0

In Schwartz "Quantum field theory and the standard model" pag 160, the generators of the rotation are Hermitian, while the generators of boosts are anti-Hermitian, as an example:

$ J_1 = \left( \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ \end{matrix}\right)\,, K_1 = \left( \begin{matrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{matrix}\right) $

So you have $\langle \psi | \psi \rangle $ is rotation invariant since $e^{i\theta J_1}$ is Unitary an then

$\langle \psi | \psi \rangle \to \langle \psi | \left(e^{i\theta J_1}\right)^\dagger e^{i\theta J_1} |\psi \rangle = \langle \psi | e^{-i\theta J_1} e^{i\theta J_1} | \psi \rangle = \langle \psi | \psi \rangle $

But for the boost you have that $\left(e^{i\beta K_1}\right)^\dagger = e^{i\beta K_1} $.

  1. So why generator of boosts are chosen to be anti-Hermitian instead of Hermitian?

  2. Furthermore, using Hermitian generators for rotation you get the bracket $[J_i, J_j] = i\epsilon_{ijk}J_k$, so in this way the $i$ factor makes the Lie algebra $\mathfrak so(1,3)$ not closed in respect to the brackets, unless you re-define the bracket with a $-i$ factor.

Qmechanic
  • 201,751
Andrea
  • 467
  • 2
    Hi! Maybe this could be relevant – Ratman Oct 10 '21 at 20:44
  • 2
    Do you understand how so(1,3) differs from so(4), its compact brother? How do their respective generators correspond? For noncompact groups like SO(1,3), unitary reps are infinite dimensional, but finite dimensional reps are nonunitary. Have you gone through WuKi Tung's book? – Cosmas Zachos Oct 10 '21 at 21:09
  • $[J_m,J_n] = i \epsilon_{mnk} J_k ~, ~ [J_m,K_n] = i \epsilon_{mnk} K_k ~, ~ [K_m,K_n] = -i \epsilon_{mnk} J_k $ is the Lie algebra of so(1,3). Check your statements... – Cosmas Zachos Oct 10 '21 at 21:23
  • Shouldn't the $J_i$ and $K_i$ be $4\times 4$-matrices instead of $3\times 3$ ? – Thomas Fritsch Oct 10 '21 at 21:29
  • Sorry, just fixed the errors in the post – Andrea Oct 10 '21 at 21:41
  • 2
    Related/possible duplicate: https://physics.stackexchange.com/q/99051/50583 – ACuriousMind Oct 10 '21 at 21:42
  • Still trying to figure out why SO(1,3) can't have unitary finite dimensional representation, cause it seems $K_i$ is anti-Hermitian cause to get the group we exponentiate the Lie algebra as $e^{i\beta K_i}$ but it seems we can get it also with $e^{\beta K_i}$ where $K_i$ is hermitian, actually usually mathematician uses $e^{\lambda M}$ instead of $e^{i\lambda M}$ – Andrea Oct 10 '21 at 22:02
  • I see that boost in that way is not unitary since the generator is not Hermitian. But It seems I can merely remove the $i$ in both the generator and the exponentiation to get the generator hermitian and the transformation unitary – Andrea Oct 10 '21 at 22:39
  • No you cannot. To prevent confusion, take real parameters and extract the algebra out of the CBH expansion of group element products. – Cosmas Zachos Oct 10 '21 at 23:24

1 Answers1

2

These $+$ and $-$ signs in the generators originate from the signs in the Lorentz transformation matrices $\Lambda_{\mu\nu}$ (used for transforming $4$-vectors), which in turn originate from the signs in the Minkowski metric of spacetime.

Consider two examples:

Rotation

A rotation around the $x$-axis (with angle $\theta$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&\cos\theta&-\sin\theta\\ 0&0&\sin\theta&\cos\theta \end{pmatrix}$$ Especially notice the opposite signs of the two off-diagonal coefficients (which assures $y'^2+z'^2=y^2+z^2$).
This matrix can be generated by $$\Lambda=e^{-iJ_x\theta}$$ with the generator $$J_x=\begin{pmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&-i\\ 0&0&i&0 \end{pmatrix}$$

Boost

A boost along the $x$-axis (with rapidity $\beta$) has the Lorentz transformation matrix $$\Lambda=\begin{pmatrix} \cosh\beta&\sinh\beta&0&0\\ \sinh\beta&\cosh\beta&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}$$ Especially notice the same signs of the two off-diagonal coefficients (which assures $c^2t'^2-x'^2=c^2t^2-x^2$).
This matrix can be generated by $$\Lambda=e^{iK_x\beta}$$ with the generator $$K_x=\begin{pmatrix} 0&-i&0&0\\ -i&0&0&0\\ 0&0&0&0\\ 0&0&0&0 \end{pmatrix}$$

Thomas Fritsch
  • 39,317
  • Why I can't set $\Lambda=e^{K_x\beta}$, where $$K_x=\begin{pmatrix} 0&-1&0&0\ -1&0&0&0\ 0&0&0&0\ 0&0&0&0 \end{pmatrix}$$ ? In this way $K_x$ would be Hermitian and $\Lambda$ unitary – Andrea Oct 10 '21 at 22:31
  • 1
    @Andrea this is just a redefinition, but it will still net you a non-unitary group element, as you may check! – Cosmas Zachos Oct 10 '21 at 22:36
  • @Andrea You could. It seems just like a convention to me. Likewise for the rotation you could set $\Lambda=e^{J_x\beta}$ with $$J_x=\begin{pmatrix} 0&0&0&0\0&0&0&0\0&0&0&-1\0&0&1&0 \end{pmatrix}$$ – Thomas Fritsch Oct 10 '21 at 22:39
  • @CosmasZachos you are right, may bad – Andrea Oct 10 '21 at 22:51