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?


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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ZeroTheHero, Jon Custer, stafusa, AccidentalFourierTransform
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ 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
  • 2
    $\begingroup$ When taking the complex conjugate $i \rightarrow i^*=-i$ and $\psi \rightarrow \psi^*$ $\endgroup$ – pp.ch.te Sep 18 '18 at 10:31
  • 2
    $\begingroup$ 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

The conjugate of the product of two numbers is the product of their conjugates:

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


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


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

  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.