# Question in String Theory / Mass of States / Number Operator

1. The problem statement, all variables and given/known data

I have the following definition of the space-time coordinates

1. Relevant equations

Working in a certain gauge we can also do:

From which we can find:

Where $$N_{lc}$$ sums over the transverse oscillation modes only.

1. The attempt at a solution

MY QUESTION:

I don't understand the RHS of the first excited state, how does $$N$$ take integer values? what exactly are the $$\alpha$$?

So (here the summation is over all modes, whereas above it is over the transvere oscillation modes only but ignoring this)

so for the first excited state do we have all alpha modes are zero except $$\alpha_{-1}$$ ?

So Then $$N= \alpha^{j}_{-1}$$ ? how is this $$N=1$$?

So for the next excited state I expect:

$$m^2 \alpha_{-1}^i \alpha_{-1}^j |p^i\rangle =m^2\alpha^{j}_{-2}|p^i\rangle=(2-a) \alpha_{-1}^i \alpha_{-1}^j |p^i\rangle=(2-a) \alpha_{-2}^j |p^i\rangle$$ right?

And I am told in my notes that this can be achieved by either $$\alpha_{-1}^i \alpha_{-1}^j |p^i\rangle$$ or $$\alpha_{-2}^i$$

How does this get $$N=2$$? since, well looking at $$\alpha_{-1}^i \alpha_{-1}^j$$, the sum in $$N$$ is over $$n$$ not $$i$$ or $$j$$ so am I looking at two different number operators here?

For the other expression So Then $$N= \alpha^{j}_{-2}$$ ? how is this $$N=2$$?

I'm just really confused as you can tel..

Not sure why you didn't ask this on Physics SE...

First off, calling the ground state $\vert p^i\rangle$ is bad notation since it makes it seem like $i$ is a free index here and there is a vector of states we're talking about (gets worse when you hit it with $\alpha^i$). How about $\vert\vec p\rangle$ instead?

I don't understand the RHS of the first excited state, how does N take integer values?

It's an operator whose eigenvalues are integers, so when it is applied to an eigenstate it becomes an integer. It's a sum of number operators for the individual normal modes of the string.

What are the $\alpha$?

Raising and lowering operators descending from the classical normal modes (i.e. Fourier coefficients) of the string. For positive $n,$ $\alpha_{-n}$ are raising operators and $\alpha_{n}$ are lowering. $\alpha_n^{i}$ is the lowering operator corresponding to the $n$-th mode in the $i$-th transverse direction. Like in the quantum mechanics of a single harmonic oscillator, the eigenstates of the number operators $\alpha^i_{-n}\alpha^i_n$ are the ground state with some number of raising operators applied to it.

so for the first excited state do we have all alpha modes are zero except $\alpha_{-1}$

Yes, basically, though I don't see how it connects to what you said just above. There are many first excited states but they are all linear combinations of states of the form $\alpha_{-1}^i\vert \vec p \rangle$ where $\vert \vec p\rangle$ is the ground state. Note we read this as 'the state with one quantum of the first mode in the $i-$th direction."

So Then $N= \alpha^{j}_{-1}$ ? how is this $N=1$?

No idea what you mean here, but hopefully what I said above clarified. $N$ is not equal to $\alpha^{j}_{-1},$ it is equal to $\sum_{i,n} \alpha_{-n}^i\alpha_{n}^i.$ It becomes one when applied to the eigenstate $\alpha_{-1}^j\vert \vec p \rangle$ because $\alpha_{-1}^j$ commutes with every lowering operator $\alpha_{n}$ except $\alpha_{1}^j$ so they all pass through it to annihilate the vaccuum. So the only term that matters is the one, and using the commutation relation $[\alpha^i_{n},\alpha^j_{-n}]=n\delta_{i j},$ we have $$N\alpha_{-1}^j\vert \vec p \rangle=\alpha_{-1}^j\alpha_{1}^j\alpha_{-1}^j\vert \vec p \rangle = \alpha_{-1}^j(1+\alpha_{-1}^j\alpha_{1}^j)\vert \vec p \rangle = \alpha_{-1}^j\vert \vec p \rangle$$ so $\alpha_{-1}^j\vert \vec p \rangle$ is an eigenvector of $N$ with eigenvalue one.

So for the next excited state I expect: $m^2 \alpha_{-1}^i \alpha_{-1}^j |p^i>=m^2\alpha^{j}_{-2}|p^i>=(2-a) \alpha_{-1}^i \alpha_{-1}^j |p^i>=(2-a) \alpha_{-2}^j |p^i>$ right?

All but the last equation. $\alpha_{-1}^i\alpha_{-1}^j \vert \vec p \rangle$ is not the same state as $\alpha_{-2}^j\vert \vec p \rangle$ although they have the same energy. The first has an excitation in the first mode in the $i$ direction and the first mode in the $j$ direction while the second has one excitation in the second mode in the $j$ direction. These are very different things even though they happen to have the same energy.

How does this get $N=2$? since, well looking at $\alpha_{-1}^i \alpha_{-1}^j$, the sum in $N$ is over $n$ not $i$ or $j$ so am I looking at two different number operators here?

Don't understand this either but I expect it's an offshoot of your earlier confusion. Again, the state $\alpha_{-1}^i \alpha_{-1}^j \vert \vec p \rangle$ is an eigenstate of $N$ with eigenvalue $2.$ The sum in the definition of $N$ is over both direction and normal mode indices. It represents the sum (weighted by $n$) of the quanta in all normal modes in all transverse directions. Thus the sum over $i$ in my expression above (perhaps this is obscured in your reference by the Einstein summation convention).

For the other expression So Then $N= \alpha^{j}_{-2}$ ? how is this $N=2$?

Even though as I mentioned before, $\alpha_{-2}^j\vert\vec p\rangle$ is not the same (or even a very similar) state as $\alpha_{-1}^i\alpha_{-1}^j\vert\vec p\rangle,$ they both happen to be eigenstates of $N$ with eigenvalue $2$ (this is 'accidental degeneracy,' if you're familiar). Since normal mode $n=2$ has twice the frequency, it has double the energy of the $n=1$ mode, so a state with one excitation of one of the $n=2$ modes has the same energy as a state with two excitations of (possibly but not necessarily different) $n=1$ modes. This can be confirmed by short calculations similar to the one I did above for $N=1.$