Malament's theorem posits that, given a few assumptions on relativistic QM, it is impossible to have localized particles. For $E_\Delta$ the proposition that a particle is certain to be found within a bounded region $\Delta$ of space, the assumptions are

  • Localizability : If $\Delta$ and $\Delta'$ are disjoint regions, then $E_\Delta E_{\Delta'} = 0$ (the particle can only be found in one region)
  • Translation covariance : For some translation by $a$, we have $U(a) E_\Delta U(a)^\dagger = E_{\Delta + a}$
  • Energy bounded from below
  • Microcausality : if $\Delta$ and $\Delta'$ are disjoint regions, separated by a non-zero distance, then there is an $\varepsilon$ such that $[E_\Delta ,E_{\Delta' + ta}] = 0, t\in [0,\varepsilon)$ for all translations $a$.

With those conditions, $\forall \Delta, E_\Delta = 0 $. There is no bounded region in which we can say that a particle can be detected. This is akin to forbidding a position operator in QFT, requiring the wavefunctionals to depend on distributions rather than positions.

On the other hand, the quantization of the basic string action (such as in the first chapter of Polchinski) seems to be pretty much that. I know that the situation in actual string theory is more complicated, but for the basic quantization of the Polyakov string, we seem to be able to get quantum states with measurements of the string's presence in a specific region.

For instance, in the Polyakov theory, we have the operator $\hat X^i(\sigma)$, which, as far as I can tell, associates a point in $\mathbb R^{d-1}$ for every value of $\sigma$. Then we can have a state with initial condition at $\tau = 0$ such that, outside of a compact support $\Delta$, $\hat X^i(\sigma) \vert \omega \rangle = 0$. Not quite sure if the mathematical details are totally exact, since it's not quite directly a position operator, but I get the impression that we can define a state where the "string" should be limited to a compact region.

Am I misunderstanding the string's quantization here, is one of Malament's assumption broken or is it an approximation that does not withstand closer scrutiny?

  • 2
    $\begingroup$ It would help if you explicitly wrote down where you think string theory violates what you've written - and in particular, what $E_\Delta$ is supposed to be. You say that the quantization seems to be "pretty much that", but I don't see at all where in string theory such an operator would arise. $\endgroup$
    – ACuriousMind
    Jun 6 '16 at 17:42
  • $\begingroup$ There is something missing in your description of microcausality. The condition following "such that" has no verb. $\endgroup$
    – WillO
    Jun 6 '16 at 19:14
  • $\begingroup$ Related question by OP: physics.stackexchange.com/q/337039/2451 $\endgroup$
    – Qmechanic
    Jan 4 '18 at 21:28

I do not know almost anything about string theory, but I can say something from the general quantum theory viewpoint.

First of all you stated Malament's theorem hypotheses into a not very precise form. The sets $\Delta$ are assumed to be subsets of a 3D spacelike surface $\Sigma$ (the rest space of an observer) whereas $a$ in the third requirement is a four dimensional vector, so the statement includes both spatial (along $\Sigma$) and temporal (normal to $\Sigma$) translations. Regarding the last requirement $a$ is supposed to be timelike.

Well, already in non-relativistic quantum theories these hypotheses are violated. The problem, the same as for relativistic theories concerns the last requirement and the temporal case ($a$ timelike) of second requirement. Omitting these requirements everything goes right. In non-relativistic theories you have the standard position operators for a particle, in relativistic theories you have the theory of Newton-Wigner position operators relying on Mackay's imprimitivity theory.

The reason for the failure of the second part of axiom II (I mean $a$ timelike) and axiom IV is physically clear in my opinion. For every natural definition of position operator $X_k$ in a rest $3$-space we must have either $$U^*_t X_k U_t = X_k + t P_k/m\:,$$ or $$U^*_t X_k U_t = X_k + t P_k/P^0\:,$$ in the non-relativistic or relativistic case respectively. As $X_k$ and $P_k$ satisfy the standard canonical commutation relations (CCRs), $U^*_t X_k U_t$ and $X_k$ do not commmute. A closer scrutiny proves that this non commutativity remains if focusing on the spectral measures: The spectral measure of $U^*_t X_k U_t$ is non-local respect to the one of $X_k$ even for arbitrarily small $t$. This is nothing the operatorial version of the fact that a localized free wavefunction at $t=0$ ceases to be localized for every $t>0$ arbitrarily small and relies on the CCRs.

Maybe a more physically sound approach consists of imposing a notion of spacetime localization. In other words replacing the spatial regions $\Delta$ for spacetime regions. In this case projection valued measures cannot be exploited (think of Pauli's theorem for the non-existence of time operator), but some result can be grasped replacing PVMs for POVMs.


The string action $$ S_p = -T\int d\tau d\sigma (-\gamma)\gamma^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu, $$ or Poyakov action, is evaluated in a path integral $Z = \int{\cal D}[X]e^{-iS}$, or defines states $$ |\psi\rangle = \int{\cal D}[X]e^{-iS}|n\rangle $$ on a Fock basis. In the light cone gauge, $X^\pm = \frac{1}{2}(X^0\pm X^i)$ the Hamiltonian for canonical momenta $p_\mu = \partial L/\partial\dot X$ is $$ H = p_-\dot X^- + p_iX^i - L $$ such that $X^+$ is gauged away. The action is then $$ S = \frac{1}{2}\int d\sigma p_-(2\dot X^- - \dot X^i\dot X^i) $$ The momentum $p_-$ is not independent and is a constraint on the light cone with $p_+ = -p_-$. The $p_-\dot X^-$ is the bare action that corresponds to a units of action that in quantization leads to $[p_-, X_-] = i$.

So far we have only derivatives of position, or the string position. The quantization of this system is over the momentum eigenstates $|p_-,~p^i\rangle$. The string action is set up with the idea of a string world sheet, but the actual form of the action is such that we are interested in the momentum states. As such there is nothing here explicitly that violates Malament's theorem.

  • 1
    $\begingroup$ You just wrote down some generalities about string theory, but how do you address the question? $\endgroup$
    – ACuriousMind
    Jun 6 '16 at 20:02
  • $\begingroup$ As I say at the end, this is all in a momentum representation. $\endgroup$ Jun 6 '16 at 22:08

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