3D Dirac operator - Quantum Calculus

The three dimensional space $\mathbb{R}^3$ is important because we live in it. With the scalar Laplacian $-\Delta$ in dimension q=3, the Hydrogen operator $-\Delta + 1/r$ (leaving out constants) essentially explains the periodic system of elements and so the starting point of chemistry. The eigenvalue difference $1/n^2 - 1/m^2$ explain spectral lines (like Lyman (UV light) , Balmer (visible light) , Paschen (infrared light) ). The integral operator expressions for the inverse of the Laplacian explain why the electromagnetic or gravitational potentials have the form $1/r$ . The three dimensional integral calculus is taught in multi-variable calculus courses, where the exterior derivative grad, curl and div appear. One usually does not write down the Dirac operator on all differential forms. There are 0 forms, 1 forms, 2 forms and 3 forms. We usually identify 0 and 3 forms and 1 and 2 forms. The curl of a vector field is still considered to be a vector field. The Dirac operator in flat 3 dimensional space is $D = \left[ \begin{array}{cccc} 0 & -{\rm grad}^* & 0 & 0 \\ {\rm grad} & 0 & {\rm -curl}^* & 0 \\ 0 & {\rm curl} & 0 & {\rm -div}^* \\ 0 & 0 & {\rm div} & 0 \end{array} \right]$ . The Hodge Laplacian $L=D^2$ has 4 diagonal blocks containing the scalar Laplacian $L_0 = -\Delta$ . For 3-forms, it looks the same just $L_3 f dxdydz = -\Delta f dxdydz$ . On 1-forms and 2-forms which in multi-variable calculus always are identified as “vector fields”, one would write there $L_1 ( P dx + Q dy + R dz) = -\Delta P dx -\Delta Q dy - \Delta R dz$ and similarly for 2-forms $P dydz + Q dxdz + R dxdy$ . The solution of the wave equation in 3 dimensional space $u_{tt} = \Delta u$ can be written down conveniently as $u(t) = \cos(Dt) u(0) + t {\rm sinc}(Dt) u'(0)$ . And now we can note that $\cos = \phi_1$ and ${\rm sinc}= \phi_3$ are Bessel functions. This triggered me to talk on Saturday about a PDE riddle. The path $u(t) = \phi_k u(0)$ is a solution of a modified wave equation $u_{tt} = - D^2 u - (k-1) u_t/t$ as the computation in the box below shows. I had hoped that $u(t) = \phi_k u(0) + t \phi_{k+2} u'(0)$ would more generally solve this wave equation; but this did not pan out: for the velocity part, it only fits for $k=1$ , where it is the usual wave equation and where $\phi_{3} ={\rm sinc}$ is the sinc function.

A PDE attempt:

Motivated by the fact that $u(t) \phi_1(tD) u(0) + t \phi_3(tD) u''(0)$ solves the wave equation $u_{tt} + D^2 u=0$ on an arbitrary Riemannian q-manifold, we attempted to see whether in general, $u(t) = \phi_k(tD) u(0) + \phi_{k+2}(tD) u'(0)$ satisfies a modified PDE. Here, $\phi_q(r)$ solves the Bessel differential equation $\phi''(r) + (q-1) phi'(r) /r + phi(r) =0$ with the initial condition $\phi(0)=1, \phi'(0)=0$ . The first cases are $\phi_1(r) = \cos(r)$ , $\phi_2(r) = J_0(r)$ , $\phi_3(r) = {\rm sinc}(r) = \sin(r)/r$ , $\phi_4(r) = 2 J_1(r)/r$ and $\phi_5(r) = \frac{\sin (r)-r \cos (r)}{r^3}$ . Since the d’Alembert type formula $u(t) = \phi_1(tD) u(0) + t \phi_3(tD) u'(0)$ satisfies the wave equation $u_{tt} = - D^2 u$ on a general Riemannian manifold $(M,g)$ with Dirac operator $D=d+d^*$ , the square root of the Hodge Laplacian $d d^* + d^* d$ , it was tempting to explore whether $\phi_k(tD) + \phi_{k+2}(tD)$ satisfies this PDE. On Saturday morning, I thought it does, but if done correctly, there is an annoying single term $\phi_{q+2} u'(0) (q-1)/t$ left, when trying with the wave type equation $u_{tt} + D^2 u + u_t (q-1)/t = 0$ . (The computation is below). The reason, why I would have liked it is because it would give more intuition about why $D_t f = \phi_{q+2}(tD) Df$ is a good deformation of the Dirac operator. What we know is that $D_t f(p)$ is completely determined by $f$ only on $W_t(p)$ the wave front. In dimension $q$ , we need $\latex \phi_{q+2}$ when doing sphere averages and I still wonder why $q+2$ and not $q$ in dimension $q$ . It is clear in 1-dimension, where $\phi_3= {\rm sinc}$ is needed to get the discrete derivative $[f(x+t)-f(x-t)] /2$ .

