1
$\begingroup$

For a non-degenerate ground state in a system with $N$ electrons, we may write the wave function as,

\begin{equation} \psi(r_1, r_2,..., r_N) \end{equation}

Where the $r_i$ represent the position of the $i^{th}$ electron. When we say the square of this wave function represents the probability of finding those $N$ electrons at $r_1,r_2,..., r_N$, it becomes the probability of finding the entire system in that particular configuration of electronic positions.

My question: what about the probability of finding an electron at any position $r$? The definition (notation) above makes me think we have a particular value of probability which is the same at each of those $r_1, r_2,..., r_N$ and $0$ elsewhere. That thought seems limited in scope.

Improve this question
$\endgroup$
2
  • $\begingroup$ Which electron are you talking about? The wavefunction you have written stands for the entire system, not just one electron. $\endgroup$
    – Yejus
    Commented Jan 2, 2021 at 10:31
  • $\begingroup$ Hi @Yejus, I was reading about the electron density in density functional theory and got curious if we could write the electron density in terms of this collective wavefunction. Weirdly, I asked it in a roundabout way 😅. $\endgroup$
    – Hitanshu Sachania
    Commented Jan 2, 2021 at 16:15

1 Answer 1

Reset to default
3
$\begingroup$

This is known as the electron density and it is obtained by integrating over the coordinates of all electrons except one.

For simplicity, let us take a 2-electorn case with wavefunction $\psi(r_1, r_2)$ and ignore the spin. Then the electron density is $$\rho(r) = \int dr_1 |\psi(r_1, r)|^2 + \int dr_2 |\psi(r,r_2)|^2 = 2 \int dr_2 |\psi(r, r_2)|^2$$ The last equality holds because the wavefunction is anti-symmetric $\psi(r_1,r_2) = - \psi(r_2, r_1)$. (In the general case, anti-symmetry involves exchanging all coordinates including the spin)

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for your answer. In fact, this question baffled me when I was reading about the electron density in density functional theory. This means we can define the electron density only in terms of single electron wavefunctions and not the collective wave function of all electrons in the system, isn't it? $\endgroup$
    – Hitanshu Sachania
    Commented Jan 2, 2021 at 16:10
  • $\begingroup$ The example I gave explicitly defines the electron density in terms of the collective wave function $\psi(r_1, r_2)$, so I'm not sure what do you mean with your comment. Are you asking about the electron density of several electrons? If so, then you can in general define the $n$-electron density by integrating over the coordinates of all electrons except $n$ ones. $\endgroup$
    – Quantum-Collapse
    Commented Jan 2, 2021 at 16:30
  • $\begingroup$ Is $\psi(r_1,r)$ the same as $\psi(r_1,r_2)$? Also, can you help me understand what does $\psi(r_1,r)$ mean? I thought a single eectron wave function would be denoted as $\psi(r_1)$ and a collective wave function as $\psi(r_1,r_2)$. $\endgroup$
    – Hitanshu Sachania
    Commented Jan 2, 2021 at 16:52
  • 1
    $\begingroup$ $\psi(r_1, r)$ and $\psi(r_1, r_2)$ are the same function evaluated at two different points...they give the same value when $r=r_2$. $\psi(r_1,r)$ is the wavefunction value when the first electron is at $r_1$ and the second electron is at $r$. $\endgroup$
    – Quantum-Collapse
    Commented Jan 2, 2021 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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