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$\newcommand{\ket}[1]{\left|#1\right>}$I've recently been studying the quantisation of the bosonic string (using GSW as my main text). However, I have some issues which I believe should be pretty straightforward, but I can't seem to work out!

We derive the spectrum of states of (open, for example) strings. The masses of these states are given by the mass operator $M^2 = 2(N-a) = 2(N-1)$ setting $\ell = 1$, $\alpha' = \frac{1}{2}$, and with $N$ the 'level/number' operator. We act with this on the space of states generated by the sequential application of any of $\alpha^i_{-n}$ onto the ground state $\ket{0;k^\mu}$ (which we understand to exist from something to do with the Stone-Weierstrass theorem). This is the tachyonic ground state.

This implies that (in light-cone gauge):

$M^2\ket{0;k^\mu} = -2\ket{0;k^\mu}$

This means that $k^\mu k_\mu=2$! That would be fine - but now I apply $\alpha^i_{-1}$ to my (string) vacuum state and create a level $N=1$ massless photon-like particle. The momentum operator commutes with the above creation operator, and so the state still has the same 26-momentum. However, it should also now have mass $M^2=0$. I don't see how these two statements can be coherent. Is it simply that if one attempts to create a state with the incorrect 26-momentum for its mass, the result is unphysical and so decouples? I have found neither hide nor hair of such a result or argument in GSW, Tong, etc.

What am I missing? Thanks very much...

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You've demonstrated that the creation operator $\alpha^i_{-1}$ can map a physical state to a non-physical one. There is no contradiction. If you've seen the BRST approach yet, a succinct way to phrase this is that the action of the oscillators does not preserve the subspace of BRST-cohomology classes (i.e. physical states).

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