I'm reading about correlators in string theory for amplitude computations. More specifically formulae 7.1.53 and 7.1.54 in Green Schwarz Witten. I don't see how they can be derived.

E.g. 7.1.53 is $$g\langle 0;k_1|V_0(k_2)|0;k_3\rangle = g$$

with $V_0(k)=Z_0(k)W_0(k)$

with $$Z_0=\exp(ikx)z^{k\cdot p +1}=z^{k\cdot p -1}\exp(ikx)$$ and $$W_0=\exp(k\cdot \sum_{n=1}^{\infty}\frac{1}{n}\alpha_{-n}z^n)\exp(-k\cdot \sum_{n=1}^{\infty}\frac{1}{n}\alpha_{n}z^{-n})$$

I obviously tried plugging $Z_0$ into the equation above. Then the second term of $W_0$ immediatly give 1 (I think) so that I'm left with

$$ gz^{k_2k_1-1}\langle0;k_1|exp(ik_2x)\exp(k\cdot \sum_{n=1}^{\infty}\frac{1}{n}\alpha_{-n}z^n)|0;k_3\rangle$$

How do I continue from here? I don't see it...?


1 Answer 1


This is simply proportional to the 3-tachyon correlation function. They warn you above the equation that the momentum conservation factor such as $$(2\pi)^D\delta^{(D)}(k_1-k_2-k_3)$$ is omitted everywhere and I am not even sure whether their normalization of the states includes the power of $2\pi$ factor. Pick Polchinski's textbook if you want every formula to be much more robust and reliable about similar details. At any rate, this delta function is what you get from all the factors of the form $\exp(ik_j\cdot x)$ for $j=1,2,3$.

You got the first power of $g$ which is a good thing, a part of the result.

One of the two last excessive things you got in your calculation is the exponential of $\sum \alpha_{-n}z^n/n$. But this exponential may simply be replaced by $1$ because all the creation operators $\alpha_{-n}$ for positive $n$ annihilate the ket vector $\langle 0;k_1|$ on the left side – the Hermitian conjugate claim to the claim that annihilation operators $\alpha_{n}$ for a positive $n$ annihilate the ket vectors on the right. So from the Taylor expansion for the exponential, only the leading term $1$ survives.

That's great and the only excessive factor you're left with is $z^{k_2 k_1-1}$. But this is also equal to $1$ because the exponent vanishes when all the physical conditions are satisfied. Note that $$k_1\cdot k_2 = \frac 12[(k_1+k_2)^2 - k_1^2-k_2^2] =\frac 12( k_3^2-k_1^2-k_2^2) $$ where, in the second $=$ step, I used the momentum conservation because everything is multiplied by the delta-function for the sum of momenta, anyway. However, the calculation is also meaningful for on-shell $k_1,k_2,k_3$ only. But the squared masses of the open-string tachyon is $$-k^2=m_T^2=-\alpha'$$ where the minus sign in front of $k^2$ arises because they use the mostly-plus convention for the metric. So in the units $\alpha'=2$ they selectively use for open strings (one has $\alpha'=1/2$ for closed strings, to make things more confusing), $$k_1\cdot k_2 = \frac 12(k_3^2-k_1^2 - k_2^2) =-\frac 12 (-2+1)m_T^2 = \frac 12\alpha' =1$$ so the exponent above $z$ is actually $k_1\cdot k_2-1=0$ and all the factors except for $g$ are equal to one. Note that the string amplitudes are on-shell (scattering amplitudes) and the calculations only simplify when the on-shell conditions are imposed. In fact, the general string amplitudes don't have any natural or canonical extension for off-shell momenta (although, of course, when you write down an effective low-energy field theory for string theory, that theory gives you formulae for the off-shell amplitudes, too)!

If you want a more pedagogical treatment that doesn't use these "inconsistent" conventions for $\alpha'=1/2$ or $2$, doesn't omit the momentum-conservation delta-functions, and is more explicit about the moments when the on-shell conditions are used and how, try e.g. Polchinski's book or one of the newer competitors. On the other hand, if you start to calculate as many amplitudes as Green and Schwarz did in the early 1980s, it may be pretty helpful to use all the "seemingly sloppy" simplifications in the notation that still capture all the physical essence.

  • 1
    $\begingroup$ You're welcome. Now I see some sign error around the $(-2+1)$ because I do seem to have forgotten that $k^2=-m^2$ in these conventions but I will leave the search to you... ;-) Update, no, I forgot one more sign, now fixed it, and things are OK now. $\endgroup$ Commented Mar 26, 2013 at 12:58

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