In classical mechanics, we have the following relation based on the kinematics with a constant acceleration (namely driven by a constant force):

$$v(t_2)^2-v(t_1)^2 = 2 a x \qquad \qquad (1)$$

and also we all know the dynamical evolution of classical objects governed by the Newtonian's law:

$$F=ma \qquad \qquad (2)$$

combining the formula $(1)$ and $(2)$ one can get the following equation

$$\dfrac{1}{2}mv(t_2)^2-\dfrac{1}{2}mv(t_1)^2 = Fx \qquad \qquad (3)$$

of course, we can generalize the $eq.(3)$ to a more general situation (driven by a variable force and living in a higher dimension).Finally one can get work-energy theorem:

$$\dfrac{1}{2}mv(t_2)^2-\dfrac{1}{2}mv(t_1)^2 = \int_A^B \textbf{F} \cdot d \textbf{X} \qquad \qquad (4)$$

If the force $\textbf{F}$ is a conservative force then one can obtain the following famous $\textbf{law of conservation of energy}$:

$$K_1+U_1 = K_2+ U_2 \qquad \qquad (5)$$

where $K=\dfrac{1}{2}mv^2$ and $\textbf{F} = -\nabla U$.Everything is so fine.

So my question is about can we follow the same logic in quantum mechanics? I mean can we derive the law of conservation of energy from Schrodinger equation? Or what's the exact meaning for the conservation of energy in quantum world ?

  • $\begingroup$ Related: physics.stackexchange.com/questions/203770/… $\endgroup$ Jan 5, 2017 at 12:06
  • 1
    $\begingroup$ The Hamiltonian operator is the energy operator, and it is the generator of time evolution as it appears in the Schroedinger equation. Therefore on any quantum state, the evaluation of the energy commutes with time evolution, or in other words energy is conserved. $\endgroup$
    – yuggib
    Jan 5, 2017 at 13:16
  • $\begingroup$ How about the Hamiltonian is time-dependent? Is it still commuting and then conservative ? $\endgroup$
    – Jack
    Jan 5, 2017 at 13:36
  • $\begingroup$ No, it is not. Nether in classical Hamiltonian mechanics the Hamiltonian is conserved along the motion of the system when it explicitly depends on time because $\frac{dH(t, q(t),p(t))}{dt} = \frac{\partial H}{\partial t}|_{(t,q(t),p(t))}$. $\endgroup$ Jan 5, 2017 at 15:02


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