Quantum Mechanics (1) · Problem 5.9

Rotations and angular momentum

Physics — Quantum Mechanics · baker-campbell-hausdorff-lie-algebra

Problem

We have discussed Lie algebras (with Lie product given by the commutator) and Lie groups, in our attempt to deal with rotations. At one point, we asserted that the structure (multiplication table) of the Lie group in some neighborhood of the identity was completely determined by the structure (multiplication table) of the Lie algebra. We noted that, however intuitively pleasing this might sound, it was not actually a trivial statement, and that it followed from the ``Baker-Campbell-Hausdorff'' theorem. Let's try to tidy this up a bit further here. Let

\mathcal{L}

be a Lie algebra, and

\mathcal{G}

be the Lie group generated by this algebra. Let

X, Y \in \mathcal{L}

be two elements of the algebra. These generate the elements

e^X, e^Y \in \mathcal{G}

of the Lie group. What we want to show is that the group product

e^X e^Y

may be expressed in the form

e^Z

, where

Z \in \mathcal{L}

, at least for

X

and

Y

not too ``large''. ((a)) Suppose

X

and

Y

commute. Show explicitly that the product

e^X e^Y

is of the form

e^Z

, where

Z

is an element of

\mathcal{L}

. (If you think this is trivial, don't worry, it is!) ((b)) Now suppose that

X

and

Y

may not commute, but that they are very close to the zero element. Keeping terms to quadratic order in

X

and

Y

, show once again that the product

e^X e^Y

is of the form

e^Z

, where

Z

is an element of

\mathcal{L}

. Give an explicit expression for

Z

. ((c)) Finally, for more of a challenge, let's do the general theorem: Show that

e^X e^Y

is of the form

e^Z

, where

Z

is an element of

\mathcal{L}

, as long as

X

and

Y

are sufficiently ``small''. We won't concern ourselves here with how ``small''

X

and

Y

need to be. Here are some hints that may help you: First, we remark that the differential equation

\frac{df}{du} = Xf(u) + g(u), \tag{5.511}

where

X \in \mathcal{L}

, and letting

f(0) = f_0

, has the solution:

f(u) = e^{uX}f_0 + \int_0^u e^{(u-v)X}g(v)\,dv. \tag{5.512}

g

is independent of

u

, then

f(u) = e^{uX}f_0 + h(u,X)g, \tag{5.513}

where, formally,

h(u,X) = \frac{e^{uX}-1}{X}. \tag{5.514}

Second, if

X, Y \in \mathcal{L}

, then

e^X Y e^{-X} = e^{X_c}(Y), \tag{5.515}

where

X_c(Y) \equiv [X,Y]

. This fact may be demonstrated by taking the derivative of

A(u;Y) \equiv e^{uX}Y e^{-uX} \tag{5.516}

with respect to

u

, and comparing with our differential equation above to obtain the desired result.

Related theory

Lie Groups And Lie Algebras

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