Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I'm reading through Seiberg and Witten's paper "String Theory and Noncommutative Geometry," and one part in $\S$2.1 isn't quite clear to me. (Sorry, in advance, for the length.)

My question is about equation (2.7) on page 9, where they say:

The coefficient $\theta^{ij}$ in the propagator also has a simple intuitive interpretation, suggested in [15]. In conformal field theory, one can compute commutators of operators from the short distance behavior of operator products by interpreting time ordering as operator ordering. Interpreting $\tau$ as time, we see that

$$ [x^i(\tau),x^j(\tau)]=T(x^i(\tau)x^j(\tau^-)-x^i(\tau)x^j(\tau^+))=i\theta^{ij}. \tag{2.7} $$


Question: How did they get the first equality?

I'm sure that there is probably some really simple argument I'm not seeing, but, as this equation is the crucial step in which noncommutativity of coordinates appears, I figure I should probably understand it. Any help?

The context is: consider a flat space (with metric $g_{ij}$) in the presence of a constant NS $B$-field with $Dp$-branes. Further, for simplicity, they assume that $B_{0i}=0$, $B_{ij}\neq 0$ for $i,j=1,\cdots,r$, and $g_{ij}=0$ for $i=1,\cdots,r$ and $j\neq 1,\cdots,r$, where $r$ is the rank of the matrix $B_{ij}$. Then one can obtain the action

$$ S=\frac{1}{4\pi\alpha'}\int_{\Sigma}\big(g_{ij}\partial_ax^i\partial^ax^j-2\pi i\alpha'B_{ij}\epsilon^{ab}\partial_ax^i\partial_bx^j\big) $$

(with $\Sigma$: the string worldsheet), which leads to the boundary conditions (along the $Dp$-branes): $g_{ij}\partial_nx^j+2\pi i\alpha'B_{ij}\partial_tx^j\big|_{\partial\Sigma}=0$ (where $n$ is the normal to $\partial\Sigma$).

Taking $\Sigma$ to be a disc, these become:

$$ g_{ij}(\partial-\bar{\partial})x^j+2\pi\alpha' B_{ij}(\partial+\bar{\partial})x^j\big|_{z=\bar{z}}=0. $$

Then, it was shown elsewhere that the propagator can be written:

$$ \langle x^i(z)x^j(z') \rangle = -\alpha'\bigg[g^{ij}\log|z-z'|-g^{ij}\log|z-\bar{z}'|+G^{ij}\log|z-\bar{z}'|^2 $$ $$ \hspace{3.4in}+\frac{1}{2\pi\alpha'}\theta^{ij}\log\frac{z-\bar{z}'}{\bar{z}-z'}+D^{ij}\bigg], $$

where $G_{ij}$ is the open string metric given by

$$ G_{ij}=g_{ij}-(2\pi\alpha')^2(Bg^{-1}B)_{ij} $$

so that

$$ G^{ij}= \bigg( \frac{1}{g+2\pi\alpha'B}g\frac{1}{g-2\pi\alpha'B} \bigg)^{ij}, $$

the constants $D^{ij}$ depend only on $B$ and "play no essential role and can be set to a convenient value," and

$$ \theta^{ij}=-(2\pi\alpha')^2\bigg( \frac{1}{g+2\pi\alpha'B}B\frac{1}{g-2\pi\alpha'B} \bigg)^{ij}. $$

Finally, they focus on open string vertex operators, restrict to real $z$ and $z'$ (which they then denote by $\tau$ and $\tau'$), and make a convenient choice of $D^{ij}$ so that the above propagator reduces to

$$ \langle x^i(\tau)x^j(\tau')\rangle = -\alpha'G^{ij}\log(\tau-\tau')^2+\frac{i}{2}\theta^{ij}\epsilon(\tau-\tau'). $$

Then the above claim about the commutator is made.

share|improve this question
I think it is generally true that $T(A(t)B(t-\epsilon)-A(t)B(t+\epsilon))=A(t)B(t)-B(t)A(t)=[A(t),B(t)]$. Here $\epsilon$ is an infinitesimally small positive number. This is a common trick used to obtain (the expectation value of) equal time commutators. –  Isidore Seville Jan 9 '14 at 4:32
Ah yes, just by definition! That was easy. It seems that I have rather selective memory when it comes to QFT. :D (which isn't surprising, as I actually study math, and only have a passing familiarity with it… perhaps I should fix that.) –  Ralph Mellish Jan 9 '14 at 4:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.