# Problem 14.2.1

A First Course in String Theory

14.2 Generating function for the unoriented bosonic open string theory.

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What is the difference between oriented and unoriented bosonic open strings?

p.268: “The theory of unoriented strings is obtained by restricting the oriented string spectrum to the set of states that are invariant under the action of  $\Omega$. Unoriented strings are not strings without orientation: they should be viewed as a quantum superposition of states that as a whole, are invariant under orientation reversal.”

An unoriented state is a superposition of 2 opposite oriented states.

— Me@2015.07.03 12:12 PM

Clue 2: “… adding a term that implements the projection to unoriented states.”

Equation (14.63):

The generating function $f_{os}$ for bosonic open string theory is

$f_{os} (x)$
$= \frac{1}{x} + 24 + 324 x + 3200 x^2 + ...$
$= \frac{1}{x} \left( 1 + 24 x + 324 x^2 + 3200 x^3 + ... \right)$

Clue 3: p.278 Problem 12.12e

$\Omega = (-1)^{N^\perp}$

An unoriented string state is a superposition of two opposite-oriented string states. An unoriented string state has twist invariant.

We can choose to keep all states $|\psi \rangle$ with $\Omega |\psi \rangle = + |\psi \rangle$.

We can also choose the states with $\Omega |\psi \rangle = - |\psi \rangle$. However, they are only valid for the basis states, not for other states, because other states are superpositions of basis states. The relative phase between basis states are physical in a superposition.

Effectively, the states $|\psi \rangle$ with $\Omega |\psi \rangle = + |\psi \rangle$ are the only possible choices. In other words, $N^\perp$ must be even.

— This answer is my guess. —

For the unoriented open strings, we should keep only the even powers of $N^\perp$:

$f_{uos} (x) = \frac{1}{x} \left( 1 + 324 x^2 + 176256 x^4 + ... \right)$

— This answer is my guess. —

— Me@2015-09-17 02:27:20 PM