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if anyone can give assistance on this question it is much appreciated! Suppose I have a Hamiltonian $$H=\frac{p^{2}}{2m}+V(r)+F(r,t) $$ where $$F(r,t) = Q(r)\Re{(e^{{iT(t)}})}:\quad Q(r), T(t),T'(t) \neq 0$$ is seperable.

Now since $$ \frac{\partial H}{\partial t}=Q(r)\Re (ie^{iT(t)}\cdot \frac{d T}{dt}) \neq 0 $$ does this mean the Hamiltonian (being time dependent) is not conserved (this is my guess, since it does not equal zero)

Also since the Hamiltonian is time dependent and we have that $$T=\frac{p^2}{2m}, \quad V=V(r),\quad E=T+V \Longrightarrow H-E \neq 0 \Longleftrightarrow H \neq E $$ i.e. because we have that extra time dependent term, I'm guessing that signifies the Hamiltonian cannot be the total system energy. Any information/places is apppreciated :)

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Your Hamiltonian has the "potential" $U = V + F$, just because $F$ is time-dependent doesn't make it not part of the potential. You are correct that the Hamiltonian is not conserved.

The Hamiltonian is the total energy of the system if the Lagrangian does not contain terms linear in the velocity or when the generalized coordinates in the Lagrangian do not depend explicitly on time. In that case, the Hamiltonian is T+V and is automatically the total energy.

However, if these conditions are not met, it does not mean that $H$ is not the total energy. The obvious example here is the vector potential.

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