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26 Jun 2024: d2h. dy22ydh. dy+ (Ẽ 1)h = 0 (2.26). You can check that this is indeed satisfied by our earlier solution (2.24) with h = 1. ... We learn that the. Gaussian wavefunction (2.24) that we guessed earlier is actually the lowest energy.

26 Jun 2024: The gluon exchange is now. 1N2. (6.24). but there are order N3 triplets of quarks, so again the total amplitude scales as N. ... 6N2. X. i 6=j 6=k. V3(xij, xjk). where xij = xi xj and the coecients in front of the potentials are taken from (6.23)and (6.24

26 Jun 2024: integral as. n 4. (2)3 eµ. Z 1. 0. dp p2ep2/2m =. mkBT. 22. 3/2eµ (2.24). Let’s try to interpret this. Read naively, it seems to tell ... anti-aligned to give 3 dierent spin 1 states. With these two amendments, our expression for the number density (2

26 Jun 2024: 0 for all S(r) and R(r) (4.24). This is clearly satisfied if R and S are finite at the origin.

26 Jun 2024: in (1.24) because E = (0,0,E) lies parallel to the side of the torus, not perpendicular. ... Instead, both (1.23) and (1.24) are best thought of as integrating the 2-form Fµν over.

26 Jun 2024: where. Kµ = µtr. A@A. 2i. 3AAA. (2.24). This means that, as in the Maxwell case, the theta term does not change the. ... derivative (2.24), the temporal component was K0 = 42W. For now, the key property of W(A) that we will need is.

26 Jun 2024: p. Figure 24:. to be conserved if we deform the theory, provided that both. ... not Grassmann-valued) four-component spinors n satisfying. i /Dn = nn (3.24).

26 Jun 2024: Zrmax(t). 0. drp1 kr2/R2. = c a(t). Zt. 0. dt0. a(t0)(1.24). ... Indeed,. mathematically it could be that the integral on the left-hand side of (1.24) does not.

26 Jun 2024: For example, we. have. m2K+. m2K0. m2. =mu mdmu md. (5.24). ... corrections, we can generalise (5.24) to. (m2K m2K0) (m2 m20). m20.

26 Jun 2024: 6.24). Of these, U(1)3 acts on V22, leaving the overall sum 1 2 3 4 unchanged.