# A Real parameters and imaginary generators

Tags:
1. Jan 7, 2017

### Heisenberg1993

I was reading some lecture notes on super-symmetry (http://people.sissa.it/~bertmat/lect2.pdf, second page). It is stated that ". In order for all rotation and boost parameters to be real, one must take all the Ji and Ki to be imaginary". I didn't understand the link between the two. What does imaginary generators have to do with getting real parameters? By parameters, do we mean the ones that appear in the group element as exponential of the generator, or the ones used when deriving the fundamental representation of these generators (see: https://www.classe.cornell.edu/~pt267/files/notes/FlipSUSY.pdf, page 3, phrase after (2.16) to understand what I mean)?

2. Jan 10, 2017

### jambaugh

The parameters parametrize the Lie group elements and for a real Lie group must be real numbers. We often for algebraic expediency consider the complexification (extension by allowing the parameters to be complex) especially when we embed the whole system within a representation algebra (typically a matrix algebra over a complex space.) The assertion here is that the exponent must be imaginary presumably in order to assure the Lie group element is within the mentioned SU(2)xSU(2) compact group. That exponent being the product of parameter and generating operator leads to the stated conclusion.

3. Jan 14, 2017

### Heisenberg1993

Let "T" be the generator. Then the group element is g=eiαT, where α is the parameter. The existence of the "i" already makes the exponential imaginary. If we make the generator pure imaginary, this will in fact require α to be pure imaginary.

4. Jan 17, 2017

### jambaugh

The exponent in your form is imaginary provided the T generator is real (has all real e-vals). You can absorb the i into the generator but it is kept separate in QM because it also is identified as an observable. So the question about the "imaginarity" of iT passes through to the question of the "reality" of T.