# Finite lorentz transform for 4-vectors in terms of the generators

One or two sets of notes (one of them by Timo Weigand) on QFT that I have come across state explicitly that a finite lorentz transformation for 4-vectors can be written in terms of the generators $J^{\rho\sigma}$ as:

$$\Lambda^{\mu}_{\,\nu}(\Omega)=\lim_{N\rightarrow \infty}(\delta^{\mu}_{\,\nu}-\frac{i}{2N}\Omega_{\rho\sigma}(J^{\rho\sigma})^{\mu}_{\,\nu})^{N}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}(J^{\rho\sigma}))^{\mu}_{\,\nu}}$$

Now, my question is that, isn't it required of the generators of a Lie group to commute(or equivalently for the group to be abelian) in order to be able to represent it as an exponential of its generators in the finite case? This follows from the condition that $e^{A+B}=e^{A}e^{B}$ only when $[A,B]=0$ (in this case, aren't we simply constructing the finite exponential by multiplying the infinitesimal exponential terms together and then assuming that the generators commute?). If so, than why can the above be written in the way it is given that the generators(J) do not commute. What am I missing?

• Note that your representation of the exponential cannot be correct - you need a full matrix product, not a power of the individual entries. – Emilio Pisanty Sep 13 '17 at 18:25
• mu and nu represent the particular entries of the exponential as a whole and not the exponential of the mu, nu-th entries of the input matrices. Is it not clear, the paranthesis defines what is being exponentianted. This is exactly Eqn 3.16 of the notes that I have shared. – Arnab Barman Ray Sep 13 '17 at 18:39
• Minor comment to the post (v2): Please consider to mention explicitly author, title, etc. of link, so it is possible to reconstruct link in case of link rot. – Qmechanic Sep 13 '17 at 18:45
• Regarding the notation, the middle of the limit is incorrectly notated (unless you explicitly acknowledge the abuse of notation, which you haven't done); just because other people use wrong notation doesn't mean you get a pass. It's claer waht yoou wnanted to sy, bot jsst beecase poelple cann undnerstnd yo doesn't mean it's not wrong. – Emilio Pisanty Sep 15 '17 at 13:22

The extreme RHS exponential is correct. This is exactly the way to write a Lorentz transformation for any size parameters $\Omega$. It is also true for making the matrices of other representations, not just the 4x4 matrices which rotate 4-vectors. For the M x M representation, each of the 16 generators $J^{\rho \sigma}$ would be a M x M matrix. Actually there are not 16 but only 6 generators for the Lorentz Group because the array of parameters is antisymmetric $\Omega _{\rho \sigma}=-\Omega_{\sigma \rho}$ and picks out only the antisymmetric set of generators $J^{\rho \sigma}=-J^{\sigma \rho}$ .

The limit expression is not written correctly. It should be written

$$\Lambda^{\mu}_{\,\nu}(\Omega)=[\lim_{N\rightarrow \infty}(I-\frac{i}{2N}\Omega_{\rho\sigma}J^{\rho\sigma})^{N}]^{\mu}_{\,\nu}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}J^{\rho\sigma})^{\mu}_{\,\nu}}$$

where I is the 4x4 identity matrix diag(1,1,1,1). Now matrices are being multiplied together and not just the $\mu \nu$ elements.

Notice that each of the N terms in the limit product is exactly the same, and therefore commutes with all the other terms and your worry is resolved.

$$\Lambda^{\mu}_{\,\nu}(\Omega)=[\lim_{N\rightarrow \infty}(e^{-\frac{i}{2N}\Omega_{\rho\sigma}J^{\rho\sigma}})^{N}]^{\mu}_{\,\nu}=\exp{(-\frac{i}{2}\Omega_{\rho\sigma}J^{\rho\sigma})^{\mu}_{\,\nu}}$$

No, it's not required. You do need the generators to commute if you want to split the exponential as a product of exponentials, but there's no need for the latter. The linear combination $\Omega_{\rho\sigma}J^{\rho\sigma}$ is a perfectly valid four-by-four matrix, no matter what is coefficients or the commutation properties of its internal components, and it's perfectly OK to explore its matrix exponential.

• I am aware that that particular linear combination is a valid matrix. My beef is with whether the exponential accurately represents the finite lorentz transformation. And I am sure that e^{A}e^{B}=e^{A+B} only when A and B commute. – Arnab Barman Ray Sep 13 '17 at 18:48
• @ArnabBarmanRay What do the two have to do with each other? Yes, $e^{A}e^{B}=e^{A+B}$ is only true if $[A,B]=0$, but why do you need the former? In what circumstances would you care about $$\prod_{\rho\sigma} e^{-\frac{i}{2}\Omega_{\rho\sigma}J^{\rho\sigma}}$$ (with no summation in the exponent)? The fact that the product is order-dependent should be a huge red flag to begin with. – Emilio Pisanty Sep 15 '17 at 13:25
• Yes, it is clear now. A passage in Weinberg's book(Part 1,chapter 2 of QFT) says that such exponential notations were only possible for abelian groups. That was what was misleading me all along. Thanks. – Arnab Barman Ray Sep 16 '17 at 14:01

The equation that you have used is based on the simple limit $$\lim_{n\rightarrow\infty}\bigg(1+\dfrac{x}{n}\bigg)^n=e^x.$$ This limit has nothing to do with the details of what $x$ is. To answer your question, usually any operator that is connected to the identity operator (${\delta^{\,\mu}}_{\nu}$ in this case) can be written as an exponential in terms of its generators because the above limit can be used.

• The equation is probably wrong, as Pisant pointed out, the limit takes the individual elements as arguments instead of composing matrices together and then taking the relevant index. Even so, what I was actually asking is whether it is possible for a finite lorentz transformation to be written in that way(extreme RHS of the equation), as an exponential. – Arnab Barman Ray Sep 13 '17 at 19:18
• The RHS is a finite Lorentz transformation built out of infinitesimals so yes it can be written as an exponential, I'm not sure what the confusion was but I hope it's clear – nomaan x Sep 14 '17 at 19:58