For this form of the equation:

$$\hat H|\psi(t)\rangle = i \hbar \frac{\partial{}}{\partial{t}}|\psi(t)\rangle.$$

For instance:

"The total energy of a quantum state at time t is equal to $i\hbar$ times the rate of change of the state w.r.t. time"?

I'm getting confused about how exactly to interpret the derivative.

  • $\begingroup$ The derivative has the exact sense from functional analysis, namely as a strong limit. SE is just the postulation that the limit exists in a separable Hilbert space and it is in the codomain/range of the Hamiltonian. In reverse, the domain of the Hamiltonian operator is the set of all vectors in a separable Hilbert space for which that limit in the right hand side exists. $\endgroup$ – DanielC Oct 16 '17 at 20:58
  • $\begingroup$ A translation I don't think should be a full answer: The Schrodinger equation is stating that $H$ is the generator of infinitesimal time evolutions. Similar to how $p = -\mathrm{i} \hbar \partial_x$ generates infinitesimal translations, so does $H = \mathrm{i}\hbar \partial_t$ generate infinitesimal 'translations in time.' $\endgroup$ – Diffycue Oct 16 '17 at 22:01

One ought to remember that the derivation of the Schrodinger equation is rather heuristic, and that rules like \begin{align} E\to \hat H&=i\hbar \frac{\partial }{\partial t} \tag{1} \\ p\to \hat p&=-i\hbar \frac{\partial }{\partial x}\tag{2} \end{align} were first “justified” using plane waves since it is empirically true that particles do exhibit wave behavior. Thus, using the Einstein and deBroglie relations $E=\hbar \omega$ and $p=\hbar k$ in the plane wave expression $$ \Psi(x,t)=A e^{i(px-Et)/\hbar} $$
one can recover the energy and momentum of the plane wave by taking the appropriate derivatives and then “factoring out” $\Psi(x,t)$: \begin{align} \hat H\Psi(x,t)&=i\hbar \frac{\partial }{\partial t}\Psi(x,t)=E\Psi(x,t)\, ,\\ \hat p\Psi(x,t)&=-i\hbar\frac{\partial }{\partial x}\Psi(x,t)=p\Psi(x,t)\, . \end{align}
In this sense the rules of (1) and (2) are just “tricks” to recover $E$ and $p$ from $\Psi(x,t)$ using derivative operators. The tricks encapsulate the observations that, for plane waves (expressed as complex exponentials), the rate of change in time of $\Psi(x,t)$ is related to the energy, while the rate of change in space of $\Psi(x,t)$ is related to the momentum.

Maybe the surprise is that the rules of Eqs.(1) and (2) obtained from plane waves remain valid even when one includes a potential $V(x)$ (and thus the solutions are no longer plane waves). In this case, the simplest thing is to extend the derivative rules of (1) and (2) to the full Schrodinger equation with potential $$ i\hbar\frac{\partial }{\partial t}\Psi(x,t)= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\Psi(x,t)+V(x)\Psi(x,t)\, . \tag{3} $$ so that this becomes compatible with the statement the total energy of the system is the sum of the kinetic plus potential: $$ E=\frac{p^2}{2m}+V(x)\, .\tag{4} $$

Using the standard separation of variables for (3), the “trick” to get the energy of the state is still an application of the time derivative to solutions of the form $\Psi(x,t)=e^{iEt/\hbar}\psi(x)$ where $\psi(x)$ now satisfies the time-independent Schrodinger equation, which ultimately is an expression of (4): $$ -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi(x)+V(x)\psi(x)=E\psi(x)\, . $$

The difficulty is that $\Psi(x,t)$ does not have a direct physical interpretation (as it is complex). The physically meaningful quantity is $\Psi(x,t)\Psi(x,t)^*$; for solutions of fixed energy, the time dependence goes away (these are steady state solutions). If $\Psi(x,t)$ is not of the factored form $e^{-iEt/\hbar} \psi(x)$ then the time derivative of $\Psi(x,t)$ is not proportional to itself, and the connection between the time derivative and the energy is lost: this is no surprise as these types of solutions are not interpreted as having definite energy.

  • 1
    $\begingroup$ Extremely minor comment: The "derivation" of the equation is heuristic, not the equation itself. $\endgroup$ – Javier Oct 16 '17 at 21:48
  • $\begingroup$ @Javier good catch. Fixed. $\endgroup$ – ZeroTheHero Oct 16 '17 at 21:50
  • $\begingroup$ The equation (3) makes people think that x and t have the same weight (up to order of partial differentiation in that PDE) in QM, whereas they don't. $\endgroup$ – DanielC Oct 16 '17 at 23:08
  • $\begingroup$ @DanielC I’m aware that this question can be answered with various levels of sophistication. I (and I will wager many in this community) would surely benefit from your contribution. $\endgroup$ – ZeroTheHero Oct 16 '17 at 23:46

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