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

I have the following definition of the space-time coordinates

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  1. Relevant equations

Working in a certain gauge we can also do:

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From which we can find:

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

  1. The attempt at a solution

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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)

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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..

Many thanks for your help in advance

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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 $N=1.$

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