I've seen it written many times that the commutation relation


is required for Lorentz invariance in the light cone gauge quantisation of the bosonic string. This follows immediately if we assume that the $M^{\mu\nu}$ must satisfy the Lorentz algebra.

But why does Lorentz invariance force the $M^{\mu\nu}$ to satisfy the Lorentz algebra as quantum operators?

Okay perhaps the Poisson brackets happen to work out this way. But surely they have nothing to do with Lorentz invariance. A priori the $M^{\mu\nu}$ are just a collection of $\frac{1}{2}D(D-1)$ numbers. Why should they have anything to do with Lorentz invariance?

Sorry if I'm confused on something really basic, and apologies for the rephrasing of this question. Hopefully it's clearer now, and someone will be able to answer!

Many thanks in advance.


I) What does it mean that a theory is invariant wrt. a group $G$? Well, in particular this means that there is a well-defined given prescription on how the constituents of the theory change under the action of the group. Often in physics (but not always), it happens that the group action is linearly realized, i.e. the fields, the matrix elements, and other objects form linear representations of the group $G$. Let us for simplicity assume this in what follows. See also e.g. this and this Phys.SE post.

II) In our case the group $G$ is the Lorentz Lie group $O(D-1,1)$. This is the set of real $D\times D$ matrices $\Lambda\in {\rm Mat}_{D\times D}(\mathbb{R})$, such that

$$ \Lambda^t \eta \Lambda ~=~\eta.$$

Here $\eta_{\mu\nu}$ is the Minkowski metric in $D$ spacetime dimensions. Thus we want to study a linear representation $\rho$ of the Lorentz Lie group. Note that the representation $\rho(\Lambda)$ of a Lorentz matrix $\Lambda$ is typically not a $D\times D$ matrix. It could be an operator. Here the representation (map) $\rho$ is a Lie group homomorphism.

III) Then, at the infinitesimal level, the $D\times D$ Lorentz matrices

$$\Lambda~\in~ O(D-1,1)~\subseteq ~{\rm Mat}_{D\times D}(\mathbb{R}) $$

are generated by a $D(D-1)/2$-dimensional basis of $D\times D$ matrices $J^{\mu\nu}\equiv-J^{\nu\mu}$ in the corresponding Lie algebra

$$so(D-1,1)~\subseteq ~{\rm Mat}_{D\times D}(\mathbb{R}).$$

Similarly, at the infinitesimal level, the Lie group action on a physical object (in the theory) is generated by the corresponding Lie algebra generators $M^{\mu\nu}=\rho(J^{\mu\nu})$ in the pertinent linear representation $\rho$.

As a consequence of the group action, the Lorentz Lie algebra generators $M^{\mu\nu}$ in the relevant linear representation $\rho$ have to satisfy the explicit form of the $so(D-1,1)$ Lorentz Lie-algebra.

IV) Finally, use (i) this explicit form and (ii) the fact that the Minkowski metric in light-cone coordinates is given as

$$\eta_{\mu\nu}~=~\left[ \begin{array}{cc} 0 & -1 & 0 \\-1 & 0 & 0 \\ 0 & 0 &\delta_{IJ}\end{array} \right], \qquad \mu,\nu=+,-,2,3,\ldots, D-1,$$ $$\qquad I,J=2,3,\ldots, D-1,$$

to conclude that


V) At the classical level, one may use Noether's Theorem to find the classical representation of the angular momentum generators $M^{\mu\nu}$.

Concretely how the classical bosonic string is quantized, and concretely how the corresponding quantum operators $M^{\mu\nu}$ are explicitly realized as operators, is e.g. explained in Barton Zwiebach, A First Course in String Theory, Chapter 12. The construction is non-trivial due to operator ordering issues.

  • 2
    $\begingroup$ very concise! Good job @Qmechanic $\endgroup$ – Dylan O. Sabulsky Nov 20 '12 at 1:37
  • $\begingroup$ @Qmechanic - thanks for your answer. Maybe I wasn't clear enough. I understand why the Lorentz Lie Algebra has that commutation relation property. I just don't see why it has to be satisfied for the theory to be Lorentz invariant. Could you possibly explain why a Lorentz invariant theory has to have Lorentz generators which satisfy the Lorentz algebra? $\endgroup$ – Edward Hughes Nov 20 '12 at 13:00
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    $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Nov 20 '12 at 18:06
  • $\begingroup$ Thank you @Qmechanic. Sorry to be so demanding but just one more question. How do you know that the generators $M^{\mu\nu}$ are precisely the charges you get from Noether's theorem? This is the point I'm really struggling to understand! Have upvoted and will accept if you have an explanation. Very many thanks once again. $\endgroup$ – Edward Hughes Nov 20 '12 at 22:00
  • $\begingroup$ Better phrased: why does the classical Noether charge turn into a quantum generator? Is there some version of Noether's theorem that will give me a proof of this? Or is it an axiom? Thanks once again. $\endgroup$ – Edward Hughes Nov 20 '12 at 22:08

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