susy and sustrings

• Pauli Exclusion and pre-susy:

  
  > The Esteemed Kevin Scaldeferri wrote:
  >
  >psi(r1,s1,r2,s2) = - psi(r2,s2,r1,s1)
  >
  >Where r is the position and s is the spin of the particle.
  
  It is a cute application of the transform principle of quantum
  to show how this applies in non-coordinate bases.  We go to a
  ket notation, and let me make the spins implicit (e at r1 has s1) :
  
  Psi = Psi(r1,r2)|r1>|r2>
  Phi = Psi(r1,r2)|r2>|r1>
  
  With a generalized Einstein convention, integrals on r1 and r2 are
  implied.  The exclusion principle states that Psi = - Phi.  Now let's
  transform to a general orthogonal basis |n>
  
  Psi = |n>|m> |r2>
  Psi = Psi(n,m) |n>|m>
  Psi(n,m) = Int(r1,r2) |r2>
  Psi(n,m) = Int(r1,r2) n(r1)m(r2)Psi(r1,r2)
  
  Phi = Phi(n,m) |n>|m>
  Phi(n,m) = Int(r1,r2) |r1>
  Phi(n,m) = Int(r1,r2) n(r2) m(r1) Psi(r1,r2)
  Phi(n,m) = Int(r1,r2) n(r1) m(r2) Psi(r2,r1)
  Phi(n,m) = - Psi(n,m)
  
  This is not at all amusing for the experts; we've only
  shown that the antisym rule is preserved under all basis
  transforms, as it should be for consistency.  However,
  this has direct application : if n & m are orbital
  wavefunctions, for example, we've derived the orbital
  structure.  Also, we've shown Phi(n,n) = 0
  (undergrads should understand the antisym rule as corresponding
  to the anticommutation of fermion creation operators, as opposed
  to boson (harmonic oscillator) creators).
  
  Of course the best way to work with this is as a matrix:
  
  M = |r1> Psi(r1,r2)  , of course.
  
  Finally, we should've guessed supersymmetry from this long
  ago : if I add an antisym matrix (a fermion) + a symm matrix
  (a boson) I get an unconstrained matrix = "normal" (classical)
  statistics.  This is equivalent to the cancellation of loops
  in Quantum Field Theory by (-1) vs. (1) factors.
  
  > selipsky@wuphys.wustl.edu says :
  >>  I (Charles) said :
  >> Finally, we should've guessed supersymmetry from this long
  >> ago : if I add an antisym matrix (a fermion) + a symm matrix
  >> (a boson) I get an unconstrained matrix = "normal" (classical)
  >> statistics.  This is equivalent to the cancellation of loops
  >> in Quantum Field Theory by (-1) vs. (1) factors.
  >
  >   Yikes!  The matrices operate on different state spaces, so it's
  >not clear what you mean by adding their "corresponding (?) entries"
  >by brute force.  And does this mean classical statistics is somehow
  >broken by the SUSY breaking scale? Or will stop working if we fail to
  >find the lightest Higgs state below 130 GeV?
  
  Here it is explicitly : let b_n be a bosonic creator with some 
  index n, and f_n a fermion creator.  Consider the two particle
  state space.  We can make general states :
  
  X = x_mn b^m b^n |vac>
  
  or
  
  Y = y_mn f^m f^n |vac>
  
  where the x matrix is constrained to be symmetric, and the y
  matrix is constrained to be antisymmetric (we can also
  make mixed states bf , but their matrices are unconstrained ("normal")
  so we won't bother with them).
  
