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The definition of the inner product is not important here, but it is (v, o) = f dªx*(x)o(x). The relevant bit here is th
Posted: Wed May 25, 2022 7:21 am
by answerhappygod
- The Definition Of The Inner Product Is Not Important Here But It Is V O F Dax X O X The Relevant Bit Here Is Th 1 (111.2 KiB) Viewed 17 times
The definition of the inner product is not important here, but it is (v, o) = f dªx*(x)o(x). The relevant bit here is that terms that look like Wi2 are normalized to one and cross terms like B*W¹ go to zero. Consider the W+ boson. We know that it is W+ = N (W¹ + iW²), (1.2) we can determine the normalization N by fixing |W+2 = N² (|W¹|² + |W²|²) = 1. This means N= √2 (1.3) The linear combination for the Z boson is proportional to the combination that gets a mass. 1.1 Inserting the vev Show that iv D₁ (H): iv g (W¹ - iW²) 2√/2 g'B - gW³ (9W 1/²) = 2/12 (²1²+). (1.4) 2√2 where g = g² + g² is the characteristic Z-boson interaction strength. In the last step we just defined the properly normalized W and Z bosons. Based on the above note, convince yourself that g'BgW³ = -9₂Z with respect to the properly normalized Z. The minus sign is conventional. 1.2 Masses When you take D₁ (H)2, you end up with masses M²W+W+M²Z² +/M²7² (1.5) The factor of 1/2 is convention for terms with two identical particles.³ Show that the masses are MW g²v² 4 M² = 922 (1.6) 4 Which particle is heavier, the Z or the W? =