PoS(TASI2018)004 https://pos.sissa.it/ vacua. Next, I review θ ∗ hook at umd.edu Speaker. These TASI lectures where given inand 2018 the and . provide an I introduction start by tomechanical introducing the perspective, the Strong calculating Strong CP the CP problem problem eDM from and both discussing a the classical and a quantum the various solutions and discuss thevarious active experiments areas proposed to of look axion for model these building. solutions. Finally, I summarize ∗ Copyright owned by the author(s) under the terms of the Creative Commons c Attribution-NonCommercial-NoDerivatives 4.0 International License (CC BY-NC-ND 4.0). Theoretical Advanced Study Institute Summer School4 2018 - ’Theory 29 in June, an 2018 EraBoulder, of Colorado Data’(TASI2018) University of Maryland E-mail: Anson Hook TASI Lectures on the Strong CP Problem and PoS(TASI2018)004 is θ . 8 Anson Hook has 1 loop RG running θ . I discuss non-axion solutions 4 . The various dark matter aspects 6 value. It is sometimes said that the Strong ) 1 ( O 1 , which when sent to zero does not have an enhanced 3 down . Asking a student to draw the neutron θ / 1 − and axionic solutions in Sec. 5 . Finally I give a theorist’s overview of experiments in Sec. 7 , I present a classical description of the Strong CP problem and some of its solutions. 2 3 up and two charge / , I present the Strong CP problem at the quantum level. I then discuss the vacuum structure 3 In some sense it is also one of the least robust problems of the , because if Partly due to the robust nature of the Strong CP problem, solutions to it have always been of Finally, a hot new topic is designing experiments to look for the axion or its variants. Each In Sec. At its heart, the Strong CP problem is a question of why the neutron In these lectures, I will review the Strong CP problem, its solutions, active areas of research The Strong CP problem is in some sense both one of the most and least robust problems of the to the Strong CP problem in Sec. In Sec. of QCD and how it resolves various historical confusions in Sec. interest. There are several standardsolution. symmetry-based While solutions none to of the these problemfield have as particularly theories well convincing (EFTs) as UV are the completions, very axion some economicalvariations of and of the simple. the effective axion Most and model its building dark these matter days aspects. focuses on experiment has different sensitivities toreviews different with regions excruciating of amounts parameter of detail space.endeavor about to As past, do current there in and exist these future good notes experiments, what is I to will explain the basic idea behind each experiment. (eDM) is so small.described at It the turns classical level. outcharge Classically, that the 2 neutron both can the be problem thought of and as all composed of of a the single common solutions can be CP problem is even more robust thanfor the which hierarchy there problem is because no it anthropic iscan solution. the constitute Thus only a even puzzle problem people of need with the to the SM have most an extreme opinion position of on some what sortset on to be the small Strong at CP some problem. sense, scale, then one it can stays just small by setminimal renormalization it SM group to and (RG) evolution. be doesn’t In small hold some from in and the most forget gluino of about mass its phase it. and extensions. this However, In this “set the it property and MSSM, is forget unique it" approach to fails. the of axions are discussed in Sec. 2. The Strong CP problem and its solutions2.1 at the The classical Strong level CP problem symmetry. Thus it natural to expect it to have an and the various experimentsprovide searching a beginning for graduate the student with axion.related all project. of the As The requisite with background any hope introduction need isbias to to a start on that topic, a these the this Strong notes CP subject set willattempt have so of much to it of provide lectures my as is own will many personal highly referencesthat encouraged as I I for miss. can, readers As but a given to result my develop an own exhaustive their laziness literature there own search will opinions. is bestandard left references as model I an (SM). will exercise Unlike for the theStrong reader. flavor CP problem, problem but involves like a the parameter Higgs mass hierarchy problem, the TASI Lectures on the Strong CP Problem and Axions 1. Introduction 2 (32) (33) (34)

PoS(TASI2018)004 (2.3) (2.4) (2.1) Anson Hook cm. Because e 13 − . By taking the difference − ν D cm (2.2) . If asked to calculate the eDM of the e 2 . . θ | cm D dE . cos , the student would then arrive at the classical e r ~ π ± q − 26 m B 1 − 2 / ∑ µ 1 √ | 10 = 2 13 ∼ ~ − d U n = | ≤ r n 10 ± d | ν ✓ | ≈ qd n d | = p ] 3 , 2 , , the experimentalist turns off the electric and magnetic fields and measures how 1 3 1 t . The experimentalist then redoes the experiment with anti-parallel electric and mag- + ν A classical picture of the neutron. From this picture, an estimate of the neutron eDM may be Many experiments have attempted to measure the neutron eDM and the simplest conceptual 2 3 many of the have precessedfrequency into the spin-down position. This determines the precession After some time netic fields. This new experiment determines theof precession these frequency two frequencies, theneutron neutron eDM eDM is can [ be bounded. The current best measurement of the way to do soprepared is a bunch via of a spinapplies up precession a neutrons experiment. set all of pointing parallel in Imagine electricat and the that a magnetic same rate fields an direction. of to the unspecified The system, experimentalist experimentalist which then causes has Larmor precession made. usually ends up with something similar to that in Fig. Thus we have the natural expectation thateDMs the are neutron a eDM vector, they should need be to ofbreaks point order Lorentz in 10 some symmetry, and direction. that The is neutron itsspin has only spin. (possibly a with Thus single a the vector minus eDM which sign). will point in the same direction as the Figure 1: neutron, the student would simply take the classical formula TASI Lectures on the Strong CP Problem and Axions Using the fact that the neutron has a size estimate that + PoS(TASI2018)004 (2.6) (2.7) (2.8) (2.5) 3. / π to be true n Anson Hook ˆ d , we find that 0. The critical n − ˆ d . Remembering = n = ˆ d = ? Phrased another s θ s 13 = − s and ˆ 10 n ˆ d ≤ = θ s . . s s → or in terms of the angle s → − . . π t x s ~ . By the same reasoning, this again means that n ˆ d , under . We have studied the neutron and it → − → − 3 − d n , t x ˆ ~ d d = − s : : → → − = to ˆ T P d d s n ˆ d to ˆ : : = , we have under parity, n ˆ P T p s ~ d × = r ~ s = is dynamical and can change. It can be proven that the minimum will s ~ θ 0, it quickly relaxes to 0. Motivated by this example, the axion solution = ] and the Strong CP problem is solved. 4 θ 0 [ = θ The second classical solution is time-reversal (T) symmetry, typically called charge parity The last solution that can be seen at the classical level is the axion solution. The situation of There are three solutions to the Strong CP problem that can be described at the classical level. . The plus charged carbon is exactly between the two oxygens with the equilibrium condition 2 is an experimental fact thatis there for is the only neutron a to single go neutron to whose itself spin under is parity. 1/2. The only Thus way the for only both option ˆ (CP) symmetry due to theUnder time fact reversal, that the combined CPT symmetry is a good symmetry of nature. Thus a neutron is taken from ˆ is if the dipole momentexperimentally is we zero. have observed This thatit parity is is is the maximally a parity broken bad solution by symmetry tocomplicated. the of the weak nature Strong interactions. and CP any Thus problem. application to However, the Strong CP problem is necessarily more Considering again a neutron whose spinunder and time eDM reversal, point in the same direction, ˆ As before, a neutron is taken from ˆ the neutron eDM must beand zero. is in As fact with maximally parity, broken CP since or the equivalently CP-violating T phase is in the not CKM a matrix symmetry is of about nature having two negative charges on opposite sides of a positive charge seems very natural, just look at We first consider a neutron whose spin and eDM point in the same direction, ˆ being that the angle between theidea two bonds for is making exactly this situationinitial work angle is is that not theis angle the between idea the that two the bondsalways angle be is at dynamical. If the The first requires thatitself. parity be a good symmetry of nature. Under parity, space goes to minus TASI Lectures on the Strong CP Problem and Axions We have thus arrived at theway, Strong the Strong CP CP problem, problem orquarks is why on simply is the the the same statement angle line! that the student should have2.2 drawn all Solutions of the that angular momentum is CO PoS(TASI2018)004 . A ) 1 (3.2) (3.1) (3.3) ( U × B Anson Hook ) 1 ( ] provide good 6 U × . If the angle between R 2 ) ), left-handed quarks 2 µ ( are Weyl . For A c A q SU ) ] and Ref. [ 1 O 5 -2 × ! ( d L and 0 U ) m q 2 ( u B C 0 m ) , SU 1 1 1 -1 1 (

U = O M R ) , 2 0 ( ). The fermions = 4 , ) SU c c i 6 = c d c qq L u h ) qMq 2 D ( + = ( ˜ c G SU q G 2 2 s U ) π g 3 θ ( 32 plays no roll in this subsection and will be ignored for now. This adj θ SU ⊃ D . L c µ ρσ q q M A G gauge group and 4 global symmetries ) 3 ( µνρσ ε 2 1 SU = ) and right-handed quarks ( A axion solution to the Strong CP problem is treating the neutron like CO ) µν ˜ G ud At low energies, this theory becomes strongly coupled and we have no analytic traction on We now formulate the Strong CP problem at the quantum level. As the Strong CP problem We consider QCD with two light flavors. This theory has ( = ( q Under these symmetries, the particles and spurions transform as theory has an what happens. Instead, what we willfield do theory is of use the various pions. inputsit from The has experiment starting been to point determined build is an experimentally the that effective measured fact that QCD confines. In particular, introductions to the topic. Aside from the kinetic terms, the theory has the Lagrangian Figure 2: is a question aboutlow-energy the QCD. properties In of this section, thelow-energy we neutron, QCD, follow aka we a the need semi-historical theory of route. toanomalous pions, symmetries. develop We incorrectly. Next, first a we We describe then describe theory how how fixneutrons to of it to into get via neutrons get the the a theory theory and better of and pions understanding calculate correctly. of the Finally, neutron we eDM. add 3.1 Low-energy QCD done incorrectly TASI Lectures on the Strong CP Problem and Axions the up and down quarksdipole is moment. dynamical, This dynamical it angle will is relax called itself the to axion. the minimum3. energy configuration The that Strong has CP no problem at the quantum level ( those unfamiliar or in search of a review of Weyl fermions, both Ref. [ where PoS(TASI2018)004 0 η . (3.7) (3.6) (3.8) (3.4) (3.5) 2 has the π i 2 U ± √ Anson Hook 1 . As with any π A ) = 1 ( . At this point, we ± 0 U π η is associated with the m 960 MeV. This clearly preserved the diagonal 0 + ≈ 0 Π U , 0 π η ! m 0 m 0 ., η π A c = ··· . ) is the identity matrix. h + 1 2 + π ( † a ! + m U d d π , and also breaks µ m m UU D ∂ ) MU B − + a 2 ) u u ( π 1 Tr ( µ m m , 3 π ∂ a SU U d d 140 MeV while 2 1 σ a f . m m π is actually not a good symmetry and that the f † ≈ a 2 = R Π + is the identity matrix. R 0 A + − 5 √ ) † ) † i π u u 0 2 1 e = ( U m U ( m m σ µ µ L =

