A nice way of arriving the the SE is by looking at time evolution of a quantum state' date=' and modeling this through a unitary operator.
Let the state [imath']| \alpha \rangle [/imath] evolve with time as :
[math]| \alpha \rangle ~ \xrightarrow{time~evolution} | \alpha,t \rangle [/math]
We create an opertor [imath]{\cal U} (t,0) [/imath] that provides this time-evolution, so that :
[math] | \alpha ,t \rangle = {\cal U} (t,0) ~| \alpha \rangle [/math]
Now, this time evolution operator is required to satisfy a whole bunch of properties. For instance, one must have
[math] \lim _{dt \rightarrow 0} {\cal U} (dt,0) = \mathbf {1} [/math]
All these requirements are satisfied by writing [imath] {\cal U} (dt,0) = \mathbf {1} - i \mathbf{\Omega} dt[/imath], where [imath]\mathbf{\Omega} [/imath] is a Hermitian operator. Now borrowing from classical mechanics, the idea that the Hamiltonian is the generator of time-translation (just as momentum is the generator of spatial translations), we choose [imath]\mathbf{\Omega} [/imath] to be [imath] {\cal H} / \hbar [/imath]. And the Schrodinger Equation follows automatically from playing with the composition property of the time-evolution operator. This property requires
[math] {\cal U} (dt_1 + dt_2, 0) = {\cal U} (dt_2,dt_1) ~{\cal U} (dt_1, 0) [/math]
The need for this property is obvious : if you evolve a state through time [imath]dt_1[/imath] and then through [imath]dt_2[/imath], you expect the final state to be the same as that caused by a time-evolution through [imath]dt_1 + dt_2[/imath].
Now, we can use this composition property to write
[math] {\cal U} (t+dt, 0) = {\cal U} (t+dt,t) ~ {\cal U} (t,0) = \left( \mathbf{1} - \frac{i {\cal H} dt}{\hbar} \right) {\cal U} (t,0) [/math]
Multiplying out, and rearranging terms, this can be written in the differential form :
[math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0) = {\cal H} {\cal U} (t,0) [/math]
Oprating this on the initial state ket [imath] | \alpha \rangle = |\alpha, t=0 \rangle [/imath] gives
[math]i \hbar \frac {\partial}{\partial t} {\cal U} (t,0)~| \alpha \rangle = {\cal H} {\cal U} (t,0)~| \alpha \rangle [/math]
But from the definition of the time-evolution operator, this gives
[math]i \hbar \frac {\partial}{\partial t} | \alpha , t \rangle = {\cal H} | \alpha ,t \rangle [/math]
ta dah !