# Baker-Hausdorff Lemma from Sakurai's book

I'd like to show that, given to hermitian operators $A,G$ on a Hilbert space $\mathscr{H}$, the following identity holds: $$e^{iG\lambda}A e^{-iG\lambda} = A + i\lambda [G,A] + \frac{\left(i\lambda\right)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots]+\ldots$$ where $\lambda$ denotes a real parameter and $[\ \!,]$ indicates the commutator.

This is a proof left to the reader by Sakurai in his books on Modern Quantum Mechanics.

• alternatively, if $f(\lambda) = e^{iG\lambda}Ae^{-iG\lambda}$ then look at the Taylor expansion; $f(0)+f'(0)\lambda+\frac{1}{2}\lambda^2f''(0)+ \cdots$ . Note, $f(0)=A$, $f'(0)=iGA-iAG =i[G,A]$ etc... I think I worked this out from Greiner's Quantum Mechanics and symmetries text. Anyway, nice work! +1 Commented Mar 14, 2014 at 7:24
• Thanks, I hadn't thought of that. It may prove useful in the future, perhaps I will try to show it that way as well... Commented Mar 14, 2014 at 8:41
• You can also see, with $L_B$ the left multiplication with $B$, $R_B$ the right and $\mathrm{ad_B}(A):=[B,A]=(L_B-R_B)(A)$, that $L_B, R_B$ commute and then: $$\exp(B)A\exp(-B)=\left(\exp(R_{-B})\exp(L_B)\right)(A)=\exp(L_B-R_B)(A)=\exp(\mathrm{ad_B})(A)\\ =A+[B,A]+1/2[B,[B,A]]+1/6[B,[B,[B,A]]]+....$$ of course this proof, like yours, needs all operators to be bounded. This certainly is not the case in most applications of Quantum mechanics! Of course, this proof is also incredibly short. Commented Nov 20, 2016 at 18:59
• @s.harp I am not sure on how we justify the first equal sign in your proof: $e^B A e^{-B} = e^{R_{-B}}e^{L_B} A$. The other steps are clear and quite clever though. Commented Nov 22, 2016 at 9:09
• $\exp(B)A=\sum_n \frac{B^nA}{n!}=\sum_n\frac{L_B^n(A)}{n!}\overset*=\left(\sum_n\frac{L_B^n}{n!}\right) (A)=\exp(L_B)(A)$. $L_B$ is a bounded linear map $L(\mathcal H)\to L(\mathcal H)$, so $\exp(L_B)$ exists as the norm limit of the sum (since norm convergence implies SOT convergence, $*$ is true). Do the same with $A\exp(-B)=\exp(R_{-B})(A)$ and then use that $R_{-B}$ and $L_B$ commute. Commented Nov 22, 2016 at 10:27

Using the series definition of exponential:

$$e^{iG\lambda}A e^{-iG\lambda} = \sum_{p=0}^\infty\frac{(iG\lambda)^p}{p!}A\sum_{q=0}^\infty\frac{(-iG\lambda)^q}{q!} = \sum_{p=0}^\infty\sum_{q=0}^\infty(-)^q\frac{(i\lambda)^{p+q}}{p!q!}G^pAG^q=\\ \sum_{s=0}^\infty\sum_{d=0}^s(-)^d\frac{(i\lambda)^s}{d!(s-d)!}G^{s-d}AG^d=\\ A+i\lambda[G,A]+\frac{(i\lambda)^2}{2!}[G,[G,A]]+\ldots+\frac{(i\lambda)^n}{n!}\sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^k+\ldots$$ So we are left with the following relation which we have to verify, and which would prove the statement: $$\mathscr{F}(n): \sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^k=\underbrace{[G,[G,[G,\ldots[G}_{n\ times},A]]]\ldots].$$ Proceeding by induction, since the first terms shown above are compatible with the formula, we have to show that, if $$\mathscr{F}$$(n) holds then $$\mathscr{F}$$(n+1) is true as well.

