Monads in Haskell can be thought of as composable computation descriptions. The essence of monad is thus separation of composition timeline from the composed computation’s execution timeline, as well as the ability of computation to implicitly carry extra data, as pertaining to the computation itself, in addition to its one (hence the name) output, that it will produce when run (or queried, or called upon). This lends monads to supplementing pure calculations with features like I/O, common environment or state, etc.

2015.09.21 Monday ACHK

# Problem 14.2.1

A First Course in String Theory

14.2 Generating function for the unoriented bosonic open string theory.

~~~

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