Existence and uniqueness of spurious states in bosonic string theory

Question

Green, Schwarz and Witten introduce so called spurious states $\phi$, that fulfill $$ (L_0 -a )\phi = 0\quad \text{and} \quad (\phi,\psi)=0 \text{ for all physical states }\psi $$ where the $L_n$ are the Virasoro generators, that can be expressed through the exitation creation/annihilation operators of the String: $$ L_m=\frac{1}{2} \sum_{-\infty}^\infty \alpha_{m-n} \cdot \alpha_n $$ Now for $m=0$ this has an odering ambiguity which is resolved by defining $$ L_0=\frac{1}{2} \alpha_0^2 + \sum_{n=1}^\infty \alpha_{-n} \cdot \alpha_n $$ and adding this parameter $a$, which is just a c-number, in every formula containing an $L_0$.

The suprious states however can be written as $$ \phi = \sum L^\dagger_n ~\chi_n = \sum L_{-n} ~\chi_n $$ with $$ (L_0 -a +n) \chi_n = 0 $$

Does anybody know whether the $\chi_n$ are unique and how do I know that such states exist in the first place?

Answer 1

A state, |$\phi$>, is said to be spurious if it satisfies the mass-shell condition,

$$(L_{0}-a)|\phi>=0 $$

and is orthogonal to all other physical states,

$$<\phi|\psi> = 0, \ ∀\ physical\ states\ |\psi>$$

In general, a spurious state can be written as

$$|\phi> = \sum_{n=1}L_{-n}|\chi_{n}>$$

where $|\chi_{n}>$ is some state which satisfies the, now modified, mass-shell condition given by

$$(L_{0}-a+n)|\chi_{n}>=0 $$

-Proving the equation above:

Following from the definition of a spurious state, since if

$$<\phi|\psi>=0$$

then

$$\sum_{n=1}L_{0}L_{-n}|\chi_{n}>- a\sum_{n=1}L_{-n}|\chi_{n}>=0$$ $$\sum_{n=1}([L_{0},L_{-n}]+L_{-n}L_{0})|\chi_{n}>- a\sum_{n=1}L_{-n}|\chi_{n}>=0$$ $$\sum_{n=1}(nL_{-n}+L_{-n}L_{0})|\chi_{n}>- a\sum_{n=1}L_{-n}|\chi_{n}>=0$$ $$\sum_{n=1}(nL_{-n}+L_{-n}L_{0}- aL_{-n})|\chi_{n}>=0$$ $$\sum_{n=1}L_{-n}(L_{0}- a+ n)|\chi_{n}>=0$$

which holds for all states, and thus

$$(L_{0}- a+ n)|\chi_{n}>=0$$

Answer 2

Classically, the negative energy eigenvalues of the string hamiltonian $L_{0}-a$ are the unwanted states on the string spectrum, it is to say, all the states $\chi_{n}$ which obey $$(L_{0}-a)\chi_{n}=-n\chi_{n}.$$

The key observation is that all the spurious states are organized in the confomal families to which the states $\chi_{n}$ are the highest-weight representants. A general spurious state then defined as a sum of Virasoro descendants of the states $\chi_{n}$ for a finite set of integers $n$.

Let's verify that if $\chi_{n}$ verify the negative mass-shell condition, then the negative Virasoro descentant $L_{-n}\chi_{n}$ satisfy the condition of being spurious.

Proof: Physical states should obey the condition of being highest weight representations of the Virasoro algebra. In particular, if $\psi$ is physical then for all $n>0$ $$L_{n}\psi=0.$$

Now recalling that if $\psi$ is physical and $\chi_{n}$ spurious then $$(\chi_{n},\psi)=0,$$ in particular the vaccum $L_{n}\psi=0$ is physical, $$(\chi_{n},L_{n}\psi)=(\chi_{n},L_{n}\psi)^{*}=0$$ where the superscript $*$ denotes complex conjugation. Now it's easy to recognize that if $$(\chi_{n},L_{n}\psi)^{*}=(\psi,L_{n}^{\dagger}\chi_{n})=(\psi,L_{-n}\chi_{n})=0$$ with $\psi$ an arbitary physical state, then it follows that $L_{-n}\chi_{n}$ is spurious.

Now it's easy to verify that $\phi=\sum_{n} L_{-n}\chi_{n}$ where $n$ runs over a finite subset of the integers is spurious.

How do you we now that the states $\chi_{n}$ exist? That's because the operator equation $(L_{0}-a)\chi_{n}=-n\chi_{n}$ can be solved in a given representation because any linear equation can be solved in principle.

Are the states $\chi_{n}$ unique? the answer is no, you could verify that if $\chi_{n}$ obey the negative mass-shell condition then $\chi_{n} + \phi$ with $\phi$ spurious also does.