Comments on a generalisation of Heisenberg's ... - CEUNES

<strong>Comments</strong> on a generalisation of Heisenberg’suncertainty principle and eventual consequences tophenomena in low energy physicsJune 15, 2012AbstractIn these notes we discuss a generalization of the Heisenberg’s uncertainty principlemotivated by some attempts to reconcile quantum mechanics and general relativity.1 IntroductionFollowing the traditional way for new developments in physics, quantum theory has appearedas an explanation for the data from experiments probing Nature at a distance of the order of10 −10 meters. Its huge success made the quantum aspects of the laws of physics a fundamentalpart to understand the World: the phenomena we observe and things that we measure arethough to have their roots in the micro-scale, and even theories describing Nature in biggerlength scales (> 10 −10 m) are expected to have their “quantum versions”.Quantum mechanics (QM) was extrapolated to a quantum theory of fields (QFT) which,with the standard model (SM) produce spectacularly good predictions, being the most reliabletheories we have when gravitational effects are negligible.On the other hand, our understanding of gravity is expressed by Einstein’s general relativity(GR) which establishes the relation (equality) between the gravitational field and thespacetime geometry.The picture of the World we have is sustained by these three pillars: QFT, SM and GR.All the rest is speculation. However, this picture is both (a) fragmented and (b) incomplete,from the theoretical point of view. Since Newton’s observation that the force that keeps ourMoon in orbit is the same that makes an apple falls down, physicists are always looking (and1

adius associated to the mass ∆M, R S = 2G∆M , otherwise we would not be able to observec 2the photon, since it cannot scape the black-hole. This leads us to conclude that ∆x L pwhere L p =√Gc 3is called the Plank’s length.Notice that the HUP, as said before, does not predict any minimum length, i.e., there is nolimitation in ∆x. However the above calculation shows that when gravity enters in the gamesuch a limitation appears naturally: HUP breaks down for higher energies, or equivalentlysmaller distances (of the order of Plank’s length) where quantum gravity effects would berelevant. Of course, the HUP must be recovered when gravity is negligible, so such quantumgravity effects should be treated as corrections.In 1989 Amati, Ciafaloni and Veneziano suggested [2] a generalized uncertainty principle(GUP) in the context of string theory of the form ∆x∆p 2 + β(∆p)2 . The fact that thisGUP implies a minimum length scale can be seen as follows. Without the second (new) termone can make the localization in position as small as one likes by making the uncertainty inthe momentum large. However, when doing that the RHS grows faster (quadratic in ∆p) thanthe LHS (linear in ∆p), and for large enough ∆p, the inequality sign would be contradicted.To avoid this a minimum and finite ∆x appears, implying a maximum value of ∆p, preservingthe inequality.Solving this inequality for ∆p the condition 3 ∆x √ 2β must be imposed in order toguarantee real roots. This is, in fact, what defines the minimum value of ∆x. Then bycomparison with our previous discussion we set β = α L2 p where α is dimensionless.Notice that the “correction” to the HUP is proportional to the square of the Plank’s length,so, it is related to Plank’s area instead. On the other hand, it is the square of the Plank’slength that is proportional to the gravitational constant G, so this correction does look like acorrection dua to gravitation. Nevertheless one could consider a generalization such that thecorrection is in Plank’s length instead. In fact this issue was discussed in [3] and the lessonwe learn from there is roughly that since these terms constitute corrections to a relation thatone still cannot verify empirically, any leading order is acceptable.Modifications of HUP appear in many places, one being slightly different from another, andthe reasons for these changes are also diverse: from string theory, cosmic rays related issues[4],modifications of Lorentz symmetry [5], and so on. As we mentioned in the beginning, QM andGR are very well tested theories and therefore their predictions are very reliable. So, amongall the possible places a generalization of the uncertainty principle can come from, probablya model independent circumstance based on these two theories would be very interesting.Indeed, Maggiore showed [6] that a different version of the Heisenberg’s microscope gedanken3 β 0.3

experiment where gravitational effects were included gives rise to a GUP just like that one wediscussed above (see also [7]): ∆x∆p 2 + β(∆p)2 .The laws of the quantum World produce results that are very distant from what we canintuitively guess, being completely disconnected from our sensitive experience as humans. Inparticular the implications from the uncertainty principle (whose mathematical structure issurprisingly very simple) require reflection, since we would like to understand from where oneshould expect some “testable” modification of it.The stability of the atom was a huge problem before quantum theory: electromagnetismpredicts that the electron moving around the proton (forming the hydrogen atom) shouldradiate, and therefore its motion would be a spiral going towards the nucleus, making ourexistence a mystery. The fact that there is the HUP in quantum systems is what makes theatom stable. First of all, it’s the uncertainty principle together with the fact that the electronis confined that makes the electron moves! This can be seen as follows: take the H-atom tobe a box of size 10 −10 meters. Then, if we want to localize the electron within this precision(∆x ∼ 10 −10 m) we have an uncertainty in its momentum of ∆p ∼ 10−24 Kg·ms. The mass ofthe electron is approximately 10 −30 Kg, so that its velocity is more or less one third of thevelocity of light. Basically when we try to confine (localize) quantum particles they start tomove like crazy. From here we see immediately that it’s just not possible for the electron tobe at r = 0, i.e., to fall into the proton, and be at rest there (p = 0).Using the HUP it’s not difficult to estimate the minimum energy the electron must havein the hydrogen atom (the binding energy).So, here it is a good place to estimate howthe GUP would change that. The calculation is simple. First, by symmetry we expect that〈r〉 = 0 = 〈p〉, which means that (∆r) 2 = 〈r 2 〉, and analogously for the momentum. The totalenergy is given by E = p2− e2(∆p)2, and the average of that is therefore 〈E〉 = − 2m r 2m e2 〈 1 〉. ∆rNow, HUP implies that ∆p , so the mean energy becomes 〈E〉 2 − e2 , and the∆r 2m(∆r) 2 ∆rRHS of this expression is exactly the minimum energy, so that its derivative with respect to∆r must vanish, which give us the value of the radius (for which the energy is minimized)as 2Finally, plugging this result into the expression for the minimum energy we getme 2 .〈E〉 min = − 1 2me 4 2≈ −13.6 eV.We proceed in the same way but now using the GUP 4 ∆x∆p 1 + αL 2 p(∆p) 2 . Solvingthis for the momentum uncertainty we get4 Take = 1.∆p ∆x2αL 2 p(1 −√)1 − 4αL2 p.(∆x) 24

