Problem 14.1a

Posted on July 26, 2015 by rpflee

A First Course in String Theory

14.1 Counting bosonic states

~~~

$n$ is the number of $a$ ‘s.

$k$ is the number of different $a$ ‘s.

For $a^{i_1} a^{i_2}$ , $n=2$ .

(Indices $i_1$ and $i_2$ are not powers. Instead, they are just upper indices for representing different $a$ ‘s.)

When all $a$ ‘s commute, $a^1 a^2$ and $a^2 a^1$ , for example, represent the same state. So we have to avoid double-counting, except for the same $a$ ‘s states, such as $a^3 a^3$ .

By direct counting, without using the formula, the number of products of the form $a^{i_1} a^{i_2}$ can be built is

$\frac{k(k-1)}{2} + k$
$= \frac{(k + 1)k}{2}$

—

Let $N(n, k) = {n + k - 1 \choose k - 1}$ , the number of ways to put $n$ indistinguishable balls into $k$ boxes.

By using the formula, the number of products of the form $a^{i_1} a^{i_2}$ can be built is

$N(2,k)$

$= \frac{(2+k-1)!}{2!(k-1)!}$

$= \frac{(k+1)k}{2}$

…

— Me@2015-07-26 08:43:22 AM

2015.07.26 Sunday (c) All rights reserved by ACHK

Problem 14.1 | Problem 12.11

Posted on July 14, 2015 by rpflee

A First Course in String Theory

14.1 Counting bosonic states

12.11 Counting symmetric products

~~~

$a^{I_1} a^{I_2} ... a^{I_n}$

$i = 1, 2, ..., n$

$I_i = 1, 2, ..., k$

$n$ is the number of $a$ ‘s.

$k$ is the number of different $a$ ‘s.

For $k=6$ and $n=9$ , one of the possible configurations is

$a^1 a^2 a^2 a^4 a^4 a^4 a^5 a^6 a^6$

The problem is equivalent to that of assigning $n$ balls to $k$ boxes.

Box 1 collects the $a$ ‘s with index 1.

Box 2 collects the $a$ ‘s with index 2.

…

— Me@2015-07-13 03:34:32 PM

2015.07.14 Tuesday (c) All rights reserved by ACHK

Quick Calculation 14.8

Posted on July 5, 2015 by rpflee

A First Course in String Theory

What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

~~~

— This answer is my guess. —

What is the meaning of “left-moving NS”?

p.322 “Conventionally, the first input in $(\cdot, \cdot)$ is the left-moving sector and the second input is the right-moving sector.”

What is the meaning of “+” in “NS+”?

p.321 “… we truncate the NS sector to the set of states with $(-1)^F = +1$ . The resulting states comprise the so-called NS+ sector.”

“The NS+ sector contains the massless states and throw away the tachyonic states.”

What is the meaning of “consistent” closed string sector?

p.322 “… guarantees that the left and right sectors give identical contributions to the mass-squared:”

$\alpha' M_L^2 = \alpha' M_R^2$

If NS+ is the left sector, the corresponding candidate right sectors are NS+, R+, R-.

$M$ in NS-sector, Equation (14.37):

$M_{NS}^2 = \frac{1}{\alpha'} \left( - \frac{1}{2} + N^\perp \right)$

$M$ in R-sector, Equation (14.53):

$M_{Ramond}^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_n^I + n d_{-n}^I d_n^1 \right) = \frac{1}{\alpha'} N^\perp$

Equation (14.78):

$\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2$

NS+ equations of (14.38):

$\alpha'M^2=0,~~N^\perp = \frac{1}{2}:~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle$ ,

$\alpha'M^2=1,~~N^\perp = \frac{3}{2}:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$ ,

$...$

NS- equations of (14.38):

$\alpha'M^2=-\frac{1}{2},~~N^\perp = 0:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$ ,

$\alpha'M^2=\frac{1}{2},~~N^\perp = 1:~~~~\{ ... \} |NS \rangle \otimes |p^+, \vec p_T \rangle$ ,

$...$

Mass levels of R+ and R- (Equations 14.54):

$\alpha'M^2=0,~~N^\perp = 0:~~~~|R_a \rangle~~||~~|R_{\bar a} \rangle$

$\alpha'M^2=1,~~N^\perp = 1:~~~~...$

$...$

There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

(NS-, NS-)

— This answer is my guess. —

— Me@2015-07-05 11:09:32 PM

2015.07.05 Sunday (c) All rights reserved by ACHK

Quick Calculation 14.7

Posted on June 29, 2015 by rpflee

A First Course in String Theory

Count the number of graviton, Kalb-Ramond, and dilation states in ten dimensions. Add these numbers up and confirm you get 64.

