About fermionic particles and bosonic particles in RQFT, ST In relativistic quantum field theory, we can observe that the Dirac equation is a square root of Klein-Gordon equation. But, we can get the Dirac equation by defining Dirac spinor as $(1/2,0)\oplus(0,1/2)$ representation of $\mathrm{Spin}(3,1)$ like Klein-Gordon equation and Maxwell action (Weinberg, Schwartz). So, I have a question.

Why must Dirac operator be a square root of Klein Gordon operator? Is there a fundamental reason?

Weakly,

Why do the fermionic particles admit first order partial differential
  equation even though the bosonic particles admit second order partial
  differential equations? Is there a fundamental reason?


This is a simple question. 

Why do people think the relativistic quantum field theory is the theory of point particle? Is there a formalism which I can get whole quantum field theory from point-particle action (einbein field formalism)?


In bosonic string theory, we can only have the bosonic states from Polyakov action.

Is there a principle which includes the fermionic states like 'prinicple of string' without supersymmetry naturally?


If we quantize the point-particle action (einbein field formalism), we can get the state of spin-$0$ particles only unlike Polyakov action.

Why?


I want the fundamental reasons what we called 'higher perspective'.
 A: *

*Dirac equation is an equation for the relativistic quantum ave function of a single electron.\

*we shall regard the dirac wave function as a field, which will subsequently be quantized along with the electromagnetic field.\

*On quantisation, the electromagnetic fields $ A_{\mu}(x) $, $ F_{\mu}\nu $ become spece and time dependent operators, the dirac fields $ \psi(x) $ also.\
Dirac invented his equation in seeking to make Schrodinger equation for an electron compatible with special relativity.  $ i\frac{\partial \psi}{\partial t} = H\psi $
to sewre a symmetry between spacetime, dirac postulated the hamiltonian for a free electron to be of the form:\
\begin{equation}
H_{D} = \alpha.p + \beta m = -i\alpha .\nabla + \beta m
\end{equation}
where $ m $ is the mass of the electron, $ p $ its momentum,\
\begin{equation}
\alpha = (\alpha_{1}, \alpha _{2}, \alpha_{3})  
\end{equation}
Thus, $ \alpha_{1} $, $ \alpha_{2} $, $ \alpha_{3} $ and $ \beta $ are matrices. $ \psi $ is a column vector, and the Schr¨odinger equation becomes the multicomponent 
\begin{equation}
( i\frac{\partial}{\partial t} + i\alpha \nabla - \beta m)\psi = 0
\end{equation}
If this equation is to describe a free electron of mass m, its solutions should also
satisfy the Klein–Gordon equation. Multiplying by the operator\
$ ( i\frac{\partial}{\partial t} - i\alpha \nabla + \beta m)\psi $ we get:
\begin{equation}
\left[ - \frac{\partial^{2}}{\partial t^{2}} + \sum_{i < j}(\alpha_{i}\alpha_{j} + \alpha_{j}\alpha_{i})\partial_{i}\partial_{j} + im\sum_{i}(\alpha_{i}\beta + \beta\alpha_{i})\partial_{i} -\beta^{2}m^{2}\right]\psi = 0
\end{equation}
\textbf{Note:} for particle with mass m and spin zero we find that:\
$ (\alpha_{i}\alpha_{j} + \alpha_{j}\alpha_{i}) = 0 $ , $ i \neq j  $ \
$ (\alpha_{i}\beta + \beta\alpha_{i}) = 0 $\
$ \beta^{2} = 1 $, $ \alpha_{1}^{2} = \alpha_{2}^{2} = \alpha_{3}^{2} = 1 $\
Then the last Dirac equation becomes identical to the Klien-Gordon equation\
Dirac equations are satisfied by the set of $ 2 \times 2 $ Pauli spin matrices 
\begin{equation}
 \sigma = (\sigma^{1}, \sigma^{2}, \sigma^{3}) , where:\\
\end{equation}
$ \sigma^{1} = $
$                              
\begin{pmatrix}             
0 & 1\\                        
1 & 0 \\                       
\end{pmatrix}                 
$
$ \sigma^{2} = $
$
\begin{pmatrix}
0 & - i\\
i & 0\\
\end{pmatrix}
$
$ \sigma^{3} = $
$
\begin{pmatrix}
1 &  0\\
0 & - 1\\
\end{pmatrix}
$
$ \sigma^{0} = $
$
\begin{pmatrix}
1 &  0\\
0 &  1\\
\end{pmatrix}
$
\
\
We now represent the $ \alpha^{i} $ and $ \beta $ by $ 4 \times 4 $ matrices, shall use the so-called chiral representation:\
\
$ \alpha^{i} = $
$
\begin{pmatrix}
- \sigma^{i} &  \textbf{0}\\
\textbf{0} &    \sigma^{i}\\
\end{pmatrix}
$
$ \beta = $
$
\begin{pmatrix}
\textbf{0} &  \sigma^{0}\\
\sigma^{0} &  \textbf{0}
\end{pmatrix}
$
$ I^{0} = $
$
\begin{pmatrix}
\sigma^{0} &  \textbf{0}\\
\textbf{0} &    \sigma^{0}\\
\end{pmatrix}
$
$ \textbf{0} = $
$
\begin{pmatrix}
0 &  0\\
0 &  0\\
\end{pmatrix}
$
\
Dirac equation has signification of relativistic Schr\"{o}dinger equation, the solutions of Dirac equation represent by wave function $ \psi $ which is a fourcomponent column matrix, where exist ‘negative energy’ solutions, which Dirac interpreted as antiparticles.\