U ≈ ∂ ∂ SU  + U 0 U U π µ µ η . Meanwhile the other pions are typically referred to L ∂ m ∂ 0 ) 0 2 η π ( Tr Tr  2 2 π π f is a unitary matrix so that f SU π down to its diagonal 2 R a f U = = ) is expanded in terms of the pion fields, the kinetic term is canonically . 2 U + 2 ( , L L U 1 − π π SU + × = π L 2 ) , ) and is called the d 1 2 A m ( Π ) 1 + ( SU u while breaking the axial group are the Pauli spin matrices and and U m 3 R ( 0 − π π 1 = a f σ L = 3 As we know nothing of how we got to this theory, we will write down all renormalizable = Π V where a is an arbitraryLagrangian in constant terms that of the will pion be fields, determined one obtains by the matching mass with matrix data. Expanding this boson obtains a large mass from another source. normalized. We now include theother mass words, of we the perform quarks, ahas keeping series transformation only expansion properties the in under leading-order the small operator. flavor masses. symmetries, In we Remembering write that the the leading-order mass operator as matrix spontaneous symmetry breaking, there will existare Goldstone expressed : in These terms are of called the the matrix pions and symmetry transformation properties All other terms in the potentialthe are requirement higher-dimensional that operators when and their coefficients are fixed by We see that there are fouragain light turn particles to whose experiment masses and obey finddoes 2 that not obey the sumdiscuss rule in that the the next section, EFT it just turns derived, out so that something has gone wrong. As we will as where breaking of TASI Lectures on the Strong CP Problem and Axions which breaks I’ll leave it as angroup exercise to the reader to demonstrate that the vev of operators consistent with symmetries withone arbitrary can write coefficients. down is The leading order operator that PoS(TASI2018)004 (3.9) in the (3.12) (3.11) (3.10) is now M θ . Anson Hook θ as a spurion. 0, the pseudo- M = θ . To obtain a spurion 0 does not convert the U by shifting = θ 0 limit so that the Goldstone . 3.10 α 1, so even if → 2 small and apply a Taylor series. as a spurion and = . − M | θ M θ i θ θ i e , e c | . → u ˜ G α θ i , e G ˜ G 2 symmetry discussed above is actually not a 2 π → acts on the pion matrix G g because 2 c , A A 16 ) u 2 ) π d θ 1 g 1 6 α α ( i ( 32 e + U U θ , → L ⊃ u d α i → e L → L , u u appear in the Lagrangian as α i e θ realizes the symmetry non-linearly, i.e. it shifts under the symmetry → θ . Despite this, it still has its uses, as we will see in the next subsection. u 1 , the masses of the pseudo-Goldstone bosons are suppressed by M . 3.10 0 then the anomalous symmetry would be a good symmetry, but then confinement does not occur. = s ], one finds that if one rotates the quarks by g 0 limit. The reason is that the symmetry is restored in the 7 For the spurion Note that there are several important differences between If By remembering that there was a term in the Lagrangian that is In this subsection, we discuss how the Because there is no symmetry, there should be no Goldstone boson. However, explicitly bro- 1 → that transforms linearly, we let M then under this rotation, the Lagrangian also changes as A major difference is that boson masses must go toHowever, zero this sort in of this expansion limit. is impossibleGoldstone Thus for boson we mass can is take still non-zero.anomalous This symmetry is into reflected a in true the symmetry.a fact The symmetry that anomalous of symmetry the never theory was and never will be rather than changing multiplicatively the way good symmetry of the theoryRef. and [ the implications of it. From any number of QFT textbooks, e.g. we see that we canour cancel spurion. the For piece 2-flavor QCD, added the to proper the anomalous Lagrangian symmetry in is Eq. ken symmetries are stillbuilding useful. a theory of After pions even all,star when of in there the are previous the show explicit were previous mass spurions,to terms subsection, constants find that that a we break transform constant under the under showed symmetries. symmetries. which how Thus, weThis The we can to particular wish take example this start non-symmetry is and usually turnanomaly called it in into an Eq. a anomalous spurious symmetry symmetry. due to the association with the TASI Lectures on the Strong CP Problem and Axions 3.2 Anomalous symmetries The reason for this anomalous symmetrytion. is that the measure is not invariant under this transforma- PoS(TASI2018)004 , A ) 1 ( (3.16) (3.14) (3.15) (3.13) (3.20) (3.17) (3.18) (3.19) U Anson Hook isn’t a good . But that is . If the quark A A d ) ) 1 m 1 ( ( + U U u m . . transforms under ∼ M c . as 0 α π i h . η − m . + M e α 3.16 α i U 2 → − e − det M 4 π θ → ··· b f symmetry is not a good symmetry of . + M → 2 A + , ! . It is left as an exercise to the reader to θ ) θ  . π α i 1 is still valid. Thus we see that the phase of 0 a f 2 ( + but not invariant under η 2 σ 2 , 0 MU φ f . i π U − f B a √ θ α √ − 2 i θ ) , π θ Tr 2 3.15 e √ e 1 φ d , we find that | i ( 7 3 − π α e b θ − i 0 i | U → = 2 θ U e i + θ a f η 0 i θ e = φ × 0 i transforms. Thus there is an anomalous symmetry  + → e 0 b π → R c = behaves, we can finally go back and redo the theory of is not a good symmetry of nature, we can write down a † h ) 2 η d

0 θ 2 A U m qq U ( ) η , µ = 1 2 0 1 ∝ ∂ η ( SU f U = , U U U , u α µ × 2 α ∂ U L i L α ) e i √ , transforms under this symmetry, the corresponding pseudo-Goldstone 2 Tr e ( θ + 2 → π 0 f → u boson, this means that the following is a good symmetry of the theory : SU is fixed to be η 0 = U b η → tells us how 0 L q was never a true symmetry to begin with. Note that while η boson can be obtained by Taylor expanding Eq. α i 0 A e ) η has an expectation value 1 ( → 0 U q π , obtains a mass even in the limit where the quark masses goes to zero. 0 η Now armed with the fact that As in the case of non-zero quark masses, broken symmetries are still useful in constraining Now that we understand how the As mentioned in the previous subsection, the symmetry, the anomalous symmetry given in Eq. fine because new term in the effective Lagrangian : which is invariant under we note that Written in terms of the Because a constant of nature, boson, how their corresponding pseudo-Goldstone boson appears. To see how the complex coefficient show that the charged pions do not obtain an expectation value. Thus we are expanding about nature. Recall that the anomalous symmetry is The mass of the Plugging this expectation value into the matrix pions carefully. The first stepnon-trivial. is The to easiest find way the to vacuum see about this which is to to expand. look at This the vacuum pion can masses, be TASI Lectures on the Strong CP Problem and Axions 3.3 The theory of pions and neutrons done properly masses were negative, then theassume pion that mass would also be negative. To find the vacuum state, we PoS(TASI2018)004 .  c ρ (3.26) and . UN u , 0 ). As (3.24) (3.22) (3.23) (3.25) (3.21) µ θ  R η ∂ ) ) † invariant 2 term, and 2 θ ( A U ) Anson Hook θ µ − 1 is necessarily SU σ ( ( φ † , a ( U c L . ) N , lending credence . 2 cos 2 0 θ ) ( d + d π 2 0. As a fun aside, I m m N SU † sin ≈ + and 2 U ) + u θ ) µ + 2 d θ d m ∂ π ( m m + U π u µ + must be real. As the reader can φ m a f u ( σ 4 a † m = ( N cos ±  u 2 π ) − m 1 m 1  . The particle that cuts off this divergence − , 3 doublet. π , s + A 2 d π ) ! f g a f θ π symmetry. The mass term is 2 p n ( , 2 ( 2 π i appears due to the expectation value of A + 2 θ

− m ) u θ 1 SU − θ − 8 = ( = c cos . + U = d N c . † θ m qqq h V u U ) transforms as a doublet under = , while † = + m c 0 2 θ M N π N ! , giving the first indication that † ( 0 , + U π 2 d 2 θ N m ! NU m d 2 θ to express the potential in a clean form. Note that c ≈ i tan + e 0 2 u d d + φ − d π m c m m m m u + − q . Thus all CP-violating effects in QCD itself come from the θ π u u i θ e 0 m m NMN u a f 1 m c dependence and thus the arbitrary constant = = 0 − θ φ is positive or negative has no physical effect, so for simplicity we take it to be 2 π c . I have not written down how the indices are contracted. I leave as a fun exercise

. c a m N N q tan † c Tr q 3 π c NU q a f N = meson. The standard quadratic divergence estimate for the mass difference cut off by the − m appears is restricted by the anomalous c ρ − = N θ With a little bit of effort, the minimum of this potential can be found to be After this long and arduous trek, we finally have a theory of pions that gives the correct pion Working through the indices, The potential comes from the term = can be non-zero) and V comes from the expectation value of d so that there is no check, whether It is an observed fact that positive. On the otherwritten them hand, out the explicitly. masses can break CP with their non-trivial phases so we have θ encourage the reader to attribute thequadratic mass divergence due difference to between the the electric chargedis charge and the of neutral the pions to the meson should reproduce the measured difference in mass between the before, we now write down all of the leading-order termsL with arbitrary coefficients : to the reader to contract thequark, indices and and that show the that proton the and proton neutron is fall made into of a two up quarks and a down masses. We can now incorporate protonsment, we and know neutrons that into protons and the neutrons theory.a are Again nucleon each composed appealing field of to three experi- quarks. We can thus construct where we have used a shiftreal of because the QCD Lagrangian is CP conserving up to thehow mass terms of the quarks (i.e. Expanding about this minimum, we find that the pion masses are to standard arguments for quadratic divergences. with TASI Lectures on the Strong CP Problem and Axions φ ) p ( u ) (3.28) 2 π  ) m N + m 2