~~~

Equation (13.69):

$\sum_{I,J} \hat S_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

p.292 “… the states (13.69) represent one-particle graviton states … ”

Equation (13.64):

In the closed string state space, the general state of fixed momentum at the massless level is

$\sum_{I,J} R_{IJ} {a_1^I}^\dagger {\bar a}_1^{J \dagger} | p^+, \vec p_T \rangle$

Equation (13.68):

$R_{IJ} = \hat S_{IJ} + A_{IJ} + S' \delta_{IJ}$

$S' = \frac{S}{D-2}$

Equation (10.108):

For the symmetric traceless part, $\hat S_{IJ}$ (with the size $(D-2) \times (D-2)$ ), the number of independent components is

$n(D) = \frac{1}{2} (D - 2) (D - 1) - 1$

The $-1$ at the end reflects the fact that if the trace is zero, the last component of the trace is not independent anymore.

In ten dimensions ( $D = 10$ ),

$n(D) = \frac{1}{2} (8) (9) -1 = 35$

p.292 For $A_{IJ}$ (a skew-symmetric matrix), the number of independent components is

$n(D) = \frac{1}{2} (D - 3) (D - 2) = 28$

p.293 “The oscillator part of (13.71) has no free indices ( $I$ is summed over), so it represents one state. It corresponds to [an] one-particle state of a massless scalar field. This field is called the dilation.”

— Me@2015-06-29

2015.06.29 Monday (c) All rights reserved by ACHK

Quick Calculation 14.4

Posted on June 13, 2015 by rpflee

A First Course in String Theory

How many states are there at $N^\perp = \frac{3}{2}$ ?

~~~

This answer is my guess.

There are $(D-2)$ transverse (i.e. non-lightcone) coordinates. When $D = 10$ , the number of transverse coordinates is 8. So there are 8 states with $N^\perp = \frac{1}{2}$ .

For $N^\perp = \frac{3}{2}$ , there are 3 indices $I$ , $J$ , and $K$ . So there are totally 24 states.

— Me@2015.05.26 03:56 PM

The above answer is incorrect.

Equation (14.38) $\left(\alpha'M^2 = 1,~~N^\perp = \frac{3}{2}\right)$ :

$\{ \alpha_{-1}^I b_{-1/2}^J, b_{-3/2}^I, b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8 \times 8$ states for

$\alpha_{-1}^I b_{-1/2}^J |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $8$ states for

$b_{-3/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle$

There are $\frac{8 \times 7 \times 6}{3!}$ states for

$b_{-1/2}^I b_{-1/2}^J b_{-1/2}^K |NS \rangle \otimes |p^+, \vec p_T \rangle$ .

So the total number of states is 64 + 8 + 56 = 128.

You can check this answer against Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

— Me@2015-08-21 08:15:50 AM

2015.06.13 Saturday (c) All rights reserved by ACHK

Quick Calculation 14.6

Posted on June 4, 2015 by rpflee

A First Course in String Theory

Construct explicitly all the states with $\alpha' M^2 =2$ and count them, verifying that there are indeed a total of 3200 states.

~~~

$\alpha' M^2 = 2~~~\to~~~N^\perp = 3$

The first possibility group includes the cases of one $a_3^\dagger$ . Since there are 24 species, this group has 24 possibilities.

The second possibility group includes the cases of one $a_1^\dagger$ with one $a_2^\dagger$ . (Their order is not important because interchanging the order would create the same state anyway.) Since there are 24 species for each of $a_1^\dagger$ and $a_2^\dagger$ , this group has $24 \times 24$ possibilities.

The third possibility group includes the cases of three $a_1^\dagger$ ‘s. We need to further divide this group into 3 sub-groups.

For the first sub-group, three $a_1^\dagger$ ‘s are all different:

$a_{1A}^\dagger a_{1B}^\dagger a_{1C}^\dagger$

Since there are 24 species for each $a_1^\dagger$ , there should be $24 \times 23 \times 22$ possibilities.

However, since their order is not important, there are repetitions among those possibilities. To eliminate repetitions, we divide the number with $3!$ . So, this subgroup has

$\frac{24 \times 23 \times 22}{6}$

possibilities.

For the second sub-group, two of the three $a_1^\dagger$ ‘s are identical:

$a_{1A}^\dagger a_{1B}^\dagger a_{1B}^\dagger$

~~There should be $24 \times 24$ possibilities.~~Me@2018-04-20 12:18:38 PM

There should be $24 \times 23$ possibilities.

For the third sub-group, all three are identical. In other words, there are 24 possibilities.