To do this we exploit: $$\underbrace{[G,[G,[G,\ldots[G}_{n+1\ times},A]]]]\ldots] = \underbrace{[G,[G,[G,\ldots[G}_{n\ times},[G,A]]]\ldots]$$

Then substituting $$\mathscr{F}(n)$$ yields: $$\underbrace{[G,[G,[G,\ldots[G}_{n+1\ times},A]]]]\ldots]= \sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}(GA-AG)G^k =\\ \sum_{k=0}^n(-)^k \binom{n}{k}G^{n+1-k}AG^{k}-\sum_{k=0}^n(-)^k \binom{n}{k}G^{n-k}AG^{k+1}=\\ G^{n+1}A+\sum_{k=1}^n(-)^k \binom{n}{k}G^{n+1-k}AG^{k}-\sum_{k'=1}^{n}(-)^{k'-1} \binom{n}{k'-1}G^{n+1-k'}AG^{k'}+(-)^{n+1}AG^{n+1}$$ where in the last passage we changed summing index in the second sum, and took out the first term from the first and the last from the second. Now: $$\binom{n}{k}+\binom{n}{k-1} = \binom{n+1}{k}$$ which gives $$\ldots=G^{n+1}A + \sum_{k=1}^n(-)^k \binom{n+1}{k}G^{n+1-k}AG^{k} + (-)^{n+1}AG^{n+1}= \sum_{k=0}^{n+1}(-)^k \binom{n+1}{k}G^{n+1-k}AG^{k}.$$

And therefore $$\mathscr{F}$$(n+1) holds.

• Wow! Phew! What grind! +1! Commented Mar 13, 2014 at 21:24
• I'm finding the notation in the formula after "To do this we exploit:" a little difficult to understand. Could you clarify it a bit? Commented Apr 27, 2016 at 13:16
• @AlexDB I'm simply stating that $n+1$ commutators can be regarded as $n$ commutators with a commutator Commented Apr 27, 2016 at 19:50
• Ah of course, gotcha. Commented Apr 29, 2016 at 7:02

Let $A$ and $B$ be any two operators on the Hilbert space $\mathscr H$, hermitian or not. We assume $A, B \in L(\mathscr H)$, the Banach algebra of bounded linear maps from $\mathscr H$ to itself. Consider the linear operator ordinary differential equation

$\dfrac{dX}{d \lambda} = [B, X] \tag{1}$

with initial condition

$X(0) = A. \tag{2}$

We observe that

$X(\lambda) = e^{\lambda B}Ae^{-\lambda B} \tag{3}$

is the unique solution to (1), (2), for from (3) it follows that

$\dfrac{dX}{d \lambda} = \dfrac{e^{\lambda B}}{d \lambda}Ae^{-\lambda B} + e^{\lambda B}\dfrac{dA}{d \lambda}e^{-\lambda B} + e^{\lambda B}A\dfrac{e^{-\lambda B}}{d \lambda} =$ $Be^{\lambda B}Ae^{-\lambda B} - e^{\lambda B}Ae^{-\lambda B}B = [B, e^{\lambda B}Ae^{-\lambda B}], \tag{4}$

where we have used the fact that $dA / d \lambda = 0$ and the Leibniz product rule for derivatives in (4), and furthermore it is evident from (3) that $X(0) = A$.

We next recall that for any $B \in L(\mathscr H)$ the adjoint linear operator $\text{ad}_B: L(\mathscr H) \to L(\mathscr H)$ may be defined via

$\text{ad}_B(A) = [B, A] \tag{5}$

for all $A \in L(\mathscr H)$. Denoting by $\Vert T \Vert _L$ the standard operator norm on $L(\mathscr H)$, we see that

$\Vert \text{ad}_B(A) \Vert_L = \Vert [B, A] \Vert_L = \Vert BA - AB \Vert_L \le \Vert BA \Vert_L + \Vert AB \Vert_L$ $\le \Vert B \Vert_L \Vert A \Vert_L + \Vert A \Vert_L \Vert B \Vert_L = 2 \Vert B \Vert_L \Vert A \Vert_L, \tag{6}$

which shows that

$\Vert \text{ad}_B \Vert_L \le 2 \Vert B \Vert_L, \tag{7}$

i.e. that $\text{ad}_B \in L(\mathscr H)$ is itself a bounded linear operator of norm at most $2\Vert B \Vert_L$. Furthermore, we have

$\text{ad}_B^2(A) = \text{ad}_B (\text{ad}_B(A)) = \text{ad}_B([B, A]) = [B, [B, A]], \tag{8}$

