Can anyone please explain me in details about what are the assumptions made for the derivation of Schrödinger Wave Equation?
1 Answer
You can read further about this in the postulates of quantum mechanics.
The assumptions are as follows about
- The Hamiltonian operator
#hatH# is a linear operator (it distributes through addition, like all other quantum mechanics operators):
#hatH(A(r) + B(r)) = hatHA(r) + hatHB(r)# It must also be Hermitian, in order to correspond to a physical quantity known as an observable.
This allows us to write things like:
#int_"allspace" psi^"*"hatHpsid tau = int_"allspace" (hatHpsi)^"*"psid tau# ,
sometimes called the turnover rule
#hatH^† = (hatH^"*")^T = (hatH^T)^"*" = hatH#
(the adjoint of#hatH# is itself.)We are used to using scalar Hamiltonians
#hatH# , so typically we just say#hatH^"*" = hatH# and don't worry about the transposition part of the adjoint operator#†# .We also typically use time-independent Hamiltonians, though time-dependent Hamiltonians do exist.
- The wave function, or state function
#psi# , belongs to a complete set of eigenfunctions#{psi_i}# such that:
#int_"allspace" psi_n^"*"psi_nd tau = 1# ,
for the#n# th state of the particle having a probability of#100%# for the particle being found in the entire universe.
#int_"allspace" psi_m^"*"psi_nd tau = overbrace(delta_(mn))^"Kroenecker Delta" -= {(0, m ne n),(1, m = n):}# ,
for the#m# th and#n# th states overlapping in the relevant boundaries. The condition of#m = n# is called normality, and the condition of#m ne n# is called orthogonality. Together they form the orthonormality conditions of#psi# .
#hatHpsi= Epsi# follows, i.e.#psi# must be prepared in a state that is an eigenstate of#hatH# that gives the energy#E# (operating with#hatH# must give#E# ). Otherwise, either#hatH# or the wave function is not well-chosen.Each
#psi_i# in this complete set must be linearly independent, i.e. they can be written as linear combinations#psi_(i) = sum_k c_(ik) phi_k# that do not equal another#psi_i# in the set.
- The wave function
#psi# also must be "well-behaved", so it must follow these conditions:
#psi# is finite in the relevant boundaries.
#psi# goes to zero at#pmoo#
#psi# is single-valued and both#psi# and#(dpsi)/(dr)# continuous. (#r# is#sqrt(x^2+y^2+z^2)# .)
#psi# is square-integrable (so that the differential equation can be solved in the first place), i.e.#int_"allspace" psi^"*"psid tau# has a solution.