33-652: String Theory Assignment 6

Peter Battaglino 7. April, 2005

Problem 13.6: Orientifold Op-planes.

a. For an O23-plane the two normal directions x^{24}, x^{25} can be represented by a plane. A closed string at a fixed τ appears as a parametrized closed curve X^{a}(τ, σ) in this plane. Draw such an oriented closed string that lies fully in the first quadrant of the (x^{24}, x^{25}) plane. Draw also the string X̃^{a}(τ, 2π − σ).

b. For any operator O the action of Ω_{p} is defined by O → Ω_{p}OΩ_{p}^{−1}. Use the expansion (13.24) to calculate the action of Ω_{p} on the operators x_{0}^{a}, p^{a}, α_{n}^{a}, ᾱ_{n}^{a}, x_{0}^{i}, p^{i}, α_{n}^{i}, and ᾱ_{n}^{i}.

The closed string coordinates are represented by

X^{μ}(τ, σ) = x_{0}^{μ} + \sqrt{2α′}α_{0}^{μ}τ + i\sqrt{\frac{α′}{2}} ∑_{n≠0} \frac{e^{−inτ}}{n} (α_{n}^{μ}e^{inσ} + ᾱ_{n}^{μ}e^{−inσ}). (1)

The action of Ω_{p} on each of the tangential coordinates is the same as that of the twist operator in problem 13.5. Thus,

Ω x_{0}^{i} Ω^{−1} = x_{0}^{i}  (2)
Ω p^{i} Ω^{−1} = p^{i}  (3)
Ω α_{n}^{i} Ω^{−1} = ᾱ_{n}^{i}  (4)
Ω ᾱ_{n}^{i} Ω^{−1} = α_{n}^{i}.  (5)

For the transverse dimensions, we have

Ω x_{0}^{a} Ω^{−1} = −x_{0}^{a}  (6)
Ω p^{a} Ω^{−1} = −p^{a}  (7)
Ω α_{n}^{a} Ω^{−1} = −ᾱ_{n}^{a}  (8)
Ω ᾱ_{n}^{a} Ω^{−1} = −α_{n}^{a}.  (9)

For the + and − coordinates, Ω simply reverses the orientation. This is obviously the case for X^{+}. For X^{−} we know that the only non-trivial actions will involve products like α_{n}^{a}α_{q−n}^{a} and their barred equivalents. The negative signs produced by the Ω action will cancel, and we will obtain the same result as in problem 13.5. The orientifold transformations are symmetries of closed string theory because they leave the closed string Virasoro operators invariant for exactly the same reason that they leave the X^{−} coordinate invariant. The closed string Virasoro operators are the operator-ingredients for the closed string Hamiltonian, so

Ω H Ω^{−1} = H ⇒ [Ω, H] = 0  (10)

c. Denote the ground states by |p^{+}, p^{i}, p^{a}⟩. Assume that the states |p^{+}, p^{i},0⟩ are invariant under Ω: Ω|p^{+}, p^{i},0⟩ = |p^{+}, p^{i},0⟩. Prove that

Ω|p^{+}, p^{i}, p^{a}⟩ = |p^{+}, p^{i}, −p^{a}⟩.

We know that p^{a} transforms as Ω p^{a} Ω^{−1} = −p^{a}, so Ω and p^{a} anticommute. Hence,

p_{op}^{a}Ω|p^{+}, p^{i}, p^{a}⟩ = −Ωp_{op}^{a}|p^{+}, p^{i}, p^{a}⟩  (11)
= −p^{a}Ω|p^{+}, p^{i}, p^{a}⟩,  (12)

so we see that

Ω|p^{+}, p^{i}, p^{a}⟩ = |p^{+}, p^{i}, −p^{a}⟩.  (13)

d. The massless states of the oriented closed string theory are given by

|Φ⟩ = ∫ dp^{+} d^{p−1}p^{i} d^{d−p}p^{a} Φ_{IJ}^{±}(τ, p^{+}, p^{i}, p^{a})(α_{−1}^{I}ᾱ_{−1}^{J} ± α_{−1}^{J}ᾱ_{−1}^{I})|p^{+}, p^{i}, p^{a}⟩.

