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

In summary, the statement that "p operator equals to zero" means that the momentum quantity has gone to zero. This happens when we make the value of h smaller in classical mechanics.
  • #1
the_pulp
207
9
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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  • #2
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.
 
  • #3
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.
 

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

What is the H=0 and momentum operator in quantum mechanics?

In quantum mechanics, the H=0 and momentum operator is a mathematical representation of the energy and momentum of a particle in a given system. It is used to calculate the probability of finding a particle at a certain position and time.

How is the H=0 and momentum operator used in quantum mechanics?

The H=0 and momentum operator is used to solve the Schrödinger equation and determine the energy states of a particle in a given system. It is also used to calculate the expectation value of a particle's position and momentum.

What is the significance of the H=0 and momentum operator in quantum mechanics?

The H=0 and momentum operator is a fundamental concept in quantum mechanics as it helps us understand the behavior of particles at the atomic and subatomic level. It allows us to make predictions about the energy and momentum of particles in a given system.

How is the H=0 and momentum operator related to the uncertainty principle?

The H=0 and momentum operator is related to the uncertainty principle, which states that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa. The H=0 and momentum operator is used to calculate the uncertainties in position and momentum of a particle.

Can the H=0 and momentum operator change over time?

Yes, the H=0 and momentum operator can change over time in quantum mechanics, as it is dependent on the energy states of the particle. When a particle interacts with its environment, its energy and momentum may change, causing the H=0 and momentum operator to change as well.

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