Notice that the root with the plus sign is ignored. This is because we want to recover theHUP in the limit L 2 p → 0, which is only possible for the root with the minus sign.G):Next we consider a Taylor expansion of the RHS of the above expression for small L 2 p (orUp to first order in L 2 p we have 〈E〉 1∆p 1∆x + α G(∆x) + 2G23 α2(∆x) + . . . 5− e2 +2m∆x ∆xαGm(∆x) 4≡ E(∆x) then dE(∆x)d∆x= − 1m(∆x) 3 +e 2(∆x) 2 − 4αGm(∆x) 5 = 0 which give us the polynomial equation me 2 (∆x) 3 − (∆x) 2 − 4αG = 0.Notice that for α = 0 this is the same equation one gets (in the same step of calculations) forthe HUP. So, we expect the solution to be that solution, i.e., ∆x ∝ 1 plus something withme 2α. On the other hand, looking at the cubic equation above it is clear that the solution mustbe a linear combination of the two sets of constants appearing: ∆x = Ame 2 +BαG, otherwisethere is no hope to arrange things on the LHS to give zero. Of course, by simple comparisonwe can guess that A = 1m 2 e 4 . This, and B = 4me 2 are found 5 when we plug this ansatz intothe equation and look separately to each order in α (i.e., α 0 and α 1 ). So, ∆x = 1me 2 +4me 2 αG.Now, put this into 〈E〉 and expand up to first order in α to get〈E〉 − me42 + αGm3 e 8from where we recognize the usual binding energy of −13.6 eV and a correction which is foundto be of the order of 10 −48 eV.This is very small. In [3] it is showed that a contribution of the order of 10 −23 eV can beobtained if one considers a GUP whose leading order is in L p instead of L 2 p. However, this isstill 12 orders above any laboratory capacity.Indeed, it is very important to search for a physical system that could give any measurablecorrection predicted by a GUP. A series of papers by Saurya, Elias and others are exactlyin this direction: using a GUP that, according to them, fits in many of the possibilitiesalready discussed by other authors, they have calculated the modifications in some observablequantities in well known quantum systems.In the rest of these notes we shall review particularly two papers, namely, “Plank scaleeffects on some low energy quantum phenomena” [8] and “Discreteness of Space from theGeneralized Uncertainty Principle”[9].5 This equation has two more solutions, but imaginary ones, which are not relevant here.5

2 The generalized canonical relationsSo far we saw that apparently to reconcile gravity with quantum mechanics one needs in fact tochange what is the cornerstone of the latter: the uncertainty principle. From theorem (A.19)it is possible to relate the uncertainty principle between two observables with the canonicalcommutation relation of the associated operators, which is needed for the calculations. Theproblem we discuss now is that of finding the commutation relations that give rise to a GUP,but before that let us take a look in the origin of the standard relation [ˆx, ˆp] = i.Consider a system 6 whose physical state is described by an observer, say O 1 . A secondobserver O 2 is displaced a distance a from O 1 .One can imagine that the description 7 ofthe physical system given by the two observers will be different in certain ways. In order tocommunicate, the first observer must make the boost x → x + a to reach the second observer,and the second observer makes x → x − a to reach the first. The crucial point here is that thespace being homogeneous implies that the results of the measurements made on the systemmust be the same for the two observers 8 | 〈ψ 1 | ψ ′ 1〉| 2 = | 〈ψ 2 | ψ ′ 2〉| 2 .This last statement together with the fact that both vectors |ψ 1 〉 and |ψ 2 〉 are in the sameHilbert space lead us to use Wigner’s theorem to relate them by a unitary (or anti-unitary)operator. Since the only property we are interested in is the position, let us label the vectorin Hilbert space described by O 1 as |x〉, then, in O 2 we have |x + a〉 = ̂T (a) |x〉. Now, asimple calculation (see appendix (B)) shows that the translation generator ̂T (a) is unitaryand therefore can be written as ̂T (a) = e −iaˆk. Consider now the position operator ˆx, whosetransformation from one reference frame to the ] other is given by ˆx → ˆx a = ̂T (a) −1ˆx ̂T (a). ] Foran infinitesimal translation: ˆx a = ˆx + ia[ˆk, ˆx + O(a 2 ). Since ˆx a = ˆx + a we get[ˆk, ˆx = −i.As showed in appendix (B), consistency with the results from wave mechanics implies that 9ˆk = ˆp , so that the standard canonical relation between position and momentum operators isobtained.What we conclude from this discussion is that HUP is intimately related to the homogeneityproperty of the physical space. Specifically, it is this property that implies that measurements(of the same observable) done by different observers would agree, and as a consequence, themomentum operator is the generator of the translation transformation.Apparently it is not trivial to do such a construction based on symmetries to go beyond6 For simplicity we consider only one dimension.7 The two observers use the same rules to associate the physical state with vectors in the Hilbert space.The Hilbert space will be the same for both, but the vectors will not.8 The notation is obvious, but to be sure no doubts remain: the index stands for which observer is describingthat state.9 Where ˆp = −i∂ x .6

the HUP in the direction we pointed out in the introduction. What we can do concretely isto find a commutation relation between position and momentum operators that can give thedesired GUP.As a warm up exercise consider the commutation relation [ˆx, ˆp] = i (1 + αˆp 2 ) in the RHSof (A.19). What one gets is the uncertainty principle ∆x∆p 2 (1 + α(∆p)2 + α〈p〉 2 ) which(up to this last term that is a constant) is of the kind of GUP we were discussing so far. In [8]the canonical relation proposed between position and momentum contains probably the mostgeneral polynomial expression in the momentum up to second order:()p i p j[x i , p j ] = i δ ij + δ ij α 1 p + α 2p + β 1δ ij p 2 + β 2 p i p j ; i, j = 1, 2, 3 (2.1)where p 2 = ∑ 3i=1 p ip i . The constants α 1 , α 2 , β 1 and β 2 can be fixed using the Jacobi identity[[x i , x j ] , p k ] + [[x j , p k ] , x i ] + [[p k , x i ] , x j ] = 0as we now show. First, notice that space is still commutative (sorry Paulo!) and therefore thefirst term above vanishes ([x i , x j ] = 0) leaving us with[[x j , p k ] , x i ] − [[x i , p k ] , x j ] = 0.Then, using (2.1) we calculate the first term, and for the second we just exchange i and j.Lets do it bit by bit:⎧⎫⎨[[x j , p k ] , x i ] = i⎩ δ [ ] [ ]⎬jkα 1 [p, x i ] +α} {{ } 2 pj p k p −1 , x i +β 1 δ jk p 2 , x i +β 2 [p j p k , x i ]} {{ } } {{ } } {{ } ⎭ .1234Each of these commutators is calculated under approximations to first order in momentum.Also, the first two of them require some clever tricks. For the first one we use the following:from one hand we have 10 [p 2 , x i ] = 2[p j , x i ]p j = −2i (p i + α 1 p i p + α 2 p i p)+higher order terms,simply by using 11 p 2 = p i p i . On the other hand, one can write p 2 = p · p, so that [p 2 , x i ] =2 [p, x i ] p. Then, we compare these two results, which must be the same thing, to get [p, x i ] p =−ip i (1 + (α 1 + α 2 )p). Multiplying by p −1 results in[p, x i ] = −ip i p −1 − i(α 1 + α 2 )p i . (2.2)10 Notice that [x i , p j ] is a function of momentum according to (2.1), so it will commute with any otherfunction of momentum.11 In most places we use Einstein’s sum convention.7

