How Do You Differentiate e^(A*t) When A is a Constant Operator?

[sunrah]

Homework Statement

calculate $\frac{d}{dt}e^{\hat{A}t}$ where $\hat{A} \neq \hat{A}(t)$ in other words operator A doesn't depend explicitly on t.

Homework Equations

The Attempt at a Solution

$\frac{d}{dt}e^{\hat{A}t} = (\frac{d}{dt}(\hat{A})t + \hat{A})e^{\hat{A}t} = (\sum^{n}_{i=0}\frac{d\hat{A}}{dx_{i}}\frac{dx_{i}}{dt}t + \hat{A})e^{\hat{A}t}$

if the x_i ≠ x_i(t) we get $\hat{A}e^{\hat{A}t}$

but is this correct I know how to define the derivative of an operator if it is explicitly dependent on the variable of differentiation but not in this case.

 

[conquest]

First off are you sure this isn't just a partial differntiation in which case there is no problem. Otherwise this looks quite allright.

 

[dimension10]

Homework Statement

calculate $\frac{d}{dt}e^{\hat{A}t}$ where $\hat{A} \neq \hat{A}(t)$ in other words operator A doesn't depend explicitly on t.

Homework Equations

The Attempt at a Solution

$\frac{d}{dt}e^{\hat{A}t} = (\frac{d}{dt}(\hat{A})t + \hat{A})e^{\hat{A}t} = (\sum^{n}_{i=0}\frac{d\hat{A}}{dx_{i}}\frac{dx_{i}}{dt}t + \hat{A})e^{\hat{A}t}$

if the x_i ≠ x_i(t) we get $\hat{A}e^{\hat{A}t}$

but is this correct I know how to define the derivative of an operator if it is explicitly dependent on the variable of differentiation but not in this case.

Yup, there's nothing wrong with your solution.

 

[dextercioby]

It makes a world of difference if the operator in the exponent is bounded or not. Either way, there's a strict definition of such a derivative in terms of limits which can be found in almost all books on functional analysis.

 

[paranormal]

Yes, your approach is correct. When the operator A does not depend explicitly on t, it can be treated as a constant and the derivative of e^{\hat{A}t} is simply \hat{A}e^{\hat{A}t}. This can also be seen by using the definition of the derivative as a limit and noting that the limit of a constant is just the constant itself.