A: Look, Klein-Gordon equation drive from Schr¨odinger equation but it relativistic case for particle with mass m and spin zero, the different is the Hamiltonian, we replace the Hamiltonian in Schr¨odinger equation by Hamiltonian in relativistic case, so this Hamiltonian becomes quadri-dimensional vector $\P^{\mu}$.
you know that Schr¨odinger equation is second order partial so the bosons characterized by Klein-Gordon remain second order partial differential.
due to the Einstein equation of energy $ E^{2} = p^{2} + m^{2}$ that give negative solution, Dirac interprated this nigative solution by the existence of anti-matter and proposed new Hamiltonian H = p + m. without the squar so from this Hamiltonian the equation  characterized the fermion seem as first   order partial differential. you must understand the role of spin in solution of dirac equation because dirac us Schr¨odinger equation that study the electron but with his new Hamiltinian. the photon as boson can separat to electron and positron (matter and anti-matter), Dirac do just this separation for the Hamiltonian to find two equation one of electron(matter) the other for positron(anti-matter).
by other side the fermions enter in the frame of U(1) transformation, you see that the solution of dirac equation give a vector (one colon).
A: This answer is from my lecture in QFT, I hope that you find what you ask
\subsection{Dirak equation solution}
Thus, we solve the Dirac equation by consider that $ p_{\mu} $ is the four momentum that satisfies the relativistic energy-momentum relation
\begin{equation}
p^{\mu}p_{\mu}  =  m^{2} 
\end{equation}
if we take 
\begin{equation}
p_{\mu} = (E, \textbf{p}) 
\end{equation}
we find two solutions : one concerns positive energy and the other concerns negative energy, 
\begin{equation}
p^{\mu}p_{\mu}  = m^{2} 
\end{equation}
then 
\begin{equation}
E = \pm\sqrt{m^{2} + \textbf{p}^{2}} 
\end{equation}
\
We express the solutions to Dirac equation with two independent solutions, for a free fermion the wavefunction is the product of a plane wave and a Dirac spinor $ u(p) $ and ( $ v(p) $ for antiparticle), thus 
$ \Psi(x^{\mu}) $ = $ u(p^{\mu})e^{-ipx} $:
Substituting the fermion wavefunction, ψ, into the Dirac equation:\
\begin{equation}
(i\gamma^{\mu}\partial_{\mu} - m)u(p) = 0
\end{equation}
For a particle at rest, $ p $ = 0, we find the following equations:
\begin{equation}
(i\gamma^{0}\dfrac{\partial}{\partial t} - m)\Psi = (\gamma^{0} E - m)u = 0
\end{equation}
So we write:
$
\begin{pmatrix}
m & 0 & 0 & 0\\
0 & m & 0 & 0 \\
0 & 0 & - m & 0\\
0 & 0 & 0 & - m\\
\end{pmatrix}
$
$
u = Eu 
$