$\text{ad}_B^3(A) = \text{ad}_B (\text{ad}_B^2(A)) = \text{ad}_B([B, [B, A]]) = [B, [B, [B, A]]], \tag{9}$

and so on:

$\text{ad}_B^n(A) = [B, [B, [B, . . . [B, A]]] . . . ], \tag{10}$

where the operator $\text{ad}_B = [B, \cdot]$ occurs a total of $n$ times on the right-hand side of (10). We see that in fact (1) may be written in terms of $\text{ad}_B$ as

$\dfrac{dX}{d \lambda} = \text{ad}_B(X). \tag{11}$

Now set

$Y(\lambda) = A + \lambda [B, A] + \dfrac{\lambda^2}{2!}[B, [B, A]]$ $+ \ldots + \dfrac{\lambda^n}{n!}\underbrace{[B, [B, [B, \ldots [B}_{n \; \text{times}}, A]]]] \ldots ] + \ldots; \tag{12}$

from the above we see that $Y(\lambda)$ may be written

$Y(\lambda) = A + \lambda \text{ad}_B(A) + \dfrac{\lambda^2}{2!} \text{ad}_B^2(A) + \ldots + \dfrac{\lambda^n}{n!} \text{ad}_B^n(A) + \ldots$ $= \sum_0^\infty \dfrac{\lambda^n}{n!}\text{ad}_B^n(A) + \ldots = e^{\lambda \text{ad}_B}(A); \tag{13}$

since by (7) $\text{ad}_B$ is a bounded operator on $L(\mathscr H)$, all the series occuring above converge absolutely and uniformly on compacta for all $\lambda \in \Bbb R$, in fact for all $\lambda \in \Bbb C$. We thus have, exactly as in the case of ordinary calculus, that the derivative $Y'(\lambda)$ is given by

$\dfrac{dY}{d\lambda} = \text{ad}_B(e^{\lambda \text{ad}_B}(A)) = [B, e^{\lambda \text{ad}_B}(A)], \tag{14}$

and furthermore

$Y(0) = A, \tag{15}$

which follows trivially from (12) and/or (13). Comparing (1), (2), (11), (14) and (15), we see that $X(\lambda)$ and $Y(\lambda)$, satisfying as they do the same ODE with identical initial conditions, must by uniqueness etc. be identical for all $\lambda$: $X(\lambda) = Y(\lambda)$. Using (3) and (12), (13) we thus see that

$e^{\lambda B}Ae^{-\lambda B} = e^{\lambda \text{ad}_B}(A)$ $= A + \lambda [B, A] + \ldots + \dfrac{\lambda^n}{n!}\underbrace{[B, [B, [B, \ldots [B}_{n \; \text{times}}, A]]]] \ldots ] + \ldots; \tag{16}$

if we now set $B = iG$ we obtain

$e^{i\lambda G}Ae^{-i\lambda G} = e^{i\lambda \text{ad}_G}(A)$ $= A + i\lambda [G, A] + \ldots + \dfrac{(i\lambda)^n}{n!}\underbrace{[G, [G, [G, \ldots [G}_{n \; \text{times}}, A]]]] \ldots ] + \ldots, \tag{17}$

where we have used the fact that $\text{ad}_{iG} = i\text{ad}_G$, a consequence of the linearity of the bracket $[G, A]$ in each of its variables $A, G$. Equation (17) is the desired result. QED.

Note: The technique used here, based on uniqueness of ODEs, is similar in spirit to that used in my answers to several other questions; in particular see this one and this one.

Another Note: A couple of interesting formulas related to the above: $[B, e^{\lambda B}Ae^{-\lambda B}] = e^{\lambda B}[B, A]e^{-\lambda B}$ and $A = e^{-\lambda B} e^{\lambda \text{ad}_B(A)} e^{\lambda B}$.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