Commutator labelled as 2 is[pj p k p −1 , x i]= pj p k[p −1 , x i]+ pj [p k , x i ] p −1 + [p j , x i ] p k p −1 .The second and third terms on the RHS are given by (2.1). The first one must be found,again, using a trick, which consists in writing 1l = p · p −1 , then [1l, x i ] = 0 becomes p [p −1 , x i ] +[p, x i ] p −1 = 0, from where we find using (2.2)[ ]p −1 , x i = ipi p −3 + i(α 1 + α 2 )p i p −2 . (2.3)At this point the commutator number 2 is[pj p k p −1 , x i]=(ip j p k p i p −3 + (α 1 + α 2 )p −2) −−()p i p ki δ ik + δ ik α 1 p + α 2p + β 1δ ik p 2 + β 2 p i p k p j p −1 −−()p i p ji δ ij + δ ij α 1 p + α 2p + β 1δ ij p 2 + β 2 p i p j p k p −1 . (2.4)Now the commutator number 3 is direct: [p 2 , x i ] = −2p j [x i , p j ]. Using (2.1) and organizingits terms we get[p 2 , x i]= −2i(pi + (α 2 + α 2 )p i p + (β 1 + β 2 )p i p 2) . (2.5)Finally the last of the four commutators is also direct:[p j p k , x i ] = −ip j(δ ik + δ ik α 1 p + α 2p i p kp + β 1δ ik p 2 + β 2 p i p k)− ip k(δ ij + δ ij α 1 p + α 2p i p jp + β 1δ ij p 2 + β 2 p i p j). (2.6)The next step is to plug results (2.2), (2.4), (2.5) and (2.6) back where they belong and rewritethe Jacobi identity:(α1 p −1 + α 1 (α 1 + α 2 ) − α 2 p −1 − α 2 α 1 − α 2 β 1 p + 2β 1 + 2β 1 (α 1 + α 2 )p+ 2β 1 (β 1 + β 2 )p 2 − β 2 − β 2 α 1 p − β 1 β 2 p 2) (δ jk p i − δ ik p j ) = 0.Ignoring the higher order terms we find(α 1 − α 2 ) p −1 + α1 2 + 2β 1 − β 2 = 0.Now, taking α 1 = α 2 = −a and (for dimensional reasons) β 1 = a 2 this equation fixes β 2 = 3a 28

and we are left with equation (1) of [8]:with a = a 0L p( ([x i , p j ] = i δ ij − a δ ij p + p )ip j+ a ( ) ) 2 δ ij p 2 + 3p i p j , (2.7)pwhere a 0 is dimensionless.As discussed previously in “ordinary” quantum mechanics the momentum operator is foundto be the generator of translations in space, and in the space of wave functions it acts as aderivative, while the position operator acts multiplicatively. In fact this can be easily verifiedfrom the HUP: [x, −i∂ x ] ψ = iψ. Such a simple representation is quite unlikely to be foundfor the momentum operator satisfying the GUP (2.7). On the other hand we have to keepin mind that when gravitational effects become negligible we need to recover HUP from thisGUP, and in the same way we need to recover that usual representation of the momentumoperator as a derivative.consider the momentum operator to beSo, thinking in the context of perturbations what we 12 do is toˆp j = ˆp 0j + Ap 0ˆp 0j + Bp 2 0ˆp 0jwhere ˆp 0j = −i∂ j , is the usual momentum operator and p 2 0 = ∑ 3i=1 p 0ip 0i . The positionoperator is the same as the usual (notice that the RHS of the GUP contains no position), so[ˆx i , ˆp 0j ] = iδ ij . Naturally we expect the coefficients A and B to be related to a, so that whenG → 0 we get ˆp i → ˆp 0i . In fact, from dimensional analysis we need A ∼ a and B ∼ a 2 .In order to fix the coefficients we plug the above formula into [x i , p j ], and then compareto (2.7). So,For the first term on the RHS we have[x i , p j ] = iδ ij + A [x i , p 0 p 0j ] + B [ x i , p 2 0p 0j].[x i , p 0 p 0j ] = iδ ij p 0 + [x i , p 0 ] p 0jand we need this last commutator. In order to find it we use a similar trick as used to find(2.2). First of all we have [x i , p 2 0] = 2 [x i , p 0k ] p 0k = 2iδ ik p 0k = 2ip 0i (p 0 p −10 ). But the LHS} {{ }1lcan be written as 2 [x i , p 0 ] p 0 , which implies that [x i , p 0 ] = ip 0i p −10 . Then[x i , p 0 p 0j ] = iδ ij p 0 + ip 0i p −10 p 0j .Now, this last term can be improved. Up to first order in a we have p j ≈ p 0j (1 + Ap 0 )+O(a 2 ).12 The reader has to remember that most of the “we” in these notes refer to the work of Saurya and others.9

Then p 2 = p j p j = p 2 0 + 2Ap 2 0 + O(a 2 ), so that to order zero in a we get p ≈ p 0 and also, to firstorder in a p j ≈ p 0j (1 + ap). This last relation can be inverted and once again approximated upto first order in a giving p 0j ≈ p j (1 − ap). With this result we can calculate p 0 , which appearsin the first term on the RHS of the commutator we are working in: p 0 = √ p 0j p 0j ≈ p(1 − ap).Also, the product we are improving becomes p 0i p −10 p 0j ≈ p i p j p −1 (1 − ap). Finally[x i , p 0 p 0j ] = iδ ij p(1 − ap) + ip i p j p −1 (1 − ap). (2.8)The second commutator is[xi , p 2 0p 0j]= 2p(1 − ap)ipi p j p −1 (1 − ap) + p 2 (1 − 2ap + a 2 p 2 )iδ ij . (2.9)Considering the canonical relation up to second order in a and in the momentum we have[x i , p j ] ≈ i ( δ ij + Apδ ij + Ap i p j p −1 + ( 2B − A 2) p i p j + (B − A 2 )p 2 δ ij)and when compared to (2.7) gives A = −a and B = 2a 2 , leading us to equation (5) of [8]:p i = ( 1 − ap 0 + 2a 2 p 2 0)p0i . (2.10)3 The generalized Schrodinger equationThe quantum state evolves accordingly to the Schrodinger equation Ĥψ = i∂ tψ, where thehamiltonian operator is Ĥ = ˆp22m + ̂V . Following this perturbative philosophy we should expectour hamiltonian to be something like Ĥ = Ĥ0 + aĤ1 + a 2 Ĥ 2 . Indeed, this is obtained whenwe square the momentum given by (2.10): p 2 = p 2 0 − 2a|p 0 |p 2 0 + 5a 2 p 4 0.Here we point out that the operator with third power in momentum is not well defined 13 .Once the Schrodinger equation is a scalar equation, all momentum operators appearing thereare in fact related to the norm |p 0 |. Naturally, the occurrence of a square-root in odd powersof this quantity make them untreatable by the usual means.Ignoring the fact that we don’t know how to act with this third power in momentumoperator, and for convenience writing it as ∇ 3 , the generalized Schrodinger equation (GSE)readsi∂ t ψ =(− 22m ∇2 + V − ia3m ∇3 + 5a2 42m ∇4 )ψ. (3.11)We might think of this equation as a correction to the Schrodinger equation when gravitational13 Only in one dimension we can in fact define it since there a scalar and a vector are the same thing.10