Upshot: the form $u(t) = \phi_q(tD) f$ satisfies the PDE $u_{tt}+D^2 u + (q-1) u_t/t=0$ with $u(0)=f$ . The path $u(t)=t \phi_{q+2}(tD) f$ does not satisfy this PDE with initial velocity $u_t(0)=f$ however. The question had been triggered by the fact that the expression $d_t f = \phi_{q+2} (tD) df$ is a nice deformed exterior derivative that uses the wave front at distance t.

Just to recall, the starting point had been that $d_t f = \phi_{q+2} (tD) df$ is the flux of $f$ through $W_t(p)$ if $f$ is a $(q-1)$-form. This was a consequence of the Jeffrey Ovall formulas for sphere or ball averages. One can deduce from this that the Huygens property holds for all differential forms and not only q-1 forms. Using the magical Cartan formula for the Lie derivative $L_X = i_X d + d i_X$ , one can in generalsee by taking inner derivatives that also for general k-forms, the formula $d_t f$ produces an exterior derivative that only uses $f$ on the wave front: one just has to write every differential form as a linear combination of decomposable forms i_X g, where g is invariant under the flow of X like $i_{e^1} g(y) dx= g(y)$ or $i_{e^2} g(x) dy = g(x)$ or $i_{e^1+e^2} g(x-y) (d(x+y)) = g(x-y)$ in the case $q=2$ and derive the Huygens principle for $k=0$ forms if it is known for $k=1$ forms. Why do we see the Bessel case $q+2$ in the sphere averages in dimension $q$ ? The PDE $u_{tt} + D^2 u + u_t (q-1)/t$ almost works for the velocity. It does for position, but for velocity there is a term left. If the initial velocity $u_t(0)=0$ is zero, then $u(t) = \phi_q(tD) u(0)$ satisfies this modified wave PDE. It would have been nice (and was wishful thinking triggered by the 1- dimensional case) to explain better the q+2 Bessel solution appearing in the exterior derivative $d_t f = \phi_{q+2}(tD) df$ that is the center of attention as it has the property that $d_t f(p)$ only depends on the wave front $W_t(p)$ . There is a cancellation explaining a bit the shift from the q-Bessel equation $f''+(q-1) f/r + f=0$ to the (q+2)-Bessel equation $f''+(q+1) f/r = f=0$ solved by $\phi_{q+2}$ , but it is not enough.

Attempt: is it true that u = f_q u(0)  + t f_{q+2} u'(0)  = f X + t g Y
solves the PDE           u_tt + D^2 u + u_t (q-1)/t   where f_q  solves
the Bessel equation f''(r) + (q-1) f'(r)/r  + f(r) = 0,  f(0)=1, f'(0)=0         ?

Write u(0) = X and  u'(0) = Y   for the initial position and initial velocity.
Write simply f = f(tD) =f_q(tD)  and     g = g(tD) = f_{q+2}(tD) and use

Bessel  f''(tD) + (q-1) f'(tD)/(tD) + f(tD) = 0                (*)
Bessel  g''(tD) + (q+1) g'(tD)/(tD) + g(tD) = 0                (**)
chain rule :   d/dt f(tD) = f'(tD) D  and   d^2/dt^2  f(tD) =    f''(tD) D^2  

       POSITION                          MOMENTUM
------------------------------------------------------------------------------
u    = f       X                   + t g     Y    
u_t  = f'  D   X                   + t g'  D Y               +   g    Y     
u_tt = f'' D^2 X                   + t g'' D^2 Y  +  g' D Y  +   g' D Y
------------------------------------------------------------------------------            

Multiply the first equation by D^2, the second by (q-1)/t and switch q-1 to q+1
with      t g' D^2 Y (q-1)/(tD)  =   t g' D^2 Y (q+1)/(tD)    - 2 g' D Y    

------------------------------------------------------------------------------            
D^2 u       = f   D^2 X            + t g   D^2 Y    
u_t (q-1)/t = f'  D^2 X (q-1)/(tD) + t g'  D^2 Y (q+1)/(tD) + g Y (q-1)/t - 2g' D Y
u_tt        = f'' D^2 X            + t g'' D^2 Y            + 2g' D Y  
------------------------------------------------------------------------------

              Bessel  (q-1)          Bessel (q+1)   
 Adds to 0    Add to 0               Add to zero              g Y (q-1)/t    
 by PDE       by (*)                 by (**)                  Remains

A PDE attempt:

Author: oliverknill