# Zero Planck constant

• fricke

#### fricke

What is the effect in quantum if Planck constant is zero?

Here are some points that I could think of:
if Planck constant is zero, the Heisenberg Uncertainty Principles will become zero, therefore either momentum or position of a particle can be known exactly.

if Planck constant is zero, both momentum and position can be known simultaneously.

if Planck constant is zero, all quantum behavior will become classical behavior, where energy will become continuous, momentum will become zero (which means it is known exactly) and etc.

So, are my points correct? and please do tell if there are other points for "if Planck constant is zero".
Thank you.

if Planck constant is zero, the Heisenberg Uncertainty Principles will become zero, therefore either momentum or position of a particle can be known exactly.
Either momentum or position can be known exactly even if the Planck constant is not zero.

if Planck constant is zero, both momentum and position can be known simultaneously.

if Planck constant is zero, all quantum behavior will become classical behavior, where energy will become continuous, momentum will become zero (which means it is known exactly) and etc.
Those two statements are correct.

Either momentum or position can be known exactly even if the Planck constant is not zero.
Hmm, so, if Planck constant is zero means that there is no uncertainty in both momentum and position, right? Does it mean if either one of momentum or position can be known exactly, therefore the other one is not known exactly?

Hmm, so, if Planck constant is zero means that there is no uncertainty in both momentum and position, right? Does it mean if either one of momentum or position can be known exactly, therefore the other one is not known exactly?
Note the bolded "not" in my sentence that you quoted.

Note the bolded "not" in my sentence that you quoted.
Yes, I noted that.
I was asking the understanding of if the Planck constant is zero. :)

Hmm, so, if Planck constant is zero means that there is no uncertainty in both momentum and position, right? Does it mean if either one of momentum or position can be known exactly, therefore the other one is not known exactly?
If h=0, then both momentum and position can simultaneously be known exactly.

The idea of quantization of most physical quantities breaks down if ##h=0## ; the deBroglie wavelength associated with any particle becomes 0, the canonical commutation relation becomes ##[x,p] = 0## (the commutation of the position and momentum operators removes the "weird" effects in QM, and reduction to classical mechanics follows), and the wave equation of any particle takes the trivial and non-renormalizable solution of ##Ψ (r,t) =0 ## (there is no wave equation for any particle if h=0; this of course agrees with the fact that the deBroglie wavelength becomes 0). From this, you can also derive the fact that the Heisenberg 'uncertainty' vanishes.

the wave equation of any particle takes the trivial and non-renormalizable solution of Ψ(r,t)=0
That is actually not quite true. If the limit h=0 is applied in the appropriate way, the complex Schrodinger equation reduces to two decoupled real classical equations. One of them is the classical Hamilton-Jacobi equation, which is a wave equation for a classical particle. The other is the continuity equation, which can be interpreted in terms of classical statistical mechanics.

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• Demystifier
If the limit h=0 is applied in the appropriate way
I don't think the OP is talking about a limit. As I read it, the OP is asking what would happen if ##h## assumed the exact value of 0, not what would happen if it tended to 0. If ##h=0## , then the terms ##iħ \frac {∂Ψ}{∂t}## and ##\frac {-ħ^2}{2m} ∇^2## (in the Hamiltonian) will drop out, and in that case, Schrodinger's (time-dependent) wave equation would become ##V Ψ (r,t) = 0## , which can only be true for any potential function if ##Ψ (r,t) = 0 ## for all r and t.

• Derek Potter