— Me@2015-06-04 10:19:30 PM

Quick Calculation 14.3

Posted on May 26, 2015 by rpflee

A First Course in String Theory

To test the $N^\perp$ includes the fermionic contribution to the number, show that the eigenvalue of $N^\perp$ on $b_{-r_1}^I b_{-r_2}^J | NS \rangle$ , with $r_1, r_2 > 0$ , is $r_1 + r_2$ .

~~~

$N^\perp b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \left[ \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^I b_r^I \right) \right] b_{-r_1}^I b_{-r_2}^J |NS \rangle$

$= \sum_{I'} \left( \sum_{p=1}^\infty \alpha_{-p}^{I'} \alpha_p^{I'} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} \right)$ $|n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r b_{-r}^{I'} b_r^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= \sum_{I'} [ r_1 b_{-r_1}^{I'} b_{r_1}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I'} b_{r_2}^{I'} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle]$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{I} b_{r_2}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$+ r_1 b_{-r_1}^{J} b_{r_1}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

$= r_1 b_{-r_1}^{I} b_{r_1}^{I} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle + 0 + 0 + r_2 b_{-r_2}^{J} b_{r_2}^{J} |n_{r_2}^J = 1, n_{r_1}^I = 1 \rangle$

Quick Calculation 14.2

Posted on May 3, 2015 by rpflee

A First Course in String Theory

$\sum_{s_1, s_2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

Prove that “only the part of the wavefunction $\psi$ that is antisymmetric under the simultaneous exchange $p_1 \leftrightarrow p_2$ and $s_1 \leftrightarrow s_2$ contributes…”

~~~

We divide the proof into two parts. In the first part, we prove that being antisymmetric under $p_1 \leftrightarrow p_2$ is a must. Then, in the second part, we use the same method to prove that being antisymmetric under $s_1 \leftrightarrow s_2$ is also a must.

First, we separate $\psi$ into two parts:

$\psi (p_1, s_1; p_2, s_2) = \psi_{sp} + \psi_{ap}$ ,

where $\psi_{sp}$ is symmetric under $p_1 \leftrightarrow p_2$ and $\psi_{ap}$ is anti-symmetric.

Wavefunctions $\psi_{sp}$ and $\psi_{ap}$ can be defined as

$\psi_{sp} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) + \psi (p_2, s_1; p_1, s_2) \right]$
$\psi_{ap} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) - \psi (p_2, s_1; p_1, s_2) \right]$

Now, we consider the contribution of the symmetric part, $\psi_{sp} (p_1, s_1; p_2, s_2)$ .

$\int d \vec p \psi_{sp} (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_2, s_1; p_1, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

By interchanging the dummy variables $p_1$ and $p_2$ in the second term, we have

$...$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger | \Omega \rangle$
$=\frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) \left[ f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger + f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger \right] | \Omega \rangle$
$= 0$

So we have proved that the part symmetric under $p_1 \leftrightarrow p_2$ does not contribute.

Using the same method, we can also prove that the part symmetric under $s_1 \leftrightarrow s_2$ also does not contribute.

Commutator product rule

Posted on March 31, 2012 by rpflee

[A,BC] = [A,B]C + B[A,C]

This commutator product rule works just like the differentiation product rule:

assume that A is independent of C, you get the first term;

assume that A is independent of B, you get the second term.

— Me@2012-03-27 7:24:30 PM

Sage (mathematics software)

Posted on March 14, 2010 by rpflee

Design Philosophy of Sage

William Stein realized several important facts when designing Sage.

* To create viable alternatives to Magma, Maple, Mathematica, and MATLAB, would take hundreds, or thousands of man-years if one started from the beginning.
* There was a large collection of open-source mathematics software already written, but which was written in different languages (C, C++, Fortran and Python being the most common).

So rather than start from the beginning, Sage which is written in Python and Cython integrates all their specialized mathematics software into a common interface. A user needs to know only Python.

Where no open-source option exists for a particular class of problem, then this would be written in Sage. But Sage does not reinvent the wheel. The same design philosophy is used in commercial mathematics program such as Mathematica, but Sage can use a much wider range of open source software solutions than nonfree software, since some open source licensing imposes restrictions on commercial use of code.

— Wikipedia on Sage (mathematics software)

2010.03.14 Sunday ACHK

Octupole Expansion

Posted on April 6, 2008 by rpflee

I have used a whole month to do this derivation.

$\displaystyle{V(\mathbf{R}) = ... + \frac{1}{4 \pi \varepsilon_0} \frac{3}{2 |\mathbf R|^4} \sum_{ijk} \hat R_i \hat R_j \hat R_k Q_{ijk} + ... }$

$\displaystyle{Q_{ijk} = \int_{\mathbf r} \left[ \delta_{ij} ( r^2 - 5 r_i r_j + \frac{10}{3} r_i r_j \delta_{jk} ) r_k - |\epsilon_{ijk}| 10 r_i r_j r_k \right] \rho(\mathbf r) d^3 r}$