Multiplying two function operators

[Pencilvester]

TL;DR

I’m having trouble working out how $e^{\hat A} e^{\hat B} = e^{\hat A + \hat B} e^{[\hat A , ~ \hat B ]/2}$

I am reading Zettili’s “Quantum Mechanics: Concepts and Applications” and I am in the section on functions of operators. It starts with how $F(\hat A)$ can be Taylor expanded and gives the particular and familiar example: $$e^{a \hat A} = \sum_{n=0}^\infty \frac{a^n}{n!} \hat A^n \tag{2.109}$$ Later it says how if $[\hat A , ~ \hat B] \neq 0$ then $e^{\hat A} e^{\hat B} \neq e^{\hat A + \hat B}$. Then here’s where I am having trouble: It says, “using (2.109) we can ascertain that $$e^{\hat A} e^{\hat B} = e^{\hat A + \hat B} e^{[\hat A , ~ \hat B ]/2},\\
e^{\hat A} \hat B e^{- \hat A} = \hat B + [\hat A , ~ \hat B] + \frac{1}{2!} [\hat A , ~ [\hat A , ~ \hat B ]] + \frac{1}{3!} [\hat A , ~ [\hat A , ~ [\hat A , ~ \hat B ]]] + \cdots”$$ I guess I am not a savvy enough mathematician to figure out how they got either of these from eq. 2.109. Can someone help me out?

 

[fresh_42]

It isn't very funny to do, but at least the first couple of terms can be calculated: Just use distribution and multiply the sums:
$$
\left( 1+A +\dfrac{1}{2}A^2+\dfrac{1}{6}A^3 +\ldots \right)\cdot \left( 1+B +\dfrac{1}{2}B^2+\dfrac{1}{6}B^3 +\ldots \right) = \ldots
$$
and compare it to the right hand side. Just do not forget that matrix multiplication isn't commutative, so $AB$ is different from $BA$ and e.g. $(A+B)^2= A^2+AB+BA+B^2$.

 

[Keith_McClary]

A special case of the Baker–Campbell–Hausdorff formula.
Proof?
Also:
https://www.physicsforums.com/threads/proof-of-exp-a-exp-b-exp-a-b-exp-1-2-a-b.644881/

 

[vanhees71]

Summary: I’m having trouble working out how $e^{\hat A} e^{\hat B} = e^{\hat A + \hat B} e^{[\hat A , ~ \hat B ]/2}$

I am reading Zettili’s “Quantum Mechanics: Concepts and Applications” and I am in the section on functions of operators. It starts with how $F(\hat A)$ can be Taylor expanded and gives the particular and familiar example: $$e^{a \hat A} = \sum_{n=0}^\infty \frac{a^n}{n!} \hat A^n \tag{2.109}$$ Later it says how if $[\hat A , ~ \hat B] \neq 0$ then $e^{\hat A} e^{\hat B} \neq e^{\hat A + \hat B}$. Then here’s where I am having trouble: It says, “using (2.109) we can ascertain that $$e^{\hat A} e^{\hat B} = e^{\hat A + \hat B} e^{[\hat A , ~ \hat B ]/2},\\
e^{\hat A} \hat B e^{- \hat A} = \hat B + [\hat A , ~ \hat B] + \frac{1}{2!} [\hat A , ~ [\hat A , ~ \hat B ]] + \frac{1}{3!} [\hat A , ~ [\hat A , ~ [\hat A , ~ \hat B ]]] + \cdots”$$ I guess I am not a savvy enough mathematician to figure out how they got either of these from eq. 2.109. Can someone help me out?

Using (2.109) you get
$$\mathrm{d}_a \exp(a \hat{A})=\hat{A} \exp(a \hat{A})=\exp(a \hat{A}) \hat{A}.$$
Now define
$$\hat{F}(a)=\exp(a \hat{A}) \hat{B} \exp(-a \hat{A}).$$
We (formally) expand $\hat{F}$ in a power series
$$\hat{F}(a)=\hat{F}(0)+a \hat{F}'(0) +\frac{a^2}{2} \hat{F}''(0)+...$$
We have $\hat{F}(0)=\hat{B}$,
$$\hat{F}'(a)=\exp(a \hat{A}) [\hat{A},\hat{B}] \exp(-a \hat{A}).$$
Now use this formula again, but instead of $\hat{B}$ for $[\hat{A},\hat{B}]$, which leads to
$$\hat{F}''(a)=\exp(a \hat{A})[\hat{A}, [\hat{A},\hat{B}]] \exp(-a \hat{A}).$$
It's easy to see that the $j$-th derivative is
$$\hat{F}^{(j)}(a)=\exp(a \hat{A}) [\hat{A},\hat{B}]_j \exp(-a \hat{A}),$$
where
$$[\hat{A},\hat{B}]_{j}=[\underbrace{\hat{A},[\hat{A},\ldots}_{j \quad \text{times}},B].$$
Setting $a=0$ in the power series, you get the 2nd formula you have trouble with.

