Commutator Algebra Homework: Analyzing Functions of Operators

[helpcometk]

Homework Statement

Analytic functions of operators (matrices) A are defined via their Taylor expansion about A=0 .Consider the function

g(x) = exp(xA)Bexp(-xA)

Compute : dⁿg(x) /dxⁿ |x=0
for integer n

and then show that :exp(A)Bexp(-A)= B+[A,B] +1/2 [A,[A,B]] +1/6[A,[A,[A,B]]]+ ...

Homework Equations

for the first part of the question i don't understand what x=0 means in the formalism and it is unclear how one should proceed ,becuase we are dealing with operators.

about the second part i would like to note that :

g(1) =g(0) +g'(0) +1/2 g''(0)+ ...
if A=0 e^0 =1 and we start with B in the expansion formula

The Attempt at a Solution

About the first part i cannot imagine how i could proceed ,but about the second question i think we must use somehow a taylor expansion to resemble the monsterous expression of the RHS. But what i can't see is how one should manipulate [A,B], how could this come from a taylor expansion of the exponential ?

 

[D H]

x is a scalar. Some hints:

• Matrix multiplication is transitive, but not necessarily commutative. That the latter is the case is the point of the commutator.
• The chain rule still works as expected with your dg(x)/dx. You will either need to show this or cite the relevant theorem from your text/lectures.
• For any square matrix A, exp(A) commutes with Aⁿ for all nonnegative integers n. Once again, you will either need to show this or cite the relevant passage.
• What is $\frac{d}{dx} \exp(xA)$?

 

[helpcometk]

If chain rule is correct ,then i would have that the derivative is equal to 0 ,right?
about the second part do i just take the taylor expansions of both exponentials and mmultiply them ?

 

[D H]

If chain rule is correct ,then i would have that the derivative is equal to 0 ,right?

No! You apparently don't understand the notation.

Compute : dⁿg(x) /dxⁿ |x=0

This means you are to compute the value of the n^th derivative of g with respect to x at x=0. It says nothing about the derivative being zero.

The chain rule still works as expected with your dg(x)/dx. You will either need to show this or cite the relevant theorem from your text/lectures.

Ooops. My bad. Sorry about that. I meant product rule, not chain rule.

 

[helpcometk]

I understand the notation ,but I am very tired becuase i don't sleep overnight and i make a mistake ,
so we have :
dng(x) /dxn = -2xAg(x) ,do you agree with this ?

 

[D H]

I understand the notation ,but I am very tired becuase i don't sleep overnight and i make a mistake ,
so we have :
dng(x) /dxn = -2xAg(x) ,do you agree with this ?

Sorry for the very late response. I didn't see that you had added a post until just now.

No, I don't agree with that. Let's start with $dg/dx$. Applying the product rule (which is still valid because matrix multiplication is associative),
$$\begin{align} \frac {d\,g(x)}{dx} &= \frac d {dx} \bigl(\exp(xA)\,B\,\exp(-xA)\bigr) \\ &= \frac{d\exp(xA)}{dx}\,B\,\exp(-xA) + \exp(xA)\,B\,\frac{d\exp(-xA)}{dx} \end{align}$$
The key challenge is to evaluate $d\exp(xA)/dx$. Using the fact that $A$ and $\exp(xA)$ commute (show this!), you can reduce the derivative to something of the form
$$\frac {d\,g(x)}{dx} = \exp(xA)\,C\,\exp(-xA)$$
where $C$ is some matrix. The question itself suggests a form for this matrix: It has to somehow involve the commutator. (Hint: It is the commutator.) See if you can get to that. With this, getting $d^ng/dx^n$ should be a snap.