PoS(TASI2018)004 ) + 2 2 / / 0 q p / − l − 2 )(( / 2 π p / m (3.30) (3.29) (3.27) − gives a l + / 2 2 c − as well as ) . There is ( 2 Anson Hook 5 3 θ + / γ 1 q c + + 5 = l γ ) . + N )(( c d 7 using the measured 2 N m . d m 1 m m + + u ≈ 2 + u m / 2 0 + m ) ]. The matrix element of the p c / 2 8 = / − 0 p µ 2 . n / ) + p / p 2 ( − / ± u . , p 5 l 2 / c ⇡ . γ i n − + N ν ( 5 27 a l . q γ τ i 1 (( µν N ) µν 0 a ≈ γ γ p ) π 0 A ( n p 9 p g N u , ( µν ) µ π m u f l p is the UV cutoff of our theory of pions. F A matrices seen in Dirac notation. ) 2 n ( g q π 5 4 d u i f ( ± ) l γ 5 ∗ µ 4 γ π π ⊃ − ⇡ i ε d 4 2 c n is small, we have performed a Taylor series in ν ( d q N L ∼ θ 2 a Z µν τ Λ ) = γ . N q ) 0 a ( q n p ∗ µ iM π ( ε u µ π ) µ f + q + c gives the leading-order interaction between protons and neutrons, so by ( c π ∗ µ θ A f θ ε g 2 − 2 2 π Λ m √ = N 2 The Feynman diagram giving the leading-order contribution to the neutron eDM. π log L m f is better related to the decay of the neutron, but that would be a long digression all by itself. A π µ 2 f g A + π 2 g c 4 ) is the incoming (outgoing) momentum of the neutron, and q is the incoming momen- 0 A √ p g ( ie θ Figure 3: p e Actually, Let us now pretend that the neutron has an eDM in the Lagrangian, To obtain the neutron eDM, we calculate the Feynman diagram shown in Fig. ≈ − ≈ 2 mass splitting between various nucleons andvalue can of be their determined to masses. be scattering protons off neutrons, we can measure not much to beusing learned Dirac from notation. the For computation those interested itself, in so more details, I see will Ref. only [ briefly sketch the procedure where taken the leading-order piece in tum of the photon. Anticipating that Feynman diagram is iM This would correspond to a diagram with the matrix element TASI Lectures on the Strong CP Problem and Axions Expanding these terms outleading to CP-preserving and leading violating order interactions are in pions and integrating by parts, we find that the Note that in Weylcoupling notation, is the imaginary or difference real, between and CP not the preserving and violating is whether the PoS(TASI2018)004 (4.1) (4.2) (3.31) (3.32) (3.34) (3.35) (3.33) Anson Hook . While I will θ . People (this author θ . θ ! We already know that all . versus θ α α . . θ ˜ + + when they mean G is the only physical quantity, we θ θ G θ e cm 2 θ 2 s θ → → π g θ 16 θ θ ˜ 32 G 10 G + 2 c 2 s × π , g , 3 dd . θ α α 10 d 32 i ∼ θ i − i − − should not affect any physical quantities. This is x . To see that e + 2 2 π u | 10 d d 10 2 θ Λ θ m θ m G . iHt 3.22 4 1 e → + → log θ c u d θ µ , then θ 2 uu π ⊃ − + θ u f c θ i 2 A e L π g u , , 8 , we find that it is independent of θ u m d e α α i i H ⊃ e = e , we see that n . L → d → θ as defined in Eq. u 3.28 d θ and thus any physical answer can only depend on to the bounds on the neutron eDM gives n θ or just d θ vacua. The first is the following statement : If we start with the Lagrangian θ vacua θ In this section, we are motivated by two confusing puzzles whose solutions lie in what is To show that these results are correct, we can perform an important check that the potential Unfortunately, the Strong CP literature is often not clear about try to be careful in thisthe review, the author reader means should be alert in general and use context to determine if and calculate the Hamiltonian Thus, all we needHamiltonian to does know not about depend a at system all on are its initial conditions and the Hamiltonian. If the 4. The known as the the first puzzle. depends exclusively on Finally, comparing and These anomalous symmetries are simplyphysical a quantity reflection is of invariant how underinvariant you’re quantity these defining is your anomalous quarks, symmetries. so any It is easy toincluded) see will often that be the sloppy in only their notation and simply write of physics can be boiledalways down given to by asking how a given state evolves in time, and that the answer is TASI Lectures on the Strong CP Problem and Axions Comparing this with Eq. remind the reader of the QCD Lagrangian Two anomalous symmetries constrain the theory : PoS(TASI2018)004 3 is r / 1 (4.8) (4.3) (4.4) (4.5) (4.6) (4.7) 4.1 cannot ∼ θ K Anson Hook . This is the θ . Thus, we discover 3 S . . i to ∞ c σ ) → A r 3 , | b ρ ( 2 ˆ r 1 A r K SU 2 abc < 2 s f π g E g 3 . θ . 32 . 4 1 † iS x r − 3 e ] U goes to at infinity are all equivalent, or d a < x , ρσ µ [ F 2 . In particular, one can either prove or look ∂ µ ∞ Z d 3 h f A G U scales can be obtained using the fact that we appears in super selection rules in the Hamil- a ν x < i = x A ∼  θ ˜ K Z → G 2 ˜ ] but there exist other more recent references such G µ B G = 11 9 A G 2 , and the action to be independent of , we need the gauge field to become pure gauge so i µνρσ 2 s + i 3 π g ε x ∞ 2 r ∼ | θ / E r 32 = / → x , µ x 4 iHt r K 3 ∞ e d K d | ∼ Z f → at infinity, then this quantity will integrate to zero and Z , are all equivalent to have finite energy. Via handwaving logic, x K r 3 h 3 ⊃ 2 ∂ S r ∼ , r / µ / L x K 4 µ d ∂ Energy Z = . The 3 in the subscript indicates mapping ˜ = G Z S G to decrease faster than 1 fields vanish : K )) = 3 B ( SU ]), so I will only sketch a brief outline. vanishes faster than 1 and ( 3 10 K E π , and the sphere at infinity, ) For the more mathematically minded, this is a question of homotopy. Our starting point is asking what sort of boundary conditions to impose on our theory. We The second puzzle is related to the first and stems from the fact that the coupling in Eq. The resolution of these puzzles lies in the instantons and the vacuum structure of QCD. In 3 3 ( to phrase it in theSU parlance of a mathematician,up whether that the mappings between the gauge group, Then we ask whether the gauge configurations that that the An alternative way to calculate physical quantities is to use the principle of least action, What is of interest isdivergence the theorem, action, which is the integral over all space of the Lagrangian. Using the want a system with finite energy, so as are dealing with a system at finite energy : as Ref. [ a total derivative : Thus, if have any physical effect. An expectation for how TASI Lectures on the Strong CP Problem and Axions where E decreases faster than 1 Thus we expect particular, the first puzzle is solved by the fact that second puzzle. tonian formalism and thus playsis a solved role by in the the presence specification of finite-energy of field the configurations initial known states. as instantons The where second puzzle and the surface integral does notown vanish entire at review infinity. (the Instantons classic are reference a is Ref. gnarly [ subject and require their PoS(TASI2018)004 (4.9) (4.12) (4.11) (4.10) , or how contains is just the ) )) states have 3 3 1 Anson Hook ( ( S n 1 state. Since U to ( SU 1 = ) π 2 n ( and the second circle SU 1 φ is the winding number at 2 0 state. In the path integral, , n τ = ~ · A x ~ i r + . 1 2 4 n x − = 1 0. We see that this mapping is characterized g . n , iS 2 = = e ] φ 2 ˜ n works in much the same way. As A G φ [ 12 sphere, we see that mapping , G 1 d Z 3 → f 2 − S i 1 A g π 0 to 1 A shift symmetric. When mapping one circle to the other φ µ integrates to, we can dictate the asymptotic behavior of Z ∂ 32 )) = = π is a very important fact and more details can be found in ˜ g 3 x 1 G ( 4 2 φ can go to at spatial infinity are characterized by an integer. G d ρ 2 µ SU 0 and we are looking for tunneling into the 4.10 r Z + ( A 3 2 = states are specified by how the gauge field falls off at infinity. The r π n n = are energy eigenstates. The first thing to note is that all µ n A 0, we can say that our initial state is the = n 0 initial state. Rather than deriving the answer, we shall take a guess-and-check = , which is the symmetry of a , typically called the winding number, which indicates how many times the first ) . Being angles, they are 2 n A 2 2 ( φ is the winding number of the gauge field at infinity and SU 1 ]. n 10 Since the vacuum is an eigenstate of energy which has the lowest energy, let’s check if the field We have now discovered that the asymptotic behavior of the gauge field is characterized by an For a better understanding of this result, we consider the simpler example of our initial state has we integrate over all finite-energydeforming gauge-field the configurations that can beapproach. reached We by simply continuously guess the following gauge-field configuration, which is called an instanton : the same energy. Each of these lowest energy state in each ofall these points sectors in is space when time. the Because gaugethus the field the gauge takes field same is on energy. pure Next its gauge, we asymptotic they askTo value all answer whether at have that these zero states question, E mix we and with first B one fields remind another and ourselves under that time physical evolution. quantities are obtained from Imagine that our initial state is Ref. [ configurations specified by circle winds around the second. integer. It turns out thatby the winding number of any given gauge configuration can be determined by the integer the group higher dimensional version of thenumber. previous example, which is again characterized by the winding TASI Lectures on the Strong CP Problem and Axions that the gauge configurations that They cannot be smoothly deformed intosmoothly each connected. other without leaving pure gauge, as integers are not to map a circle to anotherhas circle. an angle Imagine that the first circle has an angle where for simplicity we have mapped the gauge field configurations. Eq. circle, we have Where the origin. Thus, by specifying what PoS(TASI2018)004 i n states | (4.17) (4.18) (4.13) (4.15) (4.14) (4.16) term in n O | θ . Anson Hook ∆ ˜ G G + 2 π n θ h 32 ˜ G + G 2 L . π x 1 4 32        d x 1 R 4 i d 1 2 3 − R ··· , though its eigenvalues do θ th root of unity. We can now i D E dAe e D .        n , are the Pauli spin matrices, and D Z · / ∆ ∑ τ and i j      π = ) = ε 2 E i ˜ e G n 1 1 | G − − = , . 2 j ··· D D , mixing with each other. The matrix that i also appears in the Lagrangian. We note O 1 ε ε π w n 1 | D θ | 2 = 32 ∆ ε n ··· ··· ˜ x G θ 4 i + 2 1 2 e G d , ε ε ε n 2 h 13  and throw out all other values when we define our 1 Z ∆ ∑ π 1 E ε 1 1. In fact, we can connect any two different θ i − θ N − 1 32 e j D = n 1 ∆ − x ..... , = w E ( ε 4 n D ∆ ∑ i ε d δ ˜ θ G = is completely orthogonal to these other states, so there is no ···      Z | G i · j 2 θ 2 n π 0 and  θ | w 32 1 j = + O − n w | L x D 1 4 m d h  ) R i n ··· appears as part of the state in the Hamiltonian formalism. Now we show − θ m vacua in the Hamiltonian is completely equivalent to having the limit, we weight each of the states by a phase and sum. The eigenstates of ( dAe θ i θ ∞ e Z 1 2 3 n ,  ∆ → ∑ ∑ m D = = i θ | is just some normalization factor that doesn’t matter. is an arbitrary constant that is the size of the instanton, is what is called a super-selection rule. Because it is impossible to transition from one value ρ N O | to another, we choose a single value of Now we will find the eigenstates of the Hamiltonian. To get a simple intuition for the answer, We have seen that θ θ θ h is pure gauge. It is simple, but annoying, to check that this gauge-field configuration has finite Hence choosing the the Lagrangian. energy and that TASI Lectures on the Strong CP Problem and Axions where g Hence this field configurationoverlap has between the all two of states the properties we need to show that there is non-zero we first pretend that there are a finite number of states, by superimposing these gauge-field configurations. we are diagonalizing is called a circulant matrix and is of the form They are simply the sum ofsee each that state weighted in by the a multiple of the that the Hamiltonian are that when we transition to path integral formulation, reason to include them in our theory. Its eigenvectors are completely independentdepend on of them. the The values eigenvectors of are theory. Our sector, with it’s value of of where PoS(TASI2018)004 does . We (5.1) (5.2) (5.3) (5.4) (5.5) θ θ Anson Hook ] but has been 11 . As mentioned before, d θ . . α α α and + + + u θ θ θ θ , θ → → → α θ θ θ + θ d θ , , , → α + α α u θ − − − θ u 14 d d θ + θ θ θ , → → → should be zero making the action independent of u = u d d α 2 i θ θ θ θ e π 32 → ]. The neutron dipole moment is a CP-odd quantity. Taking / ˜ u , , , G 14 u d d , α α α i xG i i e 13 4 e e , d → R → → 12 u d d have an effect when it does not appear in the Hamiltonian. The answer is that it θ , and the new anomalous symmetries are