# Problem 14.5c9

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) Calculate the total number of states in heterotic string theory (bosons plus fermions) at $\displaystyle{\alpha' M^2 =4}$.

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— This answer is my guess. —

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spacetime bosons:

$\displaystyle{NS'+ \otimes NS+}$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-2}^I, \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \right) \otimes \left( \{ \alpha_{-1}^{I'} b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^{I'}, b_{\frac{-1}{2}}^{I'} b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

\displaystyle{\begin{aligned} I, J, I' &= 2, 3, ..., 9 \\ A, B, C, D &= 1, 2, ..., 32 \\ \end{aligned}}

Number of states:

Let $\displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}$, the number of ways to put n indistinguishable balls into k boxes.

\displaystyle{\begin{aligned} &\left( 8+ N(2,8) +8 \times {32 \choose 2} + 32^2 + {32 \choose 4} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\ &= 40996 \times 128 \end{aligned}}

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$\displaystyle{R'+ \otimes NS+}$

\displaystyle{\begin{aligned} \left( |R_\alpha \rangle_L \right) \otimes \left( \{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

Following the same logic:

Postulating a unique vacuum $\displaystyle{|0 \rangle}$, the creation operators allow us to construct $\displaystyle{2^{16}}$ degenerate Ramond ground states.

Therefore, there are $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha \rangle_L}$.

— Me@2018-10-29 03:11:07 PM

\displaystyle{\begin{aligned} I, J, K &= 2, 3, ..., 9 \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned} &\left( 2^{15} \right) \times \left( 8^2+8+{8 \choose 3} \right) \\ &= 32768 \times 128 \end{aligned}}

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spacetime fermions:

$\displaystyle{NS'+ \otimes R-}$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \right) \otimes \left( \alpha_{-1}^{I'} |R_{a} \rangle,~d_{-1}^{I'} |R_{\bar a} \rangle \right) \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned} &\left( 40996 \right) \times \left( 8^2 + 8^2 \right) \\ &= 40996 \times 128 \end{aligned}}

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$\displaystyle{R'+ \otimes R-}$

\displaystyle{\begin{aligned} \left( |R_\alpha \rangle_L \right) \otimes \left( \alpha_{-1}^{I} |R_{a} \rangle,~d_{-1}^{I} |R_{\bar a} \rangle \right) \\ \end{aligned}}

Number of states:

\displaystyle{\begin{aligned} &\left( 2^{15}\right) \times \left( 8^2 + 8^2 \right) \\ &= 32768 \times 128 \end{aligned}}

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— This answer is my guess. —

— Me@2019-01-03 05:26:59 PM

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# Photon dynamics in the double-slit experiment, 5

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What is the relationship between a Maxwell photon and a quantum photon?

— Me@2012-04-09 7:38:06 PM

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The paper Gloge, Marcuse 1969: Formal Quantum Theory of Light Rays starts with the sentence

Maxwell’s theory can be considered as the quantum theory of a single photon and geometrical optics as the classical mechanics of this photon.

That caught me by surprise, because I always thought, Maxwell’s equations should arise from QED in the limit of infinite photons according to the correspondence principle of high quantum numbers as expressed e.g. by Sakurai (1967):

The classical limit of the quantum theory of radiation is achieved when the number of photons becomes so large that the occupation number may as well be regarded as a continuous variable. The space-time development of the classical electromagnetic wave approximates the dynamical behavior of trillions of photons.

Isn’t the view of Sakurai in contradiction to Gloge? Do Maxwell’s equation describe a single photon or an infinite number of photons? Or do Maxwell’s equations describe a single photon and also an infinite number of photons at the same time? But why do we need QED then at all?

— edited Nov 28 ’16 at 6:35
— tparker

— asked Nov 20 ’16 at 22:33
— asmaier

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Because photons do not interact, to very good approximation for frequencies lower than $\displaystyle{m_e c^2 / h}$ ($\displaystyle{m_e}$ = electron mass), the theory for one photon corresponds pretty well to the theory for an infinite number of them, modulo Bose-Einstein symmetry concerns. This is similar to most of the statistical theory of ideal gases being derivable from looking at the behavior of a single gas particle in kinetic theory.

Put another way, the single photon behavior $\displaystyle{\leftrightarrow}$ Maxwell’s equations correspondence only holds if you look at the Fourier transform version of Maxwell’s equations. The real space-time version of Maxwell’s equations would require looking at a superposition of an infinite number of photons — one way to describe the taking [of] an inverse Fourier transform.

If you want to think of it in terms of Feynman diagrams, classical electromagnetism is described by a subset of the tree-level diagrams, while quantum field theory requires both tree level and diagrams that have closed loops in them. It is the fact that the lowest mass particle photons can produce a closed loop by interacting with, the electron, that keeps photons from scattering off of each other.

In sum: they’re both incorrect for not including frequency cutoff concerns (pair production), and they’re both right if you take the high frequency cutoff as a given, depending on how you look at things.

— edited Dec 3 ’16 at 6:28

— answered Nov 27 ’16 at 23:08

— Sean E. Lake

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Maxwells equations, which describe the wavefunction of a single noninteracting photon, don’t need Planck’s constant. I find that remarkable. – asmaier Dec 2 ’16 at 14:16

@asmaier : Maxwell’s equations predate the quantum nature of light, they weren’t enough to avoid the ultraviolet catastrophe. Note too that what people think of as Maxwell’s equations are in fact Heaviside’s equations, and IMHO some meaning has been lost. – John Duffield Dec 3 ’16 at 17:45

— Do Maxwell’s equations describe a single photon or an infinite number of photons?

— Physics StackExchange

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2019.01.03 Thursday ACHK