Understanding Exponential of Operator: Expansion and Eigenvalues Explained

[KFC]

Hi there,
I learn from the text that the exponential of an operator could be expanded with a series such that

e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}
So if the eigenvalue of the operator \hat{A} is given as a_i
e^{\hat{A}}|\psi\rangle

will be a matrix with diagonal elements given as \exp(a_i), is that right?

So I am wondering what happen if \hat{A} is now written as a power form, i.e. \hat{A}^n, can we conclude that
So if the eigenvalue of the operator \hat{A} is given as a_i

e^{\hat{A}^n}|\psi\rangle

gives a matrix with diagonal elements as \exp(a_i^n) ?

 

[bhobba]

That's correct - for operators that can be diagonalised.

Its basically how functions of diagonalisable operators are defined - when diagonalised its a function of the eigenvalues.

Non diagonalisable operators are however a bit trickier - there are a few definitions I have seen and even a paper showing they are basically all equivalent - but the point is they are a more difficult proposition.

Thanks
Bill

 

[lugita15]

If \hat{A}|\psi\rangle = a_i|\psi\rangle, then e^{\hat{A}}|\psi\rangle= \exp(a_i)|\psi\rangle and e^{\hat{A}^n}|\psi\rangle = \exp(a_i^n)|\psi\rangle. You should be able to prove this pretty easily using the Taylor series definition you mentioned.

 

[tom.stoer]

afaik the proof of

f(A) = \sum_n f_n A^n

where f_n are the Taylor coefficients is rather involved mathematically. What you need is the spectral theorem for self-adjoint operators. For us physicists it reads

f(A) = f(A) \sum_n |a\rangle\langle a| = \sum_n f(a) |a\rangle\langle a|

where the states |a> are the eigenstates of A and form a complete set.