Can Zero Be a Valid Eigenvalue for an Eigenstate?

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In quantum mechanics, an operator acting on a function can yield a scalar multiple of that function, defining an eigenstate. The discussion centers on whether zero can be a valid eigenvalue for an eigenstate. It is confirmed that eigenvectors can indeed have an eigenvalue of zero. An example provided is the operator d2/dx2 acting on the function 5x, which can be considered an eigenstate with an eigenvalue of zero. Therefore, zero is a valid eigenvalue in the context of eigenstates.
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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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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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