Since, classically we define force as rate of change of momentum and by weyl rule we can map classical functions to quantum operators. Then, why is it that we don't often hear the word 'force operator' in quantum mechanics?
Dear Gaurav,
It is a very interesting question. Searching for the reason/s behind not using a force operator quantum mechanics I found a very interesting dialog which fully covers the answer to your question:
Force through quantum mechanics
In classical physics force is:
F = dp / dt
How about quantum mechanics? In Old Quantum Mechanics momentum is: p=ℏ⋅k so force will be:
tF=ℏ dk / dt
what does dk / dt mean?
Answers
It's still true in quantum mechanics – it's a "Heisenberg equation of motion" – but both sides are operators. However, because we often want to find the spectrum of the Hamiltonian in quantum mechanics – a set of possibilities – and the properties of the energy eigenstates, the importance of the force is diminishing. The force may be written as i[H,p]/ℏ, so it's the commutator of the momentum with the Hamiltonian, but we often need to study the whole Hamiltonian and not just its commutators with other operators. That's why the potential energy V appears much more often in quantum mechanical calculations than the force – even though the latter may be written as −V′(x) if a potential is known. Note that −V′(x) is the result of the calculation of the commutator i[H,p]/ℏi[H,p]/ℏ for Hamiltonians H=p2/2m+V(x).
For more answers to your question, please use the following link:
http://physics.stackexchange.com/questions/46544/force-through-quantum-mechanics
Hoping this will be helpful,
Rafik
Well, yes we can have a force operator in the mathematical sense as outlined by Rafik Karaman and in his referred stack-exchange discussion by Lubos Mottl. The only thing is that its not much useful or even 'appropriate' to use it in the context of quantum mechanics. This is because as an operator, it cannot give observable eigen values which can be experimentally detected with accuracy.The reason behind this is the uncertainty principle. If one elevates force from the classical definition to quantum, then in the practical sense, one is actually making a measurement of the momentum at some initial time and subsequently at a later time which is infinitesimally close to the initial time.But whenever the measurement of momentum is performed on a state,it would yield an eigen value corresponding to its measurement(by collapse if it is not a momentum eigenstate initially).Since fundamentally completely isolated states are not realistic(fluctuations of quantum fields in the ground state can also affect other states,as happens in atomic transitions often) the state will evolve in time and any subsequent measurement of momentum cannot give an accurate value since there would be uncertainty in momentum measurements.Therefore force as an operator will be not much of a help from a realistic point of view.Even one can argue on the grounds of relativity that position is not a 'good' operator in quantum mechanics as well(multiparticle states being created at length scales < compton wavelength). But yes, as seen from the viewpoint of expectation values(via Ehrenfest's theorem), the mathematical way of stating the 'force' as is fine.But that is just the expectation value, not pertaining to a single measurement of a system.
Dear all,
I would like to add to the discussion and point out that the concept of force operators do occasionally appear in the context of quantum Langevin equations. As an example, attached is a link to the paper by Ford and Kac, J. Stat. Phys. 46, 803 (1987). But as you have already discussed, force operators have not become a staple of quantum mechanics classes for the reasons given.
https://deepblue.lib.umich.edu/bitstream/handle/2027.42/45153/10955_2005_Article_BF01011142.pdf?sequence=1&isAllowed=y
In quantum mechanics (QM) we don't "hear" of the force operator because we may have problems with applying the Newtonian 2nd law
(1) F = dp/dt .
For instance in a conservative field where we define classically the force as -∇V(r), Newton's law gives
(2) dp/dt = -∇V(r).
The QM disagrees with this equality. Indeed, by the quantum rule of derivation of operators in time we have
(3) dp̂/dt = i[Ĥ,p̂]/ħ + ∂p̂/∂t = i[Ĥ,p̂]/ħ,
because the linear momentum operator is not explicitly time dependent. Now, if the Hmiltonian has the simple form
(4) Ĥ = p̂2/2m + V(r),
there results
(5) dp̂/dt = i[Ĥ,p̂]/ħ = i[V(r),p̂]/ħ = V(r) ∇ - ∇V(r).
Compare this with (2) !
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