Can Zero Be a Valid Eigenvalue for an Eigenstate?

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If an operator (O) acts on a function ψ and transforms the function in a scalar manner as described below, it is said to be in an eigenstate:
Oψ=kψ
in this case, O is the operator and k some scalar value.

My question is essentially if k=0, can this still be a valid eigenstate?

for example, O could be d2/dx2 and ψ could be 5x; would that be an eigenstate?
 
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fapyfapy said:
If an operator (O) acts on a function ψ and transforms the function in a scalar manner as described below, it is said to be in an eigenstate:
Oψ=kψ
in this case, O is the operator and k some scalar value.

My question is essentially if k=0, can this still be a valid eigenstate?

for example, O could be d2/dx2 and ψ could be 5x; would that be an eigenstate?

Sure it would. You can have eigenvectors with eigenvalue 0.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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