We now look to the LC quantisation of the string.

The first thing to note is that the commutation relation translates on the coefficients of the Fourier mode expansion in such a way that: you must substitute in the F-mode expansion for $X^{i}$, $prod^{i}$, and also for the delta function. This calculation is quite involved.

In the meantime, recall from Polchinski (p.20):

[ [X^{-}, P^{+}] = -i eta^{-+} = -i ]

[ [X^{i}(sigma), prod^{j} (sigma^{prime})] = idelta^{ij} delta(sigma – sigma^{prime}) ]

With the latter, we have the commutation relations for zero-mode coefficients plus higher harmonics. We can focus on this a bit more, noting that:

[ [X^{i}, frac{P^{+}}{l} frac{P^{i}}{P^{+}} l] = idelta^{ij} = 1 ]

Here $X^{i}$ is the zero-mode. We then see that from an infinite set of Fourier harmonics (in which $tau$ dependence remains isolated), everything essentially cancels and we are left with what follows:

[ [X^{i}, P^{i}] = idelta^{ij} rightarrow a point particle ]

And, if you work through all modes and not just the zero-mode, note that you will discover the following identity:

[ [a_{m}^{i}, a_{n}^{i}] = mdelta^{ij}delta_{m, -n} ]

Again, the computation is quite involved and will not be included in this text. In essence, however, one should understand that this identity is representative of the simple harmonic oscillator (SHO) algebra. Moreover, we can show the SHO algebra by, for example, fixing $m$. Notice,

[ a_{n}^{i} = sqrt{n}a ]

[ a_{n}^{i} = sqrt{n}a^{+} ]

Where $[a, a^{+}] = 1$, $n = 1, …, infty$, and $i = 2, …, D-1$. par

A few extra comments are necessary. In short, it turns out to be the case that the Polyakov string is:

1) Based on centre of mass motion;

2) Comprised of an infinite set of harmonic oscillators.

Indeed, and it may be helpful to at least write this explicitly. To do so we should note: if $m,n$ have opposite signs in the identity $[alpha_{m}^{I}, alpha_{-n}^{J}] = mdelta_{m,n}eta^{IJ}$, then the right-hand becomes zero.

In the case where $m>0$ and $n<0$, then $-n = n^{prime} > 0$. And so you will find, $[alpha_{m}^{I}, alpha_{n^{prime}}^{J}] = 0$. For the case where $m<0$ and $n>0$, $-m = m^{prime} > 0$. So, we have $[(alpha_{m}^{I})^{dagger}, alpha_{n}^{J})^{dagger}] = 0$. This means the $a$ and $a^{dagger}$ terms commute.

Now, when $m, n$ are both positive: $[sqrt{m}alpha_{m}^{I}, sqrt{n} (alpha_{m}^{J})^{dagger}] = mdelta_{m,n}eta^{IJ}$. Therefore,

[ [alpha_{m}^{I}, (alpha_{m}^{J})^{dagger}] = frac{m}{sqrt{mn}} delta_{m,n} eta^{IJ} = delta_{m,n}eta^{IJ} = delta_{m,n}delta^{IJ} ]

So we see very clearly that what we have is indeed an infinite set of perfectly normalised harmonic oscillators.

At this point, too, everything has been discretised.

In the next post, we’ll proceed deeper into an analysis of the quantised free bosonic string, where we will study the string mass. We will also come across one of the truly great early results in bosonic string theory, in which, to give a hint, a discussion of the Riemann-zeta function will come directly into focus.

References

Joseph Polchinski. (2005). “String Theory: An Introduction to the Bosonic String“, Vol. 1.

Timo Weigand. (2015/16). “Introduction to String Theory” [lecture notes].