Momentum operator in Schrodinger equation

Because, in (non relativistic) quantum mechanics, all observable quantites are replaced by operators.

http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/qmoper.html

http://www.cobalt.chem.ucalgary.ca/ziegler/Lec.chm373/lecture6.html

http://vergil.chemistry.gatech.edu/notes/quantrev/node1.html

http://en.wikibooks.org/wiki/Quantum_Mechanics/Operators_and_Commutators

etc.

 

A key basic idea in QM is that the commutator of p(momentum) and q(position),

qp - pq = i (The tradition in high energy physics is to use units in which c=1, and h bar=1)

For his equation to hold, both q and p must be operators. Note also that this commutator is responsible for the Heisenberg Uncertainty Principle.

In fact this is an assumption, but a good one, and as basic as it gets. You would benefit from a bit of homework, as suggested by malawi_glenn above.
Regards,
Reilly Atkinson

 

He took one part de Broglie and one part Einstein and mixed it together... on a serious note, [tex]p \mapsto -i \hbar \tfrac{\partial}{\partial x}[/tex] follows from de Broglie's work and is based on analogies between classical mechanics of particles and waves.

 

That is not strange, every field has its postulates (axioms) that can't be derived nor be prooved.

However, you may postulate something, then from those axioms derive formulas and relations which can be verified and thus one can tell if you can build up a coherent system (a paradigm) or not.

Here you can read Schrodingers first publication about his new equation:
http://home.tiscali.nl/physis/HistoricPaper/Schroedinger/Schroedinger1926c.pdf

Hanve fun :-)