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I'm working through the book "Lie Groups: An Introduction Through Linear Groups", by Wulf Rossmann. In the first section, the author introduces the matrix exponential and derives its basic properties. I have done most of the exercises that come right at the end of this section, but the last one has given me trouble, and I would appreciate any help to solve it.

The problem is posed like this: if $P$ and $Q$ are operators such that $$PQ-QP=k\mathbf{1},\tag{1}$$ for some scalar $k$ (here $\mathbf{1}$ denotes the identity operator) then they are said to satisfy Heisenberg's Commutation Relation. The problem then asks to show that $P$,$Q$ satisfy this relation if and only if $$\exp(\sigma P)\exp(\tau Q)=e^{\sigma\tau k}\exp(\tau Q)\exp(\sigma P) \tag{2}$$ for all real $\sigma$,$\tau$.

I understand that this should follow from the basic properties of the exponential function, and that it is in a way a generalization of the result (proved in the book), that two matrices $X$ and $Y$ commute if and only if $\exp(\sigma X)$, $\exp(\tau Y)$ commute for all $\sigma$,$\tau$, but I've been unable to prove either direction.

Qmechanic
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    Be warned that there are in fact no matrices with this property (over, say, R or C). – Qiaochu Yuan Sep 20 '17 at 23:55
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    @QiaochuYuan why? – Hugo Sep 20 '17 at 23:57
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    Take the trace of both sides. – Qiaochu Yuan Sep 20 '17 at 23:59
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    @QiaochuYuan I forgot to mention it, but I have actually found a pdf by the author with corrections and comments on the book and there he says that "such matrices only exist in infinite dimensions when $k \neq 0$", but that "the result is a formal consequence of properties of the exponential ". From this I presume that it is provable and is in fact a useful result when working not with matrix exponentials, but exponentials of operators on infinite dimensional spaces. Am I right? –  Sep 21 '17 at 00:05
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    Yes, that's right. – Qiaochu Yuan Sep 21 '17 at 00:06
  • My guess is this follows from the Baker-Campbell-Hausdorff formula (https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula) although perhaps you can give a direct proof by looking at the proof of that formula and use only the parts that are useful for proving your result. But I didn't check the details – Vincent Nov 22 '19 at 20:48
  • If you figure it out, please post it here, I am quite curious honestly. – Vincent Nov 22 '19 at 20:48
  • @Vincent In the book, this exercise is found two sections before BCH, so I believe there must be a more direct solution. I haven't tried it in a long time, though (this question is from 2yrs ago). But I'm still curious too! –  Nov 22 '19 at 21:03
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    It turns out one of the two directions of the equivalence is proven on Wikipedia: see section 'An important lemma and its application to a special case of the Baker–Campbell–Hausdorff formula'. The special case alluded to is the case where both $X$ and $Y$ commute with $[X, Y]$ which is obviously the case in your situation with $X = \sigma P$, $Y = \tau Q$ and $[X, Y] = \sigma \tau k I$. – Vincent Nov 22 '19 at 21:54
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    The other direction can probably be handled by taking derivatives but I have to think about it a bit more. – Vincent Nov 22 '19 at 21:57
  • @bsd "I have actually found a pdf by the author with corrections and comments on the book" -- could you pls share the pdf? i'm reading Prof Rossmann's book now and it could be quite helpful. – athos Nov 06 '22 at 10:33

1 Answers1

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Sketched proof:

"$\Leftarrow$": Differentiate both sides of eq. (2) wrt. both $\sigma$ and $\tau$. Then put $\sigma=0=\tau$ to deduce eq. (1). $\Box$

"$\Rightarrow$": Use the truncated BCH formula

$$ e^Ae^B~=~e^{A+B+\frac{1}{2}C} \tag{i}$$

where the commutator

$$ C~:=~[A,B]\tag{ii}$$

is assumed to commute with both $A$ and $B$,

$$[A,C]~=~0\quad \text{and}\quad [B,C]~=~0, \tag{iii} $$

to deduce that

$$e^Ae^B~=~e^Ce^Be^A. \tag{iv} $$

$\Box$

Qmechanic
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