Sunday, 16 December 2018

quantum mechanics - Solution of the coupled non-linear oscillators by using perturbation theory



The integration shown here, $$∫_{-\infty}^{+∞}x^r\mathrm{Exp}[−x^2]\mathrm{H_n}^2[x]\mathrm{d}x,$$ appears when we try to calculate the spectrum of the perturbed non-linear oscillators by using perturbation theory in quantum mechanics. Is there any direct way to perform the definite integration of the form shown above. I want the solution of the integral for $r≥4$. I hope there may exist some techniques which can be used to calculate the integration of above integral. Please i need suggestion from this forum to solve this integration. Highly appreciated!



Answer



$\textbf{Note :}$ Below a generating function is derived for calculating the following matrix element (useful in perturbative computations) : $$O_{m,n,k}^{}=\frac{2_{}^{k}}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{} H_{m}^{}(x) x_{}^{k} H_{n}^{}(x) e_{}^{-x_{}^{2}}.$$ where $\{H_{r}^{}(x)\}$ are hermite polynamials and $k$ is a non-negative integer.



(i) Consider the generating function of the Hermite polynomials :


$$G[z,x]=\sum_{n=0}^{\infty}\frac{z_{}^{n}}{n!}H_{n}^{}(x)=e_{}^{2 x_{}^{} z_{}^{}-z_{}^{2}}.$$


(ii) Define a generating function $Z[z_{1}^{},z_{2}^{},z_{3}^{}]$ :


$$Z[z_{1}^{},z_{2}^{},z_{3}^{}]=\frac{1}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{}e_{}^{-x_{}^{2}}G[z_{1}^{},x]G[z_{2}^{},x]e_{}^{2x z_{3}^{}}=\frac{e_{}^{-[z_{1}^{2}+z_{2}^{2}]}}{\sqrt{\pi}}\int_{-\infty}^{+\infty}dx_{}^{}e_{}^{-x_{}^{2}+2x[z_{1}^{}+z_{2}^{}+z_{3}^{}]}$$ $$\Rightarrow Z[z_{1}^{},z_{2}^{},z_{3}^{}]=e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}.$$ (iii) Now notice : $$O_{m,n,k}^{}=\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}Z[z_{1}^{},z_{2}^{},z_{3}^{}]\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}.$$ (iv) Hence : $$O_{m,n,k}^{}=\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}$$ $$\therefore \int_{-\infty}^{+\infty}dx_{}^{} H_{m}^{}(x) x_{}^{k} H_{n}^{}(x) e_{}^{-x_{}^{2}}=\frac{\sqrt{\pi}}{2_{}^{k}}\left(\frac{\partial}{\partial z_{1}^{}}\right)_{}^{m}\left(\frac{\partial}{\partial z_{2}^{}}\right)_{}^{n}\left(\frac{\partial}{\partial z_{3}^{}}\right)_{}^{k}e_{}^{-[z_{1}^{2}+z_{2}^{2}]+[z_{1}^{}+z_{2}^{}+z_{3}^{}]_{}^{2}}\Big|_{(z_{1}^{},z_{2}^{},z_{3}^{})=(0,0,0)}^{}$$


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...