instead θ However, if the is massless, there are only two CP-violating terms in the Lagrangian, Thus we have finally resolved the two puzzles presented in the introduction. The first question In this section, we briefly discuss non-axion solutions to the Strong CP problem. Axion-type The massless up quark is the simplest solution to the Strong CP problem [ and d and Physical quantities such asinvariant the under these neutron anomalous eDM symmetries. The cannot only depend physical quantity on is field thus redefinitions and must be and With these anomalous symmetries, it isunder impossible the to anomalous construct a symmetries, quantityanother and that way, does thus if not it the transform isviolating up parameter impossible quark to to is zero and massless, write the it down theory has a is a possible neutron CP symmetry. using eDM. field Put redefinitions to set every CP θ various anomalous symmetries can be used to shift these constants around : experimentally ruled out [ the 2-flavor QCD Lagrangian, the only CP-odd constants are was why did appears instead as a super-selectionwas the rule naive upon estimate that the states we are considering. The second puzzle solutions will be explored in theirsociological own section. statements based While presenting on these my solutions, ownrankings I biases. independent will of I make what a strongly other few encourage peoplethink students consider so to interesting, we make since should their nature explore own doesn’t all care options. what we 5.1 The massless up quark TASI Lectures on the Strong CP Problem and Axions showed that with an instanton background, itappear does in not the vanish action and and is instead does an have integer. an Thus effect. 5. Non-axion solutions to the Strong CP problem PoS(TASI2018)004 . θ 5.8 (5.7) (5.8) (5.6) . As 4 ] . If 19 ) particle is d , 0 Anson Hook m η 18 as spurions of + u m CKM ( θ / comes from the only d has no phase. It turns θ † m u and u flavor symmetries of the Y u m c θ Y d ) = 3 ( µ SU × c . u , † ) 2 d d 3 Y needs to see both CP and P violation. It Y ( † 2 u u θ Y Y d d SU 4 d Y m Y will be important. In particular, these solutions u × 4 u 10 Y θ Q Y − ) 15 3 ∏ 10 ( because 2 < arg Tr g SU u 2 . This can be seen by treating g arg Tr m 0 at a scale larger than the QCD scale and then utilizes the small RG = = CKM = θ θ θ θ β β is a topological parameter that does not appear in perturbation θ , and to note that it is proportional to is set to 0 at some high scale and the EFT to low energies is just the θ 3.31 ], but other than that, the massless up quark solution is resting peacefully 0 at some RG scale and utilize the fact that in the SM, RG running of 16 , = ]. If θ 15 θ 17 will still be very small at low energies, thus it can be effectively ignored. θ By UV, what we mean is that it sets The first “UV" solution to the Strong CP problem that we discuss is parity [ It is important to note that In the following two solutions, RG running of As no one in his or her right mind would try to perform a 7-loop computation, we will instead An alternate way of understanding the massless up quark solution is to calculate the neutron 4 running to result in a small neutron eDM. other CP-violating phase in the SM, will attempt to set theory. Instead, it the phase of the quark masses that5.3 evolves with Parity RG. occurs at 7-loops [ SM, then the CP symmetry. It mustSM. also Thus, respect the A fun exercise is todoes write not have down a a vanishing product phase. of For example, Yukawa the matrices simplest that product respects Tr the symmetries, but where we have included anis extra left factor as of an exercise forTwelve Yukawas the and reader two to gauge specify couplings the meanto that non-trivial see contractions we any RG of need running. to the write indices down in a Eq. 7-loop contribution mentioned above, if parity were a good symmetry of nature, then the neutron eDM would be zero. use symmetries to show that allstandard 6-loop lore diagrams is and that below anything cannotloops generate not there any forbidden will RG probably by be running. symmetries a The non-vanishing will correction. be The non-zero, RG running so of we say that at 7- eDM explicitly using Eq. in its grave. 5.2 RG running of TASI Lectures on the Strong CP Problem and Axions out that a non-vanishing phase requires then the neutron eDM willquark satisfy is the an current axion bounds. solution It to is the amusing Strong to CP note problem that where the the massless already up discovered the axion. However, current latticeout results this show solution. that the There massmassless have up of been quark the a [ up few quark attempts is to non-zero, build ruling models that are similar in spirit to the PoS(TASI2018)004 ]. θ 27 (5.9) , (5.13) (5.12) (5.11) (5.10) 26 3 [ ], although / Anson Hook 1 23 , ± , 22 is observed to be † R , L θ 21 , ↔ L 20 L .. c . , . h † † R . d q Y H , with hyper charge + c i c , † R d = q d ↔ a Y H R L η R d v , the right-handed leptons/neutrinos into ia H and d Λ Q which obtain complex expectation values, that need to be addressed. Additionally, R d y † A L q a Q Λ θ H + η = arg det L j c d , so that the UV completion is not far away from + Q d Y i † R d u R y Y Q v 16 HQ + ∼ ↔ i j , R u to make the electroweak sector parity invariant. The L † Y u H Λ R . Many models that UV complete the Yukawa couplings Y Q arg det R + H θ c Q = + u L θ qq R Λ 0, but their individual elements can be complex. Thus we see angle much due to the flavor structure of the SM, other tree- u H v and µ , L u Λ = ) appear after the right-handed Higgs obtains its vev. Under = θ R R y Q ) ) = Y θ u 2 Y 2 y = ( ( L u ⊃ Y SU SU L ↔ L ) 2 ( SU are small. : ]. They posit that CP is a good symmetry in the UV, so that both the CKM and θ P 25 , 24 Parity can be broken softly by giving the right-handed Higgs a larger bare mass than the left- There are many models that move beyond the minimal parity solution [ The simplest example uses new vector-like quarks, Solutions to the Strong CP problem that utilize CP symmetry are typically called Nelson-Barr and introduced a gauged angles vanish. However, because the CKM is observed to be large, while . Due to these issues and other sociological factors, parity solutions are not as popular as the R R handed Higgs. Allbreaking of and the the Strong nice CP properties problemproblem of is is on solved. the par Thus parity with the the solution simplest axion parity are in solution its preserved to simplicity. with the Strong this CP type of θ vanishes. small, these models are carefullycorrections constructed to so that CP breaking gives a large CKM angleAdditionally, while there are more than one complex scalars breaking CP. The tree-level Lagrangian under consideration is obtaining a large top Yukawa requires that term is P and CP odd and is forbidden by parity, while the Yukawas are of the form Hermitian matrices obey arg det that the CKM matrix can have a non-trivial phase, but the neutron eDM, which is proportional to they quickly run into newbased issues solutions that is need that to tree-level bethe phases, addressed. CKM such phase as The doesn’t the beauty effect CKM andlevel the phase, annoyance phases are can of allowed cause parity- at 1-loop tree effects on level. While involve a bi-fundamental Higgs field.allowed, and Once there this are field again is 1-loop added, contributions new to CP-violating couplings are models [ TASI Lectures on the Strong CP Problem and Axions To see this in field theory language, we note that under parity where we have collected theL right-handed quarks into The standard Yukawa matrices ( v axion solution. 5.4 CP parity, the Yukawa matrices obey PoS(TASI2018)004 ) c a f (6.2) (6.1) HQq (5.14) and Anson Hook c qq η , we can replace θ ] just to name a few. Even 31 , ), and a single new coupling ( 30 a , , . By fiddling with the size of various 29 α . η , ˜ G − A 28 G ] is typically considered to be the simplest θ . 2 , 35 π µ 1 → , ! d 32 θ η 34 m A ,  17 θ 0 µ 33 can be made small. , , +

a θ a f 32 a f = α  M + ⊃ a L → a term and the axion coupling to demonstrate a simple trick that shows θ 3 down quark mass matrix. It is simple to check that at tree level arg det appears at tree level. For these reasons, CP-based solutions have fallen out × θ is the 3 Yv = d 4 mass matrix for the quarks is then m QCD Axion : Solves the Strong CP problem ALP : Does not solve the Strong CP problem Axion : Figure it out yourself × 0, while the CKM phase is non-zero and large if Aside from this simplest of CP-based solutions, there are also a plethora of fun things you can Axions and their variants are by far the most popular solution to the Strong CP problem. As After the massless up quark, the axion [ • • • = must be forbidden or which dictates how the axion can coupleof to the particles. axion For example, can every non-derivative be interaction obtained by observing that wherever we have a coupling solution to the Strong CP problem,a though run the for minimal parity-based its solutionpopularity. money. The gives EFT the consists axion The of EFT EFT a single of new particle, the the axion axion ( is extremely simple and is the main reason for its do when building models with CP symmetry, e.g. Refs [ more so than parity-based solutions, CP-basedscales solutions are are needed very for fragile, the as CKM many angle coincidences to of be large. Additionally, couplings such as such, I’ll dedicate a wholeterminology surrounding section axions to can be describing slightly axions annoying and and confusing variations : on the axion theme. The If the reader encounters the wordit “axion", solves he/she the will Strong have CP to problem. determine from the context6.1 whether The QCD axion of fashion. 6. Axions M TASI Lectures on the Strong CP Problem and Axions The 4 where Yukawa couplings, loop-level corrections to : We have written both the how the axion couples. Asmetry is apparent from this interaction, the axion obeys an anomalous sym- PoS(TASI2018)004 in θ (6.5) (6.6) (6.3) (6.4) ]. 37 Anson Hook ]. It is of some 38 [ 0, so they are not 20 = − 10 ∂θ − . , you can quickly convince θ  18 − 2 θ 10 + . , there is no symmetry associated a ∼ ˜ f a W 2 6.1 θ W  2 2 i− π 1 a . f sin 0 32 / . 2 a ) W = Q a h f d d µ θ m m σ + u † + 18 ˜ + B m a Q u a f B 4 a 2 m Q µ f ( ∝ π ∂ 1 n − 32 d 1 B a f s coupling. 