Find the conditions that must be satisfied by Φ_{ab}^{±}, Φ_{ia}^{±}, and Φ_{ij}^{±} to guarantee Ω invariance. All of the conditions are of the form

Φ_{IJ}^{±}(τ, p^{+}, p^{i}, p^{a}) = ⋯ Φ_{IJ}^{±}(τ, p^{+}, p^{i}, −p^{a}),

where the dots represent a sign factor that you must determine for each case.

We can write the action of Ω on |Φ⟩ as

Ω|Φ⟩ = ∫ d^{D}p/dp^{−}Φ_{IJ}^{±}(…)(Ωα_{−1}^{I}Ω^{−1}Ωᾱ_{−1}^{J}Ω^{−1}
± Ωα_{−1}^{J}Ω^{−1}Ωᾱ_{−1}^{I}Ω^{−1})Ω|p^{+}, p^{i}, p^{a}⟩.  (14)

When (IJ) = (ab) the operator sum in the expression for |Φ⟩ is going to undergo the transformation α ↔ ᾱ, while the action of Ω on the basis states will invert their transverse momenta. The operator sum is symmetric under the α ↔ ᾱ transformation when the ± is a +, and antisymmetric when the ± is a −. Thus we need a factor of − to cancel the asymmetry and nothing to cancel the symmetry, so we need a factor of ±:

Φ_{ab}^{±}(p^{a}) = ±Φ_{ab}^{±}(−p^{a}).  (15)

When (IJ) = (ia), we pick up negative signs in the operator sum under the action of Ω, and upon reordering terms we obtain, schematically:

∓Φ_{ia}^{±}(p^{a})(α_{−1}^{i}ᾱ_{−1}^{a} ± α_{−a}^{a}ᾱ_{−1}^{i})|p^{+}, p^{i}, −p^{a}⟩.  (16)

So the action of Ω contributes a factor of ∓ to the operator terms and inverts the basis states, and we thus require, for (IJ) = (ia), that

Φ_{ia}^{±}(p^{a}) = ∓Φ_{ia}^{±}(−p^{a}).  (17)

The action of Ω yields the same results when (IJ) = (ij) as for when (IJ) = (ab) so we need a factor of ± in this case as well:

Φ_{ij}^{±}(p^{a}) = ±Φ_{ij}^{±}(−p^{a}).  (18)

Problem 13.7: Massive level in the open superstring.

a. Consider eight anticommuting variables b^{i}, with i = 1, …, 8. Ignoring signs, how many inequivalent products of the form b^{i_{1}}b^{i_{2}} can be built? How many b^{i_{1}}b^{i_{2}}b^{i_{3}}? How many b^{i_{1}}b^{i_{2}}b^{i_{3}}b^{i_{4}}?

Ignoring signs, the question we are trying to answer is “How many different ways can we choose N out of 8 objects?”. This is the case because permutations don’t matter when we are ignoring signs, and we can’t choose the same object twice, or else we get zero (since the b’s are Grassmannian). Thus, for b^{i_{1}}b^{i_{2}} we can choose \binom{8}{2} = 28 different products, for b^{i_{1}}b^{i_{2}}b^{i_{3}} we have \binom{8}{3} = 56, and for b^{i_{1}}b^{i_{2}}b^{i_{3}}b^{i_{4}} we have \binom{8}{4} = 70 choices.

b. Consider the first excited level of the open superstring (α′M^{2} = 1). List the states in the NS sector and the states in the R sector. Confirm that you get the same number of states (the number is more than 64).

In the NS sector we have the following allowed states:

\begin{matrix}b_{−3/2}^{I}|NS⟩ ⊗ |p^{+}, p⃗_{T} ⟩ & 8 \\ b_{−1/2}^{I} b_{−1/2}^{J} b_{−1/2}^{K}|NS⟩ ⊗ |p^{+}, p⃗_{T} ⟩ & \binom{8}{3} = 56 \\ α_{−1}^{I} b_{−1/2}^{J}|NS⟩ ⊗ |p^{+}, p⃗_{T} ⟩ & 8 × 8 = 64 \end{matrix}

yielding a total of 128 states. In the R sector we have the following allowed states:

\begin{matrix}d_{−1}^{I}|R_{1}^{a}⟩ ⊗ |p^{+}, p⃗_{T} ⟩ & 8 × 8 = 64 \\ d_{−1}^{I}|R_{2}^{a}⟩ ⊗ |p^{+}, p⃗_{T} ⟩ & 8 × 8 = 64 \end{matrix}

also yielding a total of 128 states. Thus we see that the NS and R sectors have the same number of α′M^{2} = 1 states.

