Cadabra
a field theory motivated approach to computer algebra

Bosonic string oscillator algebra

The computation below shows how you can do oscillator algebra for the (open) bosonic string. We illustrate this by the classic computation which shows that you can construct a state which satisfies all the constraints but has negative norm when $d>26$. This is not a terribly difficult computation with pen-and-paper, but it is tricky to do with many other computer algebra systems (and if they handle it, their output will not look as nice as what you see below).
\alpha_{n}^{\mu}::SelfNonCommuting; \vac::LaTeXForm("|k\rangle");
\(\displaystyle{}\text{Attached property SelfNonCommuting to }\alpha_{n}\,^{\mu}.\)
\(\displaystyle{}\text{Attached property LaTeXForm to {\tt $\backslash{}$}vac}.\)
def post_process(ex): substitute(ex, $\sqrt{2} \sqrt{2} -> 2$) collect_terms(ex)
We first need to define the various objects, mostly non-commuting operators.
{\alpha_{n}^{\mu}, \vac}::NonCommuting; {\mu,\nu,\rho}::Integer(0..d-1). {\mu,\nu,\rho}::Indices. {m,n}::Integer. {m,n}::Symbol. \delta{#}::KroneckerDelta. \delta{#}::Diagonal.
\(\displaystyle{}\text{Attached property NonCommuting to }\left[\alpha_{n}\,^{\mu}, |k\rangle\right].\)
The trial state is given by $|\chi\rangle = |\chi_1\rangle + |\chi_2\rangle + |\chi_3\rangle$ with the three terms given by the following expressions:
chi1:= \alpha_{-1}^{\mu} \alpha_{-1}^{\mu} \vac; chi2:= \alpha_{0}^\mu \alpha_{-2}^\mu \vac; chi3:= \alpha_{0}^{\mu} \alpha_{-1}^{\mu} \alpha_{0}^{\nu} \alpha_{-1}^{\nu})\vac;
\(\displaystyle{}\alpha_{-1}\,^{\mu} \alpha_{-1}\,^{\mu} |k\rangle\)
\(\displaystyle{}\alpha_{0}\,^{\mu} \alpha_{-2}\,^{\mu} |k\rangle\)
\(\displaystyle{}\alpha_{0}\,^{\mu} \alpha_{-1}\,^{\mu} \alpha_{0}\,^{\nu} \alpha_{-1}\,^{\nu} |k\rangle\)
The main ingredient to verify that the $L_1$ and $L_2$ Virasoro constraints hold is, of course, the commutation relations between the $\alpha_{m}^\mu$ operators. What we want to do is commute all annihilation operators to the right until they hit the momentum eigenstate $|k\rangle$. Note that in the rule below, you need to write the indices on $\delta_{m, -n}$ using two separate subscripts, otherwise you get $m-n$ as index.
rl:= { \alpha_{m}^{\mu} \alpha_{n}^{\nu} | m > n -> \alpha_{n}^{\nu} \alpha_{m}^{\mu} + m\delta_{m}_{-n} \delta^{\mu\nu}, \alpha_{m}^{\mu} \vac | m > 0 -> 0, \alpha_{0}^{\mu} \vac -> \sqrt{2} k^{\mu} \vac, k^{\mu} k^{\mu} -> -1 };
\(\displaystyle{}\left[\alpha_{m}\,^{\mu} \alpha_{n}\,^{\nu}\quad\text{with}\quad{}m > n \rightarrow \alpha_{n}\,^{\nu} \alpha_{m}\,^{\mu}+m \delta_{m -n} \delta^{\mu \nu}, \alpha_{m}\,^{\mu} |k\rangle\quad\text{with}\quad{}m > 0 \rightarrow 0, \alpha_{0}\,^{\mu} |k\rangle \rightarrow \sqrt{2} k^{\mu} |k\rangle, k^{\mu} k^{\mu} \rightarrow -1\right]\)
Obviously we need to do this substitution a number of times, and distribute any factors over sums. We will see shortly how to automate this with converge.

