6
$\begingroup$

Is there any simple proof of the no-ghost theorem in string theory?

$\endgroup$
  • 1
    $\begingroup$ It depends whether the demonstration of the equivalence with the light-cone gauge spectrum - which is manifestly ghost-free - is simple in your opinion. $\endgroup$ – Luboš Motl May 19 '11 at 7:21
  • $\begingroup$ A couple of references: 1. P. Goddard and C. B. Thorn, Phys. Lett. B40 (1972) 235; 2. Green, Schwarz and Witten, "Superstring Theory, Vol 1"; 3. Wikipedia. $\endgroup$ – Qmechanic May 19 '11 at 18:57
1
+50
$\begingroup$

The proof using DDF formalism involves constructing a set of operators that commute with the Virasoro operators, and when applied to the ground state, they give all possible physical states. These operators $A^{i}_{n}$, where $i$ runs over $d-2$ transverse dimensions of spacetime and $n$ is an arbitrary integer generate among themselves what is called the spectrum generating algebra. These operators are in one-to-one correspondence with the transverse components of $\alpha^{\mu}_{n}$, which arise as coefficients in the mode expansion of the string with appropriate boundary conditions and are promoted to operators upon quantization.

The proof is broadly (and nicely) sketched out for the open bosonic string case using the DDF formalism, given in [1]. Using this formalism, it is shown that there exists no ghosts in the Hilbert space after implementing the old covariant quantization scheme in $d = 26$. The way this is shown is by making contact with states resulting from implementing light cone quantization scheme, which is manifestly ghost-free. The same formalism can be used to prove the no-ghost theorem in $d=10$ for a string with world-sheet supersymmetry, as given in [2].

The two references do a very nice job in presenting the same, so there is no use replicating the proof.

[1]: Section 2.3.2, Superstring Theory, Volume I; Green, Schwarz, Witten

[2]: Section 4.3.2, Superstring Theory, Volume I; Green, Schwarz, Witten

$\endgroup$

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.