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∂
I've some doubts in the basic quantum mechanics.I know these are too simple. But can't figure out the exact answer.
1. Ordinary wave equation relates second order time derivative to second order space derivative.But Schrödinger equation connects first order time derivative with second order space derivative.Why this difference?
2. I've seen in a book that wavefunction should have a dimension of L-3/2(because |ψ|2dx should be dimensionless).Then why can't we measure wavefunction?
3.From Schrödinger equation, ψ should be continuous.But how can we say that ∂ψ/∂x should be continuous also?
1. A book says this is to make wave function independent of the properties of the particle(if wave function for free particle is considered).But i can't understand it properly.
3.I think (may be wrong) when ψ is a continuous function of x, ∂ψ/∂t should be a continuos function of x. ie, LHS of Sch. equation is continous. Then RHS also need to be continuous. But it don't imply that ∂ψ/∂x should be continuous.
Homework Statement
I've some doubts in the basic quantum mechanics.I know these are too simple. But can't figure out the exact answer.
1. Ordinary wave equation relates second order time derivative to second order space derivative.But Schrödinger equation connects first order time derivative with second order space derivative.Why this difference?
2. I've seen in a book that wavefunction should have a dimension of L-3/2(because |ψ|2dx should be dimensionless).Then why can't we measure wavefunction?
3.From Schrödinger equation, ψ should be continuous.But how can we say that ∂ψ/∂x should be continuous also?
The Attempt at a Solution
1. A book says this is to make wave function independent of the properties of the particle(if wave function for free particle is considered).But i can't understand it properly.
3.I think (may be wrong) when ψ is a continuous function of x, ∂ψ/∂t should be a continuos function of x. ie, LHS of Sch. equation is continous. Then RHS also need to be continuous. But it don't imply that ∂ψ/∂x should be continuous.