# What is the $d/dt$ in the Schrödinger equation? [duplicate]

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.

• I've deleted some comments that responded impolitely to an earlier version of the question. – rob Apr 6 at 15:55
• Does this answer your question? Time derivative in Schrödinger equation – jacob1729 Apr 6 at 16:19

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.

In Schrödinger equation $$\hat{\mathcal{E}}=i\hbar\frac{\partial}{\partial t}$$ is the generator of translation in time which, according to the Noether's theorem corresponds to the energy. This is quite similar to the momentum operator components $$\hat{p}_j=i\hbar\frac{\partial}{\partial x_j},$$ which are the generators corresponding to the infinitesimal translations in space. The Schrödinger equation is thus simply the non-relativistic relation between the energy an the momentum, $$\hat{\mathcal{E}}= \frac{\hat{\mathbf{p}}^2}{2m},$$ transformed to quantum mechanics via the correspondence principle - i.e. by replacing the classical quantities with their operators.

Note that Klein-Gordon and Dirac equations are obtained in a similar fashion, but using the relativistic relation $$\hat{\mathcal{E}}^2=\hat{\mathbf{p}}^2c^2 + m^2c^4,$$ and taking the necessary care of the spin.