  So, consider the combination :
  
  X+Y = (x_mn b^m b^n + y_mn f^m f^n) |vac>
  
  We can write this with a single matrix z_mn , as :
  
  X+Y = Z = 1/2( (z_mn + z_nm) b^n b^m + (z_mn - z_nm) f^m f^n ) |vac>
  
  	= z_nm (b^n b^m + f^n f^m ) |vac>
  
  Now we can add a general mixed b,f state :
  
  I = i_nm (f^n b^m + f^mb^n) |vac>
  
  We can thus obviously make simple states:
  
  J = j_nm (f+b)^n (f+b)^m |vac>
  K = k_nm (f-b)^n (f-b)^m |vac>
  
  J = j_nm (f^nf^m + b^nb^m + f^nb^m + f^mb^n) |vac>
  K = k_nm (f^nf^m + b^nb^m - f^nb^m - f^mb^n) |vac>
  
  if we put z = j + k , and i = j - k , we have
  recovered the full state space of "fermionic" (antisym
  matrix) ff states, bosonic bb states, mixed fb/bf states
  though only a pair of commuting (="ordinary") particles
  (f+b) and (f-b) .
  
  If you draw the pictures, you'll see how we get ordinary
  statistics from bose + fermi : if ---- is a boson and ..... is a fermion
  
  ----------
  ----------
  
   +
  
  -----\/-----
  -----/\-----
  
  +
  
  ...........
  ...........
  
  +
  ....\/....
  ..../\....
  
  
      =
  
  
  ----------
  ----------
  
   +
  
  -----\/-----
  -----/\-----
  	
      =
  
  ----------
  ----------
  
   +
  
  ...........
  ...........
  
  In article <6gb79g$i0q$1@agate.berkeley.edu>, selipsky@wuphys.wustl.edu says...
  >   But why can we use the same dummy indices m,n for both f and b
  >operators?  Do we have to have a one-to-one correspondence between
  >f-created states and b-created states, in a not necessarily
  >supersymmetric theory?
  
  Yes, it is true that I assumed that the "internal" index _n running
  on b_n and f_n was the same, so that in a sense I put in the
  supersymmetry by hand, and have just derived some results of it
  (all be-it in a different way).  As for your other questions, I have
  not explored advanced topics in susy from a picture like this (really,
  this picture is too primitive, but the idea of fermi + bose = normal
  could be applied to things like Witten's index (perhaps)).  I should
  note that the "normal" statistics (that is, (b+f) commutes with
  itself) means there is no normal ordering problem, which means the
  vacuum energy is zero, which reproduces that old susy result in
  this picture.
      
  

• Kaku's simple susy:

  
  I'll try to answer some of your questions at the end, but
  first a very nice susy example from Michio Kaku.
  (you must know elementary quantum mechanics)
  
  Consider the Hamiltonian :
  
          H = b* b + f* f
  
  (where the * should really be a dagger)  Where b and f are
  "bosonic" and "fermionic" oscillators, respectively.  So that
  
          [b*,b] = b*b - bb* = 1
          {f*,f} = f*f + ff* = 1
          [f,b] = 0 , etc.
  
  Then there exists a SUSY generator:
  
          Q = f*b + b*f
  
  It is easy to show :
  
          [Q,H] = 0
          {Q,Q*} = 2H
          <0|H|0> = 0
  
  (where the state |0> is defined by b|0> = f|0> = 0 )
  These are all the basic properties of susy.  Here it's very nice
  and simple because:
  
          Q|f> = Q f*|0> = (f*bf* + b*ff*)|0> = b*ff*|0> 
                   = b*(1 - f*f)|0> 
                   = b*|0> = |b>
  
  so Q just turns a fermionic basis into a bosonic.
  (the other propositions are proved similarly).  Now, the vacuum |0>
  has zero susy charge:   Q|0> = 0 , and QH = HQ , so
          <0|QH|0> = <0|HQ|0> = 0
          so H|0> = 0
  
  the vacuum has zero energy.  This does not occur without susy (or
  fine tuning) and is one of the main attractions.
  
  So, this basic algebra is the foundation of susy, but you really
  wanted the path-integral or geometric viewpoint.
  
