Quantum Field Theory III

Quantum Field Theory III Prof. Erick Weinberg March 28, 2011

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Lecture 18

Last time we consider solitons in 2D. We need to develop a language which describes what we want more efficiently. Let’s consider a manifold M which is the set of vacua and φ0 is a vacuum. Let the gauge group be G and it spontaneously breaks down to H. So if φ0 is a vacuum then gφ0 is a vacuum, and hφ0 = φ0 . In particular we have gφ0 = ghφ0 (1) So the manifold is essentially M = G/H. Let’s consider 2D, then the infinity is essentially a circle S 1 . If we going around the circle in the soliton solution, then this gives us a loop in the space of vacua. The argument that a vortex solution can’t be deformed continuously into a vacuum solution is that a simple loop in the vacua manifold can’t be continuously deformed into a point. The mathematical language which describes this is homotopy. Let’s consider a space M and a point x0 in M .

x0

Figure 1: Space M with a part cut out We define a loop as a map f (s) : [0, 1] −→ M such that f (0) = f (1) = x0 . Suppose we have two loops f (s) and g(s) we can define the continuous deform k(s, t) where 0 ≤ s, t ≤ 1 such that k(s, 0) = f (s) and k(s, t) = g(s). If such a continuous k exists then we say f and g are homotopic at x0 . We can define the

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product of two loops f ◦ g as ( f (2s) f ◦ g(s) = g(2s − 1)

0 ≤ s ≤ 1/2 1/2 ≤ s ≤ 1

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We can define the identity as I(s) = x0 and the inverse of f can be found as f −1 (s) = f (1 − s), the loop traversed inversely. In order to consider this as a group, instead of considering loops we consider homotopy classes of loops and call [f ] the set of loops homotopic to f . Then the group operations are [f ] ◦ [g] = [f ◦ g],

[f ]−1 = [f −1 ],

i = [I]

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This group is π1 (M, x0 ) which is either called the fundamental group of M at x0 , or first homotopy group of M at x0 . If M is connected, then the group does not depend on x0 , however obviously if it has multiple connected pieces then the group will depend on x0 . If π(M, x0 ) = 1 then we say the manifold is simply connected. Let’s consider a few examples. Suppose we have two dimensional Euclidean space. It is connected so we just write π1 (R2 ). Obviously it is trivial because we can shrink everything. Same for π1 (Rn ). Now suppose M = R2 − disk as our example above. If a loop does not go around the disk then it can be shrunk into a point. Another observation is that loops that circle the disk different number of times can’t be deformed into each other, and their product goes around the disk the n1 + n2 times. So the group is π1 (M ) = Z. Similarly we have π1 (S 1 ) = Z. How about S 2 ? We can punch a hole on it and stretch it to R2 and obviously we can shrink any loop there. So π1 (S 2 ) = 0. This argument goes for any n > 2. Now consider the space with two holes. Let a be a loop goes around one hole once, and b the other hole once. The loop aba−1 b−1 can’t be contracted to a point, however aa−1 bb−1 can. So the fundamental group for this space is non-Abelian. Now we are interested in the fundamental group of Lie groups, and for any Lie group G we have π1 (G) is Abelian. Let’s consider SU (2) and SO(3) and this is the place where π1 becomes important. The group SU (2) is the group of 2 × 2 unitary matrices with determinant 1. We can write the group element as u = a0 + ia · σ,

a20 + a21 + a22 + a23 = 1

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From the constraint we know that SU (2) ∼ = S 3 so that π1 (SU (2)) = 0. Now consider SO(3). It is described by axis of rotation n ˆ and angle ϕ. But we have the additional identification (ˆ n, π) = (−ˆ n, π). We can think of this as a 3D ball of radius π, with antipodal points identified. There are two kinds of loops. One is just inside the ball like an ordinary loop. The other is a loop that goes to the boundary and by antipodal identification goes to the other end of the ball and comes back. These loops can’t be deformed into each other. However if the loop hits boundary twice we can move the two points closer and closer so they meet and we reduce to the first case. So loops with even number of jumps are contractable and loops with odd number of jumps is not so π1 (SO(3)) = Z2 (5) What is the relation between SO(3) and SU (2)? For (ˆ n, ϕ) in SO(3) we can identify the element exp[i(ˆ n · σ)ϕ/2. Now if we change ϕ → ϕ + 2π then there is no effect on rotation, but in SU (2) we change U → −U . So the full range of rotation in SO(3) is 2π and in SU (2) is 4π. We can interpret this using the center of the group, which is the set of elements that commute with all elements. In SU (2) the center is {I, −I} = Z2 . Starting from SU (2) we can define an equivalence g ∼ = −g and this equivalence

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preserves group multiplication because −I commutes with everything. Now the group that we get from this equivalence is just SO(3) SU (2)/Z2 = SO(3) (6) We can also think of this as S 3 with antipodal points identified, which will just be a hemisphere with the antipodal points on the boundary identified, which is exactly the 3D ball with antipodal points on boundary identified. Now from any Lie algebra we can get a unique simply connected group. We call this the universal covering group. Suppose G is the universal covering group and K the center of G. We could take G/K and get some other group. However we can also take K as some subgroup of the center. Now G/K is another Lie group which is not simply connected. In fact π1 (G/K) = K