effects are taken into account.3.1 Discretization of spaceIn 2009 Ahmed, Saurya and Elias [9] considerer non-perturbative solutions of the one-dimensionalgeneralized Schrodinger equation (3.11) up to first order in a for a free particle of mass Mconfined to a box of size L. As usual they showed that for the stationary case energy is quantized,but also a new phenomenon takes place: the size of the box the particle lives cannotbe anyone; it must be a multiple of the Plank’s length. In what follows we reproduce thesesresults.Consider the GSEi∂ t ψ = − 22M ∂2 xψ + V ψ − ia3M ∂3 xψ (3.12)where V is that of a square-wellV ={0 if 0 ≤ x ≤ L;∞otherwise.Notice 14 that although here one can define the cubic momentum operator, the parity transformationis no longer a symmetry of the equation. This is quite unpleasant since space is notisotropic in that case.For definite energy configurations we plug the ansatz ψ(t, x) = ϕ(x)e − i Et in (3.12) obtaining− d2 ϕdx 2 − i2ad3 ϕdx 3 = 2ME 2 ϕ.Define k 2 ≡ 2ME and consider ϕ(x) ∼ e mx . The above differential equation becomes a 2polynomial equation in m:2iam 3 + m 2 + k 2 = 0.For a = 0 this problem is the well known quantum-well problem, and the solution is m = ±ik.Now, for a ≠ 0 we expect some generalization of that.equation and expanding up to first order in a we getCalculating the solutions of thism = ik(1 + ak) m = −ik(1 − ak) m = i2a − i2ak2 ,ϕ(x) = Ae ik′x + Be ik′′x + Ce ( i2a −i2k2 a)x ,14 This remark appeared in one of the many fruitful discussions with Rodrigo Pereira.11

As we said, for a → 0 we must recover the result of ordinary Schrodinger equation. Since thelast term of the expression above cannot be expanded, we require C → 0 when a → 0.The boundary condition ϕ(0) = 0 implies a constraint between the constants A+B+C = 0,so that we take B = −(A + C). Then, the above solution is written asϕ(x) = 2iA sin (kx) e ik2 ax − Ce −ikx e ik2 ax + Ce i2a x e −i2k2 ax .Lets take A = −i ˜B , so that for a = 0 we recover the usual solution of a quantum particle in2a box ϕ(x) = ˜B sin ( πn x). Then we expand what we can in the above equation, obtainingLϕ(x) ≈ ˜B( )sin (kx) + Ce i2a x − Ce −ikx + ik 2 ax ˜B sin (kx) − Ce −ikx − 2Ce i2a x .Now, since |C| → 0 for a approaching zero, we expect that |C| depends at least linearly in a.This means that the two last terms on the RHS of the above expression can be neglected forthey are at least of second order in a.Then, we consider the other boundary condition ϕ(L) = 0 using C = |C|e −iθ C. This givesthe relation()˜B sin (kL) = |C| e −i(kL+θC) − e i( L2a −θ C)− ik 2 aL ˜B sin kL.For a → 0 we have k → πn (n ∈ Z) and |C| → 0. Then, the LHS above vanishes as well asLthe RHS. For the case a ≠ 0 we take kL = nπ + ε with ε ∈ IR going to zero for a → 0. Thenthe factor k 2 a falls to zero faster than a, and we can ignore that last term in the expressionabove. We remain with()˜B sin (kL) = |C| e −i(kL+θC) − e i( L2a −θ C),whose imaginary part is( ) Lsin (kL + θ C ) + sin2a − θ C = 0that is satisfied for L2a − θ C = 2πq − (nπ + θ C ) and L2a = (2q + 1)π + nπ + 2θ C, with q ∈ Z.This shows that the length of the box obeys a discretisation rule 15 and therefore cannotbe anyone! This is in agreement with our first discussion in the introduction of these notes:we expect that quantum gravity implies discretisation of geometry.15 Notice that the parameter θ C is not determined but is fixed.12

3.2 Effects of GUP in the quantization of the magnetic flux in asuperconductorHere, as in [8], we shall not go deep in the theory of superconductivity (it is pointless). Theonly feature we will need is the quantization of magnetic flux that happens in superconductingmaterials. Then, we shall evaluate how the generalization of Heisenberg’s uncertainty principle(2.7) would change that.The superconductivity phenomena is characterized, among other things, by the fact thatthe sample behaves as if it had no measurable electrical resistivity. Due to interactions withthe atoms of the lattice, a tiny attraction appears between two electrons moving in the sample,and they can form a bound state 16 which will be described as a Bose particle of charge 2e.These pairs, called Cooper’s pairs, exist for very low temperatures, and any increase of it candestroy them. The wave function of the Cooper pair is defined by the Schrodinger equationof charged particles moving in a magnetic field 17 (see appendix (C)):i∂ t ψ = 1 (−i∇ − qA) (−i∇ − qA) ψ. (3.13)2mThe local U(1) symmetry of this theory implies, due to Noether’s theorem, a conserved chargegiven by (the details are presented in appendix (C)) ∫ dx |ψ| 2 , whose associated current iswith Dψ = −i∇ψ − qψ.J = 12m (ψ∗ Dψ + ψ(Dψ) ∗ )Now if we write the wave function as ψ = √ ρe iθ a direct calculation givesJ = m(∇θ − q A )ρ.Let us consider a sample which is a lump. This (the topology of the sample) is very important.Notice that the density ρ must be uniform. If this was not the case a net charge would producean electric field pushing the electrons apart and destroying the Cooper pairs.Then, taking the divergence of the above equation, and using the gauge where ∇ · A = 0,one finds that in the steady state ∇ 2 θ = 0, which is satisfied for a constant phase θ. This alsogives J ∼ −A, with proportionality constant q m .16 Thanks to the remaining N − 2 electrons as showed by Cooper.17 We wrote q = 2e instead.13

Locally the magnetic field can be written as B = ∇ × A. From the equationswith J = − q A we get London equationm∇ × B − µ 0 ε 0 ∂ t E = µ 0 J ∇ × E + ∂ t B = 0∇ 2 A =( qρε 0 mc 2 )A.Just to get a hint about what is happening here one might consider this equation in onedimension. The solution is of the form e − x λ , with 18 λ −2 =qρε 0 mc 2and we see that the fielddecreases when we go inside the material. Thus, the magnetic field does not penetrate thesample very deeply. This is called the Meissner effect: if one takes a magnetic material at hightemperature, then apply an external magnetic field to it, and finally lowers the temperaturebelow the critical temperature 19 , the field is expelled.Now, suppose that the sample is, instead of a lump, a ring with thickness larger than λ.What we are going to do is the following. First, at high temperatures, a magnetic field isapplied. Then, we cool down the sample below the critical temperature, which makes thematerial a superconductor, and as we saw, the magnetic field is expelled from it. The finalstep is to remove the source of magnetic field. However, from ∇ × E + ∂ t B = 0, and fromthe fact that there is no electric field inside the superconductor we conclude that the fluxdΦof magnetic field cannot change after we remove its source: = 0. In order to solve thisdtapparent contradiction Nature bends the lines of field trapping them to the ring. After that,a current in the surface of the ring is generated to keep the flux constant.Remember that for a sample which is a lump, we took θ to be a constant. However, this isnot the case for our new sample, which is a ring. Since inside the superconductor J = 0 onegets ∇θ = qA. The flux of magnetic field inside the superconductor is Φ(Σ, B) = ∮ A · dl,then ∆θ = qΦ. Due to the external gauge field one cannot expect to get ∆θ = 0. Imposing,however, that the wave function is single valued (and not θ), i.e., that ∆θ = 2πn we find outthatnh = qΦn ∈ Zconcluding that the trapped flux is quantized, as multiples of h/q.It is clear now that since the GUP implies a modification of the Schrodinger equation theelectrical current predicted from it will be different and ultimately we might have a contributionto the quantum of flux. This is what we need to calculate now.18 This characteristic length is called London penetration depth.19 The critical temperature is that for which below it some materials become superconductors. Measurementsgo from miliKelvins to 20 Kelvins.14

