The operator e A has a meaning if it is expanded as a power series: Show that if

Question

The operator eA has a meaning if it is expanded as a power series:

Show that if

a (or in Dirac notation /a>) is an eigenstate of A with eigenvalue a, then it is also an eigenstate of eA, and find its eigenvalue (Remeber A2/a> is interpreted as AA/a> and so has eigenvalue a2)

So lost!!! Please help!!!

a (or in Dirac notation /a>) is an eigenstate of A with eigenvalue a, then it is also an eigenstate of eA, and find its eigenvalue (Remeber A2/a> is interpreted as AA/a> and so has eigenvalue a2)

So lost!!! Please help!!!

Explanation / Answer

Once you accept the power series.

A|a> = a|a> because |a> is the eigenstate of operator A.

e^A|a> = 1/n! A^n |a>

A^n |a> = A^(n-1) A|a> = A^(n-1) a|a>.

The important thing here is that A^(n-1)a = aA^(n-1) because a is just a number. Repeating this procedure, you can see that

A^n|a> = a^n|a>

Therefore e^A|a> =1/n! a^n |a>

but 1/n! a^n =e^a. This is the very definition of e^a. Therefore |a> is also an eigenvector of e^A, and its eigenvalue is e^a.

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