Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Specific Question about Susskind's Lecture on Tachyons

  1. Aug 16, 2013 #1
    I may have poorly titled this post, since the lecture I'm talking about isn't just about tachyons, really. What I'm referring to is what Leonard Susskind says between the 50-60 minute mark:


    In the video, Susskind concludes that there are the only two polarization states of the first excited state of the open string, so then this string must be massless, like a photon. For this to happen, the ground state must have a negative mass squared.

    Why must [tex]m_0^2 +1 =0[/tex] and how do we know that the a's and the b's represent the only two polarization states? I understand that if there are only two polarization states, then the particle must be massless to preserve Lorentz invariance, but the rest is confusing to me.

    Can someone clear this up for me? I have no formal exposure to QM (and I apologize for this); regardless, I will try to decipher any technical answer given. Thanks so much.
    Last edited: Aug 16, 2013
  2. jcsd
  3. Aug 26, 2013 #2
    The example that Susskind takes is the simplified case of massless photons in [itex]4[/itex] dimensions. A massless photon in [itex]4[/itex] dimensions has only [itex]2[/itex] polarizations (so are [itex]a_i/a_i^+[/itex] and [itex]b_i/b_i^+[/itex], which are the anihilation/creation operators, for different excitation of energy [itex]i[/itex], corresponding to the 2 possible polarizations X and Y), but in a [itex]D[/itex] space-time dimension, a massless photon has [itex](D-2)[/itex] polarizations. For instance, in the bosonic open string, the coherent dimension is [itex]D=26[/itex], so you have [itex]24[/itex] possible polarizations. But the logic is the same, the excitations are vectors, that is from the ground state [itex]|0\rangle[/itex], and applying creation operators for the lowest energy excitation, we have states [itex](\alpha^\mu)^+_1 ~|0\rangle[/itex] with [itex]\mu = 1...D-2[/itex]. Susskind notations correspond to [itex]a = \alpha^1, b= \alpha^2[/itex]. The subscript [itex]_1[/itex], in [itex](a^\mu)^+_1[/itex] means that we consider only the lowest energy excitation. The energy of these excited states is [itex]m_0^2+1[/itex], and it is also the squared mass of this state, which must be zero, so we have [itex]m_0^2+1=m^2=0[/itex]
    Last edited: Aug 26, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook