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I came across this equation (identity) (Eq. 4 in this paper):

$\int(-i d\psi/dx)^*\psi dx = \int \psi^*(-i d\psi/dx) dx + id(\psi^*\psi)/dx\mid_{-\infty}^{+\infty}$

I have trouble proving it. I tried to use integration by parts but could not reach there. How do we take complex conjugate (Hermitian Adjoint) of the differential operator that occurs in this equation and also of any general function.

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Notice that \begin{align} i\frac{d(\psi^*\psi)}{dx} &=\frac{d\big[(-i\psi)^*\psi\big]}{dx} \\ &= \frac{d(-i\psi)^*}{dx}\psi + (-i\psi)^*\frac{d\psi}{dx} \\ &= \left(-i\frac{d\psi}{dx}\right)^*\psi + \psi^*\left(i\frac{d\psi}{dx}\right) \\ \end{align} Now subtract the second term on the right from both sides to get \begin{align} \psi^*\left(-i\frac{d\psi}{dx}\right)+i\frac{d(\psi^*\psi)}{dx} &= \left(-i\frac{d\psi}{dx}\right)^*\psi \end{align} and finally integrate both sides from $-\infty$ to $\infty$ to obtain (as Stan Liou pointed out in the comments) $$ \int_{-\infty}^\infty\psi^*\left(-i\frac{d\psi}{dx}\right)+i\psi^*\psi\Big|_{-\infty}^\infty = \int_{-\infty}^\infty\left(-i\frac{d\psi}{dx}\right)^*\psi $$ Notice that the boundary term you wrote in the identity has an erroneous derivative that goes away when you actually evaluate the integral and use the fundamental theorem of calculus.

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    $\begingroup$ You're right, except for that it's not the desired identity, because the desired identity is just plain false. I suspect a typo in the article. $\endgroup$
    – Stan Liou
    Apr 26, 2013 at 6:18
  • $\begingroup$ @StanLiou Hmm really? Is there a sign missing perhaps? I can't see where there's an error. $\endgroup$ Apr 26, 2013 at 6:21
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    $\begingroup$ When you integrate, you get an $i(\psi^*\psi)\mid_{-\infty}^{+\infty}$ term, rather than $id(\psi^*\psi)/dx\mid_{-\infty}^{+\infty}$ as the OP has. $\endgroup$
    – Stan Liou
    Apr 26, 2013 at 6:23
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    $\begingroup$ Oh of course. Thanks for pointing that out! I somehow kept seeing the correct boundary term even though i was staring at it for a few minutes. Thanks Stan. $\endgroup$ Apr 26, 2013 at 6:25
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    $\begingroup$ @DurgaDatta Once the extra derivative is removed (since it's inconsistent for dimensional reasons as Josh said), the formula is just a particular case of integration by parts with some $i$'s sprinkled in, whose conjugates change some signs. So you can prove it by IbP formula without moving any terms across sides. But IbP is itself just a re-arrangment of the integral of the Leibniz rule for differentiation, so really you'd be doing the same thing as this answer anyway. $\endgroup$
    – Stan Liou
    Apr 26, 2013 at 22:06

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