Andrea Shindler

Wilson twisted mass in the ε regimeL = F 24 Tr ∂ µ U(x)∂ µ U † (x) − Σ 2 Tr M † U(x)+U † (x)M ,M = m1 + iµ q σ 3 = m P e iω 0σ 3 m P = iU = U 0 expF ξa σ a ,m 2 + µ 2 qBär, Necco, A.S: 2010O = 1 ZD [U] O [U]e −S χ[U]The ground state U VU = U V U 0 exp iF ξa σ a D [U] =d [U 0 ] D [ξ] A (ξ)The Haar measure d [U 0 ] is invariant d [U 0 ]=d [U V U 0 ]The ground state of Wilson fermions does not play a special rôle in the ε regimeBecause of the integration over the zero-modes one does not need to find thevacuum angleMonday, January 17, 2011