I'm wondering, in the Schrödinger equation, $$ {\displaystyle i\hbar {\frac {d}{dt}}\vert \Psi\rangle ={\hat {H}}\vert \Psi } $$ what is the $ {\displaystyle {\frac {d}{dt}}} $ ? I understand it has to do with time, t, but I guess i'm hoping for some more understanding into what d is and how this derivative works, and what it means to the whole.
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$\begingroup$ I've deleted some comments that responded impolitely to an earlier version of the question. $\endgroup$– rob ♦Commented Apr 6, 2020 at 15:55
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1$\begingroup$ Does this answer your question? Time derivative in Schrödinger equation $\endgroup$– jacob1729Commented Apr 6, 2020 at 16:19
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1 Answer
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1) We are in the Schrodinger picture. Physical states are labelled by a state vector $|\psi(t)\rangle$, which evolves in time as the state evolves in time. The time $t$ is a parameter-it labels different state configurations. This is a 'ket'-it stores all possible information about the system, at the time $t$.
2) We have, by definition, $$\frac{d}{dt}|\psi(t)\rangle\equiv \lim_{\epsilon\to 0}\frac{|\psi(t+\epsilon)\rangle-|\psi(t)\rangle}{\epsilon}$$ i.e. it's a measure of how different does the state look at $(t+dt)$ from $t$.
3) The schrodinger equation, then, is an equation that tells you the CAUSAL time evolution of the state of the system-if it were, so to speak, left to itself. It doesn't tell you what happens to the system when you measure it-that is a complicated issue. All it says is, if I specify what my system was in the beginning, $|\psi(0)\rangle$, I have here a first order equation that will allow me to predict what my system will be like at some $t$, i.e. I can solve this equation for $|\psi(t)\rangle$. And this is precisely the aim-predict how an initial state evolves.
4) It is crucial to note here that we only have a $d/dt$, and not also something like a $d/dx$ here. This is because the only parameter of the theory here is $t$-position $x$ etc come in as dynamical variables (i.e. evolve in time). There is a meaning to things like 'velocity=$dx/dt$'. Time is treated as SPECIAL, in both quantum and classical mechanics.
In QFT, which is relativistic by construction, all $x,t$ are treated as parameters and not dynamical variables. This is a significant departure, and this is why the equations of motion there have all $4(3+1)$ derivatives. In this sense, regular QM is called $(0+1)$ dimensional QFT, with only one parameter-$t$.
The differential equation in $t$ now describes your physics.
