Question

QUUUU
• The momentum operator p/ for
the particle along x axis is​

Answer 1

Answer:

n this case f(x) is called an eigenfunction of ˆA and a the corresponding eigenvalue. and since by the de Broglie relation ¯hk is the momentum p of the particle, we have pf(x) = pf(x) Note that this explains the choice of sign in the definition of the momentum operator!

Answer 2

The momentum operator p/ for the particle along the x-axis is​ the fundamental operator.

• Neither the operator nor the wave function has any meaning.
• An operator is a mathematical commodity that enco des a "quantum version of observable" (while the observable itself is a parameter), and a wave function is an abstract "thing" that represents quantum states (often in the form of a complex vector).
• If we square that wave function or apply an operator to it, we will receive the parameter's average/expectation value.
• As a result, the relationship is essentially mathematical because the operator and wavefunction are mathematical objects.