• Thanks very much! I'm going to have a look at this awesome answer as soon as I have some time! Commented Mar 14, 2014 at 23:44
• It truly is a much more elegant solution, @Robert Lewis. I'm not very much into the theory of series of linear operators, however you state that proving the boundedness and thus continuity of ad is sufficient to ensure convergence. Instinctively I'd say you'd need norm less than 1 as well, what's the point? Commented Mar 15, 2014 at 8:16
• @ Brightsun: Sorry it took me so long to get back to you on this. Please note that in my answer the only operator series referenced are of the form $e^{\mu T}$ for some bounded operator $T$ and scalar $\mu$. Indeed, this series is referenced both implicitly (as $e^{i \lambda G}$ etc.) and explicitly as $A + i \lambda [G, A] + \frac{(i\lambda)^2}{2!}[G, [G, A]] + \ldots = e^{\lambda \text{ad}_{iG}}A$ in your question. The series $e^{\mu T} = \sum_0^\infty \frac{(\mu T)^n}{n!}$ converges for all bounded $T$, no matter what $\Vert T \Vert_L$ may be; (continued in following comment) Commented Mar 15, 2014 at 19:25
• @ Brightsun: (continuation of previous comment) . . . indeed $e^{\mu T}$ is majorized in norm, term by term, by $\sum_0^\infty \frac{\Vert \mu T \Vert_L^n}{n!}$, which converges absolutely and uniformly (for $\mu$ contained in a compact subset of $\Bbb C$, in any event) no matter what value $\Vert \mu T \Vert_L$ may take; it is simply of the form $e^x$ for $0 \le x \in \Bbb R$. I merely took the trouble to show $\text{ad}_B$ is bounded to ensure the readers understood $e^{\lambda B}$ exists. The restriction $\Vert \mu T \Vert_L < 1$ is not necessary for the series $e^{\mu T}$. Commented Mar 15, 2014 at 19:38
• @ Brightsun: even more: however, certain other ubiquitous series such as $\sum_0^\infty T^n = (I - T)^{-1}$ require $\Vert T \Vert_L < 1$ to make sense. OK, enough, I hope. Sorry about the terse response but I've a lot on my plate at present and so I'm in a bit of a rush. Hope these remarks clarify. Regards. P.S. In the previous comment, the final occurance of $e^{\lambda B}$ should read $e^{\lambda \text{ad}_B}$ Commented Mar 15, 2014 at 19:41

Let

$$f(i\lambda) = e^{iG\lambda}Ae^{-iG\lambda}$$

We Taylor expand

$$f(i\lambda) = f(0) + i\lambda f'(0) + \frac{(i\lambda)^2}{2} f''(0) + ... + \frac{(i\lambda)^n}{n!}f^{n}(0) + ...$$

We see that

$$f'(i\lambda) = \frac{d}{di\lambda}(e^{iG\lambda})A e^{-iG\lambda} + e^{iG\lambda} A \frac{d}{di\lambda} (e^{-iG\lambda})$$

$$= e^{iG\lambda}GAe^{-iG\lambda} - e^{iG\lambda}AGe^{-iG\lambda}$$

$$= e^{iG\lambda}[G,A]e^{-iG\lambda}$$

$$f'(0) = [G,A]$$

In general, if

$$f^{n}(i\lambda) = e^{iG\lambda}\underbrace{[G,...,[G}_{\text{n times}},A]...]e^{-iG\lambda}$$

Then

$$f^{n+1}(i\lambda) = \frac{d}{di\lambda}(e^{iG\lambda})\underbrace{[G,...,[G}_{\text{n times}},A]...]e^{-iG\lambda} + e^{iG\lambda}\underbrace{[G,...,[G}_{\text{n times}},A]...]\frac{d}{di\lambda}(e^{-iG\lambda})$$

$$= e^{iG\lambda}G\underbrace{[G,...,[G}_{\text{n times}},A]...]e^{-iG\lambda} - e^{iG\lambda}\underbrace{[G,...,[G}_{\text{n times}},A]...]Ge^{-iG\lambda}$$

$$= e^{iG\lambda}\underbrace{[G,...,[G}_{\text{n+1 times}},A]...]e^{-iG\lambda}$$

$$f^{n+1}(0) = \underbrace{[G,...,[G}_{\text{n+1 times}},A]...]$$

Putting this all together we get

$$e^{iG\lambda} A e^{-iG\lambda} = A + i\lambda[G,A] - \frac{(i\lambda)^2}{2}[G,[G,A]] + ... \frac{(i\lambda)^n}{n!} \underbrace{[G,...,[G}_{\text{n times}},A]...] + ...$$