Can Zero Be a Valid Eigenvalue for an Eigenstate?

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SUMMARY

The discussion confirms that zero can indeed be a valid eigenvalue for an eigenstate. Specifically, when an operator O acts on a function ψ, resulting in Oψ = kψ, where k = 0, the eigenstate remains valid. An example provided is the operator d²/dx² acting on the function ψ = 5x, which demonstrates that eigenvectors can exist with an eigenvalue of zero.

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  • Understanding of linear operators in quantum mechanics
  • Familiarity with eigenvalues and eigenstates
  • Basic knowledge of differential operators, specifically d²/dx²
  • Concept of scalar multiplication in vector spaces
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  • Explore the implications of zero eigenvalues in quantum mechanics
  • Learn about the role of differential operators in functional analysis
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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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