Units, dimensions and regularization

Transcription

1 Chapter Units, dimensions and regularization Instructor: Sudipta Mukherji, Institute of Physics, Bhubaneswar. Natural Units Like other courses in physics, we start with units and dimensions by first listing down values and units of certain quantities that will be appearing frequently in the lectures. ev = J, h = h 2π = Js = evs = Mevs, c = m/s, hc = MeVs m/s = MeVm = MeVfm = 97 MeVfm. (. Various formulae will simplify considerably in natural units where c and h are set equal to. In natural units lengths, mass and time all are expressed in GeV. Let us see how this works. For h = c =, since [c] = [L][T], we have [L] = [T]. The bracket denotes the dimensions of the quantities. Now using Einestein s relation E 2 = p 2 c 2 + m 2 c 4, we find [E] = [M] = [p]. (.2 Nothing is original in this note except perhaps the organization.

2 2 CHAPTER. UNITS, DIMENSIONS AND REGULARIZATION Further since [ h] = [M][L] 2 [T], setting h = c = gives [M] = [L] = [T]. (.3 Therefore [M] can be chosen as the single independent dimension in natural units. Let s now see how we can use this. In (., we have hc = 97 MeVfm. Therefore in natural units fm = 97 MeV, so m = MeV = GeV. (.4 We also see from (. that h = MeVs. Since now h =, we have s = MeV = GeV. (.5 Finally, note that kg of mass has equivalent energy equals to J which is /( ev = 9/ ev. In gev this becomes GeV. Putting all these together we get meter = GeV, second = GeV, kg = GeV. (.6 Problem : A 2.00 g insect moves at 0.00 cm/s. Find its momentum and kinetic energy in TeV. Problem 2: The mass of an electron is 0.5 MeV. Convert it into kilogram. Problem 3: Cross sections are often expressed in millibarns, where mb = 0 3 b = 0 27 cm 2. Using gev units, show that GeV = 0.389mb..2 Dimension and dimensional regularization Dimensional analysis will be important in this course as well and let us start by highlighting its usefulness. We will start with a proof of Pythagoras theorem. Consider the figure (.. The area of the big right angled triangle can be written in terms of the length c and the two dimensionless angles α, β. Since area has the unit of length 2, we write A = c 2 f(α, β, (.7 where f is a function of the two angles. This is because c is the only dimensionful quantity involved. Now the big triangle can be broken in to

3 .2. DIMENSION AND DIMENSIONAL REGULARIZATION 3 β b c α β a α Figure.: Big right angled triangle has arms a, b, c two right angled triangles as shown in the figure. The other two angles of each triangle are shown in the figure. Since sum of the areas of the smaller two triangles must be equal to A, we can write a 2 f(α, β + b 2 f(α, β = c 2 f(α, β. (.8 Dropping f(α, β from both the sides we get the Pythagoras theorem! As we will see later, in field theory, dimensional analysis is essential in making certain calculation tractable. In the next part of this section, we illustrate this with two simple problems one from electrodynamics and the other from quantum mechanics. Consider the figure (.2, an infinite straight string along the y axis with linear charge density. Suppose we want to find out the electric field at a point located at a perpendicular distance x from the line. We can do it in several ways. Simplest is to use Gauss law. Cosinder a cylinder of radius x and length l around the line charge. Using Gauss law (from symmetry consideration, electric field should be radially outward from the line charge 2πlxE = l ǫ 0, so E = 2πǫ 0 x. (.9 We see that ǫ 0 E has the dimension charge per length 2. Now suppose we want to construct the potential V. Since ǫ 0 V must have the dimension of

4 4 CHAPTER. UNITS, DIMENSIONS AND REGULARIZATION x y r Figure.2: Infinite line with constant charge density running along y. charge per length, it has to go as and no other length dependent parameter can appear (on dimensional ground. This means ǫ 0 V is independent of x, hence by differentiating, the electric field must vanish! We can see this happening also by exploiting a scaling symmetry. Consider an element dy on the line situated at a distance r from the point. The potential due to this charge is dv = dq r = dy r. (.0 Total potential is then obtained by integrating over the line V (x = + dy x2 + y 2. (. Clearly the integral is divergent for large y going as logy. Note that the above integral has the property V (ax = + dy a2 x 2 + y 2 = + dy/a x2 + (y/a 2

5 .2. DIMENSION AND DIMENSIONAL REGULARIZATION 5 = + dz x2 + z 2 = V (x. (.2 This is funny! Since V (x = V (x 2, E = V = 0 or the work done W = q(v (x V (x 2 = 0. Sounds unphysical. We will come back to it but let us first regularize the integral in (.. Consider, instead, to be non-zero over a length 2L. giving 2 V (x = V (x = log +L L dy x2 + y 2, (.3 ( L 2 + x 2 + L L2 + x 2 L. (.4 Note that the result depends on L which we will call the regulator. Divergence appearing in the limit L. However, the electric field following from (.4 V (x E(x = = L (.5 x 2πǫ 0 x L2 + x 2 is finite for L. In fact E(x = Further, since we see that in the limit L, we have ( L log 2 + x 2 + L L2 + x 2 L V (x 2 V (x 2πǫ 0 x. (.6 ( 4L 2 log x 2, (.7 ( x log. (.8 2πǫ 0 x 2 So work done in taking one unit charge from x to x 2 is indeed free of L. Note that by introducing the length L, we have broken the scaling symmetry along y. For later convenience, let us write the above equation as V (x 2 V (x + ( x log. (.9 2πǫ 0 x 2 ( 2 dy = log y+ x 2 +y 2 x 2 +y 2 x.