Imposing the constraints

The $L_m$ Virasoro generators are given by infinite sums. The function below constructs part of that sum explicitly. Of course, you would be a bit more clever when doing things with pen and paper:
def L(m): ins='' for n in range(-10,10): if n!=-10: ins+='+ ' ins += r'1/2 \alpha_{'+str(m-n)+r'}^{\mu} \alpha_{'+str(n)+r'}^{\mu}' ex=Ex(ins) return ex
Here is what $L_0$ looks like; the other operators are similar:
L0=L(0); L1=L(1) L2=L(2)
\(\displaystyle{}\frac{1}{2}\alpha_{10}\,^{\mu} \alpha_{-10}\,^{\mu}+\frac{1}{2}\alpha_{9}\,^{\mu} \alpha_{-9}\,^{\mu}+\frac{1}{2}\alpha_{8}\,^{\mu} \alpha_{-8}\,^{\mu}+\frac{1}{2}\alpha_{7}\,^{\mu} \alpha_{-7}\,^{\mu}+\frac{1}{2}\alpha_{6}\,^{\mu} \alpha_{-6}\,^{\mu}+\frac{1}{2}\alpha_{5}\,^{\mu} \alpha_{-5}\,^{\mu}+\frac{1}{2}\alpha_{4}\,^{\mu} \alpha_{-4}\,^{\mu}+\frac{1}{2}\alpha_{3}\,^{\mu} \alpha_{-3}\,^{\mu}+\frac{1}{2}\alpha_{2}\,^{\mu} \alpha_{-2}\,^{\mu}+\frac{1}{2}\alpha_{1}\,^{\mu} \alpha_{-1}\,^{\mu}+\frac{1}{2}\alpha_{0}\,^{\mu} \alpha_{0}\,^{\mu}+\frac{1}{2}\alpha_{-1}\,^{\mu} \alpha_{1}\,^{\mu}+\frac{1}{2}\alpha_{-2}\,^{\mu} \alpha_{2}\,^{\mu}+\frac{1}{2}\alpha_{-3}\,^{\mu} \alpha_{3}\,^{\mu}+\frac{1}{2}\alpha_{-4}\,^{\mu} \alpha_{4}\,^{\mu}+\frac{1}{2}\alpha_{-5}\,^{\mu} \alpha_{5}\,^{\mu}+\frac{1}{2}\alpha_{-6}\,^{\mu} \alpha_{6}\,^{\mu}+\frac{1}{2}\alpha_{-7}\,^{\mu} \alpha_{7}\,^{\mu}+\frac{1}{2}\alpha_{-8}\,^{\mu} \alpha_{8}\,^{\mu}% +\frac{1}{2}\alpha_{-9}\,^{\mu} \alpha_{9}\,^{\mu}\)
The $L_1$ operator acting on the three terms in the trial state gives the following three terms, which we will need to reduce using the commutation relations (2nd and 3rd term suppressed; you get the idea).
L1c1:= @(L1) @(chi1); L1c2:= @(L1) @(chi2). L1c3:= @(L1) @(chi3).
\(\displaystyle{}\left(\frac{1}{2}\alpha_{11}\,^{\mu} \alpha_{-10}\,^{\mu}+\frac{1}{2}\alpha_{10}\,^{\mu} \alpha_{-9}\,^{\mu}+\frac{1}{2}\alpha_{9}\,^{\mu} \alpha_{-8}\,^{\mu}+\frac{1}{2}\alpha_{8}\,^{\mu} \alpha_{-7}\,^{\mu}+\frac{1}{2}\alpha_{7}\,^{\mu} \alpha_{-6}\,^{\mu}+\frac{1}{2}\alpha_{6}\,^{\mu} \alpha_{-5}\,^{\mu}+\frac{1}{2}\alpha_{5}\,^{\mu} \alpha_{-4}\,^{\mu}+\frac{1}{2}\alpha_{4}\,^{\mu} \alpha_{-3}\,^{\mu}+\frac{1}{2}\alpha_{3}\,^{\mu} \alpha_{-2}\,^{\mu}+\frac{1}{2}\alpha_{2}\,^{