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I read from Griffith's Introduction to Quantum Mechanics that:

Now the Schrodinger equation says that, $$\frac{\partial\Psi}{\partial t}=\frac{i\hbar}{2m}\,\frac{\partial^2\Psi}{\partial x^2}-\frac{i}{\hbar}V\Psi\tag{1.23}$$ and hence also (taking the complex conjugate of Equation 1.23) $$\frac{\partial\Psi^*}{\partial t}=-\frac{i\hbar}{2m}\,\frac{\partial^2\Psi^*}{\partial x^2}+\frac{i}{\hbar}V\Psi^*.\tag{1.24}$$

How is 1.24 arrived from 1.23? What is the math rule that was used?

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closed as off-topic by ZeroTheHero, Jon Custer, stafusa, Kyle Kanos, AccidentalFourierTransform Sep 20 '18 at 13:50

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    $\begingroup$ Like it says, take the complex conjugate of each side of the equation $\endgroup$ – By Symmetry Sep 18 '18 at 10:13
  • $\begingroup$ @BySymmetry why was there a sign change on the right side and not on the left side? $\endgroup$ – Taenyfan Sep 18 '18 at 10:14
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    $\begingroup$ When taking the complex conjugate $i \rightarrow i^*=-i$ and $\psi \rightarrow \psi^*$ $\endgroup$ – pp.ch.te Sep 18 '18 at 10:31
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    $\begingroup$ The complex conjugate is an operation that puts a minus sign only in the imaginary part of a complex number. $\endgroup$ – user171780 Sep 18 '18 at 10:44
  • $\begingroup$ Please don't paste images of text and formulae, instead you should copy it into the post so it can be properly indexed by search engines. $\endgroup$ – Kyle Kanos Sep 18 '18 at 16:44
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The conjugate of the product of two numbers is the product of their conjugates:

$$ (uv)^* = u^* v^* $$

So:

$$ \left( \frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2} \right)^* = i^* \left( \frac{\hbar}{2m} \right)^* \left( \frac{\partial^2\psi}{\partial x^2} \right)^* = -i \frac{\hbar}{2m} \frac{\partial^2\psi^*}{\partial x^2} $$

Likewise:

$$ \left(\frac{iV}{\hbar}\psi\right)^* = i^* \left(\frac{V}{\hbar}\right)^* \psi^* = -i \frac{V}{\hbar}\psi^* $$

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  • $\begingroup$ For completeness you might need to add that complex conjugate of x-derivative is derivative of complex conjugate (applied on $\psi$). $\endgroup$ – npojo Sep 18 '18 at 12:01
  • $\begingroup$ Also, it seems that $\hbar$ and not $h$ should be in the denominator of the RHS (last line). $\endgroup$ – BowPark Sep 18 '18 at 15:23

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