$$. Second quantization is taken here for convenience - the property appears in all matrix elements, and can be viewed as selection rules on possible processes.

Conservation in scattering processes
Closely related to the previous point is the more global conservation of the initial and final momenta in scattering processes, which is often referred to as the on shell processes. Momentum and energy summation rules are also imposed for nodes of Feynmann-Dyson expansion.

• So it means that expectation value of some observable being time independent does not necessarily means it of conservation ? Then what is the statement of law of conservation of momentum ? I think that I have misinterpreted Generalized Ehrenfest Theorem. Could you clarify that a bit ?
– mum
Commented Oct 19, 2021 at 7:56
• In a scattering process the total some of the initial momenta of a system is the same as the total sum of the final momenta. In this the initial and the final particles are understood to be in momentum eigenstates. More generally, a quantity is conserved, if its operator commutes with the Hamiltonian - you may put the system in the eigenstate of thsi quantity and it will remain in this eigenstate. Commented Oct 19, 2021 at 8:16