What Happens to the Momentum Operator as Planck's Constant Approaches Zero?

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The discussion centers on the behavior of the momentum operator in quantum mechanics as Planck's constant (h) approaches zero. The momentum operator is defined as p = i*h*nabla, and as h approaches zero, the operator itself tends to zero, which raises questions about its physical interpretation. It is clarified that the momentum operator is a mathematical construct that does not directly translate to classical momentum. Instead, the transition from quantum to classical mechanics is understood through the limit process, where reducing h leads to classical behavior by confining wave interference effects to smaller regions.

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the_pulp
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I have a doubt making a little of thinking of basic notions of QM and I think that the answer should be very simple but I can't make up my mind. So, here I go:
I usually hear that when we want to go from to Quantum Mechanics to Classical Mechanics, one have to make h go to 0 and then the magic happens. However, when I think of the momentum operator:
p=i*h*nabla

when we make h go to 0, p goes to 0. What does "p operator equal to zero" means?
 
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Hello the_pulp,

the question how and whether wave mechanics reduces to classical mechanics is very interesting indeed.

The reason why your example doesn't work is that the momentum operator is not momentum "quantity". The operator is just an additional mathematical concept that has no use in classical mechanics.

When this operator acts on the wave function of free particle [itex]e^{i\frac{p_x x}{\hbar}}[/itex], it gives the same function multiplied by [itex]p_x[/itex], which we interpret as momentum quantity, and this need not go to zero with h -> 0.

More generally, the statement of reduction of wave mechanics into classical mechanics is usually understood in following sense:

By means of substitution [itex]\psi = e^{iS/\hbar}[/itex], Schroedinger's equation can be written as

[tex] -\partial_t S = \frac{-i\hbar}{2m} \Delta S + \frac{(\nabla S)^2}{2m} + V.[/tex]

When h - > 0, the term with the imaginary unit and h goes to zero and the equation approaches the Hamilton-Jacobi equation for the action [itex]S[/itex] from classical mechanics.
 
the_pulp said:
I have a doubt making a little of thinking of basic notions of QM and I think that the answer should be very simple but I can't make up my mind. So, here I go:
I usually hear that when we want to go from to Quantum Mechanics to Classical Mechanics, one have to make h go to 0 and then the magic happens. However, when I think of the momentum operator:
p=i*h*nabla

when we make h go to 0, p goes to 0. What does "p operator equal to zero" means?

Note that taking the limit as h goes to zero is not the same as setting h to zero in all equations. We should look at what changes when we make the value of h smaller, and extrapolate these changes to their limit as h goes to zero. What we find is that if we make h smaller, things get "more classical."

We can look at what happens to the momentum operator as h gets smaller. Suppose h is very small. Then for a particle to have a reasonably large momentum it must have an extremely short wavelength. As we make h smaller, the wavelength of a particle with a given momentum becomes smaller, because this wavelength is given by h/p. Then for instance wave interference effects, which happen in regions whose size is comparable to one wavelength, become confined to very small regions. If we make h small enough that the momentum we care about corresponds to an extremely tiny length that is way below anything we can detect, then we no longer observe interference effects and recover classical mechanics.
 

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