2 π . We can calculate the neutron eDM using the same trick to find , these couplings must be there initially, or they are not generated f ⊃ . Due to the anomalous symmetry structure of the axion and the θ ]: a θ f 2 π L 41 6.1 m θ , − − 40 = = , i V 39 a h . The other couplings generated by RG evolution are derivative interactions with 5 . Derivative couplings are more complicated because a f ] for a more detailed symmetry based analysis of the RG running of axion couplings. / 36 a + . Using our trick from before, we find that the axion potential is θ See Ref. [ Because all of the CP breaking in QCD is dictated by the spurion The whole point of introducing the axion was to hopefully solve the Strong CP problem, The axion quality problem relates to the fact that while there is a very well-defined anomalous UV completions of the QCD axion will occasionally generate other couplings, such as 3.23 5 0. This effect has been estimated to be of the size Thus, the axion vev is Thus, once the axion relaxes toto the zero. minimum As of claimed, its the potential, QCD it axion dynamically solves sets the the Strong neutron CPyourself eDM problem. that all higher-order corrections toaxion the vev. axion However, once potential you coming include fromviolating all QCD phase that three do can not generations break shift of CP the quarks, and= move the the CKM axion matrix potential away has from a having CP- the neutron eDM with it. This lack ofaxion quality proper problem symmetry [ properties generates two issues that together are called the so let’s show thatand the expectation axion value. sets We already the calculated neutron how the eDM vacuum to energy of zero. QCD depended First on we calculate the axion mass symmetry associated with the axion coupling shown in Eq. Eq. that comfort that in the asymptotic future,CP physicists problem will regardless be of able to whether test they the ever find axion solution the to axion6.2 the particle Strong itself. The axion quality problem topological nature of the spurion quarks : TASI Lectures on the Strong CP Problem and Axions it with accompanied by a corresponding Axions with these additionalhas couplings the are coupling still shown in called Eq. the QCD axion asby long RG as evolution the axion still Even if these couplings are zero at tree level, they are still generated by RG evolution [ PoS(TASI2018)004 , ) 1 46 ( (6.7) (6.8) (6.9) U (6.11) (6.10) symme- ]. PQ Anson Hook ) 45 symmetry. In 1 , ( 44 PQ U ) 1 ( U without also including a . The resulting IR theory c q ] and DFSZ [ that are only charged under c 6.1 43 . q , c and . 42 symmetry, of which the axion is h q GeV for dark matter reasons [ and ) + 1 q 12 c ( . U 10 qq can be seen to arise via the anomalous  Φ ∼ . y φ a f a 6.1 f / + + , . ia 2 a a , which break the anomalous f 4 ) a f e needs to be. This term generates a potential for 2 / † n ) − 2 n p r ia Φ  Φ 2 Φ 19 M + qe ε 2 ΦΦ a ( cos f m ∼ = λ 2 a 0 f and V q + 2 = ( coupling due to the anomaly. c symmetry is traditionally called the † ε ˜ GeV and see how severe the axion quality problem is. ) G Φ and the now heavy quarks qq with an approximate ∼ 1 ε r G 12 ( ΦΦ 2 V Φ U 10 m is an anomalous symmetry and thus by the standard rules of an − = q a f = V and Φ , we have an additional pair of vector-like quarks Φ GeV, and there is a preferred region around 8 . We have a complex scalar ]. As such, we will set imposes the anomalous symmetry, gravitational effects willa break separate it mass and term the axion that will is obtain problem. not centered around a zero neutron eDM, which reintroduces the host of other couplings. writing down every couplingproperties, allowed by there is symmetry. no Because way the to form axion the has axion no coupling symmetry in Eq. 6 10 We can now see the axion quality problem in this context. The first issue is that the The first issue is how small the coupling The two most well-known UV completions of the axion are called KSVZ [ To see the axion quality problem in action, we consider the simplest UV completion of the 48 6 2. Quantum gravity is believed to break all symmetries that aren’t gauged. Thus, even if one 1. EFTs are built by specifing the particle content and symmetries of the problem, and then & obtains an expectation value , a f EFT, we cannot forbid the couplings symmetry imposed on try. The second issue isdimensional that operators since of gravity the breaks form this anomalous symmetry, we can expect higher QCD. The Lagrangian is Φ We can integrate out the radial mode which makes the axion appear in the the axion that is the pseudo-Goldstone boson. This addition to is that of the axion.field redefinition The coupling of the axion in Eq. As we will describe in the following section, the experimental constraints on the axion require that 47 axion TASI Lectures on the Strong CP Problem and Axions PoS(TASI2018)004 10 − (6.14) (6.12) (6.13) Anson Hook and 3M copies ]. I will discuss 3 confines there is Q ) 49 by more than 10 , as well as two new M 4 ( , θ symmetry could result 3 , 2 SU , 1 PQ × ) Q PQ 1 ) ) ( 1 1 N 1 -1 -1 . ( ( U , M copies of could prevent all of the danger- 2  is an accidental symmetry of the U SU Q φ Φ L − ) + . B ) a M a  f 1 ( n ( φ  U SU + cos a ]. a f ) 4 is an accidental symmetry, theories where the − 51 N  M ]. The pseudo Goldstone boson whose breaking ( M 3 20 a PQ 3 p f ) 50 cos [ 1 SU M ( 4 ) − n a U ∼ symmetry is n p c M f ) ( 4 M 4 M 3 − PQ ( ) Q SU M ∼ 1 3 p M 2 ( SU V Q M discrete symmetry acting on U and 5, we can suppress higher-dimensional operators enough that the ⊃ 4 14 ) 3 2 ≥ shown below : Q Z N Q Q ( 1 L 14. We would need to prevent Planck-suppressed operators to a ridicu- M × × PQ Q × & ) SU M 3 symmetries can come from discrete symmetries. For example, as shown 1 M n 3 ( U PQ ) 1 ( . Thus the anomalous symmetry must be extremely good. The Planck-suppressed U GeV 19 − comes from 5D gauge symmetries, and examples from string theory [ 10 . After some amount of work, it can be shown that after PQ 4 People have developed many ways to solve the axion quality problem. These broadly fall Accidental . ) 1 Q ε ( confining gauge groups, a light pseudo Goldstone bosondown that that violates is this the accidental axion. The lowest-dimensional operator we can write By taking the gauge group gives the axion is the axion still solves the Strong CPaxion problem. quality factor Of in course this this manner, is see not e.g. the Ref. only [ model that does solves the Again, requiring that this potentialgives does the not constraint that shift the minimum away from where we have indicated that there areof three copies of the quark lously high order to solve thethis Strong axion CP quality problem! problem This is toy and UV why completion significant highlights effort how has6.3 serious been devoted Solving to the it axion over quality the problem years. into the categories of theories where the in the previous subsection, a TASI Lectures on the Strong CP Problem and Axions Requiring this potential to be small enoughin that the axion still solves the Strong CP problem results the first two approaches inmore background this that section is beyond and the skip scopeto of the the these string elegance lectures. of theory None the of exampleselegant these axion as EFT approaches EFT. is comes they The the close require fact dirt a that that typically such lot gets complicated swept models under are the needed rug in to axion justify discussions. an ous higher-dimensional operators. Morefrom plausibly, chiral the gauge accidental theories inrenormalizable much SM. the A same model way that realizes that this has four sets of quarks, operators give a potential of the form U PoS(TASI2018)004 a f / ]. All (6.16) (6.17) (6.15) charged 55 , PQ Anson Hook 54 ) 1 ( U before the Landau a f ] such that / 52 much smaller than 1 1 , can be much larger or ], and can happen if one γγ γγ ∼ ]. Another 5D approach is 58 a a ]. A way of obtaining para- symmetries can result if the g g γγ 53 2 [ a 61 ]. / g PQ , a ) , as RG evolution tends to bring f 57 a 1 60 f , ( , = , U . 56  59 ) Q UV f 4 µ ( a σ f 92 [ . . † 92 5 , and photons, . 1 of a 5D gauge field [ Q Q 1 / 5 f ) that we integrate out to get the axion- a a c Q f A µ dyA − f q ∂ , ∼ Z 21 q 1 UV f = UV  much smaller than f a f ˜ F π Q / α f 2 F a a = are only achievable if the two terms cancel each other. This γγ 4 a a γγ f g a g . Similarly, it is very difficult to make a f . For example, the simplest way to increase the axion-photon coupling a f in the bulk. Accidental anomalous / PQ without fine tuning is not known. ) 1 γγ ( a g U , is too low for the EFT to make any sense. a f Y ) is the model-dependent coupling and the second term comes from QCD. Photon cou- 1 ( UV U f There has been a recent jump in popularity of trying to make the axion coupling to fermions The EFT of the QCD axion is very elegant and simple. Its coupling to the gluons and its The coupling of the QCD axion to fermions, On the other hand, the extra-dimensional approach solves the problem with Planck-scale to within a loop factor of Q f because mixing with the pions couples the axion to photons [ is to dial up the electric charge of the quarks ( chooses Casmirs of the UVmetrically quarks small to give or photons larger than 1 mass are intrinsically tieddependent. together, As while with the any simple couplings andTwo basic to elegant variations theory, the of people the have fermions QCD started axion and to havecouplings push been photons to explored. various fermions are boundaries. The and first model photons. results in The larger-than-expected second breaks the mass6.4.1 to neutron Large coupling relationship. and photon couplings particles in the bulk, thenof the these accidental 5D PQ solutions symmetries can canmotivate be be the of particular dimensionally structure very deconstructed of high into the quality 4D theories. [ solutions, but the6.4 5D Variations versions of the QCD axion smaller than It turns out to be very difficult to make where plings parametrically smaller than cancellation accidentally happens for some GUTs where coupling. This approach canpole boost of the coupling to photons until physics by promoting the axionin shift several symmetry ways. to One a way gauge is symmetry to in make the the UV. axion This the can be done TASI Lectures on the Strong CP Problem and Axions Thus at high energies, thecoupling to axion gluons is can protected result fromto from corrections use a due a 5D to gauged Chern-Simons 5D interactionfermions [ gauge that invariance. canel anomalies The are localized on different branes. If there are no PoS(TASI2018)004 ˜ ]. G G 63 , (6.18) (6.19) (6.21) (6.20) 62 Anson Hook . A discrete symmetry, , 10 ˜ F − , ...) is taken to be large. F c . 