I am learning quantum mechics. The hypothesis is:(adsbygoogle = window.adsbygoogle || []).push({});

In the quantum mechanics, all operators representing observables are Hermitian, and their eigen functions constitute complete systems. For a system in a state described by wave function ψ(x,t), a measurement of observable F is certain to return one of the eigenvalues of the operator F.

In the 1D infinite well, the position x should be an observable, its operator x is Hermitian, and its eigen function satisfies xg(x)=x'g(x), leading to g(x)=δ(x-x'), the eigenvalue x' is any real number. But g(x)=δ(x-x') cannot be normalized, and it cannot be a wave function since a wave function must be normalizable. Then my question is:

1. Is the position x a Hermitian operator?

2. If it is, why there is no eigenwavefunction?

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# Does position operator have eigen wave function?

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