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I'm a quarter of the way towards finishing a basic quantum mechanics course, and I see no mention of force, after having done the 1-D Schrodinger equation for a free particle, particle in an infinitely deep potential well, and the linear harmonic oscillator.

There was one small mention of the fact that expectation values obey classical laws. I was wondering why we don't make more use of this fact. For example, in the linear harmonic oscillator problem, one could obtain the temporal evolution of $\langle x \rangle$ using the classical expression $\left(-\frac{dV(x)}{dx}=m\frac{d^2\langle x\rangle}{dt^2}\right)$, and if we could get the time-evolution of $\sigma$ and tack this on, we could re-create the Gaussian and get back $|\Psi(x,t)|^2$. Of course, that last part may not be very easy.

I was just wondering if anybody has tried doing something like this, or if there an obvious flaw in thinking about it this way.

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5 Answers

Because it doesn't help in interpreting the results.

The Ehrenfest equations show that the expectation values behave similar to the classical laws, but it is difficult to make use of the Ehrenfest equations in a computational context. So the usefulness of the force concept is very limited.

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So what's meant by an electric field in QM? –  Larry Harson Nov 9 '13 at 21:44
    
@LarryHarson: This is defined through QED. Nothing mathematically precise is available. However, in many applications one treats the electromagnetic field as a classical external field, ignoring its quantum fluctuations. Then it is just terms in the Hamiltonian. –  Arnold Neumaier Nov 13 '13 at 17:49
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For example, in the linear harmonic oscillator problem, one could obtain the temporal evolution of $\langle x \rangle$ using the classical expression $\left(-\frac{dV(x)}{dx}=m\frac{d^2\langle x\rangle}{dt^2}\right)$, and if we could get the time-evolution of $\sigma$ and tack this on, we could re-create the Gaussian and get back $|\Psi(x,t)|^2$.

What Gaussian?

You can't automatically assume that the wavefunction takes a Gaussian form, and that it stays in a Gaussian form over time. (In fact the eigenstates of the SHO are Hermite polynomials with a Gaussian envelope, not plain Gaussians peaks.) If you do, you're basically discarding all information about the wavefunction except for the first two moments, the mean and standard deviation, but that discarded information is necessary to predict the future behavior of the system (probabilistically of course).

Now, under some conditions, you can get away with assuming the wavefunction is and remains Gaussian, but those are more or less the same conditions under which quantum effects can be neglected, so the whole theory just reduces to classical mechanics anyway.

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Beside the reasons already provided, I will add that that because forces in there classical meaning are not fundamental, what is more fundamental? it's potentials, and what is even more fundamental are fields, and the usual classical forces, are nothing more than (roughly speaking) a sum of the effects/interaction of those fields, and the equation that you provided in your question, is nothing more than a mathematical description of that averaging those fields/quantum interactions over time, will behave very similar to the familiar Newtonian physics, thus the concept of force is nothing more than statistical averaging/approximation of fields interaction.

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In QM instead of force there is an operator of force in the momentum operator equation: $$\frac{d\hat{\vec{p}}}{dt}=\hat{\vec{F}}.$$ Operator (ordinary) equations for canonical variables are coupled and are difficult to solve for time-dependent non commuting variables. The infinite-size matrices are also generally difficult to handle in approximate calculations. So people prefer to deal with an equivalent (Schroedinger) formulation containing potentials instead of forces in a partial differential equation with time-independent operators.

EDIT: In QFT the "interaction potential" or Lagrangian is not a function, but an operator too.

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We use the concept of force in quantum mechanics. For instance the Heisenberg equations, which are the quantum analogue of Hamiltonian equations of classical mechanics, use quantum forces --the hat denote matrices--

$$\frac{\mathrm{d}\hat{\mathbf{p}}}{\mathrm{d}t}=\hat{\mathbf{F}}$$

What happens is that the Heisenberg matrix formulation is more demanding that the wavefunction formulation by Schrödinger and introductory textbooks avoid it. However, the Heisenberg formulation is very popular in several applications dealing with harmonic oscillators.

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