Is Acceleration Quantization Valid in Quantum Mechanics?

[Sangoku]

acceleration quantization ??

If [tex] x \Psi (x,t)=x \Psi(x,t) [/tex]

and [tex] p \Psi (x,t)= -i\hbar \partial _{x} \Psi (x,t) [/tex]

then should it be [tex] a \Psi (x,t) = \dot p \Psi (x,t)= \hbar ^{2} \partial _{xt} \Psi (x,t) [/tex] using usual QM

So, the direct quantization of motion equation (constraint) should it read:

[tex] \hbar^{2} \partial _{xt} \Psi (x,t)+ \frac{dV}{dx}\Psi(x,t)=0 [/tex]

 

[dextercioby]

Well, the main idea is that QM and CM can be connected using the famous "canonical quantization rules" iff the CM is presented in the Hamiltonian formulation in which there's no such thing as "generalized accelerations", since the fundamental variables are the generalized coordinates and velocities.

However, in the Heisenberg picture one can define the velocity and acceleration operators by

[tex]\hat{\vec{v}} :=\frac{d\hat{\vec{r}}}{dt} [/tex]

and

[tex] \hat{\vec{a}} :=\frac{d\hat{\vec{p}}}{dt} [/tex]

Their expressions can be found using the Heisenberg equations of motion which are known to be equivalent to the SE. But i don't see a way to represent [itex] \hat{\vec{a}} [/itex] as an operator on the [itex]L^{2}\left(\mathbb{R}^{3}, d^{3}x, \mathbb{C}\right) [/itex].

I don't think, if, let's say, it would be possible to represent it, it would have a mixed derivative in it. No reason, just a hunch.

 

[hellfire]

If [tex] x \Psi (x,t)=x \Psi(x,t) [/tex]

and [tex] p \Psi (x,t)= -i\hbar \partial _{x} \Psi (x,t) [/tex]

then should it be [tex] a \Psi (x,t) = \dot p \Psi (x,t)= \hbar ^{2} \partial _{xt} \Psi (x,t) [/tex] using usual QM

So, the direct quantization of motion equation (constraint) should it read:

[tex] \hbar^{2} \partial _{xt} \Psi (x,t)+ \frac{dV}{dx}\Psi(x,t)=0 [/tex]

I would proceed in a different way. You should find out the hamiltonian and then apply the usual substitutions [itex]p \rightarrow -i \hbar \nabla[/itex] in order to formulate the Schroedinger equation. To calculate the hamiltonian I would start with a lagrangian:

[tex]L = \frac{1}{2} m (\dot x - a t)^2[/tex]

It is interesting that you will find out that, if you are free to choose the phase of the wavefunction, then the Schroedinger equation in an accelerated frame is equal to the Schroedinger equation in a uniform gravitational potential.

See this.