0 will remain equal due to the 2 , ˜ ], a move of marketing genius b 2 G π 0 θ e , and thus contribute to the axion 65 a G 32 . , θ 2 a up to 10 a 2 f confines at a high scale, integrating , we get the IR Lagrangian  π 0 s 64 0 a 0 g θ . symmetry is only very softly broken, θ ˜ 32 + G F 2 0 − N f ˜ ≈ F  Z a G 0 0 a f = 2 ]. The simple version is θ θ 2 π G f  e 67 2 + 2 0 s , 32 π a g a f cos 66 f θ ] using UV instantons. As argued in the previous 32 f Na  22 y 79  + + , f a a f ˜ is an integer coming from the group theory of the UV ∏ ˜ G G 78 2 2 π N G , + g G 8 and defining 2 2 2 s b − b 77 π 2 s f π g f e , Nb g θ / 32 32 ∼  b 76 f a ) ,  + + a θ ( ˜ a 75 G f , V + ]: in the background of the instanton shows that the instanton potential / G a 2 2 74 2 s a f 72 , π Na g , xG  θ are pseudo-scalars. Assuming that 73 4 32 71 in the SM. The axion will still solve the Strong CP problem while the mass d b b , b f R θ 70 , and a can be kept small while the number of particles ( 15 , N 69 , , is used to make these angles identical. If the is a new confining gauge group, 2 68 0 Z G The second way to increase the axion mass is to use QCD itself to give the axion a second The other way of playing with the axion is to change the relationship between the coupling of More recent attempts have all centered around using multiple axions and mixing them in such a There are two general approaches to make the axion heavier than expected given its coupling can be taken to be large, giving us a parametrically enhanced coupling of the axion to the photons. small RG running of section, there exist instanton configurations thatpotential. are Evaluating sensitive to scales in the UV as contribution to its mass [ will be larger than expected from the neutron eDM coupling. the axion to the neutron eDM and the mass of the axion. Both of these terms come from the Although these approaches have been aroundhas for recently a renamed long them time as [ clockwork models [ where N Alternatively, 6.4.2 Changing the axion mass - neutron coupling relation usually e.g. the Higgs and mirror Higgs have different masses, then both Since QCD is perturbative in theQCD UV, becomes these strong extra in contributions are the usually UV, then negligible. they However, can if give the axion a mass that is larger than otherwise way that the coupling to gluons is small while the coupling to fermions and photons is large [ to the neutron eDM.couples The [ first is to introduce a new confining gauge group to which the axion TASI Lectures on the Strong CP Problem and Axions coupling so that the relationship isto pretty the tight in axion the mass usual must case.neutron eDM To be change is present. it, zero, another otherwise This contribution the axion extra fails contribution to has solve to the Strong be CP centered problem. around where the In order to still solve the Strong CP problem, we need completion, and out the linear combination PoS(TASI2018)004 (6.22) (6.24) (6.23) ]. Be- 80 symmetry Anson Hook N ]. However, Z 81 0 is still the minimum. The model . k = ˜ G . θ k ˜ is odd, but also makes the axion mass F G . F 2 N  / a for the QCD axion. Thus the mass and the 4 θ N f a γγ 2 f 4 a + 0 if g / k ∼ 1 = π N + ) 23 2 2 ∼ 1 θ ) a + 2 a γγ = N a ( f m a g a N 2 1 (  m a = ]. The axion can obtain its DM abundance from either the k m ∑ symmetry. The axion non-linearly realizes the 48 N L = , Z 47 L , 46 is completely unrelated to the Strong CP problem. Axion-like particles γγ a g One of the appealing qualities of the axion is that it can also be dark matter, solving two Finally, there is only a single model that can make the axion lighter than expected [ Unless one accepts complex models, problems at the same time [ and interacts with N copies of QCD given by having seen so many pretty models crashALPs and is burn, perhaps that the due strongest to motivation forof recent looking new technological for parameter advances, space, we so can we should now look cover and many see orders what of we magnitudes find. 7. Axion/ALP dark matter which does this utilizes a discrete Surprisingly, a numerical checkpotential shows not that only adding retains allexponentially a these lighter minimum different : around contributions to the axion This mass may or mayno not coupling come to gluons, from the the mass confinementother. and of the There coupling another to are gauge photons string group. are theory completely Because independent motivations from there for each is why these new particles may exist [ TASI Lectures on the Strong CP Problem and Axions expected. We can accomplish this byor by adding utilizing additional Higgsing. matter so QCD is not asymptotically free, cause the neutron eDM coupling tochy the problem axion that is is a resolved non-derivative interaction, byonly the QCD has axion dynamics. to has If solve a the the hierar- hierarchy axion problem, is but to also be ensure lighter that than expected, one not This is because the axionconverge potential exponentially in fast for this periodic case analytic can potentials. be related to a6.5 Riemann ALPs sum that is known to (ALPs) address the question :with If the we’re Strong looking for CP aALPs problem, particle are then particles whose which why couplings have have does the nothing Lagrangian the to particle do have to solve the Strong CP problem? coupling to the photon,the the tilde easiest level. of the Manycoupling, axion experiments even though have couplings been to devoted probe, to are search related, for though the only QCD at axion through this PoS(TASI2018)004 and (7.3) (7.6) (7.2) (7.4) (7.5) (7.1) 7.2 Anson Hook . We assume that the axion , t a 2 m / . 2 0 1 a  . 2 a 3 = ) m H describes an over-damped harmonic ) H a ∼ eV) gives m 7.1 . a ) . 3 a t ρ 0 , ( = T m t . R a = 2 H 2 ∼ / ( m /  a 3 p a 1 R 2 a , H m 2 p DM M m cos )( ρ 24 a 0 M respectively, ignoring the crossover regime. Using + 2 a eV A m ˙ a a 2 m / , m ∼ = H 3 0 ) ∼ 2 0 3 a < ) H a 2 p a ( + . = H ¨ M ρ 0 m a ) 2 a a t ( = H ≈ ) = R t describes an under-damped harmonic oscillator. Using the WKB H and 3 ∼ 0 ( ( A a . In the early universe, Eq. ρ R 0 SM 7.1 m . Inflation results in the same initial conditions being seen everywhere. a ρ 7 = ( > = a a H is the scale factor. When estimating the ALP number abundance, we take Eq. to apply when 3 R We note that at late time, the axion behaves like cold non-relativistic matter so that its energy This is a slight lie as misalignment is also present when PQ symmetry is restored. In this case, the initial angle The energy density of the SM and ALP when the ALP starts acting like CDM is We first study the case of ALP dark matter. In a Friedmann-Robertson-Walker (FRW) universe, The misalignment mechanism operates when PQ symmetry is broken during inflation and is 7.3 7 density red-shifts as where As long as the initial value of the ALP is this value, then it will make up all of DM. these approximations, we have is random and can be averagedphenomenology. over all Hubble patches. The inhomogeneity of the initial axion angle leads to different In the late-time universe, Eq. Eq. oscillator. Thus we approximate approximation, we find in the far future that Additionally, its energy is found viaa Fourier production transformation to mechanism be for equal coldmuch to dark its lighter matter mass. than (CDM) Thus, the we that keVbelow have works a scale. even keV when behaves This like the hot is particle dark an is matter. impressive much feat, as usually dark matter with a mass Requiring that the ALP accounts for all of DM ( the equation of motion for ALP dark matter is The topological mechanism operates when PQinflation. symmetry is restored either during inflation or after 7.1 Misalignment : ALP Dark Matter never restored after inflation TASI Lectures on the Strong CP Problem and Axions misalignment mechanism or topological defects.number abundance of In axions this from these section, various we processes. review how to estimate the People typically treat a radiation-dominatedhas universe an where initial field value PoS(TASI2018)004 , 85 , (7.8) (7.7) (7.9) (7.11) (7.10) 84 , becomes 83 θ Anson Hook . Depending 0 every Hubble θ . H , simply because )  , 1 ∼ c θ ( T + O a a f ∼ 0  θ cos isn’t small enough [ 9 8 a T Λ f s / m . d H ) . . m T 0 9 8 u ( ] and we will use dimensional analysis to a T Λ m . = ) s m c 82 a ∼ m ) T > d ( 2 a  T a f ( m θ H 2 a u m 25 + m m a + a f ) = . Because the axion is a periodic field, we work with ∼ c ˙ a , a 2 T f a  ) ( f H 0 0 T present for all instanton calculations. The QCD coupling 3 θ H ( θ cos ) a + ) = T ¨ ( = T m a 2 3 0 ( g 2 3 a a / g 2 / 2 π 8 π 8 − − e Te . The generic assumption people make is that s 0 θ m numbers. In some cases, the initial angle can be estimated in an inflationary d ) m 1 ( u m O ∼ V . 0 θ ]. We are now in a position to estimate the axion DM abundance. The equations of motion for Finally, before we start estimating the axion DM abundance, we roughly specify the initial Another effect of inflation kicking around the axion expectation value is inhomogeneities be- The estimate for axion DM is very similar to that of ALP DM, but withThe a axion critical mass difference. comes from thermal instantons [ 87 can be evaluated using the 1-loop expression for the RG running. The end result for the potential , 3 the misalignment angle of our affinity for context. Inflation will kicktime the in axion the expectation form of value a around random by walk. an This leads amount to inflation populating every value of Again we will completely ignore transition regions. At a critical temperature, the axion are g is 86 conditions for the axion. We take estimate the result. Itbetween turns 100 out MeV that and 1 we GeV. In willup this be range, quark interested QCD solution acts in like the to three-flavorunphysical. QCD. axion the As mass with Strong Thus around the CP massless temperatures theremaining problem, axion dimensions if potential can be any must made quark besuppression up using of mass proportional the the goes to temperature potential T. to the The zero product next factor then of is the the exponential masses. The As a result, the temperature dependence of the axion is on the choice of howvalue to of deal with the measure problem, this can eventween lead various Hubble to patches. predictions As for adensities, result, the different leading Hubble to patches will well-known have isocurvature different constraints dark matter if The axion mass changes withconservation. time. Let us As first estimate a the result, temperature one dependence of needs the to axion be mass. careful with the lack of energy TASI Lectures on the Strong CP Problem and Axions 7.2 Misalignment : Axion DM In the early universe, we do the same thing as before and estimate PoS(TASI2018)004 , 11 97 10 (7.15) (7.13) (7.14) (7.12) (7.16) ∼ a f 100 MeV, Anson Hook ∼ , ]. This is partly QCD 3 QCD Λ . With a changing 94 GeV and Λ < , ∼ ∼ eV 93 T c . c T . , ∼ T c for a T 92 < , DM <  T ρ 91 T ∼ , ] suggest a slightly larger value of 90 p 89 , numbers and solving the differential 3 , QCD . M ) ) t , we find that , 2 a c Λ a ) 1 ) c T a c ( m T f m T T ( ( O 7.16 >  R R cos a 4 QCD T 2 )( ˙ / m c Λ 3 2 T 0 ) ( ]. You’ll sometimes hear people quote a number 26 θ ) ) a c n , 88 T T ∼ a ( , ( f 3 ≈ a R R 0 2  m ( θ ) ) ) . Solving Eq. c and as before, we trade the GeV [ T = c  T ]. It is prohibitively difficult to simulate realistic values of T ( T c QCD ( ( a a 11 a T a H 4 QCD ) 96 Λ m , m Λ T 10  ( a ) 95 ∼ s × c a m 2 a T 2 f a ( f 0 m symmetry that is only broken by QCD. It turns out that it has strings [ a ∼ 2 a ∼ θ ) ) m m a 1 f T = ( ( a a U is a circle, and if we look asymptotically far away from a string in the trans- 4 QCD n ) Λ GeV because extrapolated lattice results [ 2 1 0 ( θ 12 U ∼ 10 a ρ ∼ a f ] because We can now compare the axion and dark matter energy densities at PQ symmetry is a After PQ symmetry breaking, each Hubble patch will settle down to its own value of the Putting everything together, we have After this relatively in-depth review of the misalignment mechanism, I will now give an un- 99 (lattice simulations have not been able to simulate physical values of quark masses yet). , a f equations exactly gives verse directions, space is alsothere a is circle. an integer As windingstrings mentioned have number. before, non-zero winding when The number. mapping vacuum has a 0 circle winding to a number, whereas circle, axion. objects As called various patches come into contact, some might form a non-trivial winding number by GeV. Using the instanton approximation with proper parameters, so all groups need to extrapolate their results. 