  The susy geometry is an extension in spinor and spinor* directions,
  which are curved over spacetime, such that parralelograms do not close
  unless you also shift in spacetime.  (this is {Q,Q*} = 2P, as above).
  Susy geometry is easy to understand if you know about BRST in gauge
  theories.  If you know 'd' in differential geometry, then (b) BRST is
  easy.  Just like 'd' is a translation in spacetime, 'b' is a translation
  along the gauge-direction.  Hence bA = Dw , for example.  Gauge Invariance
  is then the statement bS = 0 , on the action S.  On the Lagrangian,
  this means      b L = j/\*dw    so that vanishing of b S becomes the
  statement that d*j = 0 , conservation of the current generated by the
  brst local symmetry. (this is of course the ordinary electric current)
  In susy you can construct a similar d (s) which moves you along the
  fermionic direction (in a continuum between bosons and fermions). So
  we demand        s S = 0        Which determines the constraint on your
  action. 
  
  (it is interesting to note in both brst and susy that the
  physical part of the Lagrangian is in the cohomology of the symmetry:
  that is, I can always add a term to L of the form L = b X or s Y ,
  since b^2 = s^2 = 0 .  These X and Y terms just turn out to be
  gauge-fixing terms, so the physics must be in the cohomology!)
  
  Anyhoo, supergravity comes in for an interesting reason.  If we now
  try to "gauge" (make local) the susy, we try    s L = j/\*dw
  we have a problem.  Why?  because of the old susy commutator : two
  susy transforms differ by a coordinate transform.  Thus sL inherently
  contains the current for coordinate transforms as well, which is
  simply T_uv , the energy tensor, and for this to be conserved, we must
  invoke gravity.  So, we've argued that any local susy must be a theory
  of gravity.  
  
  

• supergravity as gauged susy

  
  I should note that supergravity is a lot more special than just
  "adding super-symmetry to physics" (as we do in Yang-Mills - we
  just try to add fermionic partners).  We can see why this is by
  trying to "gauge" the symmetry.
  
  First a little review of gauging symmetries in general.
  Consider some Lagrangian (density), L, which is invariant
  under a global transformation (global means shift by constant
  values), call it "q" (an operator, like a derivative), for
  example maybe q(x) = (x + a).  So
  	q L = 0
  
  Now we want to "promote" q to Q and make it a local symmetry,
  that is the shift (here a) is now a function on spacetime.  This
  means that the most general thing Q can be is :
  
  	Q L = j^u d_u a(x)
  
  since when a(x) = a(0) this must vanish.  Now, doing integration
  by parts, since L only ever appears inside an action, this is:
  
  	QL = - a(x) d_u j^u
  
  This is the standard Noether way to find currents.  Now we can
  "Gauge" the current, to keep the action invariant (we see that the
  d^u j_u above will vanish on shell (that is, when the equations of
  motion are satisfied) (actually, in 2-d it usually vanished off-shell
  as well..)).  That is, we add a term to L :
  
  	L' = - A^u j_u
  
  and define A by Q A = d^u a(x)
  so
  
  	Q(L + L') = j^u ( d_u a(x) - QA_u) = 0
  
  (actually, I assumed Qj_u = 0 here, which is not always true;
  the experts would go through a recursive procedure and possibly add
  more terms to cancel the gauging of Qj_u ).
  
  So, back to the case in point.  We have q = global supersymmetry
  transforms, and we want to gauge it.  First some counting.  q is
  parameterized by a spinor (that is, a(x) must be a spinor) so
  that j_u must be spin-3/2 :
  	j^u d_u a(x)
  since a is a spinor and d_u is a vector, j = spinor + vector = 3/2
  
  Now, it is known that the only susy-multiplets with spin-3/2 also
  contain gravitons, but I don't want to go through the representations
  of the susy algebra, so I'd like to try to see it more directly.
  
  The definition of the susy algebra is :
  
  	{q* s , s* q} = s* gamma_u s d^u
  
  (the star here should be the usual bar on spinors).
  In words : the (anti) commutator of two susy transforms is
  a coordinate shift.
  
