What is the GROUND STATE? lowest energy state What does GROUND STATE mean in Quantum Field Theory? Shouldn’t that just be the vacuum state? | 0  which has an compared to 0. Fields are fluctuations about the GROUND STATE. Virtual particles are created from the VACUUM. The field configuration of MINIMUM ENERGY is usually just the obvious   0 (e.g. out of  away from a particle’s location)

L can only carry even powers of  We’ve discussed Vector fields (spin 1 particles) , g massless Klein-Gordon (Proca) Lagrangian Dirac spinor fields (spin ½ particles) q, e, m, t Dirac Lagrangian What about Scalar fields (spin 0 particles)? By definition, to “transform like a scalar” means '(x')= (x) under a Lorentz transformation Note: a Lagrangian's dependence on space-time exists only through the field and its first derivative xm doesn't appear explicitly in any Lagrangian is a 4-vector so only appears in a Lagrangian contracted with itself! NOTE: L can only carry even powers of  i.e. *, (* )2 to itself be real and preserve U(1) and Lorentz invariance

£ £ £ £  cannot be made U(1)-invariant the simplest Lagrangian with (xm) and m  dependence is £ this is the Klein-Gordon Lagrangain! £ £ £ from which  / - m [ /  (m ) ] =0 yields the familiar Klein-Gordon equation however  cannot be made U(1)-invariant if it is real and the particles to be described are electrically charged but as you have seen (Exam 2)  cannot be made U(1)-invariant if it is real and the particles to be described are electrically charged

A U(1) transformation sends: If  is strictly real, this is not an invariant form unless q = 0 in which case we have no charge-coupling to photon fields. Insisting that  be real would mean the Lagrangian cannot be U(1) invariant. Insisting on U(1) invariance means  cannot be real.

£ £ £ -m2 So instead we try an explicitly COMPLEX FIELD: Recall: , separately satisfied conjugate field equations for a Vector Field £ to remain invariant. for a Dirac Field £ but for a simple scalar, the usual 4-vector “DOT” product becomes just the one-dimensional product  -m2 is this some kind of interaction? m2 m2 This is more like some sort of self-propagation term self-persistence… not a Feyman vertex, but a MASS term!

£=½½¼ Klein-Gordon Lagrangian for a “complex” scalar field Scalar (spin=0) particle (with “self-interaction” term) £=½½¼ Note: OBVIOUSLY globally invariant under U(1) transformations   ei “globally” means   constant Not LOCALLY invariant yet. That is introduced through a covariant derivative with new fields. A completely EQUIVALENT form found by separating out the R() and I() parts:  =1 + i2  

£=½½ ¼ Klein-Gordon Lagrangian for a “complex” scalar field Scalar (spin=0) particle (with “self-interaction” term) Then £=½½ ¼ becomes £=½11 + ½22 ½12 + 22 ¼12 + 22  Which is ROTATIONALLY invariant under SO(2)!! 1 2 Remember SO(2) U(1) are “isomorphic” Our Lagrangian yields the field equation: 1 + 12 + 112 + 22  = 0 or equivalently  2 +  22 +  212 +  22  = 0 Klein-Gordon equation some sort of interaction between the independent states

1 + 12 + 112 + 22  = 0 Obviously as 0 there are no reactions and we will have only free particle states. And as (or in regions where)   0 we have the empty state | 0  representing the lowest possible energy state and serving as the vacuum. The exact numerical value of the energy content/density of | 0  is totally arbitrary…relative. We measure a state’s or system’s energy with respect to it and usually assume it is or set it to 0. What if the EMPTY STATE did NOT carry the lowest achievable energy? = We will call 0| |0 = vev the “vacuum expectation value” of an operator state.

U U    ( , ) Clearly the Lagrangian under consideration the coefficients of our general interaction terms Now, inspired by the classical Lagrangian where L = T - V we separate the “kinetic” from the “potential” terms U U where all the dependence is carried by   (U ) And as U  0 we have free, massless particles What is the minimum U possible? Keeping just the mass term (a free particle)? Where  vanishes (empty space)? Let’s try looking at extrema in U

and I guess we’ve sort of been assuming some calculus of variations U 1 Letting  = * = 0 U 2 = 0 Extrema occur not only for but also for But since  = * > 0 This must mean -2 > 0 2 < 0 and I guess we’ve sort of been assuming  was a mass term!

defines a circle of radius 2 defines a circle of radius | | /  in the 2 vs 1 plane 1 Furthermore, from not x-y “space” U 1 we note 2U 12 which at  = 0 gives just 2 < 0 making  = 0 the location of a local MAXIMUM!

 2  1 Lowest energy states exist in this circular valley/rut of radius v = This clearly shows the U(1)SO(2) symmetry of the Lagrangian But only one final state can be “chosen” Because of the rotational symmetry all are equivalent We can chose the one that will simplify our expressions (and make it easier to identify the meaningful terms) “subtract off” the vacuum’s energy expanding the field about the ground state: 1(x)=+(x)

L=½11 + ½22 Scalar (spin=0) particle Lagrangian L=½11 + ½22 ½12 + 22 ¼12 + 22  with these substitutions: v = becomes L=½ + ½ ½2 +2v+v2+ 2  ¼2 +2v+v2 + 2  L=½ + ½ ½2 + 2 v ½v2 ¼2 +2 ¼22 + 22v+v2 ¼2v+v2

Explicitly expressed in L=½ + ½ ½(2v)2 v2 + 2¼2 +2  + ¼v4 ½  ½(2v)2 Explicitly expressed in real quantities  and v this is now an ordinary mass term!  “appears” as a scalar (spin=0) particle with a mass ½  “appears” as a massless scalar There is NO  mass term!