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Kit
Feb9-04, 09:56 AM
please refer to the attachment

in the last part, it states that...

If solutions phi.n.(r) can be found, for different values of En, then the linear combination is also a solution of the Schroedinger equation, because the Schroedinger equation is a linear equation. However, such a linear combination does not represent a particle with a well defined energy.

if it is not represent a particle with a well defined energy, then what is it represent?

does it represent all the energy states from ground state to En?

thx for answering

RedX
Feb9-04, 12:50 PM
Let i be the energy representation
|psi>=Sum{|i><i|psi>}
Transforming to the position representation j
|psi>=Sum{Sum{|j><j|i>}<i|psi>}
|psi>=Sum{u_i(x)e^(-(E_i)t/h}<i|psi>}

So a superposition of the states of definite energy represents a state with the probability <i|psi> of having energy E_i.

So the Schrodinger equation (the equation which governs the dynamics of the wave equation in the language of position space) must be linear for it to give a correct representation of the quantum system.