Question #196636

Let us consider a Hermitian operator 𝐴̂, with eigenvalues  a1=1/2a_1 = 1/2 , a2=3/2a_2 = 3/2 and a3=5/2a_3=5/2 operating in a 1-Dimensional space.


(a) Can this operator be associated with a measurable quantity? Provide a brief justification for your answer ?


(b) Describe the meaning of degeneracy of the operator and demonstrate that all the eigen-states of 𝐴̂ are non-degenerate.



1
Expert's answer
2021-05-23T16:36:15-0400

(a) Yes, this operator be associated with a measurable quantity. We actually construct a Hermitian operator to represent a particular measurable property of a physical system.


(b) A value of energy is said to be degenerate if there exist at least two linearly independent energy states associated with it. Moreover, any linear combination of two or more degenerate eigenstates is also an eigenstate of the Hamiltonian operator   corresponding to the same energy eigenvalue.


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