I found in the literature that the energy of the Hartree–Fock method is given by the following equation:

$$E_\mathrm{HF} = \int (\Psi^* \hat{H}\Psi)\,\mathrm d\tau$$

The term $\Psi$ is the wave function, but what would the term $\Psi^*$ be?

Does anyone have a simple of explanation of what this may be?

  • 1
    $\begingroup$ Something with an asterisk in QC usually refers to a complex conjugate. See Wikipedia and a related question on Physics.SE. $\endgroup$
    – andselisk
    Jul 27, 2020 at 12:06
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    $\begingroup$ I had seen it on wikipedia but thought there would be another explanation for that (physical meaning or something like that). Thanks for the references @andselisk $\endgroup$ Jul 27, 2020 at 12:16
  • $\begingroup$ I highly recommend that you review the basic math of quantum mechanical operators before tackling the HF method. It will make a lot more sense that way. $\endgroup$
    – Andrew
    Jul 27, 2020 at 21:15
  • $\begingroup$ @Andrew would you have review articles in the area to kindly recommend reading? $\endgroup$ Jul 28, 2020 at 9:44
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    $\begingroup$ Levine's Quantum Chemistry textbook has a great section just on the math that only assumes familiarity with standard calculus. $\endgroup$
    – Andrew
    Jul 28, 2020 at 12:52

1 Answer 1


For every $x$ and $t,$ $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use an example:

$$\Psi = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{-\mathrm iE_n t}.$$

We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*),$ let's do it step by step.

First we need to conjugate $\displaystyle\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\displaystyle\sin\left(\frac{n\pi x}{a}\right),$ but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $\mathrm e^{-\mathrm iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $\mathrm e^{\mathrm iE_n t}.$ So putting it all together, we have

$$\Psi^* = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{\mathrm iE_n t}.$$


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