6 6 CHAPTER. UNITS, DIMENSIONS AND REGULARIZATION Now, here are a few comments. Remember that initially we could not construct the potential having x dependence from a simple argument of dimensional analysis. However, introducing another length scale to this problem, we could immediately write down ( x ǫ 0 V f. (.20 L Clearly, f is a dimensionless function. However since we do not want to have L dependent electric field (since it is a measurable quantity, which we get from V by differentiating, f must have the property ( x f = h(x + g(l. (.2 L The only function that has this property is the logarithm. That s why we see log function appearing in (.4 - one that reduces to log(2l/x for large L. Further, by introducing the regulator L, we have also un-anbiguously extracted V (x V (x 2. Let us now digress for a moment and do the integral (. in arbitrary dimension where we replace dy = dv dv n = dω n y n dy. It 0 is known that 3 (, Ω n = dω n = 2π n 2 Γ(n/2 (.24 where Γ is the standard gamma function. The integral determining V (x is now V (x = dω n y n dy µ n x 2 + y. ( Note that we have introduced an arbitrary quantity µ with dimension of length to ensure that V (x has correct dimension. Further introducing ǫ = ( n/2, we can express the potential as V (x = ( µ 2ǫ Γ(ǫ x 2ǫ π ǫ. (.26 3 For example V 3 = dv 3 = y 2 sin θdydθdφ = dω 3 y 2 dy, (.22 with Ω 3 = 2π3/2 Γ(3/2 = 2π3/2 π /2 = 4π. (.23 /2

7 .2. DIMENSION AND DIMENSIONAL REGULARIZATION 7 Note that no where we have broken the scale invariance along y. We further note that any scaling along x can be compensated by a re-scaling of µ. Let us now examine (.26 for ǫ 0. Using the fact that, for ǫ 0 µ 2ǫ + 2ǫ log µ +..., Γ(ǫ = γ +... (.27 ǫ we arrive at V (x = ( ǫ + log (e γ + log (µ2 π x 2, (.28 where γ =.577. Physical quantities are independent under the shift of the potential by a constant. Therefore, V (x = ( log ( e γ + log (µ2 π x 2. (.29 Note that we could have equally chosen any other subtraction scheme for example we could have V (x = ( log ( µ2 x 2 (.30 as well. Now, having done that, we can get the potential in two dimensions simply by taking ǫ 0. Note that, in addition to µ, V (x also depends on the subtraction scheme. However, physical quantities are independent of such ambiguities. The fact that E and W do not depend on µ can be written as de d logµ = dw = 0. (.3 d logµ Let us refer these as the β-function equations. Here, these equations look too trivial, but in general they contain lot of information about the underlying theory. Here is one example. Scattering cross-section (σ in particle physics can generally be written as σ = FΣ, (.32 where, F, containing non-perturbative information, is hard to compute within perturbative field theory. On the other hand, Σ is a perturbatively computable object. Now, scattering matrix should be independent of unphysical scale µ. This leads to dσ d log µ = 0 = dσ d log µ F + df Σ. (.33 d log µ

8 8 CHAPTER. UNITS, DIMENSIONS AND REGULARIZATION This gives df Fdlog µ = dσ = c. (.34 Σdlog µ Since the origin of F and Σ are completely different, it is natural to assume that both terms are independently constant. If we take this fairly reasonable assumption, we get F µ c. (.35 Without knowing F, we have a prediction about its behaviour! We conclude our discussion on regularization with one more instructive example from quantum mechanics. The problem is to solve Schrödinger equation and analyze bound state for attractive δ function potential in two dimensions. The equation that we want to study is [ ] h2 2m 2 δ 2 (r ψ(r = E(rψ(r. (.36 We want to find bound states energy eigen-values of this equation. We, however, face an immediate problem. Since in two dimensions, 2 and δ 2 (r both have dimensions [L] 2 h, 2 2m and also have same dimensions h too. Energy has to be constructed out of 2 2m and. This is impossible by simple dimension arguments since both these two quantities have same dimensions. In order to see a different manifestation of this problem, let us define B 0 = 2mE h 2, 0 = 2m h 2, with B 0 > 0. (.37 We can write Schrödinger equation as [ ] δ 2 (r ψ(r = B 0 ψ(r. (.38 In Fourier space ψ(r = (2π 2 ψ(0 = (2π 2 d 2 k e ik.r ψ(k, d 2 k ψ(k, (.39 so (.38 reduces to (2π 2 d 2 k ( k 2 e ik.r ψ(k + 0δ 2 (r (2π 2 d 2 k e ik.r ψ(k = (2π 2 d 2 k B 0 e ik.r ψ(k. (.40

