- #1
KFC
- 488
- 4
Hi there,
I learn from the text that the exponential of an operator could be expanded with a series such that
[itex]e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}[/itex]
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]
[itex]e^{\hat{A}}|\psi\rangle[/itex]
will be a matrix with diagonal elements given as [tex]\exp(a_i)[/tex], is that right?
So I am wondering what happen if [itex]\hat{A}[/itex] is now written as a power form, i.e. [itex]\hat{A}^n[/itex], can we conclude that
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]
[itex]e^{\hat{A}^n}|\psi\rangle[/itex]
gives a matrix with diagonal elements as [itex]\exp(a_i^n)[/itex] ?
I learn from the text that the exponential of an operator could be expanded with a series such that
[itex]e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}[/itex]
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]
[itex]e^{\hat{A}}|\psi\rangle[/itex]
will be a matrix with diagonal elements given as [tex]\exp(a_i)[/tex], is that right?
So I am wondering what happen if [itex]\hat{A}[/itex] is now written as a power form, i.e. [itex]\hat{A}^n[/itex], can we conclude that
So if the eigenvalue of the operator [itex]\hat{A}[/itex] is given as [itex]a_i[/itex]
[itex]e^{\hat{A}^n}|\psi\rangle[/itex]
gives a matrix with diagonal elements as [itex]\exp(a_i^n)[/itex] ?
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