13
$\begingroup$

I am very confused regarding these four terms- ψ,|ψ|², radial probability and radial distribution

I know that ψ is called the wave function but is it the same as radial probability?

I also know that |ψ|² is called the probability density of finding an electron inside an atom but is it the same as radial distribution—which is the probability of finding an electron in a volume $\mathrm dV$ inside a nucleus?

$\endgroup$
1
  • $\begingroup$ You ask way too many questions at once. Radial probability is not a thing at all, and the rest are all different thing. $\endgroup$ Mar 13, 2018 at 13:39

3 Answers 3

17
$\begingroup$

All the four terms are different and they represent different concepts in quantum mechanics.
Firstly, the term $\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\Psi(x,y,z,t) = \sqrt{\frac{8}{abc}} \sin\left(\frac{n_x\pi x}{a}\right) \sin\left(\frac{n_y\pi y}{b}\right) \sin\left(\frac{n_z\pi z}{c}\right) e^{-2\pi iEt/h}$$ is an example of a wave-function.

But the $|\Psi|^2$ is mathematically defined as $\Psi\cdot\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\Psi(x,y,z,t)|^2 = \frac{8}{abc} \sin^2\left(\frac{n_x\pi x}{a}\right) \sin^2\left(\frac{n_y\pi y}{b}\right) \sin^2\left(\frac{n_z\pi z}{c}\right)$$ Here the complex part will not appear as in the previous example because $\Psi$ is multiplied with its complex conjugate.

The probability of finding the particle in a unit volume element $\mathrm dV$ is $|\Psi|^2 \mathrm dV$. In spherical polar coordinates, it is $|\Psi|^2 r^2 \, \mathrm dr \, \sin\theta \, \mathrm d\theta \, \mathrm d\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\pi$ as, $\int_{0}^{2\pi} \int_{0}^{\pi} \sin\theta \, \mathrm{d}\theta \, \mathrm{d}\phi =4\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\Pr(r) = |\Psi|^2 4 \pi r^2 \, \mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.

$\endgroup$
1
  • 4
    $\begingroup$ Nice answer. Just would change 'When you are only concerned with radial part' by 'if the radial part of the wave function is separable' $\endgroup$
    – user43021
    Mar 13, 2018 at 15:30
3
$\begingroup$

According to the Copenhagen interpretation of quantum mechanics,

$|\Psi|^2$ is the "probability density" (the probability per volume of finding a particle, such as an electron, in a given volume, in the limit the volume approaches zero).

If $\Psi(r,\theta,\phi)$ is separable as $R(r)Y(\theta,\phi)$, such as in the hydrogen atom, then:

"Radial Probability Density" is $R^2(r)$ (still proportional to probability per volume for s-states)

"Radial Probability Distribution" is $4\pi r^2R^2(r)$ (probability per incremental distance from origin, as the incremental distance approaches zero)

$\endgroup$
2
  • $\begingroup$ Actually Born's rule doesn't depend on Copenhagen interpretation. In MWI, Bohm-de Broglie theory and any other interpretation it's still valid, because it is a testable rule, unlike internals of the interpretations. $\endgroup$
    – Ruslan
    Mar 13, 2018 at 19:00
  • 1
    $\begingroup$ @Ruslan I don't think there is complete agreement on that, according to this: en.wikipedia.org/wiki/Quantum_non-equilibrium $\endgroup$
    – DavePhD
    Mar 13, 2018 at 19:17
2
$\begingroup$

I'd say the wave function is your $ψ$. This wavefunction describes your system. When you want to determine something like its energy or other operations you need to describe your system like your atom using a wave function. $|ψ|²$ is as you already said the probability density.

Your wavefunction for example in solving the hydrogen atom using the Schrödinger-equation has this step where you transform to spherical coordinates and then seperate into a radial part and an angular part. From this radial part you can get the radial probability, so to find an electron for example at the distance x. And the radial distribution is this function times the area of a sphere having the size of the radius for that distance you are looking at the moment. So this will give you the probability for the electron to be located somewhere on a sphere at the distance x.

But I can't really tell you the actual difference between taking only the radial part or the wave function itself to describe probability.

$\endgroup$
0

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.