4
$\begingroup$

Currently I'm trying to understand string theory in the light cone quantization. I just have had a look into Polchinski (Vol. 1, Introduction to the bosonic string), because – as far as I could see – GSW doesn't cover Vertex Operators in the light cone formulation (correction appreciated).

There he constructs (almost) general states on page 21 (eq. 1.3.28) via raising operators acting on Fock space vacuum states $\left|0;k \right\rangle$ with $k = (k^+, k^i)$. A few sentences later he introduces the state $\left| 0;0 \right\rangle$ for the ”ground state of a single string with zero momentum“. I think this state does not even exist as $$p^- = \frac{p^i p^i + m^2}{2 p^+}$$ diverges. This can equivalently be concluded by the fact that $\left|0;k \right\rangle$ (with arbitrary $k$) is a tachyonic state which can never have zero momentum.

This is problematic as this state is heavily used e.g. in chapter 2.8 (eq. 2.8.2 ff.).

Is there anything wrong about my arguments?

$\endgroup$
2
$\begingroup$

Comments to the question (v1):

  1. We will not discuss tachyonic states here, because they are pathological and signal an instability of the theory. Then $$\tag{1} p^{\pm}~\equiv~\frac{p^0 \pm p^1}{\sqrt{2}}~\geq~0 $$ is manifestly non-negative, since the energy $p^0\geq |p^1|$.

  2. In the light-cone formalism $p^{+}>0$ is strictly positive, since the special case $p^{+}=0$ is by definition regularized away. For the record, Ref. 1 (Ref. 2) mention that $p^{+}$ is positive on the bottom of p. 187 (on the top of p. 20), respectively.

  3. A Fock space of definite momentum is built from a ket |$N; p^+, \vec{p}_T \rangle$, where $\vec{p}_T$ is the center-of-mass transverse momentum, and $N$ are oscillator modes. Note that $p^{+}$ is often not written explicitly in the ket state notation $| 0;0\rangle$.

  4. The momenta $\hat{p}^{+}$ and $\hat{p}^I_T$ are constants of motion. In first quantized string theory [as opposed to string field theory, where strings can interact], the momentum $p^{+}$ can be assumed to be a fixed (but arbitrary) positive constant.

  5. Ref. 1 chooses light-cone gauge $$\tag{2} \hat{X}^{+}~=~2\alpha^{\prime}\hat{p}^{+}\tau,$$ and identifies $$\tag{3}\hat{H}~=~2\alpha^{\prime}\hat{p}^{+}\hat{p}^{-}$$ with the Hamiltonian, see Ref. 1 p. 238.

  6. Note that Ref. 2 at first identifies $\hat{p}^{-}$ with the Hamiltonian, cf. e.g. eqs. (1.3.6) and (1.3.30), while Ref. 2. in later chapters (implicitly?) uses the same convention as Ref. 1.

  7. Note that light-cone time $\tau$ is a parameter; not an operator. As a consequence, the CCRs
    $$\tag{4}[\tau,\hat{H}]~=~0\quad\text{and}\quad[\hat{X}^{+},\hat{p}^{-}]~=~0.$$ This is similar to ordinary quantum mechanics, cf. Pauli's objection to a time operator, see e.g. Ref. 1 p. 221; and this, this and this Phys.SE posts.

  8. On the other hand, the CCR $$\tag{5} [\hat{X}^{-},\hat{p}^{+}]~=~i\hbar {\bf 1}$$ is implemented in the light-cone formalism. Therefore the operator $e^{-ia\hat{X}^{-}/\hbar}$ becomes an intertwining operator between Fock spaces with different values of $p^{+}\to p^{+}+a$. Again, Fock spaces with non-positive $p^{+}\leq 0$ are dismissed in the light-cone formalism.

References:

  1. B. Zwiebach, A first course in String Theory, 2nd edition, 2009; p. 187; p. 221; p. 238.

  2. J. Polchinski, String Theory, Vol. 1, 1998; p. 21.

$\endgroup$
  • $\begingroup$ Isn't $p^+ > 0$ incompatible with any representation of the Poincaré group which transforms the momentum operators as a vector? Suppose a spatial rotation $D$ in the x-z-plane (x being a transversal direction) which turns z into -z and thus interchanges $p^+$ and $p^-$. For a state $\left| \psi \right\rangle$ with tachyonic momentum $p\left| \psi \right\rangle = (0, \ldots , 0, m)\left| \psi \right\rangle$ ($p^\pm =\pm m/\sqrt{2}$) we have $p U_D \left| \psi \right\rangle = (0, \ldots , 0, -m) \left| \psi \right\rangle$, thus $p^+ < 0$. $\endgroup$ – Florian Oppermann Feb 23 '15 at 10:33
  • $\begingroup$ Hi @Florian Oppermann. I assume that your new question in above comment is about the possible signs of $p^{+}$, and not so much the singular case $p^{+}=0$. Good question. It should first of all be said that light-cone formalism breaks manifest Lorentz symmetry. Ultimately the question boils down to whether it is consistent in ST to have a convention where $p^+$ and $p^-$ are only positive. This becomes more delicate in SFT, where strings can interact. $\endgroup$ – Qmechanic Feb 23 '15 at 11:27
  • $\begingroup$ I agree, the singular point $p^+=0$ might be an artefact of the light-cone coordinate choice. Nevertheless string theorists prove that there exists a unitary representation of the Poincaré group (although this is not manifest). And I assume it should transform momenta as stated above. Note that if there was no tachyon there wouldn't be any state with $p^\pm<0$ as the raising operators don't change $p^+$ (and strictly increase $p^-$ if $p^+>0$). To be honest I'm not really familiar with SFT. Back to the beginning: What does Polchinski mean with $\left| 0;0 \right\rangle$? $\endgroup$ – Florian Oppermann Feb 23 '15 at 15:55
  • $\begingroup$ Hi @Qmechanic. There is another argument which disproves the boundedness of $p^+$. It is assumed that there exists a operator $x^-$ with $\left[x^-, p^+\right] = i$. Then the operator $\exp(-i a x^-)$ shifts the eigenvalue of any $p^+$-eigenstate by an arbitrary $a$ – also to $p^+ \leq 0$. $\endgroup$ – Florian Oppermann Feb 25 '15 at 12:49
  • $\begingroup$ Thanks for your extensive answer. I sum up 1., 2. and 8. as „we just ignore the states with $p^+ < 0$“. This should at least be mentioned in some textbooks because it makes the Hilbert space much less obvious. It is often suggested that the advantage of the light cone formalism is that the Hilbert space consists only of physical states which is not true if I can leave it e.g. by acting with $e^{-i a x^-}$. Concerning your point 3: One of my initial questions remains open: Does Polchinski mean a class of states $|1\rangle$ in eq. (2.8.2) (with $p^+$ not fixed)? $\endgroup$ – Florian Oppermann Feb 27 '15 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.