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Thirty-one years ago , Dick Feynamn told me about his 'sum over histories' version of Quantum Mechanics.'The electron does anything it likes',he said, 'It goes in any direction at any speed,forward or backward in time,however it likes, and then add up the amplitudes and it gives the wave-function'. 'I said to him,'You're crazy'. But he was't.                                                             - F.J Dyson

Thomas and Fermi independently developed  a theory of complex many electrons atom or ion by choosing fermi gas of electron in ground state.The electrons confined in space by a central potential V(r) where V(r) is zero for large distance(r) from the nucleus. Apart from this assumption they also assumed the variation of potential is very slow in distance very much larger than de'Broglie wavelengths of electron and the thermal energy kT is much smaller than the potential energy in the boundary of the atom.In the near of nucleas potential remain constant .And for large distance its value is zero.Then Schrödinger equation becomes $$\frac{-\hbar^2\nabla^2\Psi}{2m}=E\Psi$$ Which solution gives $$\Psi=C\sin\left({k_x x}\right)\sin\left({k_y y}\right)\sin\left({k_z z}\right)$$ $$k^2=\frac{2mE}{\hbar^2}={k_x}^2+{k_y}^2+{k_z}^2$$ Where we are using the boundary conditions $$\Psi(x,y,0)=\Psi(x,y,L)=\Psi(x,0,z)=\Psi(x,L,z)=\Psi(0,y,z)=\Psi(L,y,z)=0$$ where the solution confined in a ...

The Einstien's mass-energy dispersion relation be$$E^2=c^2P^2+{m_0}^2c^4\space\space(1)$$ Now in Quantum mechanics $$E\rightarrow i\hbar\frac{\partial}{\partial t} \Rightarrow E^2 \rightarrow -\hbar^2\frac{\partial^2}{\partial t^2}$$and $$P\rightarrow -i\hbar\nabla\Rightarrow P^2\rightarrow -\hbar^2 \nabla^2$$ In this fashion we can write $$ -\hbar^2\frac{\partial^2}{\partial t^2}\psi=-\hbar^2c^2 \nabla^2\psi+{m_0}^2c^4$$ $$\Rightarrow  \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\psi-\nabla^2\psi+\frac{{m_0}^2c^2}{\hbar^2}\psi=0\Rightarrow \Box\psi+\frac{{m_0}^2c^2}{\hbar^2}\psi=0$$ This is the Klein-Gordon  equation. This equation solutions  is $$Ae^\frac{i p^\mu x_\mu}{\hbar}=Ae^\frac{i \left(Et-\mathbf{p\cdot x}\right)}{\hbar}$$ Where A is a normalization constant. Now substituting this solution imto Klein-Gordon  equation gives $$E^2=c^2P^2+{m_0}^2c^4\Rightarrow E=\pm\sqrt{c^2P^2+{m_0}^2c^4}$$Therefore Klein-Gordon equation gives negative energy solu...

The Maxwell's  equations are $$(1)\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$ $$(2)\nabla\cdot\mathbf{B}=0$$ $$(3)\nabla \times\mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$$ $$(4)\nabla \times\mathbf{B}=-\mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}+\mu_0\mathbf{J}$$ From (3) we get: $$\nabla\times\mathbf{E}=-\frac{\partial}{\partial t}(\nabla\times\mathbf{A})[\because \mathbf{B}=\nabla\times\mathbf{A}]\Rightarrow\nabla\times \left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right )=0$$$$\Rightarrow\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}=-\nabla\phi\Rightarrow\mathbf{E}= -\left(\frac{\partial\mathbf{A}}{\partial t}+\nabla\phi\right)$$ Now electromagnetic field tensor defined as $$F^{\mu\nu}=\partial ^{\nu}A^{\mu}-\partial ^{\mu}A^{\nu}=\frac{\partial A^{\mu}}{\partial x_{\nu}}-\frac{\partial A^\nu}{\partial x_{\mu}}$$   Electromagnetic  Four  Potential  $$ \space A^{\mu\nu}=\{\frac{\phi}{c},A_x,A_y,A_z\}$$ $$F^{00}=...