# Tag Info

## New answers tagged commutator

1

Besides the truncated BCH formula (which is also proven in this Phys.SE post and mentioned in Steven Mathey's answer), in practice one often wants to normal order the differential operators, i.e. putting all the $x$'s to the left of all the $\partial_x$. To this end the formula $$\tag{1} e^{a\partial_x}f(x) ~=~ f(x+a)e^{a\partial_x}.$$ is useful. Hence ...

3

Suresh is right, the Baker-Campbell-Hausdorff formula will save you. If $\left[A,B\right]$ commutes with $A$ and $B$ you have, $$\exp\left(A\right)\exp\left(B\right) = \exp\left(A+B+\frac{1}{2}\left[A,B\right]\right) \, .$$ Then the answer to your question is $$\exp\left(\partial_x\right)\exp\left(x\right) = 2 \exp\left(x+\partial_x\right) \sinh(1/2) \, ... 5 Here is the formal answer on your question based on particular result of Pauli theorem. Calculations are rather cumbersome, but they are general. Arbitrary fermionic field (with invariance under discrete transformations of the Lorentz group) It can be shown that each Poincare-covariant fermion field with spin s = n + \frac{1}{2} and mass m can be ... 0 OP's identity generalizes to the truncated BCH formula$$\tag{1} e^{\hat{A}}e^{\hat{B}}~=~e^{\hat{A}+\hat{B}+\frac{1}{2}\hat{C}} $$where the commutator$$\tag{2} \hat{C}~:=~[\hat{A},\hat{B}]$$is assumed to commute with both \hat{A} and \hat{B},$$\tag{3} [\hat{A},\hat{C}]~=~0\quad \text{and}\quad [\hat{B},\hat{C}]~=~0. $$[In particular, the ... 3 You have obtained several interesting answers. Here is the one I prefer, which actually is a variation of gj255's answer. Let us attack the problem by computing the function$$f(\mu) := e^{\mu X} Y e^{-\mu X}\:.$$This function verifies$$f'(\mu) = e^{\mu X} X Y e^{-\mu X} - e^{\mu X} YX e^{-\mu X} = e^{\mu X} [X, Y] e^{-\mu X} = \lambda I\:.$$... 7 There's a trick to proving this result which you would certainly be forgiven for not spotting! Consider the quantity \exp(\mu X) \exp(\mu Y) which appears on the right hand side. Now differentiate this with respect to \mu. Say what!? Yeah, bear with me, just try it:$$\frac{d}{d \mu} \exp(\mu X) \exp(\mu Y) = X\exp(\mu X) \exp(\mu Y) + \exp(\mu X) Y ...

3

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 ...

1

Here are some hints. use the definition of $\exp(x) = 1+x+\frac{1}{2}x^2+...$ on both sides. Now compare coefficients of $X^mY^n$ on both sides. Start with the $XY$ term as a warm up. Be careful to introduce commutators if you are swapping $X$s and $Y$s (Lie algebras are non-commutative in general). You should be able to write down the proof now!

1

As you have in the commutation relations, $\sigma_i \sigma_j= \sigma_j\sigma_i$ e.g. spin operators on different sites commute, so there is no minus sign to pick up.

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