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Gamma 5 matrix 
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#1
Jun2414, 08:59 PM

P: 895

I am sorry, this is a rather dump question more like a notation question since the sources I've looked into don't specify it... But in case I want to calculate (in general dimensions) the:
[itex] \Gamma^{2}_{chir} =1 [/itex] Do I have to take: [itex] \Gamma_{chir} \Gamma_{chir}[/itex] or [itex] \Gamma_{chir} \Gamma_{chir}^{\dagger}[/itex] ? 


#2
Jun2514, 03:00 AM

P: 93

[itex]\gamma_{5}[/itex] is an hermitain matrix.
so [itex]\gamma_{5}=\gamma_{5}^{\dagger}[/itex] so both options are the same 


#3
Jun2514, 05:44 AM

P: 895

is that true in general dimensions? For example D= 5 dims?



#4
Jun2514, 09:20 AM

Sci Advisor
P: 908

Gamma 5 matrix
[tex] \gamma_{ ( 5 ) } \equiv \sigma_{ 3 } =  i \sigma_{ 1 } \sigma_{ 2 } , \ \ i \sigma_{ 1 } \sigma_{ 2 } \sigma_{ 3 } = I[/tex] And in D = 5, [itex] \gamma_{ ( 5 ) } \equiv \gamma^{ 4 }[/itex] 


#5
Jun2514, 12:41 PM

P: 895

In general D dimensions you can always define a gamma 5:
[itex]\Gamma = i^{a} \gamma_{0} \gamma_{1} ... \gamma_{D1} [/itex] So far I've proven that [itex]\Gamma[/itex] anticommutes with [itex]\gamma^{M}[/itex] if D=even and commutes if D=odd... In the case of D=odd, because it commutes with all the gamma matrices (I think because of Schur's Lemma) you have [itex] \Gamma \propto 1 [/itex] And I'm trying to find the constraint on [itex]a[/itex] so that I can fulfill the requirement that [itex]\Gamma^{2}=1 [/itex] However I am not sure if by the square they mean [itex]\Gamma \Gamma[/itex] or [itex] \Gamma \Gamma^{\dagger}[/itex]... In [itex]\Gamma \Gamma[/itex] case I have: [itex] \Gamma \Gamma= (i)^{2a} \gamma_{0} \gamma_{1} ... \gamma_{D1} \gamma_{0} \gamma_{1} ... \gamma_{D1} [/itex] By doing the commutations properly I get:* [itex] \Gamma \Gamma= (i)^{2a} (1)^{\frac{D+1}{2}} 1 = (1)^{a+\frac{D+1}{2}} [/itex] So if I want to get the identity matrix, I must ask for the exponent to be even. [itex]a+\frac{D+1}{2}= 2n [/itex] [itex] a= \frac{4nD1}{2} [/itex] In the simplest case [itex]n=0[/itex] so that [itex](1)^0=+1[/itex] and we have: [itex] a=  \frac{D+1}{2}[/itex] However if [itex]\Gamma^{2}= \Gamma \Gamma^{\dagger}[/itex] and by supposing that: [itex] \gamma_{0}^{\dagger}= \gamma_{0}[/itex] [itex] \gamma_{i}^{\dagger}= \gamma_{0} \gamma_{i} \gamma_{0} [/itex] I have: [itex]\Gamma \Gamma^{\dagger}= (i)^{2a} (1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D1}\gamma_{D1}^{\dagger} ... \gamma_{1}^{\dagger} \gamma_{0}^{\dagger} [/itex] now inserting the above assumption: [itex]\Gamma \Gamma^{\dagger}= (i)^{2a} (1)^{a} \gamma_{0} \gamma_{1} ... \gamma_{D1} \gamma_{0} \gamma_{D1} \gamma_{0} ... \gamma_{0} \gamma_{1} \gamma_{0} \gamma_{0} [/itex] Now inside D points, you have D1 regions (in this case it means D1 [itex]\gamma_{0}^{2}[/itex])**. Since D is odd, D1 is even and thus the result is just a +. [itex] \Gamma \Gamma^{\dagger}= (1)^{2a} \gamma_{0} \gamma_{1} ... \gamma_{D1} \gamma_{0} \gamma_{D1}... \gamma_{1}[/itex] Since all the gammas commute with [itex]\Gamma[/itex] I can move the middle [itex]\gamma_{0}[/itex] to the 1st place, without a problem, where I'll get [itex]\gamma_{0}^{2}=1[/itex] and the rest gammas will start cancelling each other one after the other without changing anything (+)(+)(+) etc.... [itex] \Gamma \Gamma^{\dagger}= (1)^{2a+1} [/itex] again asking for the power to be even: [itex] 2a+1 =2n[/itex] [itex] a= \frac{2n1}{2}[/itex] Again in the simplest case [itex] a= \pm \frac{1}{2}[/itex] ( [itex]\pm[/itex] because I don't know if it's needed to be positive or negative, for  n=0, for + n=1) *eg [itex]D=1, ~~00 =1 [/itex] [itex]D=3, ~~ 012 012= +1 [/itex] [itex]D=5, ~~ 0123401234= 1 [/itex] etc So for general D I have [itex] (1)^{\frac{D+1}{2}}[/itex] **eg [itex]D=3, ~~ 012 2'1'0' = 012 0 200100= 012021 * (1)^{2} [/itex] [itex]D=5, ~~ 01234 4'3'2'1'0'=012340400300200100= 0123404321 * (1)^{4} [/itex] [itex]D=7, ~~ 0123456 6'5'4'3'2'1'0'= 01234560600500400300200100 = 01234560654321 * (1)^{6}[/itex] 


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