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This can be seen if we consider loops from point g. For each element z in the center, the curve that connects g to zg is a loop in the quotient group, and it is different from the curve connecting g to z 0 g. For SO(N ) the universal covering group is Spin(N ). It is always the case that SO(N ) = Spin(N )/Z2

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If we look at the center of Spin(N ), then we have   Z2 , Center(Spin(N )) = Z2 × Z2 ,   Z4 ,

N odd N = 4k N = 4k + 2

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This is because SO(N ) has nontrivial center for even N , when −I also has determinant 1. However SU (N ) has center ZN which are elements ei2πk/N I. Note that U (1) ∼ = S 1 so that π1 (U (1)) = Z. The group Sp(N ) has center Z2 , E6 has center Z3 and E7 has center Z2 . Now remember SU (2) has integer and half integer spin representations, and SO(3) only have integer spin representations. In general the universal covering group has all the representations, and G/K has some subset of the representations. So in SU (3) we can define some “triality” which is the number of upper indices minus lower indices mod 3. So the representation 3 has triality 1, where ¯3 has triality −1, and the 8 representation has triality 0. This is because SU (3) has center Z3 . For SU (N ) we can do this and classify the representations into N subsets. Now let’s get back to solitons. The loops beginning and ending at a point φ0 corresponds to loops in the space of vacuum G/H. These objects are almost identical, but not quite. In our deformation there is no reason we keep the point φ0 fixed. So we define the free homotopy. Now when we have two holes and the fundamental group is nonabelian, then free homotopy is essentially making the group abelian. Now we want to relate the group structure to topological charge. Now if we have two vortices in the space, the topological charge of one plus the other is just going around the product of two loops. Here we want abelian because we don’t want the order of going around to matter. Consider the example for U (1) completely broken by complex ϕ. The manifold for vacua is U (1) and π1 (M ) = Z. So the charge is just integers. Now consider SO(N ) broken into SO(N − 1) the space of vacua is M = SO(N )/SO(N − 1) ∼ (10) = S N −1 3

so π1 (M ) = 0. Now suppose the group is SO(3) and we have two scalar field which are in the vector representation of SO(3). The potential being V =

λϕ 2 λχ 2 (ϕ − vϕ2 )2 + (χ − vχ2 )2 + g(ϕ · χ)2 4 4

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If g > 0 then ϕ is perpendicular to χ so SO(3) is completely broken. Then π1 (M ) = π1 (SO(3)) = Z2 . So what do the solutions look like? Possible solutions look like ϕ = (0, 0, vϕ ),

χ = (cos θ, sin θ, 0)f (r),

Aj = jk x ˆk

a(r) (0, 0, 1) r

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or we can have χ = (0, 0, vχ ),

ϕ = (cos θ, sin θ, 0)f (r),

Aj = . . .

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Now the fundamental group being Z2 means that there is one kind of solution which is topologically trivial and deformable to a vacuum solution. This class of solution is just ϕ = (0, 0, vϕ ),

χ = (cos 2θ, sin 2θ, 0)f (r),

Aj = jk x ˆk

a(r) (0, 0, 1) r

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What is the other class? Suppose vϕ  vχ , then we can think of the breaking as SO(3) to SO(2) = U (1) and then broken by χ. But the above solution is stable because for it to deform into the vacuum we need to deform ϕ and it involves a lot of energy. So topology isn’t everything. Now let’s consider Weinberg-Salam model. The group is G = gSU (2) × g 0 U (1) with fields ϕ and the vacuum is ϕ† ϕ = v 2 /2. The space of vacua is S 3 and π1 (S 3 ) = 0. So there should be no vortices. However if g = 0 then SU (2) is global and U (1) is local, then we have vortices in U (1) section. These are called semi-local string and they are actually stable. Now suppose g  g 0 then there is only large finite energy barrier against decay of these strings. However what is “small” enough? It is a numerical question, and it turns out that the “small” is smaller than observed number, so there is no such things in electroweak theory. Now if there is a π1 (M ) then there is obviously π2 . The groups πn (M ) are maps from S n to M . When we consider solitons in 2D then π1 is what to look at. However when we consider solitons in 3D then π2 (M ) is what to look at. We do this by converting S 2 into a square with all sides identified. We define a “loop” as f (s, t) : [0, 1] × [0, 1] −→ M, f (0, t) = f (1, t) = f (s, 0) = f (s, 1) = x0 (15) We define homotopy by the existence of a continuous map k(s, t, u) which connects f and g. The product of two loops is roughly as before ( f (s, 2t), 0 ≤ t ≤ 1/2 f ◦ g(s, t) = (16) g(s, 2t − 1), 1/2 ≤ t ≤ 1 Note this can always be continuously deformed to g ◦ f , so π2 (M ) is always abelian. Similarly πn (M ) is abelian for n > 2. Let’s state some results. In general we have πn (S n ) = Zn ,

πn (S 1 ) = 0,

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πn (S m ) = 0 for m > n

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