9 .2. DIMENSION AND DIMENSIONAL REGULARIZATION 9 Multiplying both sides by e ik.r and integrating over r, we get d 2 re ik.r d 2 k ( k 2 e ik.r ψ(k + 0 d 2 rδ 2 (re ik.r d 2 k e ik.r ψ(k = d 2 re ik.r d 2 k B 0 e ik.r ψ(k. (.4 Performing the r integration, we get k 2 ψ(k + 0 ψ(0 = B 0 ψ(k, (.42 or We have used ψ(k = 0ψ(0 k 2 + B 0. (.43 (2π 2 δ 2 (k = e ik.r d 2 r. (.44 Now if we integrate both the sides over k we get or 0 = 4π 2 ψ(0 = 0 4π 2 d 2 k k 2 = + B 0 4π 2 (2π d 2 kψ(0 k 2 + B 0, (.45 kdk k 2 = + B 0 4π d(k 2 k 2 + B 0. (.46 Now this is the equation which is supposed to determine B 0 and hence the bound state eigen value. However, the problem is that, by redefining k k/ B 0, we can completely eliminate B 0! We also notice that the integral has a log divergence at large momentum. We introduce a large momentum cut-off and write it as 0 = Lt Λ 4π Λ 2 0 d(k 2 ( Λ 2 k 2 = Lt Λ + B 0 4π log + B 0. (.47 B 0 Like the earlier example, by introducing a cut-off we have broken the scale invariance k k/c of the initial problem. Now however we can not scale out B 0 and write it as B 0 Λ 2 e 4π 0 We define renormalized coupling constant and E = h2 2m Λ2 e 4π 0. (.48 R (µ = 0 4π log ( Λ 2 µ 2. (.49

10 0 CHAPTER. UNITS, DIMENSIONS AND REGULARIZATION We have introduced an arbitrary dimensionful constant µ. Given and Λ, we therefore have a R for each µ. This, in turn, allows us to write bound energy as E = h2 2m µ2 e 4π R (µ. (.50 Notice that the explicit dependence of Λ and 0 in E have been eliminated in favour of µ and R. The limit Λ can now be taken. Even though the form contains µ, being observable, E would be independent of µ. This determines the change of R as we change the scale µ. 4 Elaborating a bit more ( ( E µ, = E µ (, R (µ R (µ = E µ, R (µ (.5 In other words, it says the independence of the physical quantity with respect to the choice of the parametrization point. Furthermore, going from parametrization given by (µ, / R (µ to (µ, / R (µ and to that of (µ, / R (µ should be equivalent to going directly from (µ, / R (µ to (µ, / R (µ a group law. From (.49, it follows, R (µ = ( µ 2 R (µ + log µ 2. (.52 Clearly, due to the appearance of the log function, the group law would be satisfied. This is a toy example of what is known as renormalization group a topic we will discuss in detail later in this course. Let us re-write the above equation as R (µ = R (µ + R(µ 4π log( µ 2 µ 2 (.53 This equation tells us that as we increase the mass scale µ, R (µ decreases. This means that with higher and higher energy, the renormalized coupling consatnt gets weaker. This penomenon is called asymptotic freedom. Quantum chromodynamics enjoys this special property. At weak copuling, theory is almost free. Consequently, constituents of this theory interact very weakly at very high energy. Another way to see this behaviour of the coupling constant is to look at the β-function equation that follow from (.50. We know that E, being a measurable quantity, should be independent of the parameter µ. de = 0. (.54 d log µ 4 Note from (.49 that in order to have finite R, as we take the regulator Λ to infinity, 0 must go to zero in order to have a bound state.

11 .2. DIMENSION AND DIMENSIONAL REGULARIZATION This immediately implies d R (µ d log µ = R 2 (µ 2π. (.55 We see that indeed the renormalized coupling decreases with energy and this is dictated by the β-function equation. We end this chapter with some instructive problems. Problem 4: Modify the integral in (.46 by damping term e ǫk2 as = d(k 2 e ǫk2 0 4π 0 k 2 (.56 + B 0 Do the integral and expand it in powers of /ǫ. Show that this is an equivalent way of regulating the integral once we identify Λ / ǫ. What would you conclude then? Problem 5: Consider a particle of mass m in one dimensional potential V (x = { if x 0 a x 2 if x > 0. (.57 Write down the Schrödinger equation and argue on dimensional ground that it is impossible to have bound state solution. What can you say if we modify the above potential by V (x = { if x ǫ a x 2 if x > ǫ? (.58 In particular, does it have a bound state solution? What happens if we take ǫ 0 at the end of the calculation? Problem 6: Finally take a three dimensional problem. For a square well potential { V0 if r a V (r = (.59 0 if r > a, write down the equation that the radial component of the wave function satifies. Further, analyze the nature (and condition for existance of the bound state solutions.