Problem 14.1: A Dp-brane with orientifolds.

a. The Ω actions are given by

Ω X^{a}(τ, σ) Ω^{−1} = X^{a}(τ, π − σ),  (19)
Ω X^{i}(τ, σ) Ω^{−1} = X^{i}(τ, π − σ).  (20)

Give the Ω action on the oscillators α_{n}^{a} and α_{n}^{i}. What is the expected Ω action on α_{n}^{−}? Does it work out?

We have the expansions

X^{a}(τ, σ) = x̄^{a} + \sqrt{2α′} ∑_{n≠0} \frac{1}{n}α_{n}^{a}e^{−inτ} sin nσ  (21)
X^{i}(τ, σ) = x_{0}^{i} + \sqrt{2α′}α_{0}^{i}τ
+i\sqrt{2α′} ∑_{n=1}^{∞} \frac{1}{n}(α_{n}^{i}e^{−inτ} − α_{−n}^{i}e^{inτ}) cos nσ  (22)

For the transverse oscillators, we see that a factor of sin(nπ−nσ) contributes an additional minus sign to the usual (−1)^{n} for the tangential oscillators, so we get

Ω α_{n}^{i} Ω^{−1} = (−1)^{n}α_{n}^{i}  (23)
Ω α_{n}^{a} Ω^{−1} = (−1)^{n+1}α_{n}^{a}  (24)

For α_{n}^{−} we expect to pick up the same sign change as for the tangential oscillators. We just need to check the action of Ω on the Virasoro operators:

∑_{p∈ℤ}α_{p}^{I}α_{n−p}^{I} = ∑_{p∈ℤ}(α_{p}^{i}α_{n−p}^{i} + α_{p}^{a}α_{n−p}^{a})  (25)
→ ∑_{p∈ℤ}((−1)^{p+n−p}α_{p}^{i}α_{n−p}^{i} + (−1)^{p+1+n−p+1}α_{p}^{a}α_{n−p}^{a}) (26)
= (−1)^{n} ∑_{p∈ℤ}α_{p}^{I}α_{n−p}^{I},  (27)

which is what we expect.

b. Assume that the ground states |p^{+}, p⃗⟩ are Ω invariant. Find the states of the theory for N^{⊥} ≤ 2. As you will see, some massless sates survive. Interpret these states along the lines of the discussion below (14.37).

The possible N^{⊥} ≤ 2 states that we can build are

\begin{matrix}α_{−1}^{a}|p^{+}, p⃗⟩ & +1 \\ α_{−1}^{i}|p^{+}, p⃗⟩ & −1 \\ α_{−1}^{a}α_{−1}^{b}|p^{+}, p⃗⟩ & +1 \\ α_{−1}^{a}α_{−1}^{i}|p^{+}, p⃗⟩ & −1 \\ α_{−1}^{i}α_{−1}^{j}|p^{+}, p⃗⟩ & +1 \\ α_{−2}^{a}|p^{+}, p⃗⟩ & −1 \\ α_{−2}^{i}|p^{+}, p⃗⟩ & +1 \end{matrix}

If we only keep states that satisfy Ω = +1 we still end up with massless states, namely those that we get by acting with α_{−1}^{a} on the ground states. There are (d − p) of these massless scalar states. These can be viewed as excitations of the Dp-brane in its normal directions.

c. Replace the O25-plane by an Op-plane coincident with the Dp-brane at x̄^{a} = 0. Let Ω_{p} denote the operator for which this theory keeps only the states with Ω_{p} = +1. How should equations (19) and (20) change when Ω is replaced by Ω_{p}? Give the Ω_{p} action on the oscillators α_{n}^{a} and α_{n}^{i}.