98 closer to fairly brief review of the topological production of axions [ Once the axion mass stops changing around 100 MeV, energy also dilutes away like CDM. and where conservation of energy in the axions becomes a good approximation : 7.3 Topological production of axions because of my own ignoranceclear on consensus the on the subject results and [ partly because this field has not yet reached a TASI Lectures on the Strong CP Problem and Axions At this point, the axion startsthe to WKB oscillate. approximation, After which the can axion be starts used to as oscillate, long we as again want to apply Both of these are saturated around where we have used that mass, the WKB approximation tells us thatexpansion number of density the not universe energy gives density the is standard conserved volume and dilution. the Thus we find that PoS(TASI2018)004 , 11 131 10 , (7.17) ∼ and the a 0 f 130 θ ]. Anson Hook GeV. This can 122 , 11 10 121 & GeV. In this case, there , and have the potential a axion DM is parametric f ) a 11 f H ( , then some strings are stable 10 O PQ . number of strings is left in each ) to oscillate, so it starts to behave a 1 can be larger. One of the major ) f ( a 1 a ]. The axions emitted as the strings 19 ( f . m U dark matter axions using topological 1 / O completely, then the strings decay. If a  . Axions can also be DM via an entropy 106 f 0 , ]. Current estimates are that if PQ θ ) 1 a GeV 105 114 ( f , , 11 U 10 104 113 27 , ,  ]. Because the axion potential is a cosine, sitting ]. Since inflation populates all values of 2 0 θ 103 112 125 117 , , ]. Another way to get small 10, the axion starts to oscillate after its mass losses its 01 , , . / 0 p 102 111 128 124 116 M = , , , , , 1. There are several directions for generalizing dark matter 2 ∼ h ]. An entropy dump heats up the SM while leaving DM alone, so ]. These objects are even denser than miniclusters. a 101 110 ∼ 127 a 115 123 f , , [ , Ω 0 θ 120 134 π , , 100 109 126 , ∼ , 119 133 θ 95 , 108 , 118 , 107 GeV where 47 ], which is essentially Bose-enhanced decays. If another scalar has enough energy 11 1 because if the initial angle is smaller, then 10 . 129 × 0 is sub-Planckian. For θ , I will cover the typical assumptions for how axion DM behave locally. Additionally, a means that the axion takes a much longer time than 1 f few 8.2 π The energy density of axions is roughly The next variation is to have axion dark matter even when The last variation on dark matter axions explores how they show up in the galaxy today. In The ultimate fate of the string depends on whether there is a preserved subgroup of the PQ One variation is to arrange for axions to be dark matter even when ], which are typically AU in size and much lighter than the mass of a star. Note that this is still ∼ a f the interactions with QCD preserve a discreteand subgroup over-close the of the universe. 7.4 Variations of dark matter axions dump approach [ as cold dark matter muchdefects later. and One fine can tuning also [ make lower- to be dark matter, its Bose-enhanced decays into axions could result inSec. axions being dark matter. to be dark matter [ existence of life as we know it might correlate with small that the relative energy density inenergy out DM of decreases. the Finally, DM the and last dump it approach into I’ll another mention sector is viais to particle not take production enough [ dark mattersimplest in way axions is and to we tune need a way to generate even more cold axions. The people sometimes discuss formation mechanisms132 and signatures of axion miniclusters10 orders [ of magnitude more densewith than is dark Bose/axion matter stars in [ the galaxy. Another possibility people play near symmetry. If the interactions with QCD break the merge and straighten out are mostly emitted with veryGeV, low with energy, about an order ofproduction. magnitude uncertainty, then axions can be dark matter via topological resonance [ TASI Lectures on the Strong CP Problem and Axions chance. Thus on average, one stringstrings per come Hubble into volume is contact created.and with As radiating each the axions. universe other expands, The and more expectation is relax that towards approximately their ground state by closing loops when Hubble volume at all times [ temperature dependence and the results scale differently. Usually, people are interestedaxions. in a single occur if arguments for this is anthropics [ PoS(TASI2018)004 , 7 ]. . a 137 142 m and 10 excluded 8 a f Anson Hook ]. As before, 142 ], search for axions 146 ]. 143 100 MeV [ ], I will give the overall idea . , is radiated from the electron a a 136 GeV and work for masses m , , where the axion is radiated by a / a 10 135 + − n 10 + GeV and n . 8 γγ → ] and the future IAXO [ a 10 ]. g n . 145 + 141 a 28 f n , in which the axion, . a + N GeV 6 + e 100 keV, then the axion/ALP can be produced in stars [ → , where the photon is converted into an axion via the axion-photon . a 100 MeV, the axion/ALP can be probed by rare meson decays [ a N + m . + e a e m → e + γ GeV and 8 − GeV and 7 4 ]. Because the QCD axion mixes with some of the neutral pions, it can appear in of 100 keV is due to the temperature in the center of stars. 10 10 a 140 . m ]. . , a a f f is a nucleus, usually ionized so that it has a large charge. Another production mechanism is 144 139 Helioscopes, such as the current one CAST [ Supernova bounds apply when 10 If Axions in stars are produced via a variety of mechanisms depending on the couplings. One Due to the high near-nuclear densities in the center of supernovae, the dominant production Recently there has been a boom in the number of proposed experiments to look for the axion. If N , by supernovae because eventually the axionsprovide become a too good strongly cooling coupled mechanism. to escape Additionaland and bounds subsequent no come conversion from longer into axion photons production in in the supernovae galactic magnetic fields [ 8.1.4 Axion helioscopes the mass cutoff is dueused to to the constrain temperature axions of by thethe requiring supernova total that being energy the 10s in axions of produced the MeV. do Supernovae neutrinos can not emitted be carry by away the energy supernova. equal to There is only a range of neutron and the neutrons scatterunderstand via supernovae a and that pion the exchange. bounds obtainedof It in magnitude is this important in manner are to uncertainty, subject note e.g.GeV to that [ at we least the one do upper order not bound quite is probably somewhere between 10 produced in the Sun. Roughly speaking, they give coupling. 8.1.3 Supernova method is Bremsstrahlung of axions comes from nuclear Bremsstrahlung 8.1.2 Stellar cooling We understand energy transport and cooling ofthe stars axions well escape, enough or to even impose justbound constraints transport on on energy whether from the core to the outside of the star. The upper and Primakoff radiation some of the rare meson decays.in Even decays if through it its doesn’t coupling mix to with gauge the bosons pions, [ à la ALPs, it can still appear 138 As there already exist reviews that go into excruciating detail [ TASI Lectures on the Strong CP Problem and Axions 8. Experimental probes of the Axion of how each search strategysearches works will and be the categorized by region if of the parameter axion space is dark that8.1 matter. it DM probes. independent searches The various 8.1.1 Rare meson decays PoS(TASI2018)004 GeV (8.1) (8.4) (8.2) (8.3) / 7 − . It is a fun Anson Hook 10 ε . γγ a g . 2 θ E . 2 a − m 1 with a mixing term . E − 2cos2 2 2  . ± E coupling in the Lagrangian causes ) ω 2 2  B E q E 2 a · and 2 − ω + 2 ∼ 1 1 Lm 1 aE 2 E E E  E γγ ( a 2 L g =  2 sin , 2 0 1 2 , 29 E  ω sin 2 ω ∼ θ 2 a B 1 2 m , γγ E a 2 g E ) =  2 ε − 1 , ∼ → E B P 1 γγ ( = a ]. The basic idea is to shoot a laser at a wall through a region with a P g θ ∼ 147 ε tan2 1 2 eV [ 3 − ], which has a region of space near to the horizon called the ergosphere where 10 150 . , a m 149 ]. , 81 148 A fun experiment called light shining through walls gives the constraint Only the photon mode that is transverse to a B-field can mix with the axion. Thus, if a laser Black hole superradiance utilizes the interesting physics that happens around a spinning black Let’s take a Hamiltonian with two states of energy In the presence of a large magnetic field, the . The basic principle used in these experiments is axion-photon conversion. When two states travels through a region withwill a large mess magnetic with field, the a polarizationbased birefringent of effect experiments the arising photon search from in this forfect the mixing [ the transverse-to-B-field direction. change Polarization- in polarization that results8.1.7 from Black hole this superradiance birefringent ef- hole [ Taking a state to be purely stateinto 1 state at 2 some is position, after a distance L, the probability of conversion CAST utilizes a long regionproduced of in space the filled sun that with subsequently a converts large into magnetic a photon field8.1.5 inside and of searches Light CAST. shining for through an walls axion large magnetic field. Behind theby wall, there a is photon another region detector.magnetic with field. a When large The the magnetic axion field laser goes followed going through approaches through the the the wall, wall, wall, the whereas axion the it converts photon back can hits into convert it a and photon8.1.6 into which is an Polarization can blocked. be axion After measured. in the homework problem to work out that the mixing angle and eigenvalues obey whenever TASI Lectures on the Strong CP Problem and Axions eV mix, they can oscillate into each other, aphysics phenomenon see that it is very in familiar neutrino to oscillations, physicists (e.g. whereas experimentalists particle see it in Rabi flopping). Thus the axion photon conversion scales as mixing between the axion and photon. For the axion, PoS(TASI2018)004 2 r / 2 θ to be a π = Anson Hook θ force between 4 and “glue" them r 2 τ f in the Sun, and this / force. π 3 r / 6= while power experiments i θ ) t . After this time, the axion ( 2 π φ force between the two objects. 4 mv h 3 ∼ ]. As a result, an axion field profile f r ]. / τ θ 152 153 , which mediates a new Yukawa f / ψ , then it will mediate a 1 inside of the NS to 0 outside. Whenever two f am / π a ψ µ 30 θψ ψ∂ 5 γ eV, finite-density corrections would cause µ 12 for a coherence time − ψγ ] are designed to look for this new 1 mt 10 151 transitions from . ]. Experiments have measured various nuclear properties related cos a θ 2 ρ 1 eV, the number of axions per Compton wavelength is large. Hence m 2 m . 152 0 ≈ < ) t a ( m a GeV and in the Sun . Phase experiments are sensitive to the value of 16 a θ − being the minimum inside of a neutron star [ 15 π 10 0, then the axion has a coupling = . 6= θ a f θ instead of φ As the derivation is a bit tedious, we simply summarize the results : The axion acts like If the QCD axion is lighter than expected, then finite-density corrections to its potential can In this subsection, I will discuss searches for axions/ALPs that rely on them being dark matter. The first question in axion DM searches is how to treat the axion. Since we will only consider In this section, because we will use creation and annihilation operators, we will call our scalar If If a classical field with 8.1.10 Value of minimum inside of the Sun [ we will treat the axions as a classical scalar field. to the neutrinos and light coming from the Sun. Hence we know that result in I will only describe a subset of the numerous experimental8.2.1 probes available How to to us. treat axion/ALP DM very light axions with field spins. If it has both of these couplings, then there is also8.1.9 a Neutron star mergers at LIGO will surround a neutron star as can be used to place constraints on axions. 