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For example we work with 1+1D massless free boson, in canonical quantization we allow creation operators at any momentum so the Hamiltonian has continuous spectrum. But if we conformally map to a cylinder, the Hamiltonian has discrete spectrum. Why they are different?

Qmechanic
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Peter Wu
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2 Answers2

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TL;DR: The spectrum is discrete (continuous) since space is a compact circle $\mathbb{S}^1$ (non-compact line $\mathbb{R}^1$), respectively, cf e.g. this Phys.SE post.

In more details:

  1. The radially ordered punctured plane $\mathbb{R}^2\backslash\{(0,0)\}$ can (via the exponential/logarithmic map) be conformally$^1$ transformed to a time-ordered cylinder $\mathbb{R}\times\mathbb{S}^1 $ (or equivalently, to a time-ordered strip $\mathbb{R}\times [0,\ell] $ with periodic periodic boundary conditions).

  2. This is different from the time-ordered plane $\mathbb{R}^2$ with asymptotic boundary conditions.

References:

  1. J. Polchinski, String Theory Vol. 1, 1998; p.52-53.

  2. P. Ginsparg, Applied Conformal Field Theory, arXiv:hep-th/9108028; p.82.

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$^1$ Be aware that the Schwarzian derivative/central charge may shift the spectrum [1,2].

Qmechanic
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  • Thanks for reply. my question is, the free boson system possesses a conformal symmetry, then we can map the plane to a cylinder(say chapter 6 in the “yellow pages”). The mass spectrum should not change ,so what’s wrong in the argument? – Peter Wu Feb 14 '24 at 07:06
  • I updated the answer. – Qmechanic Feb 14 '24 at 08:26
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I guess I’ve figure out what’s going on, the notion of translation is different (translation on cylinder is rotation and dilation on plane), so we are working with different momentum operators in this two case, the mass spectrum is not invariant w.r.t. conformal transforms.

Peter Wu
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