Show the formula which connects the adjoint representations

In summary, the conversation discusses the use of conjugation and the adjoint representation in Lie groups and Lie algebras. The formula for the adjoint representation is shown and the use of the exponential function to relate the two representations is explained. The conversation also mentions a specific exercise in quantum mechanics that involves calculating with matrices and using the commutator operation.
  • #1
Mutatis
42
0
Homework Statement
Show that ##e^\left(-Â\right)*\hat B*e^Â = \hat B - \left[ Â, \hat B \right] + \frac {1} {2!} *\left[ \hat A, \left[ Â, \hat B \right] \right] - ... ##
Relevant Equations
## e^x=\sum_{n=0}^\infty \frac {x^n} {n!} ##
That's my attempting: first I've wrote ##e## in terms of the power series, but then I don't how to get further than this $$ \sum_{n=0}^\infty (-1)^n \frac {Â^n} {n!} \hat B \sum_{n=0}^\infty \frac {Â^n} {n!} = \sum_{n=0}^\infty (-1)^n \frac {Â^2n} {\left( n! \right) ^2} $$. I've alread tried to expand it but it leads me to a weird sum of terms...
 
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  • #2
Mutatis said:
That's my attempting: first I've wrote ##e## in terms of the power series, but then I don't how to get further than this $$ \sum_{n=0}^\infty (-1)^n \frac {Â^n} {n!} \hat B \sum_{n=0}^\infty \frac {Â^n} {n!} = \sum_{n=0}^\infty (-1)^n \frac {Â^2n} {\left( n! \right) ^2} $$. I've alread tried to expand it but it leads me to a weird sum of terms...
Where did you get this from? The proof is normally not done by direct computation; and it is usually not homework.
 
  • #3
Well, this is one exercise from my quantum mechanics class...
 
  • #4
Looks a bit troublesome to do it this way. The way I know is (details aside):

Given a Lie group ##G## and its Lie algebra ##\mathfrak{g}##.

##G## operates on itself via conjugation ##x.y :=xyx^{-1}## which gives rise to
##G## operates on ##\mathfrak{g}## by ##x.Y := \operatorname{Ad}(x)(Y) = x Y x^{-1}## which gives rise to
##\mathfrak{g}## operates on itself by ##X.Y := \operatorname{ad}(X)(Y) = [X,Y]##

Now the two adjoint representations are related by ##\operatorname{Ad}(\exp(x)) = \exp(\operatorname{ad}(X))## since the exponential function is basically the integration from the tangent space, the Lie algebra ##\mathfrak{g}##, back into the Lie group ##G##.

What you have here is exactly this formula:
\begin{align*}
\operatorname{Ad}(\exp(-\hat A))(\hat B) &= \exp(-\hat A) \hat B \exp(\hat A) \\
&= \hat B - [\hat A, \hat B] +\frac{1}{2!} [\hat A,[\hat A,\hat B]] \mp \cdots \\
&= \left(1 + \operatorname{ad}(-\hat A) + \frac{1}{2!} (\operatorname{ad}(-\hat A))^2 \mp \cdots \right)(\hat B)\\
&= \exp(\operatorname{ad}(-\hat A)) (\hat B)
\end{align*}
If it is only an exercise in calculation with matrices, then you will probably have to use ##\operatorname{ad}(-\hat A)(\hat B) = [-\hat A,\hat B] = -\hat A \hat B +\hat B \hat A## and a bit of patience multiplying those sums.
 
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Related to Show the formula which connects the adjoint representations

1. What is the adjoint representation?

The adjoint representation is a mathematical tool used in the study of Lie algebras. It maps every element in the algebra to a linear operator, allowing for the algebra to be studied through its action on vector spaces.

2. How is the adjoint representation related to the Lie bracket?

The adjoint representation is closely related to the Lie bracket, which is a binary operation on the elements of a Lie algebra. The Lie bracket of two elements is defined as the commutator of their corresponding linear operators in the adjoint representation.

3. What is the formula for the adjoint representation?

The formula for the adjoint representation is given by Ad(x)(y) = [x,y], where x and y are elements of the Lie algebra and [x,y] is the Lie bracket of x and y.

4. How is the adjoint representation used in physics?

The adjoint representation is used in physics to study the symmetries of physical systems. In particular, it is used in the study of gauge theories, which describe the fundamental interactions between particles.

5. Can the adjoint representation be extended to other mathematical structures?

Yes, the adjoint representation can be extended to other mathematical structures such as Lie superalgebras and quantum groups. It is a powerful tool for understanding the structure and properties of these mathematical objects.

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