ABSTRACT
The usual starting point for this topic is the wine bottle potential depicted in Figure 9.1. Clearly, there is an unstable position of equilibrium for a ball under “gravity” on top of the hump in the center. Once this is disturbed there is a spontaneous breaking of the symmetry and one position on the horizontal circle (shown dotted) is selected at random. Obviously we are really talking vacuum expectation values and states here. The dotted line around the lowest point of the potential is a set of massless states of equal energy, which can be transported with effectively zero force. However, the selected state hasamass, as we see by the fact that it takes energy to move it up the wall. In practice the “bottle” is represented by a quartic in scalar fields with a positive coefficient for the fourth power terms but a negative coefficient for the quadratic terms. So we can write https://www.w3.org/1998/Math/MathML"> a ( ϕ 2 ) 2 − b ϕ 2 w i t h a > 0 a n d b < 0 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/c5a9cf84-23e1-4064-9971-05f6815a4e54/content/math9_1_B.tif" xmlns:xlink="https://www.w3.org/1999/xlink"/> which has a minimum when https://www.w3.org/1998/Math/MathML"> 2 a ϕ 2 − b = 0 , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/c5a9cf84-23e1-4064-9971-05f6815a4e54/content/math9_2_B.tif" xmlns:xlink="https://www.w3.org/1999/xlink"/> https://www.w3.org/1998/Math/MathML"> r e a l l y < ϕ 2 > = b 2 a , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/c5a9cf84-23e1-4064-9971-05f6815a4e54/content/math9_3_B.tif" xmlns:xlink="https://www.w3.org/1999/xlink"/> where the lowest point is https://www.w3.org/1998/Math/MathML"> a ( b 2 ) 2 4 a 2 − b 2 2 a = b 2 4 a ( b 2 − 2 a ) , https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/c5a9cf84-23e1-4064-9971-05f6815a4e54/content/math9_4_B.tif" xmlns:xlink="https://www.w3.org/1999/xlink"/> which can be set to zero by putting b = (—)√2a. The potential is then https://www.w3.org/1998/Math/MathML"> a ( ϕ 2 ) 2 − 2 a ϕ 2 . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429184550/c5a9cf84-23e1-4064-9971-05f6815a4e54/content/math9_5_B.tif" xmlns:xlink="https://www.w3.org/1999/xlink"/>
