I know from quantum mechanics that any matter particle has both wave and particle properties. So my question is that like in Newtonian mechanics we calculate the force acting on a particle by applying Newton's 2nd law of motion, can we apply it also for waves?


4 Answers 4


Although you can't interpret force in Quantum Mechanics in the same way as we do in Classical Mechanics (which is true of all physical quantities anyway), Newtonian Force does have a perfectly good quantum analogue, thanks to a result known as Ehrenfest's Theorem. To understand this answer, you will need to understand the basic mathematics of Quantum Mechanics, as there is no (honest!) intuitive description of mechanical concepts in the quantum regime.

First, you need to recall that in ordinary old (one-dimensional) Newtonian Mechanics, for a system with a potential energy function $U(x)$, we define the force function $F=-\frac{dU}{dx}$, and this can be shown to result in our good old friend $F=ma$ or, in terms of momentum, $F=\frac{dp}{dt}$.

Now, in Quantum Mechanics we usually start with a Hamiltonian of the form $\hat H=\hat K+\hat U(x)$, where $\hat H$, $\hat K$, and $\hat U$ are operators representing the Total Energy, Kinetic Energy, and Potential Energy observables, respectively.

We can then define a Force Operator by $\hat F=-\frac{d\hat U}{dx}$, in analogy with the Newtonian definition, and it turns out that the following is true (which isn't obvious, but was shown by Ehrenfest):$$\langle\hat F\rangle=\frac {d}{dt}\langle \hat P \rangle$$or, if you prefer $$\langle\hat F\rangle=m\frac {d^2} {dt^2}\langle \hat X\rangle=m\langle\hat A\rangle$$

where $\hat X$, $\hat P$, and $\hat A$ are the Position, Momentum, and Acceleration operators, respectively, and angle brackets represent the quantum expectation values. (I'm working in the so-called Heisenberg Picture here in which the observables change in time, just like they do in Classical Mechanics.)

In other words, with the appropriate re-definitions, Newtonian Force works in terms of expectation values in Quantum Mechanics.

Again, this does not mean that you can picture the force on an electron in the same way we do the force on a chair, say, but rather that we can often use the same classical equations, as long as we think in terms of probability distributions, as we always need to do in the quantum world.


Force isn't much used in quantum mechanics.

Classical mechanics deals with point particles, which have a precise trajectory. $F = ma$ is really important. It relates forces, the cause of motion, to acceleration. If you know initial conditions and acceleration at all times, you get velocity and position.

In quantum mechanics, there are no point particles and no precise trajectories. Instead you have a wave function, a distribution of places an object might be.

This means energy and fields are more important than forces. In classical mechanics, if you have an electron in an electric field, you know the electron's energy difference between points A and B. The trajectory doesn't matter.

Likewise in quantum mechanics, if you have an initial state and a final state, you know the energy difference between them. You don't need to know a trajectory that connects these states.

The Schrodinger equation is the important equation in quantum mechanics. It is an energy equation.


The wave property of quantum mechanical particles is a wave in the probability distribution of the location of the particle in (x,y,z,t). It is not an energy wave, as sea waves, or acoustic waves , or classical electromagnetic waves.

Even with the above classical waves, the concept of force as applied in classical mechanics on particles is not a one to one correspondance. The classical waves depend on distributions of many particles ( order of 10^23) and any concept of "force" when two acoustic waves hit each other is a meta level on the classical mechanics force, and not really useful. One looks at interferences in energy transfers.

See this accumulation of single electrons displaying the probability amplitude wave interference pattern. There is no space for modeling a "force" concept. A dp/dt for individual particles could be calculated but it would be within the heisenberg uncertainty.


Yes indeed we can apply a force to a wave, though we don't generally talk about forces in this context. Instead we talk about momentum.

The key fact you need to know is that force, $F$, is the rate of change of momentum, $p$:

$$ F = \frac{dp}{dt} $$

For a particle this is easy to show because the momentum is:

$$ p = mv $$


$$ \frac{dp}{dt} = \frac{d(mv)}{dt} = m\frac{dv}{dt} = ma = F $$

because Newton's second law tells us $F=ma$. The point of all this is that waves can carry momentum just like particles, and if waves can carry a momentum that momentum can be changed and that means we can apply a force to the wave. Incidentally this is exactly why a light wave can exert a force.

For waves the momentum is given by:

$$ p = \frac{h}{\lambda} $$

This works for light waves and for matter waves.

  • $\begingroup$ @Jhon Rennie then what about the energy of the matter/light wave. How's that calculated , like in Newtonian mechanics we calculate the potential energy using the definition for the conservative force ( gravitational or electrostatic ). Is it in the same way? $\endgroup$
    – Munj Patel
    Aug 3, 2017 at 14:50
  • $\begingroup$ @MunjPatel the energy is given by $$E^2 = p^2c^2 + m^2c^4 $$ where $p$ is the momentum as described in my answer and $m$ is the mass of the particle. For light or any other massless particle $m=0$ and it simplifies to $E=pc$. $\endgroup$ Aug 3, 2017 at 14:58
  • $\begingroup$ does the above equation work for any kind of particles irrespective of their mass? $\endgroup$
    – Munj Patel
    Aug 3, 2017 at 15:01
  • $\begingroup$ does this violate the uncertainty principle? $\endgroup$
    – Munj Patel
    Aug 3, 2017 at 15:02
  • $\begingroup$ Yes, that equation works for anything. It is one of the fundamental equations in relativity. No it doesn't violate the uncertainty principle. Why do you think it would? $\endgroup$ Aug 3, 2017 at 15:37

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