A Quantum Algebra Supplement - Sample Applications

Sample Applications

The sample applications are given in the form of notebooks. For the time being there are the following:

• A notebook for the Klein - Gordon - equation
• A notebook for the Dirac - equation

In addition to local rules, they contain a series of calculations, e.g. Hamilton - operator, 3 - momentum and evaluation of charge.

Klein - Gordon - Equation

In all cases, the free fields that solve the field equations are specified.

1. The commutator of the fields, when applying the commutation relations for the coefficients of the fields, leads to the commutation relations of the fields themselves.
   

   $\begin{array}{l}\text{(* Define the wave function solutions *)}\\ \mathrm{\Phi C}:={\int }_{\infty }\left(1/\mathrm{Sqrt}\left[2\left(2\mathrm{Pi}\right)^3k_0\right]{\hat{a}}_{k_1,k_2,k_3}\mathrm{Exp}\left[-i\exp \right]\right)d^{3}k+{\int }_{\infty }\left(1/\mathrm{Sqrt}\left[2\left(2\mathrm{Pi}\right)^3k_0\right]{\hat{b}}_{k_1,k_2,k_3}^{†}\mathrm{Exp}\left[i\exp \right]\right)d^{3}k;\\ \mathrm{phi}=\mathrm{\Phi C}//\mathrm{QASimplify}\left[\right];\\ \mathrm{phicross}=\left({\mathrm{\Phi C}}^{†}//\mathrm{QASimplify}\left[\right]\right);\\ \text{(* Define the time derivative of wave function solution *)}\\ \mathrm{phicrossdot}=\left(\left({?}_{x_0}\mathrm{phicross}\right)/.\left\{k\to \mathrm{k1},x\to y\right\}\right)/.\left\{y_0\to x_0\right\};\\ \text{(* Evaluate the Commutator *)}\\ \mathrm{QAEvaluate1}[\mathrm{phi}\cdot \mathrm{phicrossdot}-\mathrm{phicrossdot}\cdot \mathrm{phi}]\\ i\mathrm{DiracDelta}\left[x_1-y_1,x_2-y_2,x_3-y_3\right]\hat{1}\\ \end{array}$

2. The Hamiltonian as an expression dependent on the fields leads to an expression in the coefficients of the fields (creation and annihilation operators).
   

   $$\begin{array}{rl}& \mathrm{\Phi }:=\sum _k(1/\mathrm{Sqrt}\left[2V{k}_0\right]{\hat{a}}_{k_1,k_2,k_3}\mathrm{Exp}[-i\exp ])+\sum _k(1/\mathrm{Sqrt}\left[2V{k}_0\right]{\hat{b}}_{k_1,k_2,k_3}^{†}\mathrm{Exp}[i\exp ])\\ & {\mathfrak{H}}_1:=({\mathfrak{D}}_{x_0}(\mathrm{\Phi }//\text{Evaluate}))\cdot ({\mathfrak{D}}_{x_0}\left(({\mathrm{\Phi }}^{†}//\text{QASimplify}[])//\text{Evaluate}\right)/.\{k\to k1\})\\ & {\mathfrak{H}}_2:=\sum _{\mathrm{i}=1}^3(\left({\mathfrak{D}}_{{\mathrm{x}}_{\mathrm{i}}}({\mathrm{\Phi }}^{†}//\text{QASimplify}[]//\text{Evaluate})\right)\cdot ({\mathfrak{D}}_{{\mathrm{x}}_{\mathrm{i}}}(\mathrm{\Phi }//\text{Evaluate})/.\{\mathrm{k}\to \mathrm{k}1\}))\\ & {\mathfrak{H}}_3:=({\mu }^{\wedge }2({\mathrm{\Phi }}^{†}//\text{QASimplify}[])\cdot (\mathrm{\Phi }/.\{\mathrm{k}\to \mathrm{k}1\}))//\text{QASimplify}[]\\ & H_1:={\int }_V{\mathfrak{H}}_1{d}^3\mathrm{x}\\ & H_2:={\int }_V{\mathfrak{H}}_2{d}^3\mathrm{x}\\ & H_3:={\int }_V{\mathfrak{H}}_3{d}^{3}x\\ & \text{hamiltonian = QAEvaluate}2[{\mathrm{H}}_1+{\mathrm{H}}_2+{\mathrm{H}}_3]\\ & \hat{1}\sum _k{k}_0+\sum _k{\hat{a}}_{k_1,k_2,k_3}^{†}\cdot {\hat{a}}_{k_1,k_2,k_3}{k}_0+\sum _k{\hat{b}}_{k_1,k_2,k_3}^{†}\cdot {\hat{b}}_{k_1,k_2,k_3}{k}_0\\ & \end{array}$$

