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Consider $$\int_{-\infty}^{+\infty} x \dfrac{\partial}{\partial x}\left(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}\right) dx $$

If I apply integraton by parts here by bringing in the x inside the derivative then its $$\int_{-\infty}^{+\infty} \dfrac{\partial}{\partial x}\left(x\cdot\left(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}\right)\right) dx $$ and taking $f = x$ and $g' = (\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x})$ the integral should be: $$-\int_{-\infty}^{+\infty} \dfrac{\partial x}{\partial x}\left(\int \left(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}\right) dx\right)dx$$. We know that $\dfrac{\partial x}{\partial x} = 1$ so integral becomes $$-\iint_{-\infty}^{+\infty} \left(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}\right) dxdx$$

Why isn't this right? The answer is just:$$-\int_{-\infty}^{+\infty} \left(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\dfrac{\partial \Psi^*}{\partial x}\right) dx$$ where did the other integral vanish?

Also, I ignored the boundary term because $\Psi \to 0$ at $\pm\infty$.

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You are not applying integration by parts correctly. You take $$ g' = \Bigl(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\,\dfrac{\partial \Psi^*}{\partial x}\Bigr), $$ and it should be $$ g= \Bigl(\Psi^*\dfrac{\partial \Psi}{\partial x} - \Psi\,\dfrac{\partial \Psi^*}{\partial x}\Bigr). $$ Then $$ \int_{-\infty}^\infty x\,\frac{\partial}{\partial x}g(x)\,dx=\text{boundary term}-\int_{-\infty}^\infty g(x)\,dx. $$

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  • $\begingroup$ But $\int f(x)g'(x) dx = \text{boundary term} - \int g(x) du$ where $u = f(x)$. According to this if $f(x) = x$ then the other term should be equal to $g'$ instead of $g$ no? $\endgroup$ Jun 20 '18 at 16:48
  • $\begingroup$ Or, are you saying to take $f' = x$ and $g$ to be the other term ? $\endgroup$ Jun 20 '18 at 16:51
  • $\begingroup$ @MohammadAreebSiddiqui Your integrand is on the form $f(x)\frac{dg(x)}{dx}$ so if you want to use integration by parts you would put $u = f(x)$ and $v' = \frac{dg(x)}{dx}$ or $v = g(x)$. This makes sure the integrand is on the form $uv'$. You have taken $v = \frac{dg(x)}{dx}$ for which the integrand is $uv$ and you can not directly apply integration by parts on this. He is saying you made the wrong choice for $v$ (or what you call $g$). $\endgroup$
    – Winther
    Jun 20 '18 at 17:00
  • $\begingroup$ The comment by Winther explains it. $\endgroup$ Jun 20 '18 at 17:05
  • $\begingroup$ Ohh right! Got it! $\endgroup$ Jun 20 '18 at 18:27

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