# Mass-mismatch in the low-lying states of the bosonic string

$$\newcommand{\ket}[1]{\left|#1\right>}$$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...

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).