- #1
nosafeway
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I am learning quantum mechics. The hypothesis is:
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 eigen wave function?
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 eigen wave function?