Why can momentum be expressed as an operator

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Momentum in quantum mechanics is expressed as an operator due to its relationship with wave functions, where the momentum operator acts on the wave function to yield momentum values. The differentiation of the wave function leads to the identification of momentum as an operator, specifically represented as -iħ (∂/∂x). This operator is crucial in the Schrödinger equation, where it interacts with the wave function to describe physical systems. While momentum is technically an eigenvector of the operator, the operator itself is commonly referred to as "momentum" for simplicity. Understanding this concept is essential for grasping the mathematical framework of quantum mechanics.
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Im having a hard time figuring out how in quantum mechanics things such as momentum can be expressed as an operator.

I know the simple algebra to get the relation. Starting with the 1D solution to wave equation\Psi=e^{i\omega x} then differentiating that with respect to x and replacing the resulting k with the de Broglie relation you get \frac{\partial \Psi}{\partial x}=i\frac{p}{\hbar}\Psi.

From here I can sort of see that p may =-i\hbar \frac{\partial }{\partial x} but I just don't get how a value such as momentum can be a operator.

Is it that to get the momentum from the wave function you need to perform this operator on the wavefunction, and since every term in the Schrodinger equation has the wavefunction in it, they simply represent momentum as the operator? Even if this is true, there has to be more to it than being a simple variable substitution right?
 
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hi bfusco! :smile:
bfusco said:
… how in quantum mechanics things such as momentum can be expressed as an operator.

technically, momentum is the eigenvector of the operator, not the operator itself

however, the operator needs a name, so we call that "momentum" too! :rolleyes:

if this was biology … where they pay such detailed attention to nomenclature … we'd probably call it "momentumase" :wink:
 
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