0
$\begingroup$

It was mentioned in http://kclpure.kcl.ac.uk/portal/files/12371620/Studentthesis-Mehmet_Akyol_2013.pdf page 28, a new concept "oscillator basis" or more precisely the author defines gamma matrices of the oscillator basis. What does it mean to have gamma matrices of oscillator basis?

The relevant equations from Akyol's paper are:

The Clifford algebra gamma matrices can also be constructed using these basis elements and are defined in the following way (1.10),$$\Gamma_0 = - e_n \wedge +i_{e_n}$$ $$\Gamma_i = e_i\wedge +i_{e_i}$$ $$\Gamma_n = e_n \wedge +i_{e_n}$$ $$\Gamma_{i+n} = i(e_i\wedge -i_{e_i})$$where $i=1, ..., n-1$ where $n=d/2$ in even dimensions. As for (1.31) it is $$\Gamma_{+} = \frac{1}{\sqrt{2}}(\Gamma_n +\Gamma_0) = \sqrt{2}i_{e_n},$$ $$\Gamma_{-} =\frac{1}{\sqrt{2}}(\Gamma_n - \Gamma_0) =\sqrt{2}e_n \wedge,$$ $$\Gamma_{\alpha} = \frac{1}{\sqrt{2}}(\Gamma_{\alpha} +i\Gamma_{\alpha+n})=\sqrt{2}i_{e_{\alpha}},$$ $$\Gamma_{\bar{\alpha}} = \frac{1}{\sqrt{2}}(\Gamma_{\alpha} - i\Gamma_{\alpha+n}) = \sqrt{2}e_{\alpha} \wedge$$ Here $d=2n$

If we take $d=4$, then $n=2$ and $i$ takes only the value $1$ we get:

$$\Gamma_0 = - e^2 \wedge +i_{e^2}$$ $$\Gamma_1 = e^1\wedge +i_{e^1}$$ $$\Gamma_2 = e^2 \wedge +i_{e^2}$$ $$\Gamma_3 = i(e^1\wedge -i_{e^1})$$ Where the oscillator Gamma matrices can be written as $$\Gamma_{+} = \frac{1}{\sqrt{2}}(\Gamma_2 +\Gamma_0) = \sqrt{2}i_{e^2},$$ $$\Gamma_{-} =\frac{1}{\sqrt{2}}(\Gamma_2 - \Gamma_0) =\sqrt{2}e^2 \wedge,$$ $$\Gamma_1 = \frac{1}{\sqrt{2}}(\Gamma_1 +i\Gamma_3)=\sqrt{2}i_{e^1},$$ $$\Gamma_{\bar{1}} = \frac{1}{\sqrt{2}}(\Gamma_1 - i\Gamma_3) = \sqrt{2}e^1 \wedge$$

$\endgroup$
4
  • 1
    $\begingroup$ 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. $\endgroup$
    – Qmechanic
    Aug 20, 2015 at 19:43
  • 2
    $\begingroup$ The link is currently dead. It's unclear what this question is asking for. Please exhibit the "gamma matrices of the oscillator basis" within the question. $\endgroup$
    – ACuriousMind
    Dec 30, 2015 at 22:47
  • $\begingroup$ @ACuriousMind please reconsider the closure, the question as been edited. $\endgroup$ Jan 2, 2016 at 15:31
  • $\begingroup$ Why exactly did you remove the details necessary to understand the question without clicking a link again? $\endgroup$
    – ACuriousMind
    Mar 5, 2016 at 21:45

1 Answer 1

1
$\begingroup$

The basis change for the Clifford algebra

$$ \{\Gamma_A, \Gamma_B\}_+~=~2\eta_{AB} \tag{1.12} $$

can be viewed as a linear coordinate transformation in the (complexified) Minkowski space$^1$ $V$ of dimension $2n$. The longitudinal and temporal coordinates go into light-cone coordinates. The transversal coordinates pair up two and two and are transformed in a similar way, see eq. (1.31) for details.

In the new "oscillator basis", the gamma matrices becomes isomorphic to standard Grassmann-odd creation and annihilation operators acting on a fermionic Fock space; or equivalently in a mathematical language, exterior and interior multiplication of degree-one elements in an exterior algebra.

--

$^1$ Since OP's link ultimately aims at describing (super)gravity in a curved spacetime $M$, we should stress that the vector space $V$ is generally not spacetime itself. For each spacetime point $p\in M$ the vielbein $e:T_pM \to V$ forms a (vector space) isomorphism with the tangent space $T_pM$ at $p$.

$\endgroup$
6
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Aug 20, 2015 at 21:04
  • $\begingroup$ @Qmechanic I would like to ask you if you may, which are considered as longitudinal,temporal and transversal coordinates? Thank you. $\endgroup$ Dec 30, 2015 at 4:56
  • $\begingroup$ @PhilosophicalPhysics please update your question because this link is dead. I am requesting this because of the discussion that is going on here (physics.stackexchange.com/questions/226780/…). $\endgroup$ Jan 2, 2016 at 13:06
  • $\begingroup$ @Fluctuations done $\endgroup$ Jan 2, 2016 at 13:41
  • $\begingroup$ The answer uses the words longitudinal coordinate, temporal coordinate, transversal coordinates & light-cone coordinates in their standard sense (as seen in e.g. E&M). Note that sign conventions, factors of $\sqrt{2}$, and number of spacetime dimensions may differ. $\endgroup$
    – Qmechanic
    Jan 2, 2016 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.