As we mentioned before the correction to first order in a to the Schrodinger equation givesa ill-defined operator (at least using the proposed representation for the momentum (2.10)).Then, we shall ignore such a term and consider the leading order to be in a 2 .However when doing that we cannot simply ignore the problematic term and use what isleft in (3.11). It is necessary to reconstruct a generalised Schrodinger equation from a GUPwhose leading order for the corrections is a 2 . This can easily be done following the samescheme used before and the result is[x i , p j ] = i ( δ ij + δ ij a 2 p 2 + 2a 2 p i p j). (3.14)Then, the momentum is given byp j = p 0j + a 2 |p 0 | 2 p 0jand the generalized Schrodinger equationi∂ t ψ = − 22m ∇2 ψ + V ψ + a 2 4m ∇4 ψ. (3.15)From this equation we get the expression for the electrical current, which now has two partsJ 0 and J 1 , like in (D.26) and (D.27), but with the minimal coupling. In particular thecontribution to the quantum flux comes fromJ gauged1 = 1 m( ( ) [ ( ) 2 ∗ [( ) ∗ ( ) 2 i ∇ − qA ψ ψ]i ∇ − qA +ψ]i ∇ − qA i ∇ − qA ψ] ∗ )+ ψ ∗ ( i ∇ − qA ) ( i ∇ − qA ) 2ψ + ψ[ (i ∇ − qA ) ( i ∇ − qA ) 2ψ.In order to estimate that we notice that after the minimal coupling is considered in (D.27) weget lots of i ∇ψ −qAψ terms. Since for the superconductor the density ρ = |ψ|2 is uniform thefirst term involving the gradient of the wave function is simply ∇θ ψ. The flux of magneticfield is given by ∫ A · dl that can be approximated to |A|2πR, with R the radius of thesample (a ring). So, we see that the flux depends on the typical size of the sample. Then, theorder of magnitude of this first term involving the change of the phase can be found taking∇θ ≈ h . For the second term we R consider20 |A| ≈ 1 |B|R, and take q = −2e, remaining2with q|A| ≈ −e|B|R. Considering the following values for the size of the sample and thestrength of the magnetic field as R ∼ 10 −3 m and |B| ∼ 10 −3 T, we see that the first term20 Since B = ∇ × A we get A = 1 2 B × r. 15

is 6 orders of magnitude smaller than the second one. Then, equation (3.16) is simplified toJ gauged1 ≈ 32e 3 |A| 2 Aρ. Then, the total current is( )J gauged 2e 32e3= ∇θ + A + a2m m m |A|2 A ρ.Since the circulation of it vanishes inside the sample we get−2e ∆θ = ( 1 + 16e 2 a 2 |A| 2) Φwhere Φ = ∮ A · dl. Imposing that the wave function is single valued fixes ∆θ = 2πn, withn ∈ Z. Then we can invert this relation by doing an approximation to order a 2 obtainingΦ = nΦ 0(1 − 16e 2 a 2 |A| 2)where Φ 0 = − h 2e .Now in order to estimate the parameter a 0 in the definition of a we notice that the deviationfrom the usual quantum flux given by δΦ 0 = 16e 2 a 2 |B| 2 L 2 Φ 0 cannot be detected ordistinguished from what we really measure. So, supposing that indeed what we measure is aflux quantum plus a correction due to effects in the Plank scale we can blame our technologicallimitation and say that the quantity 16e 2 a 2 |B| 2 L 2 is less than our experimental error δΦ 0Φ 0,which we can take to be 1 part in 10 ε . Using the values for the sample size and magnetic fieldused before we geta 0 ≤ 10 26− ε 2 ,and, of course, with better precision (much better than that we have nowadays) we could havehope in detecting the two contributions to the flux of magnetic field.3.3 The Landé g-factor in the non-relativistic approximation usinga GUPConsider Dirac’s equation for the 4-dimensional spinor ψi∂ t ψ = ( cα · p + mc 2 β ) ψwhereα i =(0 σ iσ i 0)β i =(1l 00 −1l)are 4 × 4 matrices, σ i , i = 1, 2, 3 being the Pauli matrices.16

Using equation (2.10), up to first order in a we get α · p = α · p 0 − ap 0 α · p 0 , and fromp 0 ≈ (1 − ap)p, we approximate p 0 ≈ p, and replace it by α · p getting the generalised Diracequation (GDE)i∂ t ψ = ( cα · p − c a (α · p 0 ) 2 + mc 2 β ) ψ. (3.16)Turning the gauge field on, and calling the canonical momentum Π 0 we consider the nonrelativisticapproximation writingψ(t, x) =(ϕ(t, x)χ(t, x)so that (3.16) can be written as the set of two equations)e − i mc2 tmc 2 ϕ + i∂ t ϕ = cσ · Π 0 χ + mc 2 ϕ + eΦϕ − ca (σ · Π 0 ) 2 ϕ (3.17)mc 2 χ + i∂ t χ = cσ · Π 0 ϕ − mc 2 χ + eΦχ − ca (σ · Π 0 ) 2 χ. (3.18)In (3.18) we suppose a low-varying field ∂ t χ ≈ 0 and a weak electric field so that 21 Φχ ≈ 0.Then it becomesThis can be approximated tocσ · Π 0 ≈ ( 2mc 2 + ca(σ · Π 0 ) 2) χ.χ ≈ 12mc σ · Π 0ϕ(1 − a (σ · Π )0) 22mcPlugging this result back into equation (3.17) we geti∂ t ϕ ≈( [(σ · Π0 ) 22m − a c(σ · Π 0 ) 2 + (σ · Π ])0) 4ϕ,4m 2 ci.e., a corrected Schrodinger equation due to Plank scale effects.Working on the kinetic terms using σ i σ j = 1lδ ij + iε ijk σ k , and defining the operator S = σ 2we geti∂ t ϕ =[( ) 12m − ca Π 2 −a4m 2 c Π4 − 2 e2mc− ae2 24m 2 c 3 |B|2 − aie (∇ (σ · B) ·2m 2 c 2(1 − 2acm − a mc Π2 )S · B(Π − e )c A + i∇ 2 (σ · B)) ] .We recognize from the third term the g−factor.Using the same reasoning used for the21 In fact, the electric field is irrelevant for our discussion.17

superconductivity:g = g 0 (1 − ag 1 )with g 1 = 2mc + Π2mc and g 0 = 2. Then ignoring Π2mca 0

