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Let $A$ be a Banach algebra we khow that

If $ab=ba$ then $e^{a+b}=e^{a}e^{b}$

my question is

Does $e^{a+b}=e^{a}e^{b}$ implies that $ab=ba$?

Any comment or response is appreciated.

tryst with freedom
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user62498
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  • @Ennar. You're right. I thought I had found a different duplicate target. I'm certain I've seen this question here before. – Theo Bendit Mar 06 '19 at 08:20
  • Do you mean $ab=ba$ for a particular $a$ and $b$ or for all $a$ and $b$? – MPW Mar 06 '19 at 08:46
  • Writing out $e^{a+b}-e^ae^b$, I highly doubt that the vanishing of all those terms implies the vanishing of $ab-ba$. But I might be wrong. – Arthur Mar 06 '19 at 08:46
  • Have a look at Baker-Campbell-Hausdorff formula. – N. Ciccoli Mar 06 '19 at 08:57
  • @N. Ciccoli, only one direction is obvious from the formula, right? – Ennar Mar 06 '19 at 09:02
  • Yes. On the other direction BCH tells you that if $e^ae^b=e^{a+b}$ then $\mathrm{Id}-\mathrm{ad}a/\mathrm{ad}_a(b)=b$ so your question can be reformulated as: is it possible that $\mathrm{ad}_a(b)\ne 0$ but still $\mathrm{Id}-\mathrm{ad}_a/\mathrm{ad}_a(b)=b$, so it refers to injectivity of the map $z\mapsto \sum{n=1}^\infty(-1)^{n+1}z^{n-1}/n!$. If I remember correctly this function is at least locally injective around $0$, but at moment I cannot find a ref. Hope this rewriting helps. – N. Ciccoli Mar 06 '19 at 09:35
  • @MPW, I mean for all $a, b$ – user62498 Mar 06 '19 at 18:22
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    See this: https://math.stackexchange.com/questions/349180/if-ea-and-eb-commute-do-a-and-b-commute/349382#349382 – Sungjin Kim Mar 07 '19 at 08:12
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    Does this answer your question? If $e^A$ and $e^B$ commute, do $A$ and $B$ commute? That question has a counterexample in $2 \times 2$ matrices. – GEdgar Apr 09 '21 at 11:44
  • Voting to reopen because the other question is clearly a different question. They might be equivalent but IMHO that requires a proof. – N. Virgo Apr 20 '23 at 04:10

1 Answers1

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There is a counterexample. Let the Banach algebra be the space of operators on a finite-dimensional Hilbert space. We can pick two orthonormal vectors, $|i\rangle$ and $|j\rangle$, and choose operators $\hat A$ and $\hat B$ as follows: $$\hat A = 2\pi i |i\rangle \langle i| \, , \quad \hat B = 2\pi i |i\rangle \langle i| + |i\rangle \langle j | \, .$$

Then we see that $$ \hat A^k = (2\pi i)^{k-1} \hat A \, , \quad \hat B^k = (2\pi i)^{k-1} \hat B \, , \quad (\hat A+\hat B)^k = (4\pi i)^{k-1} (\hat A+\hat B) \, . $$

Therefore $e^{\hat A} = e^{\hat B} = e^{\hat A + \hat B}$ = I, which means $ e^{\hat A + \hat B} = e^{\hat A}e^{\hat B}$. Yet $$ \hat A \hat B = (2\pi i)^2|i\rangle \langle i| + (2\pi i) |i\rangle \langle j| \not = (2\pi i)^2|i\rangle \langle i| = \hat B \hat A \, . $$

Minh Nguyen
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    Do you think using $i$ both for $\sqrt{-1}$ and for the index of your vector $|i\rangle$ is good notation? – GEdgar Apr 09 '21 at 11:43
  • @GEdgar It seems perfectly readable to me. – Pedro Apr 09 '21 at 11:50
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    I'm just saying. Why not call the orthonormal vectors $| 1 \rangle$ and $| 2\rangle$. Or, as mathematicians do, $u_1$ and $u_2$. Also, non-physicists may not know what $|i\rangle \langle i |$ means. – GEdgar Apr 09 '21 at 11:56