\mu} \alpha_{-1}\,^{\mu}+\frac{1}{2}\alpha_{1}\,^{\mu} \alpha_{0}\,^{\mu}+\frac{1}{2}\alpha_{0}\,^{\mu} \alpha_{1}\,^{\mu}+\frac{1}{2}\alpha_{-1}\,^{\mu} \alpha_{2}\,^{\mu}+\frac{1}{2}\alpha_{-2}\,^{\mu} \alpha_{3}\,^{\mu}+\frac{1}{2}\alpha_{-3}\,^{\mu} \alpha_{4}\,^{\mu}+\frac{1}{2}\alpha_{-4}\,^{\mu} \alpha_{5}\,^{\mu}+\frac{1}{2}\alpha_{-5}\,^{\mu} \alpha_{6}\,^{\mu}+\frac{1}{2}\alpha_{-6}\,^{\mu} \alpha_{7}\,^{\mu}+\frac{1}{2}\alpha_{-7}\,^{\mu} \alpha_{8}\,^{\mu}% +\frac{1}{2}\alpha_{-8}\,^{\mu} \alpha_{9}\,^{\mu}\right) \alpha_{-1}\,^{\nu} \alpha_{-1}\,^{\nu} |k\rangle\)
We can now work out what is the action of a Virasoro operator on a state by repeatedly substituting the commutation relations of the $\alpha_{m}^{\mu}$ and simplifying, until the result no longer changes. Below, we stick that logic into a function act which you can apply on an expression.
def act(ex): converge(ex): distribute(ex) substitute(ex, rl) distribute(ex) eliminate_kronecker(ex) rename_dummies(ex) return ex
Here is what happens when you apply the commutator rules to the three terms in the state $L_1 |\chi\rangle$:
act(L1c1);
\(\displaystyle{}2\alpha_{-1}\,^{\mu} \sqrt{2} k^{\mu} |k\rangle\)
act(L1c2);
\(\displaystyle{}2\alpha_{-1}\,^{\mu} \sqrt{2} k^{\mu} |k\rangle\)
act(L1c3);
\(\displaystyle{}-4\alpha_{-1}\,^{\mu} \sqrt{2} |k\rangle k^{\mu}\)
The procedure works the same for $L_2|\chi\rangle$:
L2c1:= @(L2) @(chi1). L2c2:= @(L2) @(chi2). L2c3:= @(L2) @(chi3).
act(L2c1);
\(\displaystyle{}d |k\rangle\)
act(L2c2);
\(\displaystyle{}-4|k\rangle\)
act(L2c3);
\(\displaystyle{}-2|k\rangle\)
The constraints $L_1 |\chi\rangle=0$ and $L_2 |\chi\rangle=0$ thus reduce to
tst1:=@(L1c1) + A @(L1c2) + B @(L1c3) = 0;
\(\displaystyle{}2\alpha_{-1}\,^{\mu} \sqrt{2} k^{\mu} |k\rangle+2A \alpha_{-1}\,^{\mu} \sqrt{2} k^{\mu} |k\rangle-4B \alpha_{-1}\,^{\mu} \sqrt{2} |k\rangle k^{\mu} = 0\)
tst2:=@(L2c1) + A @(L2c2) + B @(L2c3) = 0;
\(\displaystyle{}d |k\rangle-4A |k\rangle-2B |k\rangle = 0\)
These are solved by $A=\frac{d-1}{5}$ and $B=\frac{d+4}{10}$.

Computing the norm

Having pinned down the coefficients $A$ and $B$, we now want to compute the norm of the state $|\chi\rangle$ (to be added).
Copyright © 2001-2017 Kasper Peeters
Questions? info@cadabra.science