The eigenvalues are: $ E_{+} = m $ (twice) and $ E_{-} = - m $ (twice), unit[nat],
The solutions are four eigenspinors:\
$ u^{1} $
$
\begin{pmatrix}
1\\
0\\
0\\
0\\
\end{pmatrix}
$
$ u^{2} $
$
\begin{pmatrix}
0\\
1\\
0\\
0\\
\end{pmatrix}
$
$ u^{3} $
$
\begin{pmatrix}
0\\
0\\
1\\
0\\
\end{pmatrix}
$
$ u^{4} $
$
\begin{pmatrix}
0\\
0\\
0\\
1\\
\end{pmatrix}
$
and the associated wavefunctions of the fermions are:\
- for positive energy : $ \psi^{1} = e^{-imt}u^{1} $ and $ \psi^{2} = e^{-imt}u^{2} $ \
- for negative energy : $ \psi^{3} = e^{imt}u^{3} $ and $ \psi^{4} = e^{imt}u^{4} $\
Note that the spinors are $ 1 \times 4 $ column matrices, and that there are four possible states. The spinors are, however, not four-vectors: the four components do not represent t, x, y, z. The four components are a suprise: we would expect only two spin states for a spin $ - \frac{1}{2} $ fermion Note also the change of sign in the exponents of the plane waves in the states $ \psi^{3} $ and $ \psi^{4} $. The four solutions in equations (5.24) and (5.25) describe two different spin states ($ \uparrow $ and $  \downarrow $) with $ E = m $, and two spin states with $ E = −m < 0 $.\
\textbf{Negative Energy Solutions and Antimatter:}\
To describe\textit{ the negative energy states}, Dirac postulated that an electron in a positive energy state is produced from the vacuum accompanied by a \textit{hole} with \textit{negative energy}. The hole corresponds to a physical antiparticle, the positron, with charge $ + e $. Another interpretation (Feynman-St\"{u}ckelberg) is that the $ E = − m $ solutions can \textit{either} describe a negative energy particle which propagates \textit{backwards} in time, or a positive energy antiparticle propagating \textit{forward} in time:
\begin{equation}
e^{-i\left[(-E)(-t)-(-p)(-x)\right]} = e^{-i[Et - Px]}
\end{equation}
For a moving particle, $ p \neq 0 $ the Dirac equation becomes (using (5.13) and (5.17)):\
$ (\gamma^{\mu}p_{\mu} - m) $
$
\begin{pmatrix}
u_{A}\\
u_{B}\\
\end{pmatrix}
$
$ = $
$
\begin{pmatrix}
E - m & -\sigma p\\
\sigma p & - E - m\\
\end{pmatrix}
$
$
\begin{pmatrix}
u_{A}\\
u_{B}\\
\end{pmatrix}
$
$ = 0 $
\
Note that: $ \gamma^{\mu} = \gamma^{0} + \gamma^{i} $,
$ u^{1,2,3,4} = $
$
\begin{pmatrix}
u_{A}\\
u_{B}\\
\end{pmatrix}
$
\
The equations for:\ 
$ u_{A} = $ 
$
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
$
or 
$ u_{A} = $ 
$
\begin{pmatrix}
1\\
0\\
\end{pmatrix}
$
and
$ u_{B} = $ 
$
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
$
or 
$ u_{B} = $ 
$
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
$
are coupled:\
\begin{equation}
u_{A} = \frac{\sigma p}{E - m}u_{B}
u_{B} = \frac{\sigma p}{E + m}u_{A}
