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It has been said that particle identity relates to vibrational modes of its quantum string. However, I have not seen the identification of a particle related to a specific vibrational mode. At a conference I asked a physicist what in the mathematics of quantum physics led to the idea that there were actually strings. He said it was because of the harmonic nature of the results. That would lead one to believe that the fundamental mode might result in a certain particle, the next mode (three nodes and anti-nodes) would result in an another particle, four nodes and anti-nodes another, and so on. Apparently it's not so simple because all of the pictorial representations of strings show them vibrating in complex modes, rather than pure harmonics.

My questions are:

  1. Is there any correspondence between the harmonic vibration of a string and the particle's identity?

And incidentally,

  1. Do strings have different lengths, or are all closed strings, for example, the same length?
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  • $\begingroup$ The string is the particle type. The modes are the possible particle states. The amplitude of each mode is the number of particles in that state. You may find this other Physics.SE post and my answer to it interesting. $\endgroup$ – DanielSank Oct 28 '15 at 0:34
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    $\begingroup$ John I highly recommend @DanielSank's answer he cites above. $\endgroup$ – WetSavannaAnimal Oct 28 '15 at 0:41
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If a relativistic string has energy $E$ and and 3-momentum $\vec{p}$, then it corresponds to a particle of mass $$ m^2 = E^2 - |\vec{p}|^2 $$ Adding vibrational modes increases the energy $E$ of the string while keeping the 3-momentum the same. In this way, adding vibrational modes to a string changes the mass of the particle that it corresponds. Aside from mass, particles generally also have an internal spin. This arises in strings by adding different types of vibrational modes.

It is not as simple as just looking at nodes, because strings that potentially describe nature are supersymmetric and have fermionic excitations which does not have a classical analogue.

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