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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?

Update in 24/09/29: I realzed that the problem is invalid. By Fourier transform the 1+1D massless free boson on cylinder can be decoupled into harmonic oscillator, and there is a zero mode, which is a free particle on a real line, and it has a continuous spectrum. I guess those (planar wave in zero mode and ground state in other mode) corresponds to vacum $|\alpha\rangle$ created by vertex operators $:\exp(i\alpha\varphi(0)):$, eg. pp.163 in "Yellow Pages"

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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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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