\end{equation}
$ p\sigma = p_{x}\sigma^{1} + p_{y}\sigma^{2} + p_{z}\sigma^{3} $\
So, We showe the four solution of Diarc equation :\
For 
$ u_{A} = $ 
$
\begin{pmatrix}
1\\
0\\
\end{pmatrix}
$
we find
$ u^{1} = $ 
$
\begin{pmatrix}
1\\
0\\
\frac{p_{z}}{E + m}\\
\frac{p_{x} + ip_{y}}{E + m}
\end{pmatrix}
$
\
For
$ u_{A} = $ 
$
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
$
we find
$ u^{2} = $ 
$
\begin{pmatrix}
0\\
1\\
\frac{p_{x} - ip_{y}}{E + m}\\
\frac{- p_{z}}{E + m}\\
\end{pmatrix}
$
\
\
These two solutions $ u^{1} $ and $ u^{2} $ describe an electron of\ \textit{positive} energy $ E = + \sqrt{m^{2} + p^{2}} $, and momentum p\
\
For
$ u_{B} = $ 
$
\begin{pmatrix}
1\\
0\\
\end{pmatrix}
$
we find
$ u^{3} = $ 
$
\begin{pmatrix}
\frac{- p_{z}}{E + m}\\
\frac{- p_{x} - ip_{y}}{- E + m}\\
1\\
0\\
\end{pmatrix}
$
\
For
$ u_{B} = $ 
$
\begin{pmatrix}
0\\
1\\
\end{pmatrix}
$
we find
$ u^{4} = $ 
$
\begin{pmatrix}
\frac{- p_{x} + ip_{y}}{- E + m}\\
\frac{p_{z}}{- E + m}\\
0\\
1\\
\end{pmatrix}
$
\
One these two solutions $ u^{3} $ and $ u^{4} $ describe positron of \textit{negative} energy $ E = - \sqrt{m^{2} + p^{2}} $, and momentum p.\
Now we can change the spinors $ u^{4}(-p) = v^{1}(p) $ and  $ u^{3}(-p) = v^{2}(p) $
to discribe antiparticule with positive energy
\
$ v^{2}(p) = u^{3}(-p) $ 
$
\begin{pmatrix}
\frac{ p_{z}}{E + m}\\
\frac{p_{x} + ip_{y}}{ E + m}\\
1\\
0\\
\end{pmatrix}
$
and
$ v^{1} = u^{4} $ 
$
\begin{pmatrix}
\frac{p_{x} - ip_{y}}{E + m}\\
\frac{- p_{z}}{E + m}\\
0\\
1\\
\end{pmatrix}
$
\
- For a fermion with momentum p along the z-axis,describes a spin-up fermion by
\begin{equation}
 \psi = e^{-ipx}u^{1}(p^{\mu})
\end{equation}
 And describes a spin-down fermion by
\begin{equation}
 \psi = e^{-ipx}u^{2}(p^{\mu}) 
\end{equation}
- For an antifermion with momentum p along the z-axis, we describes a spin-up antifermion by:
\begin{equation}
\psi = v^{1}(p^{\mu})e^{- ipx} = u^{4}(-p^{\mu})e^{i(-p)}
\end{equation}
And we describes a spin-down antifermion by:
\begin{equation}
\psi = v^{2}(p^{\mu})e^{- ipx} = u^{3}(-p^{\mu})e^{i(-p)}
\end{equation} 
\subsection{Fermion currents}
To describe fermion current, first we define the Dirac adjoint,as
\begin{equation}
 \Psi = \Psi^{+}\gamma^{0} = (\Psi^{1}, \Psi^{2}, -\Psi_{3}^{*}, -\Psi_{4}^{*}) 
\end{equation}
and the Dirac Lagrangian can be written as 
\begin{equation}
L_{D} = \Psi(i\gamma^{\mu}\partial_{\mu} - m)\Psi
\end{equation}
using the Noether theorem, the four-vector fermion current: is 
\begin{equation}
j^{\mu} = \Psi\gamma^{\mu}\Psi = (\Psi\gamma^{0}\Psi, \Psi\gamma^{i}\Psi) = (\rho, \overrightarrow{j})
\end{equation}