The 2nd formula only holds if
$$[\hat{A},[\hat{A},\hat{B}]]=[\hat{B},[\hat{A},\hat{B}]]=0, \qquad (*)$$
To show this, we define
$$\hat{F}(a)=\exp[a(\hat{A}+\hat{B})].$$
The same series expansion technique together with the assumption (*) yields
$$\hat{F}(a) \hat{A} \hat{F}^{-1}(a)=\hat{A} + a [\hat{A}+\hat{B},\hat{A}]=\hat{A}+a [\hat{B},\hat{A}].$$
From this we get
$$\hat{F}(a) \hat{A}=\hat{A} \hat{F}(a) - a [\hat{A},\hat{B}]\hat{F}(a).$$
On the other hand with this we have
$$\mathrm{d}_a \hat{F}(a) = \hat{F}(a)(\hat{A}+\hat{B}) =\hat{F}(a) \hat{B} + \hat{A} \hat{F}(a)-a [\hat{A},\hat{B}]\hat{F}(a).$$
Taking into account again (*) it's easy to show that this differential equation is solved by
$$\hat{F}(a)=\exp(a \hat{A}) \exp(a \hat{B}) \exp \left (-\frac{a^2}{2} [\hat{A},\hat{B}] \right).$$
Setting $a=1$ gives
$$\hat{F}(1)=\exp(\hat{A}+\hat{B}) = \exp(\hat{A}) \exp(\hat{B}) \exp \left (-\frac{1}{2} [\hat{A},\hat{B}] \right).$$
Again, it's important to keep in mind that this holds only, if (*) is fulfilled. Nevertheless the formula is of some use in applications.

 

[Pencilvester]

It isn't very funny to do, but at least the first couple of terms can be calculated: Just use distribution and multiply the sums:
$$
\left( 1+A +\dfrac{1}{2}A^2+\dfrac{1}{6}A^3 +\ldots \right)\cdot \left( 1+B +\dfrac{1}{2}B^2+\dfrac{1}{6}B^3 +\ldots \right) = \ldots
$$
and compare it to the right hand side.

You’re right, this is not at all very fun to do, but I began doing it, and I’m running into a problem. If you start to expand $e^{A+B}e^{[A, ~ B]/2}$ you get $$(1+A+B+ \cdots + \frac{1}{6} A^2B + \cdots )(1+ \frac{1}{2} AB + \cdots )$$Which gives us $\frac{2}{3} A^2B + \cdots$ but the expansion of $e^A e^B$ says we should only have $\frac{1}{2} A^2B$. What am I screwing up?

 

[fresh_42]

I don't know what happens with the additional condition that $[A,[A,B]]=[B,[B,A]]=0$ has to hold, see post #4.

You can find the complete CBH formula here where the other twelfth can be found:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/
or if you don't like the proof in post #4, maybe you like this one (also with the additional conditions on commutativity of the higher order terms):
https://www.physicsforums.com/threads/commutator-identity.966227/#post-6133835

 

[Pencilvester]

The 2nd formula only holds if
$$[\hat{A},[\hat{A},\hat{B}]]=[\hat{B},[\hat{A},\hat{B}]]=0, \qquad (*)$$
...
Taking into account again (*) it's easy to show that this differential equation is solved by
$$\hat{F}(a)=\exp(a \hat{A}) \exp(a \hat{B}) \exp \left (-\frac{a^2}{2} [\hat{A},\hat{B}] \right).$$

Wow. Okay, if Zettili intended his readers to work all that out just from eq. 2.109, then my math skills must be far from where they should be to study this book. Thanks for the explanation though!

 

[vanhees71]

Hm, I don't think one gets this without knowing this (or perhaps another) simpler trick :-)).

 

[Tendex]

The 2nd formula only holds if

[^A,[^A,^B]]=[^B,[^A,^B]]=0,(∗)​

Is this premise valid for the use of this special form of the BCH in QFT when constructing field operators expanding in terms creation and annihilation operators?

 

[vanhees71]

In which context do you need it? Sometimes it's also sufficient to just use the series for the general case up to the first few terms (e.g., in the derivation of the path-integral expression for the generating functional for Green's functions etc.).

 

[Tendex]

In which context do you need it? Sometimes it's also sufficient to just use the series for the general case up to the first few terms (e.g., in the derivation of the path-integral expression for the generating functional for Green's functions etc.).

In the context of field operator construction with annihilation and creation operators. The conmutator of these(-I) commutes with either one (creation or annihilation) operator so it seems to obey (*) in your post and in the same was is done in the free fields construction.