  So, let's apply L to this formula on the right :
  
  	Q* a a* Q L + a*Q Q*a L = a* gamma_u a d^u L
  
  	Q*a (a* d^u j_u) + a*Q (d^u j*_u a) =
  
  	(d^u a*) Q*j_u a + a* Q j*_u (d^u a) = 
  
  Now, I don't know what Q*j or Qj* are, so we'll just
  use symbols :
  
  	j_u Q* = gamma^v J_uv
  	Q j*_u = gamma^v J'_uv
  
  (we've arranged these so that the left-hand side is a matrix in
  spinors : s* s is a scalar, and s s* is a matrix)
  Moving things around by parts, the lhs is :
  
  	(d^u a* gamma^v a) ( J_uv - J'_uv ) = a* gamma_u a d^u L
  
  But d^u L is just the good-old Lorentz symmetry, and local
  lorentzes creates a current (as always) which is just the
  stress-energy tensor :
  
  	d^u L = d^v T_uv
  
  So we find
  
  	(d^u a* gamma^v a) ( J_uv - J'_uv ) = d^v a* gamma^u a T_uv
  
  	 J_uv - J'_uv  = T_uv
  
  So the susy of susy-currents is nothing but the stress-energy.  This is
  actually Einstein's equation in disguise!  What we have found is that
  if you try to gauge the susy transform, because of the susy algebra, you
  inevitably get a gravity theory!  (that is, our original L MUST have
  had the Ricci scalar and whatnot in it in order to create these currents
  which satisfy this equation).
  
  (another way of saying all this is with the aside I tried to ignore :
  that when we gauge L by adding L' , we may not actually be done if
  Q(j) isn't zero.  This is the case here, and we have shown that the
  extra terms we need to add to >actually< have off-shell invariance
  are the gravity terms!)
  
  

• kappa symmetry

  
  >-----John Baez wrote:
  >I don't understand kappa symmetry, not even for superstrings, so
  >could somebody say a few words about it and why it's so crucial?
  >He gives a formula describing it yet I remain unenlightened.  He 
  >says it "always halves the number of fermionic degrees of freedom" -
  >why is that?
  
  Just like the Weyl or Majorana or Dirac conditions on spinors each
  halve the number of degrees of freedom.  For example, in the diagonal
  gamma_5 basis, the Weyl condition says the upper (lower) half of the
  spinor is zero.  The Kappa symmetry can be written as a similar
  projection form; the fermion is changed by
  
  Theta += Projector * Kappa
  
  So that any of the componenets of Theta along the Projector directions
  can be gauge-fixed to zero by a suitable choice of Kappa.
  
  It's so crucial, because it makes the number of fermionic and bosonic
  degrees of freedom equal (in 10-d a vector has 8 degrees of freedom
  (on shell) while the spinor has 2^(10/2) = 32 degress of freedom ;
  this is cut in half by the Weyl condition, then half again by the Kappa
  symmetry, to 8).  This in turn must be done so that the vacuum energy
  remains zero.
  
  I'll do an example for point particles, due to John Schwarz, as a nice
  simple warmup.  The point particle action is :
  
  S = Integral [ L dT ]
  
  L = sqrt{ p^2 } = sqrt{ p_u p^u }
  
  where T is a proper time, and p is the susy-momentum :
  
  p_u = d_t x_u - i s* g_u d_t s
  
  where s is a spinor, g is gamma matrix, and x is a spacetime coordinate
  
  So, let's see what L does under dx_u = i s* g_u ds
  
  dp_u = d_t (dx_u) - i ds* g_u d_t s - i s* g_u d_t ds
  dp_u = - 2 i s* g_u d_t ds
  
  dL = (p^u dp_u ) / L
  dL = - 2 i s* p^u g_u d_t ds / L
  
  Now we play a trick.  We put :
  
  P = p^u g_u g_11 / L
  
  So that :
  
  dL = - 2 i s* P g_11 d_t ds
  
  and note P^2 = p^up_u /L^2 = 1 , and Tr{P} = 0
  
  Now imagine some L' , with
  
  dL' = - 2 i s* g_11 d_t ds
  
  so that 
  
  d(L + L') = + 2 i ds* (1 + P) g_11 d_t s
  
  This L' is easy enough; it's L' = i s* g_11 s
  
  Now we use the kappa ; we see that (1+P)/2 is a
  projector, so that we can choose
  
  ds* = k* ((1-P)/2)
  
          (ds = (1-P)/2 k )
  
  So that the spinor varies as a projector on this Kappa.
  
