I am working through David Tong's lecture notes on string theory. The bit I am stuck on is page 41/42 in Chapter 2. The PDF file is here.

In 2.3.2 "The First Excited States" on page 41/42, he says:

We now look at the first excited states. If we act [on the vacuum state of the string] with a creation operator $\alpha^j_{-1}$, then the level matching condition (2.25) tells us that we also need to act with a $\tilde{\alpha}^i_{-1}$ operator.

The equation he is referring to is:

$$M^2 = \frac{4}{\alpha'}(\sum_{i=0}^{D-2}\alpha_{-n}^i\alpha_{n}^i -a) = \frac{4}{\alpha'}(\sum_{i=0}^{D-2}\tilde{\alpha}_{-n}^i\tilde{\alpha}_{n}^i -a).$$

The problem is, I don't understand how this tells us that we also need to act with a $\tilde{\alpha}^i_{-1}$ operator.


Define $$ N \equiv \sum_{i=0}^{D-2} \alpha_{-n}^i \alpha_n^i,~~~~{\tilde N} \equiv \sum_{i=0}^{D-2} {\tilde \alpha}_{-n}^i {\tilde \alpha}_n^i $$ The formula you have written tells us un particular that for any state, we must have $N = {\tilde N}$. This condition is known as the level-matching condition. The state $\alpha^i_{-1} |0;k\rangle$ has $N = 1$ but ${\tilde N} = 0$ (verify this). Thus, $\alpha^i_{-1} |0;k\rangle$ is not a physical state. The first excited state must have $N = {\tilde N} = 1$. The only way to have this is to have the state $\alpha^i_{-1} \alpha^j_{-1} |0;k\rangle$.


If one calculates expectation value $\langle\Psi| M^2|\Psi\rangle$ of the mass-square operator $M^2$ of the state $|\Psi\rangle~=~ \alpha^j_{-1}| 0; p\rangle $, then the level-matching condition (2.25) would be violated.


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