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The energy operator is:

$${\displaystyle {\hat {E}}=i\hbar {\frac {\partial }{\partial t}}}\tag1$$

and the momentum operator is

$${\displaystyle {\hat {p}}=-i\hbar {\frac {\partial }{\partial x}}}.\tag2$$

I know that we obtain the first by differentiating the plane wave solution of the Schrodinger equations with $t$ and the second by $x$.

Say I have the expression for one of them, can I derive the other one using the Hamiltonian equations of motion from classical mechanics?

Can I obtain $(1)$ from $(2)$ or vice versa via:

$${\displaystyle {\frac {\mathrm {d} {\boldsymbol {q}}}{\mathrm {d} t}}={\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {p}}}},\quad {\frac {\mathrm {d} {\boldsymbol {p}}}{\mathrm {d} t}}=-{\frac {\partial {\mathcal {H}}}{\partial {\boldsymbol {q}}}}.}$$

Qmechanic
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User198
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2 Answers2

1

As it stands, the question is meaningless since Eq.(1) is not the definition of an operator in the Hilbert space as, instead, Eq.(2) is.

The point is that $t$ labels different states in the Hilbert space. Contrarily, an operator acts at given time on a unique given state.

Up to factors, the Schroedinger equation equals the action of an operator $H\psi_t$ (on the state at a certain time), with the $t$ derivative of a curve taking values in the Hilbert space $\mathbb{R} \ni t \mapsto \psi_t$. Eq.(1) refers to this second type of action.

$H\psi$ does not need a $t$ parametrized curve of states to be defined.

Concersely, $\frac{d}{dt}\psi_t$ needs such a curve to be defined.

Maybe, taking the remarks above into accout,it is possible to turn the question into a meaningful one.

Valter Moretti
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There's a conceptual mistake there. What follows could find an answer with the relationship between Poisson brackets in classical mechanics and the commutator in quantum mechanics.

The conceptual mistake. The conceptual mistake you make is the same you'd make in classical mechanics defining the momentum saying that it's the time derivative is the external force $\mathbf{F}$ acting on the system.

In classical mechanics, the equation of motion tells that ime derivative of the momentum equals (and NOT is) the force acting on it.

In general, you have:

  • the state of the system
  • the set of forces acting on it
  • the equation of motion relating the evolution of the state of the system with the set of forces acting on it

Relationship between Poisson brackets and commutator in the equations of motion. Hamilton equations of motion in classical mechanics read

$$\begin{aligned} \frac{d \mathbf{q}}{dt} & = \{ \mathbf{q}, H \} \\ \frac{d \mathbf{p}}{dt} & = \{ \mathbf{p}, H \} \end{aligned}$$

while the time derivative of the operators (in Heisenberg picture, see https://en.wikipedia.org/wiki/Heisenberg_picture) reads

$$\begin{aligned} \frac{d \hat{\mathbf{q}}}{dt} & = - \frac{i}{\hbar} [ \hat{\mathbf{q}}, H ] \\ \frac{d \hat{\mathbf{p}}}{dt} & = - \frac{i}{\hbar} [ \hat{\mathbf{p}}, H ] \ , \end{aligned}$$

with the correspondence between classical and quantum equations,

$$[\hat{A}, \hat{H}] \quad \leftrightarrow \quad i \hbar \{ A, H \} \ .$$

basics
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