Rotation Operator: Interaction between Two-Level Atom in {|g>, |e>} Basis

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The discussion focuses on the interaction of a two-level atom represented in the {|g>, |e>} basis using the rotation operator R(t) = exp[i(σz + 1)ωt/2]. The Pauli matrix σz is defined in this basis, and the operator is expressed as a matrix with elements related to the exponential of its diagonal terms. The user seeks clarification on whether the exponential of a diagonal matrix applies uniformly across its elements, specifically questioning if exp(0) in one position implies the same for others. The consensus confirms that for diagonal matrices, the exponential is indeed derived from the exponential of the diagonal elements. Understanding matrix exponentiation is essential for accurately representing the rotation operator in quantum mechanics.
zDrajCa
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Hi, I'm working on the interaction between a two level atom (|g>, |e>)
In my exercise we have to use the rotation operator :

R(t)=exp[i(σz+1)ωt/2]

with σz the pauli matrix which is in the {|g>,|e>} basis :
(1 0)
(0 -1)

If i want to represent my rotation operator in the {|g>,|e>} basis. Then i can do:
σz +1 = (1 is the identity matrix)
( 2 0)
( 0 0)

Do my operator is :
(exp(iwt) 0 )
( 0 exp(0) )

Thanks for your answers.
 
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I'm not familiar with taking exponential of a matrix. But if you had exp(0) in the lower right position, would it also be exp(0) in the upper right and lower left?
 
i have read that if a matrix was diagonal (my case right) ,then the exponential of the matrix is the exponential of his diagonal term
 
scottdave said:
I'm not familiar with taking exponential of a matrix. But if you had exp(0) in the lower right position, would it also be exp(0) in the upper right and lower left?
https://en.wikipedia.org/wiki/Matrix_exponential
It can be expanded in a Taylor series.
zDrajCa said:
the exponential of the matrix is the exponential of his diagonal term
Yes, this is right.
 
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