  Then d(L + L') ~ (1-P)(1+P) = 0
  
  The really interesting thing here is that the projector P
  is not just constant, it's actually dynamical (it contains
  the p_u and whatnot), so that the kappa symmetry of the
  spinor s actually is determined by the dynamics of s itself.
  
  ====================================================
  
  In article <6dger8$7b@charity.ucr.edu>, baez@math.ucr.edu says...
  >It's a bit like reading about "minimal coupling"
  >before understanding connections on vector bundles. 
  
  Actually, that analogy is meaningful.  In a sense, we have a
  string (or particle) Lagrangian for bosonic coordinates, and we're
  trying to "minimally couple" to the susy derivatives, via
  
  	x_u -> x_u + i s* gamma_u s
  
  (this form is essentially demanded by the susy algebra; in terms of
  canonical field theory, it would be
  
  		( phi* d_u phi + i psi* gamma_u psi )
  
  is the only current conserved before & after susy "gaugings" )
  
  The problem is, this "naive" susy minimal coupling is not necessarily
  right.  You are free to add more terms to the Lagrangian that are purely
  fermionic (since you started with a pure bosonic Lagrangian).
  
  In a sense, it's the same with Gauge minimal coupling - the matter
  part (analogous to the bosonic part here) can only be couped with
  d -> D , but then I'm still free to add stuff which is purely
  gaugeonic - (eg Gauge-Fixing terms = Fadeev-Popov ghosts, whatever) -
  and I must choose the gaugeonic stuff actually in a very similar way
  that I chose the Kappa symmetry - with BRST symmetry, which cuts out
  the nonphysical degrees of freedom.
  
  

• boson <-> fermion and susy

  
  [... in response to a question...]
  
  This is a very confused sentence, but I think what you're trying to
  say is : is there a one-to-one mapping of bosonic and fermionic
  processes (in-out scattering lines).  The answer is yes.  I presume
  you don't know much field theory, but for those who do, the classic
  example is the interaction
  
  	2 g f*f b +	g^2  b^4
  
  (f is a fermion, b is a boson)
  (I may have muffed a factor of two) , where you can see the
  b+b -> b+b and f+f -> f+f processes are obviously both of order g^2 ;
  
  (btw this is just a special case of the general
  
  	V^2 + f*f dV
  
  (where dV = derivative of V wrst b)
  (for the choice of V = gb^2 )
  
  > Does it map states spaces to states spaces, in
  >such a manner that multiple bosons condensed in a single state
  >(can) be distributed to multiple fermions in different states ?
  
  I'm not sure what you mean here, but let me mention something else.
  There's a simple, cute, and mysterious technique known as bosonization.
  In it's simplest form, I can integrate out a fermion in trade for
  a boson, which is usually understood as being a meson made of
  a fermion pair.  This is a very pretty picture, especially in the
  classic Thirring - Sine-Gordon equivalence where the solitons in the
  bosonic theory are the free fermions.  However, there are some
  mysteries.  For example, the infinite zero point energy is changed by
  an infinite amount when I do this integration.  It is of course true
  that an infinite baseline for the energy is irrelevant, but I can change
  this choice of baseline when I do an integral? (this is presumably
  related to the trouble with defining path integrals).  This mystery
  becomes most troublesome when you start with a susy theory which has
  manifestly zero background energy.  The simplest case is a free fermion
  and a free boson.  I can integrate out the fermion, replacing it with
  a meson.  There is now an O(2) symmetry between the two bosons replacing
  the supersyemmtry, and my vacuum energy is no longer finite!  It is well
  known that susy is unbroken only when the vacuum energy is zero - but
  can susy be broken just by integrating out the fermions?
  
  > Is it, in any case, as well a symmetry exchanging
  >Bose-Einstein statistics with Fermi-Dirac statistics ?
  
