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.

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.