Thursday, 6 November 2014

quantum mechanics - Commutator algebra in exponents


Considering $X$ and $Y$ such that $[X,Y]=\lambda$, which is complex, and $\mu$ is another complex number, prove: $$e^{\mu(X+Y)}=e^{\mu X} e^{\mu Y} e^{-\mu^2\lambda/2}$$ My attempt (so far) is: Expand the exponent. $$\mu(X+Y)=\mu X+ \mu Y$$ and then split it. How can I introduce $\lambda$?




Taylor expansion: $$e^{\mu(X+Y)}=\sum\limits_{n=0}^\infty \frac{(\mu X+\mu Y)^n}{n!}=1+\mu X+\mu Y+\ldots$$



Answer



This is a basic example of a BCH formula. There are many ways to prove it. For example, write the exponential as $$ \exp(\mu X + \mu Y) = \lim_{N\to \infty} \left(1 + \frac {\mu X+ \mu Y}N\right)^N = \dots $$ Because the deviations from $1$ scale like $1/N$, it is equal to $$ = \lim_{N\to \infty} \left[\left(1 + \frac {\mu X}N\right)\left(1 + \frac {\mu Y}N\right)\right]^N $$ Now, we need to move all the $X$ factors to the left and $Y$ factors to the right. Each factor $1+\mu X / N$ commutes with itself, and similarly for $1+\mu Y/N$, of course. However, sometimes the $Y$ factor appears on the left of the $X$ factor and we need to use $$ \left(1 + \frac {\mu Y}N\right)\left(1 + \frac {\mu X}N\right) = \left(1 + \frac {\mu X}N\right) \left(1 + \frac {\mu Y}N\right)\left ( 1 - \frac{\mu^2 (XY-YX)}{N^2} \right ) $$ plus terms of order $O(1/N^3)$ that will disappear in the limit. The only thing we need to count is the number of such permutations of the $(1+\mu Y/N)$ factors with the $(1+\mu X/N)$ factors.


It's not hard. An average $(1+\mu Y/N)$ factor stands on the left side from $N/2\pm O(1)$ of the $(1+\mu x/N)$ factors, and there are $N$ factors of the form $(1+\mu Y/N)$, so we produce $N^2/2\pm O(N)$ factors of the form $$ \left ( 1 - \frac{\mu^2[X,Y]}{N^2} \right) $$ which is a $c$-number that commutes with everything.


Now we just collect the factors inside the limit. On the left, we see $N$ factors $(1+\mu X/N)$ which combine to $\exp(\mu X)$, then we have on the right from them $N$ factors with the $Y$ that combine to $\exp(\mu Y)$, and then there are $N^2/2$ factors of the form $(1-\mu^2[X,Y]/N^2)$ which, in the $N\to \infty $ limit, combine to $\exp(-\mu^2[X,Y]/2)$.


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