# Forces in QM?

## Main Question or Discussion Point

In QM, would you ever have to deal with forces on i.e particles?

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dextercioby
Homework Helper
Not directly, in the standard treatments (usually found in textbooks), all forces are conservative and therefore are derived from a potential. The potential, even when it's time-dependent, is crucial, since, usually, the hamilton operator is the sum b/w the KE operator and the PE operator.

ZapperZ
Staff Emeritus
In QM, would you ever have to deal with forces on i.e particles?
You really ought to make sure you understand classical mechanics before doing QM. In Lagrangian/Hamiltonian mechanics, there are no "forces" either. So this really is not a necessary concept in solving the dynamics of a system, classical or quantum.

Zz.

You really ought to make sure you understand classical mechanics before doing QM. In Lagrangian/Hamiltonian mechanics, there are no "forces" either. So this really is not a necessary concept in solving the dynamics of a system, classical or quantum.

Zz.
However, at my uni a second course in classical mechancis is not a prereq for upper QM courses. But you think it should be? If so why?

However, at my uni a second course in classical mechancis is not a prereq for upper QM courses. But you think it should be? If so why?
In mine too, however, they are taught simultaneously. Also many staff members criticize the decision to teach QM before a better coverage of Analytical Mechanics.

The question itself is interesting - I suppose one can define a force operator as the time derivative of the momentum operator, find expectation values etc. In a rather poor year and a half since I took QM1 I have never encountered it.
I've heard once that Pauli's exclusion principle actually creates pressure which can be interpreted as actual force acting on electrons. I wonder how similar it is to classical forces (can you derive it from some field? can can it be considered an elementary force - I guess not but I didn't read very deep into it)

in Carr-Parinello dynamical simulations, an extended lagrangian is applied to force the electrons into the ground-state and then a correction is made to the forces on the nuclei from this. In the Franck-Condon approximation this works well, I don't know about vibronically dominated system though.

In QM, would you ever have to deal with forces on i.e particles?
In quantum mechanics we typically don't talk about forces. It's not that the concept of force doesn't exist, but this concept is obscured behind the formalism of QM. Potential, however, is defined in terms of force (recall that the potential is some function whose gradient gives the force on a test mass), and potential is an extremely important concept in QM. It appears in the Schrodinger Equation, and greatly affects the solution to any particular problem. So in that sense, force is a very important concept in QM.

In quantum mechanics we typically don't talk about forces. It's not that the concept of force doesn't exist, but this concept is obscured behind the formalism of QM. Potential, however, is defined in terms of force (recall that the potential is some function whose gradient gives the force on a test mass), and potential is an extremely important concept in QM. It appears in the Schrodinger Equation, and greatly affects the solution to any particular problem. So in that sense, force is a very important concept in QM.
Good point, Potential and force are related by a derivative. But in QM we usually only deal with the potential in our calculations. It is also easier to measure than force? I only know how force is measured classically, that is a=F/m by using a spring balance.

In QM physicists use the SE and potential to calculate probabilities. A potential implies a force or acceleration which means a charged particle will radiate. No wonder the concept of force is obscured in the formalism. This is why many physicists believe that QM is not a completely understood theory and needs further development.

ZapperZ
Staff Emeritus
In QM physicists use the SE and potential to calculate probabilities. A potential implies a force or acceleration which means a charged particle will radiate.
Er.. come again?

We use the central potential to solve the Schrodinger Equation for a Hydrogen atom. These are stationary solutions. Where is there any "radiation"?

Want another one? Try the Kronig-Penney potential in solid state physics. Solve for that to get your Bloch wavefunction. Where is the "radiation"?

Zz.

Er.. come again?

We use the central potential to solve the Schrodinger Equation for a Hydrogen atom. These are stationary solutions. Where is there any "radiation"?

Zz.
I agree these are the stationay solutions which are in agreement with hydrogen's spectral lines. I disagree that stationary states exist in a point charge field, which is the case for the hydrogen atom. For in this field, stationary states are mechanically impossible. If the electron is stationary , it will be accelerated and fall to the nucleus. If it is in a dynamic equilibrium with the proton it will radiate and fall to the nucleus.

Is it the case that stationary here means 0 net force rather than 0 net velocity?

If I remember right, they are stationary in that the time derivative of the square of the wave function is zero.

If I remember right, they are stationary in that the time derivative of the square of the wave function is zero.
So in QM I should think about everything in terms of probability? In this case the stationary state is when the rate of change of the probability of the particle with respect to time is 0. I can see now why they saw QM is unintuitive. For one thing, probability is quite unintuitive as most people get probability type problems/puzzles wrong.

I agree these are the stationay solutions which are in agreement with hydrogen's spectral lines. I disagree that stationary states exist in a point charge field, which is the case for the hydrogen atom. For in this field, stationary states are mechanically impossible. If the electron is stationary , it will be accelerated and fall to the nucleus. If it is in a dynamic equilibrium with the proton it will radiate and fall to the nucleus.
i am confused on two levels here:

1) the situation you have described has already been disproved - it is called the Bohr model.

2) by stationary states ZapperZ is referring to solutions who's probability density has no time dependence (e.g. standing waves). This is an important result in quantum chemistry.

For example, general solutions to the time-dependent SE in the position representation are stationary states of the form (where the potential carries no time dependence):

$$\Psi(x,t) = e^{\frac{-iEt}{\hbar}} \psi(x)$$

the probability amplitude is then:

$$|\Psi(x,t)|^2 = |\psi(x)|^2$$

and the time dependence is gone.

a nice way to think about $$\Psi(x,t)$$ is that the standing wave solution $$\psi(x)$$ is rotating through the complex<->real planes with a phase factor proportional to the energy (apply Euler's trig formula to exponential propogator). It is the projection of this rotation into real position space that makes the wave appear to have time dependence. In more advanced topics, this leads to the plausibility of performing analytic continuation to go from path integral state transitions to dynamics! (e.g. setting $$\beta = \frac{it}{\hbar}$$)

in echoing what others have said, the role of force in QM is a bit ambiguous...the main difficult is in the fact that energy is quantized, so then what is the derivative $$-\frac{\partial V}{\partial x}$$? formally you are taking the derivative of a discontinous set. (Note that taking the gradient of the wavefunction is perfectly fine since the wavefunction is continuous - in fact we NEED to take the time derivative!). This, IMHO, is yet another reason why the concept of a particle in the classical sense breaks down in QM - "force" is a classical construct for particles. Having said that, in doing numerical QM dynamics, calculating forces can still be a useful idea however.

it is clear that in the formalism of QM you talk in terms of potentials.but note that there are the Ehrenfest's equation which give you the connection with classical mechanics.if you can experimetally measure an observable then theres an operator corresponding to that whose expection value is measuarble!
and as for talking of derivatives like -grad V(x) then you should have a look at what cohen-tannoudji has to say in his book(vol 1, chap 3)!

Last edited:
ZapperZ
Staff Emeritus