On the other hand, the average value of A is Ā = 1 ∑ ni=1 N i∑ ni=1 N ia i but since p i = N i ∑ ni=1 N iwe get 〈A〉 = Ā.With that we can rewrite our statistical quantities using the Born’s interpretation. Theprobability of finding the particle inside a region dx is P dx (x) = ψ ∗ (t, x)ψ(t, x)dx, so we get〈x〉 = ∫ +∞xP −∞ dx(x), and 〈x 2 〉 = ∫ +∞−∞ x2 P dx (x), and therefore(∆x) 2 = 〈(x − 〈x〉) 2 〉 = 〈x 2 〉 − 〈x〉 2so the uncertainty in the measurement of the position of the particle is defined by (the rootmean square)∆x = √ 〈x 2 〉 − 〈x〉 2When we measure the observable A, its value is determined and therefore ∆A = 0 atleast at the instant of the measurement. Define the hermitian operator 22 â ≡  − 〈A〉, andthen (∆A) 2 = 〈ψ| â |ψ a 〉 = 〈ψ a | ψ a 〉, where |ψ a 〉 = â |ψ〉. Now since ∆A = 0 and (∆A) 2 =〈ψ a | ψ a 〉 0 for all x we conclude that |ψ a 〉 = â |ψ〉 = 0, so  |ψ〉 = 〈A〉 |ψ〉, i.e., any statewith vanishing uncertainty in the observable A is an eigenstate of the corresponding hermitianoperator Â.If we want to measure A and B simultaneously, then the physical state has to be eigenstateof both  and ̂B, which is possible only if[Â, ̂B]= 0. Otherwise we have the followingtheorem:whose proof is now presented.⎛ ] [Â,(∆A) 2 (∆B) 2 ⎝ 〈 ⎞ ̂B 〉⎠2i2(A.19)To the observable A it is associated the hermitian operator  with uncertainty ∆A =√〈A2 〉 − 〈A〉 2 in the state |ψ〉. Define the hermitian operator â =  − 〈A〉 so that (∆A)2 =〈ψ| (A − 〈A〉) 2 |ψ〉 = 〈a 2 〉. Define the state |ψ a 〉 = âψ. Now, let ̂B be another hermitianoperator and use similar definitions as those for Â. Then(∆A) 2 (∆B) 2 = 〈ψ| ââ |ψ〉 〈ψ|̂b̂b |ψ〉 = 〈ψ a | ψ a 〉 〈ψ b | ψ b 〉 = |ψ a | 2 |ψ b | 2 .Using the Schwartz inequality we get(∆A) 2 (∆B) 2 | 〈ψ a | ψ b 〉| 2 = |〈âˆb〉| 2 .This last term is the norm of a complex number, which can be separated into its real and22â † = â.19

( ) 2 (2.imaginary part, giving: |〈âˆb〉| 2 = Re〈âˆb〉 + Im〈âˆb〉)Since the real part is always(2.bigger or equal to zero the following inequality holds: |〈âˆb〉| 2 Im〈âˆb〉)Finally, writingthis imaginary part as Im〈âˆb〉 = 〈âˆb〉−〈âˆb〉 ∗theorem.2i= 〈âˆb〉−〈ˆbâ〉2i=〈[Â, ˆB]〉The HUP is found from the “quantization ansatz” [ˆx, ˆp] = i directly.2iwe conclude the proof of theA.1 The time-energy uncertaintyAs a consequence of the theorem (A.19) we show that ∆E∆t 2 .Consider the observable A, and suppose that after a time ∆t, 〈A〉 changes by ∆A. Fromthe formulas showed in the previous section for the average of an observable a direct calculationmaking use of the Scrhodinger equation givesThen, from our hypothesisand thereforeddt 〈A〉 = d ∫dtdx ψ ∗ Âψ =∣ ∣∣∣ ddt 〈A〉 ∣ ∣∣∣∆t = ∆A1∣ ∣∣∣〈[Â, Ĥ]〉∣∆t = ∆A.Now, using (A.19) we get 1 ∆A∆E 12〈[Â, Ĥ]〉〈 1] [Â, 〉 Ĥ .i= 1 ∆Aso, ∆E∆t .2 ∆t 2BThe translation operatorConsider a state |x〉 and the transformation |x + a〉 = ̂T (a) |x〉.translation operator ̂T (a) is unitary. TakeNow, lets show that the〈x ′ | ̂T (a) |x〉 = 〈x ′ | x + a〉 = δ ((x + a) − x ′ ) = 〈x ′ − a| x〉where in the last step we used the fact that the delta function is symmetric. Then, 〈x ′ | ̂T (a) =〈x ′ − a| and we conclude that ̂T (a) = ̂T † (−a), so the adjoint of the translation operatortranslates backwards. Moreover it is direct to see that ̂T † (a)T (a) |x〉 = |x〉, i.e. this operatoris unitary and therefore can be written as ̂T (a) = e −iaˆk where ˆk † = ˆk.Consider the eigenvalue problem ˆk |k〉 = k |k〉, where |k〉 are also the eigenstates of20

̂T (a). Take the projection of the translation operator onto the position space:〈x| ̂T (a) |k〉 =e −iaˆkψ k (x). Doing the same thing but now using ̂T (a) = ̂T † (−a) we get 〈x| ̂T (a) |k〉 =〈x − a| k〉 = ψ k (x − a), and comparing both results: ψ(x − a) = e −iaˆkψ(x). Expanding theRHS for small a up to first order and after that taking the limit a → 0 we get −i∂ x ψ = ˆkψ.Using the de Broglie relation p = k and taking ˆp = −i∂ x we arrive to the conclusion thatˆk = ˆp .CThe minimal coupling in electromagnetic phenomenaRoughly speaking the gauge principle states that when we promote a global symmetry of thetheory to become a local one, it might be necessary the introduction of new fields - the gaugefields.We start by writing the wave function in an “internal space basis” as ψ = ψ a u a . The gaugegroup is the local U(1), whose elements are g(x) = e − i qα(x) , and it acts on the (“internalstructure” of the) wave function as ψ → gψ. Now, when we go from one point of the spaceto another as x → x + dx the wave function changes according to ψ → ψ + dψ, where dψ =ψ(x + dx) − ψ(x). On the other hand, dψ = dψ a u a + u a du a , where this change in the internalbasis elements is given by du a = g(dx)u − u. Let us make a quick (and trivial) considerationto get g(dx). First, notice that g(0) = 1l implies α(0) = 0. Then g(dx) = e − i α(dx) , andα(dx) = α(0 + dx) ≈ α(0) + ∂ µ α(x)dx µ , so α(dx) = dα, and finally g(dx) = e − i dα .considering a first order approximation in g(dx) the change in the internal basis becomes(1 −i qdα) u a = u a + du a , which gives du a = − i dαu a. Together with that we have dψ a =∂ µ ψ a dx µ and at the end: dψ = ( ∂ µ ψ a − iq ∂ µα ψ a)ua dx µ .Introducing the definition A µ (x) ≡ ∂ µ α(x) (and calling A µ (x) the gauge field or connection),so that the gauge group element reads g = e − iq ∫ Aµdx µ , and also defining what is calledthe covariant derivative as D µ ψ = ( ∂ µ ψ a − iq A µ(x) ψ a)ua , we get to the conclusion that thewave function changes from one point to the other according toψ(x + dx) − ψ(x) = D µ ψ dx µ .Another (more practical) conclusion is that the covariant derivative (and not the ordinaryderivative) transforms in the same way as ψ does under the gauge transformation if the gaugefield transforms as A µ → A µ − i q ∂ µgg −1 = A µ − ∂ µ θ, for g = e − i qθ(x) , i.e., Dψ → gDψ whenψ → gψ.Next we analyse the so called geodesic configurations, defined by D µ ψ = 0. The wavefunctions defined by this equation are such that it is the presence of the gauge field alone thatSo,21