Equations (19) and (20) change to

Ω_{p} X^{a}(τ, σ) Ω_{p}^{−1} = −X^{a}(τ, π − σ)  (28)
Ω_{p} X^{i}(τ, σ) Ω_{p}^{−1} = X^{i}(τ, π − σ).  (29)

Since Ω_{p}’s action hasn’t changed for the tangential coordinates, its action on α_{n}^{i} stays the same. However, we pick up a minus sign for the normal oscillators:

Ω_{p} α_{n}^{a} Ω_{p}^{−1} = −(−1)^{n+1}α_{n}^{a} = (−1)^{n}α_{n}^{a},  (30)

so we have

Ω_{p} α_{n}^{I} Ω_{p}^{−1} = (−1)^{n}α_{n}^{I},  (31)

which means that we have the same allowed spectrum as for the twistoperator invariant spectrum for open strings, with a D25-brane present. This excludes the massless scalar states that we did allow for the operator Ω. We are left over with only the states for which N^{⊥} is even. This means that we Battaglino 8 don’t get the same normal excitations of the Dp-brane as we did for the case with an O25-plane.

Problem 4: Yang-Mills theory.

U(N) has N^{2} independent generators. The elements of SU(N) must satisfy the additional constraint of det(M) = 1. Since M = exp(iα · J) in general, and det(M) = exp(iTr(α · J)), we require that the generators be traceless, so this additional constraint reduces the dimensionality to N^{2} − 1 for SU(N). SO(N) is just the group of rotations in N dimensions. There are \binom{N}{2} = N(N − 1)/2 dimensions in this group because there are that many planes of rotation in N dimensions.

To generalize the QED action −(1/4)F^{μν}F_{μν} to non-Abelian groups, we allow the gauge field to transform as

δA_{μ}^{a} = gf^{abc}α^{b}A_{μ}^{c} + ∂_{μ}α^{a}.  (32)

The generators of the gauge group satisfy

[T^{a}, T^{b}] = if^{abc}T^{c}.  (33)

When the group is U(1), there is only one generator for the group. Since f^{abc} can be made completely antisymmetric, the fact that [T^{a}, T^{b}] = 0 means that the structure constants are zero, and the gauge transformation reduces to the usual transformation in QED. Define

F_{μν}^{a} = ∂_{μ}A_{ν}^{a} − ∂_{ν}A_{μ}^{a} + gf^{abc}A_{μ}^{b}A_{ν}^{c}.  (34)

This then transforms as

F_{μν}^{a} → ∂_{μ}(A_{ν}^{a} + gf^{abc}α^{b}A_{ν}^{c} + ∂_{ν}α^{a}) − ∂_{ν}(A_{μ}^{a} + gf^{abc}α^{b}A_{μ}^{c} + ∂_{μ}α^{a})
+gf^{ade}(A_{μ}^{d} + gf^{abc}α^{b}A_{μ}^{c} + ∂_{μ}α^{d})(A_{ν}^{e} + gf^{ebc}α^{b}A_{ν}^{c} + ∂_{ν}α^{e})
= F_{μν}^{a} + gf^{abc}(∂_{μ}(α^{b}A_{ν}^{c}) − ∂_{ν}(α^{b}A_{μ}^{c})) + gf^{ade}A_{μ}^{d}(gf^{ebc}α^{b}A_{ν}^{c} + ∂_{ν}α^{e})
+gf^{ade}(gf^{dbc}α^{b}A_{μ}^{c} + ∂_{μ}α^{d})A_{ν}^{e} + O(α^{2})
= F_{μν}^{a} + 2gf^{abc}((∂_{μ}α^{b})A_{ν}^{c} − (∂_{ν}α^{b})A_{μ}^{c})
+gf^{abc}α^{b}(∂_{μ}A_{ν}^{c} − ∂_{ν}A_{μ}^{c}) − g^{2}f^{dce}f^{abc}α^{b}Aμ^{d}A_{ν}^{e} + O(α^{2})
= F_{μν}^{a} + gf^{abc}α^{b}F_{μν}^{c} + 2gf^{abc}((∂_{μ}α^{b})A_{ν}^{c} − (∂_{ν}α^{b})A_{μ}^{c}) + O(α^{2}) (35)

to first order in α^{i}.