8.2 Axion DM searches acquires a random phase andall amplitude time. and is Thus, no what longer onetogether acts does and as conceptually are a combined is in cosine to quadrature with divide as the time these same into separate phase slices pieces for of are size not phase coherent. objects sourcing a field approach eachaxion-mediated other, force there can is be a as force strongthis between as type them. gravity between and neutron Surprisingly, this stars either new could repulsive or be probed attractive. by A LIGO new [ force of TASI Lectures on the Strong CP Problem and Axions energies can be negative. There isis a thrown famous into process the called ergosphere. the Insidepart Penrose the of process ergosphere, the by it object, which imparts which an negative is object energyobject then to thrown leaves into some with the particle more black or energy hole than whilehole. it the Superradiance rest had is of when a the coming object similar in,cloud escapes. effect which surrounding where The extracts energy it. energy is This from transferred process the fromused spins black a to down black constrain the hole the black existence into hole of an so axions. axion that high-spin black8.1.8 holes Fifth can force be experiments force. If the axion has spin couplings, Experiments such as ARIADNE [ PoS(TASI2018)004 1  v  (8.8) (8.6) (8.7) (8.9) (8.5) ∆ n (8.11) (8.10) N Anson Hook is isotropic so f .  φ i φ . . + n n . ) θ i ) n N x φ v n · n ( v ω ip f + − . This means that all of the relative v ω 3 i e ) ) 0 ∑ t φ † n v π ( a m = 2 + i t + v f , v i 3 i x N · . = ( 0 d . In particular, for changes in velocity | ω n ip | 0 n n e a v Z † n n N † n a a † n a a n and a time state. Because we are in a classical state, cos ( n a N n 0 v n ) = n i √ ω 31 v t x i ω e ω ( , n 1 n f 2 ω ∑ α n 2 ∑ √ ω | = = dtH n 2 s i / N 2 1 N 3 H Z / h 3 l N 1 3 l | 1 ) T limit. Consider the state with an average number of particles = ∑ π n . To obtain these expectation values, we model dark matter as i ∑ = ml 2 i N N 2 ( T = ) φ | ) = i i t , t ( θ , T H i H ∑ , i x φ h v . | ( i h has a characteristic scale i H 2 φ in terms of the distribution of particles in phase space. Because particles v h N i = ) h f n t as if they commute. i ( N ) † t φ a ( h φ h and the cosine piece as approximately constant. Because of this approximation, , in terms of the creation and annihilation operators as and v n and i f a i φ ω ) t ( is the assumed classically derived distribution of axions. It can now be easily shown φ ) h v is an unimportant normalization factor : ( is the average number of particles in the f n α N Let us first express We now expand around a random position As mentioned before, people typically assume that the axion is a classical field. A classical At finite volume, the Hamiltonian is with energy , we can treat n 0 come in waves, the energy at a fixed position is not constant in time. Let us define Thus so we will treat The velocity distribution v that where Where This is exactly what we expect, as it’s just the total energyphases divided between by the the sums volume. will be basically random. For simplicity, we assume that field is a superpositionannihilation of operators in a the large large number of particles and is an eigenstate of the creation and We decompose that it is independent of the angles : N TASI Lectures on the Strong CP Problem and Axions are sensitive to the value of particles in a boxsome and energy take distribution given the by volume darkcalculates matter of simulations, the typically box isothermal, and to ask infinity. how one We assume that the axions have PoS(TASI2018)004 . For 2 (8.15) (8.16) (8.17) (8.18) (8.12) (8.14) (8.13) π 4 mv are just a ∼ φ Anson Hook i τ , θ i , b .  ) φ b , a φ , + is small. In practice, it is ) θ . , v b , v ) a ) a v r φ ∆ φ φ is constant, φ + v + v b i + , + t f a t v . t φ v a r 3 + ) ω t ω ω v ( ( ( π a 2 lm 2 ω . ( ( i as long as 2 cos cos . v e / cos v v r v v ∆  a 2 a v b 2 α ∆ α α ∆ v v + v − Re 2 only and the energy in the denominator as the ∆ π e v φ ∆ 2 i v v 4 , 2 π v a a π θ ∑ v i 4 32 vlm π α = 2 v π 4 v f ∆ π 4 vlm v 2 b v ∑ v ∆ f v π a a i ) = f v ∑ vlm v 2 a p a 2 ∆ = is no longer well approximated by a cosine leading to the m ρ q α v 2 r v n f m i ) i ( i ∑ v t 2 π a ( P ∑ = ρ 4 , so that φ r m i ρ f 2 2 p = m / is much easier to calculate and we leave it as an exercise for the ) as functions of 3 p t i ) are taken from the Rayleigh distribution, N v ( = i 2 v π i φ = ) a ω t ) h ml 2 i t ( α ( ) ( t φ v φ ( h a and ∑ h φ v h i f i ≈ to be the resolution of the experiment. We can now simulate the above field ) t v ( . We thus separate the sum into regions where ∆ ) φ h v ∆ 0 mv ( / π We are finally in the position to say how experiments should treat the expectation values There are a slew of astrophysical probes of the axion and ALP DM, which look for DM-photon reader. conversions in galactic magnetic fields, around NS,about and at astrophysics CMB and to DM name in a your few. As other you’ll lectures, hear I more won’t go into further detail here. behavior described earlier in this subsection. 8.2.2 Astrophysical probes where the random numbers The expectation value of times longer than a coherence time, values and check that they are well approximated by a cosine during a coherence time These sums converge independently ofsimplest the to value take of TASI Lectures on the Strong CP Problem and Axions we go from being abletime to 2 trust the time evolution for all time, to being able to trust it only for a mass as it varies more slowly than We will approximate Summing over this many random phases gives us The latter part of thisrandom expression walk, can wherethe be number simplified of by steps is noting the that number the of sums phases over in the volume PoS(TASI2018)004 (8.21) (8.19) (8.20) Anson Hook axis. Using axis as time . Maxwell’s z y nm δ . ) V x a = k will grow with time if + dV t t a a m m ψ m ( n ψ cos direction. The spin will precess R cos t y B so that each individually averages ext ω t ], search for evidence of DM con- B ω · . axis and stay only slightly misaligned. n ¨ . I encourage the reader to take pencil a 154 , a z ψ B ext m ˙ f a B V , so that the cavity is of order the size of the dV is the gyromagnetic ratio. In the absence of a γγ + a g m Z J g / ] searches for this non-zero B field outside of a 0 axis and small wobbles about the − + 1 a 2 n z 33 t E = ∼ 156 ω ∂ ∂ , L B , where ext 2 = B ∇ 155 gB B γγ + a = g × 2 B t B 2 ∇ = ∂ direction and an E-field in the ω ∂ and orthonormality conditions n z B m ) ψ n 2 m Γ n ω ω i = + m 2 n . Basically, each effect oscillates as cos ψ B ω ∇ ] uses NMR to hunt for the axion. In the presence of the axion, the neutron has ω − − 2 ∼ 157 a ω axis. Because the axion oscillates in time, this precession never performs a full ( m axis with a frequency y z . CASPER is designed to look for this effect. B ω Now, if the axion is present, there is also an eDM aligned with the spin that causes precession CASPER [ Haloscopes, the prototypical example being ADMX [ Axion-photon conversions are all about matching the dispersion relationships in space, so the Examining Maxwell’s equations ∼ a direction. Apply a B-field in the we notice that the axion actscurrent like running an through effective it. current along In pure B-fields.In E+M, Consider the there now presence is a a of solenoid B-field an with along axion,zero the however, B-field solenoid this and outside. B-field 0 inside ABRACADABRA B-field acts [ outside. like a small current generating a non- the axion, the slightly misaligned spins precess around the oscillation and only rocks backand and forth paper at to a sketch frequency this precession around the progresses if to zero as time progresses. However, the combined effect cos around the around the geometry, it can be seen that the spin becomes more and more misaligned with the m solenoid bent into a toroid. 8.2.5 CASPER a time-dependent eDM. Consider a spin-polarizedz block of material that is roughly polarized in the equations for the modes are TASI Lectures on the Strong CP Problem and Axions 8.2.3 Haloscopes verting into photons in athe magnetic cavity modes field. If we are looking for conversions in a cavity, we have In terms of the cavity modes, we have We see from the lastaxion term wavelength. that This generally we means want that the cavity mode to overlap as much as possible with the size of the experiment isthe about energy the of size the of axion. the Hence axion, ADMX and looks in for the time,8.2.4 energy the deposited ABRACADABRA energy by of the the axion mode into is the about cavity. axion, and that conversion is only efficient for the fundamental mode. PoS(TASI2018)004 (8.22) . These L Anson Hook . If the exper- ) L ]. This is exactly / x π 161 n , ( 160 . ˙ a f k increases. Since the integral of this wavefunc- n ± 2 34 k = 2 ] are a set of proposed experiments that use dielectrics. ω 159 , the RHS involves an integral over the wavefunctions of the 8.20 does not fall off as dx n ψ R ] and its cousin [ 158 I thank everyone involved in TASI 2018 for making it such a great experience. I thank Yvonne Hopefully these sets of lectures have been a useful introduction to the exciting field of the Again, by looking at Maxwell’s equations, we see that the dispersion relation for circularly MADMAX [ Left and right circularly polarized light traveldo at is different to phase build velocities. an Thus interferometer the to natural thing look to for the different phase velocities [ Kung and Gustavo Marques Tavarestypos for and carefully errors reviewing within it. the manuscript and fixing the many Acknowledgments Strong CP problem and axions.doing research Their in this target area. audience Oftime is course, and many graduate due topics students and to who references my haveup are lack been a interested of left lot out in energy. of due This momentum. toparity is lack On and of a the axion bustling solutions. theory field On side, of the modelfor experimental research builders how side, which are there to has expanding have search recently recently upon for been the picked these many axion basic new proposals dark proposals minimal are matter. being These built. ideas are very exciting and prototypes of many of polarized light is 9. Conclusion iment has a size muchany smaller than spatial the dependence de of Broglie thescopes wavelength axion, use of dielectrics the the to axion, higher modify so modes thesecosines that start wavefunctions and we so integrating sines can that and to ignore the zero. wavefunctionstion are over no Dielectric many longer halo- pure periods doesconversion not due vanish, to these the higher-order factexperiments modes that look for can the photons be integrals from used are axion-photon conversiondielectrics. enhanced for that are by axion-photon emitted the from length series of of stacked the material 8.2.7 Interferometers same way that we lookto for the gravity polarization waves of except the that light. gravity-wave interferometers are insensitive TASI Lectures on the Strong CP Problem and Axions 8.2.6 Dielectric haloscopes The idea is to noticecavity that mode in and the Eq. axion mode. Normally cavity wavefunctions scale as sin PoS(TASI2018)004 . 76 ]. , C77 (1991) Phys. Rev. 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