  Yes, that is absolutely true, since the susy generator Q is
  anticommuting.  if b is a boson and commuting, and f is a fermion
  and anticommuting, then
  
  	b' = Q f is commuting
  	f' = Q b is anticommuting
  
  here, b' and f' are not necessarily b and f, unless you have a
  very simple (trivial) realization of the susy.  Anyhoo, you can
  see most of the susy magic occur because of a *cancellation*
  between fermi and bose statistics which give the net system
  "null" statistics (classical).  For example, in the presense of
  a background boson, the amplitude to emit another boson goes
  up (while with a background fermion, amplitude to emit a fermion
  goes down).  Thus if you have a background of one of each, the
  net amplitude for boson or fermion is unchanged !
  
  
  

• why 11 dimensions?

  
  In article <6erq1i$76a@gap.cco.caltech.edu>, kevin@cco.caltech.edu says...
  >
  >>4) M-theory might be a rolled-up limit of a 12-d theory, F-theory,
  >>   which they're also unwilling to define
  >
  >I got the idea F-theory was a little better defined,
  
  Not to mention the fact that the fundamental theory is almost
  certainly NOT 12 dimensional, as it is well known that supersymmetry
  does not work in more than 11 dimensions (hence all the appreciation
  of supergravity and M theory in 11 dimensions).
  
  A brief review of why this is: (I don't really understand it too
  well, so maybe someone else can add to this):
  
  a spinor in D dimensions has n = 2^(D/2) components, so
  in D = 4, n = 4, but this reduces with Weyl or other conditions;
  in D = 10, n = 32, but again this reduces to 16, and in 11, n = 32,
  and since in odd dimensions you cannot have Weyl or Majorana
  conditions, this n = 32 is irreducible.  A 32-component spinor means
  the (N=1) susy must have 32 supercharges (susy is parameterized by a
  spinor). Now we argue that 32 supercharges is the most you can ever
  have. We dimensionally reduce the D = 11, N = 1, n = 32 theory down
  to D = 4, where we are more familiar with the little group (and hence,
  the helicity classifications).  Now, the dimension of spinors cuts
  down to 4, but the number of supercharges must be the same.  Thus
  we find an N = 8 theory (eight independent 4-component spinor
  charges) in 4-D.
  
  Now, since the susy generator is fermionic (anticommutes) it changes
  the spin by 1/2.  Also, it can only be applied once.  Thus, with N
  generators, we can apply them sequentially to change the spin by N/2 .
  So, if the minimum spin in the susy multiplet is (-j) , then the
  maximum is J=(N/2 - j).
  
  Finally we note that quantum theories for spin > 2 are inconsistent.
  (Weinberg in Volume 1 shows this the best, I think; basically, for
  spin >= 1 you start getting in to trouble : your demand of irreducible
  representation of lorentz gets harder and harder; with spin 1 it is
  only possible modulo gauge-transforms, so that you must couple your
  photon to a conserved current, or you lose *lorentz* invariance!  with
  spin 2 you have similar trouble, by the time you get to spin 2.5 or 3,
  the transformation properties are so ugly that you simply cannot make
  any current with the right symmetries which is also conserved; note that
  I've really specialized to the case of D=4 to make this argument, that's
  why we had to dimensionally reduce).
  
  Thus we demand j <= 2 and J <= 2 , so
  
  (N/2 - j ) <= 2
  N/2 <= 2 + j <= 4
  N <= 8
  
  So N=8 is the most you can have in 4-d, so 32 supercharges are the
  most you can have, so D=11 is the highest dimension in which you can
  have N=1 (the minimum) consistent susy.
  
  (actually, there's a caveat here.  It seems you can have N=1/2 susy
  in D=12 ; N=1/2 means you apply some further constraint, like both
  Majorana and Weyl; I haven't really thought about this carefully,
  it's a bit "post-modern").
  

• Susy-Q amusement

  When I noticed the supersymmetry (susy) charge is always called "Q", the result was inevitable (phonetic spelling) :

  You make-a fermions From-a bo-osons, You make-a bo-osons from-a fermions, Susy Q. I like-a BPS, I like your finite-ness, I like-a BPS, I like your finite-ness, Susy Q. Whoah, Susy Queue ! Whoah, Susy Queue, I can't observe you, Susy Q.

  ---

  Charles Bloom / cb at my domain
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