makes changes on them. For convenience define e ≡ q , and this equation reads∂ µ ψ − ieA µ ψ = 0.(C.20)The solution of this equation can be found iteratively and is given byψ Γ (x, x R ) = e ie ∫ Γ Aµdxµ ψ Rwhere Γ stands for a path in spacetime joining the points x R and x, and ψ R is the wavefunction calculated at x R . For completeness we discuss in (??) how the solution is found.This derivation is not needed for the rest of the discussion and can be ignored by the reader.Notice that if we turn off the gauge field then ψ(x, x R ) = ψ R , i.e., the wave function atthe final point is the same of that in the initial (reference) point. That is what one expectssince according to equation (C.20), the only thing changing the wave function is the gaugefield, which is not there in this case. So what the gauge principle is telling us is that the wavefunction of a free particle differs from that of a particle in an external electromagnetic fieldby a phase.We put a Γ under the wave function symbol in order to stress out that when the gauge fieldin on, the wave function will change by a multiplicative phase, but this phase will depend onthe path of the particle. Suppose we solve equation (C.20) for a path Γ 1 going from x R to x f ,with initial condition ψ R . Then, imagine we take the same initial condition, and the same endpoints, but now a different path, Γ 2 . The wave function in the first case is 23 ψ Γ1 = e ie ∫ Γ 1 A ψ Rand in the second case ψ Γ2 = e ie ∫ Γ 2 A ψ R . We have seen that two different solutions of equation(C.20) can be related by a gauge transformation ψ → e −ieθ(x) ψ. Then, suppose this is thecase and ψ Γ1 and ψ Γ2 differ by this multiplicative phase: ψ Γ1 = e −ieθ(x) ψ Γ2 . We then find outthat θ = − ∫ Γ 1A, where Γ ≡ Γ∪Γ −11 ∪ Γ −12 is the loop whose base point is x R . Now, Stokes2theorem implies that the integral of the gauge field over this loop equals to the integral of itscurvature F = dA over the area Σ whose boundary is Γ, i.e., the flux of the field strengththrough the area Σ: θ = − ∫ Γ 1 ∪Γ −12A = − ∫ F ≡ −Φ(Σ, F ).ΣWe have seen that when a local U(1) transformation is considered the gauge potentialenters in the game and not only the wave function changes by a (local) phase factor but alsothe ordinary derivative must be replaced by a covariant derivative. What we need now isto take all these new features into the Schrodinger equation in order to describe a chargedparticle in an electromagnetic field.The lagrangian of a non-relativistic particle of electric charge e and mass m in an electro-23 Some times we might use the differential form notation: ω = 1 n! ω µ 1···µ ndx µ1 · · · dx µn .22

magnetic field described by the gauge potentials A and φ isL = 1 2 mv2 + ev · A − eφ.(C.21)The first step to get the hamiltonian function, and therefore the Schrodinger equation, is toget the canonical momentum:p = ∂L∂v= mv + eA ≡ Π + eA(C.22)where Π = mv defines the kinetic momentum.The hamiltonian is defined by∣H = (p · v − L)∣v=v(p)and equation (C.22) can be inverted to give the velocity in terms of the momentum 24 leavingus (after a straightforward calculation) withH = 12m |(p − eA)|2 + eφ.Then, the minimal coupling discussed before has the following “practical implications”p → p − eAH → H − eφ,i.e., the ordinary (kinetic) momentum (which, in fact, should have been represented here byΠ in the LHS of the first substitution) is replaced by the momentum which has the ordinaryone plus a contribution from the external (gauge) field; and the energy gets also a contributionfrom the scalar potential.We finally arrive to the conclusion that the local gauge U(1) transformations transformthe free particle wave function by a local phase and the free particle hamiltonian asH = p22m−→ H − eφ =|p − eA|22m . (C.23)The first step to build up the quantum version of this theory 25 is to promote the variables of24 In the case the magnetic field is too strong so that the kinetic term can be neglected, the equation definingthe canonical momentum in terms of velocity cannot be inverted and we have a constrained system. In thatcase the canonical quantization procedure must be done using the Dirac-Bergmann algorithm, and we end upwith a non-commutative theory (is that right Paulo?).25 Following the canonical quantization scheme23

the phase space, position and momentum, into operators whose representation isx → ˆx = xp → ˆp = −i∇and (also/therefore) [ˆx, ˆp] = i. The gauge field acts multiplicatively and since it is in anabelian Lie algebra[µ Âν], = 0. Now, the quantum hamiltonian is obtained by “putting ahat” in the quantities at the RHS of (C.23):Ĥψ = 1 (− 2 ∇ 2 ψ − e 2mi)2e∇A ψ − A∇ψ + e 2 |A| 2 ψ + eiˆφψ. (C.24)Notice that the second term on the RHS can be made to vanish in the Lorentz gauge. TheSchrodinger equation is finally obtained taking Ĥ = i∂ t.It is possible (and nice) to formulate this as a field theory for ψ and ψ ∗ . The free theoryis described by the lagrangianL = i 2 (ψ∗ ∂ t ψ − ψ∂ t ψ ∗ ) − 22m ∇ψ∇ψ∗ .One can check that taking the variation of the action S = ∫ dtL with respect to ψ ∗ , andcomparing the result with the equation obtained from (C.24) with e = 0. For pedagogicalreasons lets rewrite the above lagrangian in an apparently inconvenient way, with the is and together with the derivatives:L = 1 2 (ψ∗ (i∂ t ψ) + ψ(−i∂ t ψ ∗ )) − 12m ((−i∇ψ)(i∇ψ)) .Although strange, with this expression it is now easier to consider the minimal coupling−i∇ → −i∇ − eAi∇ → i∇ − eAandi∂ t → i∂ t − eφ− i∂ t → −i∂ t − eφ.The “gauged” lagrangian becomesL = 1 2 ψ∗ (i∂ t ψ − eφψ) + 1 2 ψ (−i∂ tψ ∗ − eφψ ∗ )− 12m (−i∇ψ − eAψ) (i∇ψ∗ − eAψ)(C.25)and one can verify that variation of the corresponding action with respect to ψ ∗ gives exactly(C.24).24

As discussed before the minimal coupling implies that the ordinary derivatives are replacedby covariant ones, which is exactly what is happening here. One can introduceD 0 = i∂ t − eφD = −i∇ − eAand write the gauged lagrangian aswhich is a much nicer and compact form.L = 1 2 ψ∗ D 0 ψ + 1 2 ψ(D 0ψ) ∗ − 12m Dψ(Dψ)∗C.1 Solution of the holonomy equation (C.20)Let Γ be a path in spacetime parametrized by σ ↦→ x µ (σ), such that x µ (0) = x R is theinitial point - called also reference point. We consider the spacetime manifold to be piece-wisesimply-connected, so that the path is defined inside a simply-connected region. We want tosolve (C.20) for configurations on this path. Then, the first step is to project (take the innerproduct) this equation on the velocity vector dxµ in each point of Γ. Calling C dσµ = −ieA µ forconvenience we getdψdσ + C dx µµdσ ψ = 0.Integrating this equation in σ from σ = 0 to some other point we get∫ σψ Γ [σ, 0] − ψ Γ [0, 0] = − C µ (σ ′ ) dxµ0 dσ ψ Γ[σ ′ , 0]dσ ′ .′The second term on the LHS will be denoted as ψ R since it is the wave function calculated atthe reference point. Now, on the RHS we have still a wave function inside the integral. Usingagain the same equation above we can write this wave function asand plug it back gettingψ Γ [σ, 0] = ψ R −∫ σ0ψ Γ [σ ′ , 0] = ψ R −∫ σ ′0C µ (σ ′′ ) dxµdσ ′′ ψ Γ[σ ′′ , 0]dσ ′′∫ σ∫ σC µ (σ ′ ) dxµdσ ′ dσ′ ψ R + C µ (σ ′ ) dxµ′0 dσ ′ dσ′ C ν (σ ′′ ) dxν0 dσ ψ Γ[σ ′′ , 0]dσ ′′ .′′25

We notice that we still have the wave function on the RHS, so we repeat the procedure oncemoreψ Γ [σ, 0] = ψ R −−∫ σ0∫ σand in fact recurrently.0∫ σ∫ σC µ (σ ′ ) dxµdσ ′ dσ′ ψ R + C µ (σ ′ ) dxµ′0 dσ ′ dσ′ C ν (σ ′′ ) dxν0 dσ ψ Γ[σ ′′ , 0]dσ ′′ ψ ′′ RC µ (σ ′ ) dxµdσ ′ dσ′ ∫ σ ′0C ν (σ ′′ ) dxνdσ ′′ dσ′′ ∫ σ ′′0C ρ (σ ′′′ ) dxρdσ ′′′ ψ Γ[σ ′′′ , 0]dσ ′′′ .A great simplification here is that the connection is abelian andtherefore we do not need to worry about path ordering. The RHS of this expression can bewritten asψ Γ [σ, 0] =∞∑∫(−1) nn=0σσ 1 ···0dσ n · · · dσ 1dx µndσ n· · · dxµ 1dσ 1C µ1 (σ 1 ) · · · C µn (σ n )ψ R .We notice that there is an ambiguity when calculating the integrals.second term we haveFor instance, in the(∫ σ2dσ ′ C(σ )) ′ =0∫ σ ∫ σ100dσ 1 dσ 2 (C(σ 1 )C(σ 2 ))and the integration variables can be exchanged without affecting the result. Then we concludethat∫ ∫ans similarly for the other terms we getSo, the solution is written as1 dσ 2 (C(σ 1 )C(σ 2 )) =σσ 1 σ 2 0dσ 1 (∫ σ2dσ ′ C(σ )) ′ ,2 0ψ Γ [σ, 0] =(∫1 σndσ ′ C(σ )) ′ .n! 0∞∑(∫(−1) n σndσ ′ C(σ )) ′ ψ R ,n!n=0which is formally the series of the exponential function:ψ Γ [σ, 0] = e ie ∫ Γ Aµdxµ ψ R .026

DThe current for the generalized Schrodinger equationLet us consider a “practical” way of calculating the conserved current associated to the U(1)symmetry which involves a trick with the equation for ψ and its complex conjugate.Take the free (with no external gauge field) GSEi∂ t ψ = − 22m ∇2 ψ + V ψ + a 2 4m ∇4 ψand multiply it by ψ ∗ . Then, consider its complex conjugate multiplied by ψ. Subtractingthe second result from the first we get (writing it in a very pedagogical way for the minimalcoupling to be considered later)i∂ t |ψ| 2 = 1 (ψ ∗ (−i∇) 2 ψ − ψ (i∇) 2 ψ ∗) + a2 (ψ ∗ (−i∇) 4 ψ − ψ (i∇) 4 ψ ∗) .2mmBoth terms above can be written as the divergence of a gradient, as we now show. For thefirst one it is direct to check that it is − 2 ∇ · (ψ ∗ ∇ψ − ψ∇ψ ∗ ) . For the second one we firstnotice that it appears in∇ · ∇ (∇ · (ψ ∗ ∇ψ − ψ∇ψ ∗ )) = 2 ( ∇ψ ∗ · ∇ ( ∇ 2 ψ ) − ∇ψ · ∇ ( ∇ 2 ψ ∗)) + ψ ∗ ∇ 4 ψ − ψ∇ 4 ψ ∗ .Now we notice that the first term in the RHS can be written as 2∇ · (∇ψ ∗ ∇ 2 ψ − ∇ψ∇ 2 ψ ∗ ),and also we have ∇∇·(ψ ∗ ∇ψ − ψ∇ψ ∗ ) = ∇ψ ∗ ∇ 2 ψ−∇ψ∇ 2 ψ ∗ +ψ ∗ ∇∇ 2 ψ−ψ∇∇ 2 ψ ∗ . Finally,we end up withψ ∗ ∇ 4 ψ − ψ∇ 4 ψ ∗ = ∇ · (∇ψ∇ 2 ψ ∗ − ∇ψ ∗ ∇ 2 ψ + ψ ∗ ∇∇ 2 ψ − ψ∇∇ 2 ψ ∗) .Now we can write a continuity equation ∂ t ρ + ∇ · J = 0 where ρ = |ψ| 2 andwithJ 1 = 1 m( [ ( ) ] 2 ∗i ∇ψ i ∇ ψ +J = J 0 + a 2 J 1 ,J 0 = 1 ( ) (ψ ∗2m i ∇ψ( ) ∗ ) + ψi ∇ψ(D.26)[ i ∇ψ ] ∗ ( i ∇ ) 2ψ + ψ ∗ i ∇ ( i ∇ ) 2ψ + ψ[i ∇ ( i ∇ ) 2ψ] ∗ )(D.27)If one turns the gauge field on the currents above change according to the minimal coupling,as discussed before..27

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