Why does the kinetic operator depend on a second derivative?

[Steven Hanna]

The formula T = -(ħ/2m)∇² implies that T is proportional to the second spatial derivative of a wavefunction. What is the origin of this dependence?

In classical mechanics, T = p²/2m. Is it also the case in classical mechanics that p²/2m is proportional to a second spatial derivative? I have not been able to relate (d/dt)² to d²/dx² using classical mechanics.
Thanks,
Steven

 

[drvrm]

The formula T = -(ħ/2m)∇2 implies that T is proportional to the second spatial derivative of a wavefunction. What is the origin of this dependence?

check the expression for T -i think it should be hbar^2 may be a typo.
as we know the kinetic energy operator in coordinate representation is the above expression as differential operator.
it can be derived from the operator form of momentum

P(operator) = -ihbar. Del or in one dimension -ihbar. d/dx
as T= P^2/2m the operator form will be -hbar^2 /2m . d^2/dx^2

 

[Steven Hanna]

check the expression for T -i think it should be hbar^2 may be a typo.
as we know the kinetic energy operator in coordinate representation is the above expression as differential operator.
it can be derived from the operator form of momentum

P(operator) = -ihbar. Del or in one dimension -ihbar. d/dx
as T= P^2/2m the operator form will be -hbar^2 /2m . d^2/dx^2

Thanks! Let me make sure I understand your explanation: squaring the p(operator) gives (-ihbar∇)^2 = -hbar^2*(∇∙∇), and ∇∙∇ = ∇^2 because the dot product of a vector with itself is equal to the magnitude of that vector squared. Therefore, T = p^2/2m = (-hbar^2/2m)∇^2. The kinetic energy of a classical wave would also be proportional to ∇^2, because the momentum of a classical wave is proportional to ∇.

 

[blue_leaf77]

Thanks! Let me make sure I understand your explanation: squaring the p(operator) gives (-ihbar∇)^2 = -hbar^2*(∇∙∇), and ∇∙∇ = ∇^2 because the dot product of a vector with itself is equal to the magnitude of that vector squared. Therefore, T = p^2/2m = (-hbar^2/2m)∇^2.

Yes.

The kinetic energy of a classical wave would also be proportional to ∇^2, because the momentum of a classical wave is proportional to ∇.

No, the macroscopic, nonrelativistic kinetic energy is always given by $\frac{1}{2}mv^2$.

 

[bhobba]

The formula T = -(ħ/2m)∇² implies that T is proportional to the second spatial derivative of a wavefunction. What is the origin of this dependence?

Strange but true, and one of the deepest discoveries of modern physics, the answer is symmetry. You will find the detail in chapter 3 of Ballemtine:
https://www.amazon.com/dp/9814578584/?tag=pfamazon01-20

It involves advance math but is still worth taking a look at.

Thanks
Bill

 

[PeroK]

check the expression for T -i think it should be hbar^2 may be a typo.
as we know the kinetic energy operator in coordinate representation is the above expression as differential operator.
it can be derived from the operator form of momentum

P(operator) = -ihbar. Del or in one dimension -ihbar. d/dx
as T= P^2/2m the operator form will be -hbar^2 /2m . d^2/dx^2

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