3. The same applies to the expression of the 3-momentum.
   

   $$\begin{array}{rl}& {\mathfrak{P}}_{\mathrm{i\_}}:=({\mathfrak{D}}_{x_0}(\mathrm{\Phi }//\text{Evaluate}))\cdot ({\mathfrak{D}}_{x_i}\left(({\mathrm{\Phi }}^{†}//\text{QASimplify}[])//\text{Evaluate}\right)//.\{k\to k1\})+\\ & \left({\mathfrak{D}}_{x_0}\left(({\mathrm{\Phi }}^{†}//\text{QASimplify}[])//\text{Evaluate}\right)\right)\cdot ({\mathfrak{D}}_{x_i}(\mathrm{\Phi }//\text{Evaluate})//.\{k\to k1\})\\ & P_{\mathrm{i\_}}:={\int }_V{\mathfrak{P}}_i{d}^{3}x\\ & \text{Map}[\left(P_{\mathrm{\#1}}//\text{QAEvaluate2}\right)\mathrm{\&},\{1,2,3\}]\\ & \{-\sum _k{\hat{a}}_{k_1,k_2,k_3}^{†}\cdot {\hat{a}}_{k_1,k_2,k_3}{k}_1-\sum _k{\hat{b}}_{k_1,k_2,k_3}^{†}\cdot {\hat{b}}_{k_1,k_2,k_3}{k}_1\\ & -\sum _k{\hat{a}}_{k_1,k_2,k_3}^{†}\cdot {\hat{a}}_{k_1,k_2,k_3}{k}_2-\sum _k{\hat{b}}_{k_1,k_2,k_3}^{†}\cdot {\hat{b}}_{k_1,k_2,k_3}{k}_2,\\ & -\sum _k{\hat{a}}_{k_1,k_2,k_3}^{†}\cdot {\hat{a}}_{k_1,k_2,k_3}{k}_3-\sum _k{\hat{b}}_{k_1,k_2,k_3}^{†}\cdot {\hat{b}}_{k_1,k_2,k_3}{k}_3\}\end{array}$$

4. The same applies to the expression for the charge, which results from a continuity equation.
   

   $$\begin{array}{rl}& {\mathfrak{J}}_{\mathrm{i\_}}:=i(({\mathfrak{D}}_{x_i}(\mathrm{\Phi }//\text{Evaluate}))\cdot (({\mathrm{\Phi }}^{†}//\text{QASimplify}[])//.\{k\to k1\})-\\ & <malignmark></malignmark>({\mathfrak{D}}_{x_i}\left(({\mathrm{\Phi }}^{†}//\text{QASimplify}[])//\text{Evaluate}\right))\cdot (\mathrm{\Phi }//.\{k\to k1\}))\\ & J_{\mathrm{i\_}}:={\int }_V{\mathfrak{J}}_i{d}^{3}x\\ & \text{QAEvaluate2}\left[J_0\right]\\ & \sum _k{\hat{a}}_{k_1,k_2,k_3}^{†}\cdot {\hat{a}}_{k_1,k_2,k_3}-\sum _k{\hat{b}}_{k_1,k_2,k_3}^{†}\cdot {\hat{b}}_{k_1,k_2,k_3}\\ & \end{array}$$