In the context of the wave functions of electron in a hydrogen atom, I just want to get clear regarding the terms probability and probability density of an finding an electron. To be precise, we know that the value of $|\Psi|^2$ at a point gives the probability density of the electron at the point. Now, does that imply that if the value of $|\Psi|^2$ is maximum at a point, the electron is most likely to be found at that point?

Reason for my doubt: The question given by my instructor goes:

Given that the normalized wave fuction of the electron in $\mathrm{3d}_{z^2}$ orbital of an Hydrogen atom is: $$ \Psi = N \sigma^2 \mathrm{e}^{-\frac{\sigma}{3}}(3\cos^2\theta-1) \, , $$ where $\sigma={\frac{r}{a_o}}$

Now where is the electron most likely to be found?

The solution given goes like: If $P$ represents probability of finding the electron and $\mathrm{d}V$ the volume element then, $$ \frac{\mathrm{d}P}{\mathrm{d}V} = |\Psi|^2 \implies \mathrm{d}P=|\Psi|^2 \mathrm{d}V$$ Also since, $$\mathrm{d}V = r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi$$ We get, \begin{align} \mathrm{d}P &= |\Psi|^2 r^2 \sin\theta \mathrm{d}r \mathrm{d}\theta \mathrm{d}\phi \\ &= N_1 \sigma^4 \mathrm{e}^{-\frac{2\sigma}{3}}{(3\cos^2\theta-1)^2} \sigma^2 \sin\theta \mathrm{d}\sigma \mathrm{d}\theta \mathrm{d}\phi \\ & = N_1 [ \sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}} \mathrm{d}\sigma ][ {(3\cos^2\theta-1)^2} \sin\theta \mathrm{d}\theta ][ \mathrm{d}\phi] \end{align}

He later maximized the $[\sigma^6 \mathrm{e}^{-\frac{2\sigma}{3}}]$ and $[{(3\cos^2\theta-1)^2} \sin{\theta}]$ terms to get the values of ${\sigma}$ i.e. hence $r$ and ${\theta}$ at which the electron is most likely to be found. Infact, he stressed on the maximization of ${\theta}$ part which came out to be $0$. And hence, a conclusion made that for an electron in $\mathrm{3d}_{z^2}$ orbital, the electron is most likely to be found in the XY plane.

I find all this stuff confused and fishy. Can someone explain me this solution or prove it wrong via a right way of solving this question.

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    $\begingroup$ I made an attempt to improve math expressions a bit. Note to OP: the probability density is $|\Psi|^2$ and not $|\Psi^2|$. I'm not sure if the later is even defined for a complex-valued function. :) $\endgroup$ – Wildcat Oct 10 '16 at 10:17
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    $\begingroup$ I can assure you, $|\Psi|^2$ and $|\Psi^2|$ are exactly the same thing. $\endgroup$ – Ivan Neretin Oct 10 '16 at 10:23
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    $\begingroup$ @Wildcat The square of a complex number is that number multiplied by itself. You can multiply arbitrary complex numbers, can't you? $\endgroup$ – Ivan Neretin Oct 10 '16 at 10:43
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    $\begingroup$ @IvanNeretin, cause if you define the square of a complex number $c=a+b\mathrm{i}$ as $c^2:=c \cdot c$, then $|c|^2 \neq |c^2|$ since $|c|^2:=c^* \cdot c=a^2+b^2$ while $|c^2| :=c \cdot c = |a^2+2ab\mathrm{i}-b^2|$. $\endgroup$ – Wildcat Oct 10 '16 at 11:19
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    $\begingroup$ Now that's totally right, and these two expressions are in fact equal! See, $|c^2| = |a^2+2ab\mathrm{i}-b^2|=\sqrt{(a^2-b^2)^2+(2ab)^2}=\sqrt{a^4+b^4-2a^2b^2+4a^2b^2}=\sqrt{a^4+b^4+2a^2b^2}=a^2+b^2$. $\endgroup$ – Ivan Neretin Oct 10 '16 at 11:20

Now, does that imply that if the value of $|\Psi|^2$ is maximum at a point, the electron is most likely to be found at that point?

It certainly does imply something similar to what OP is saying, but not quite. As always in quantum theory one has to be very careful with his or her wordings. Namely, we are talking about a continuous random variable here, so the only probability which makes sense is that of it falling within a particular range of values, i.e. probability of finding an electron in some infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ and not on this single point. Still, if you divide the whole space into volume elements of an identical (infinitesimally small) size, the bigger the value of $|\Psi|^2$ is, the bigger is the above mentioned probability.

Now to the actual findings. So, for an electron occupying the $\mathrm{3d}_{z^2}$ orbital, the probability of finding it in an infinitesimal element of volume $\mathrm{d}V$ centered at a particular location $r$ is given as $|\Psi|^2 \mathrm{d}V$, where $|\Psi|^2$ is the probability density. We analyze the probability density $$ |\Psi|^2 = N^2 \sigma^4 \mathrm{e}^{-2\sigma/3} (3\cos^2\theta-1)^2 \, , $$ or, more precisely only its angular $(3\cos^2\theta-1)^2$ part. So, when does this expression have its maximum value? The global maxima are at $\theta = 0$ and $\theta = \pi$, i.e. alongside $z$-axis, plus there are local maxima at $\theta = \pi/2$ and $\theta = 3\pi/2$, i.e. in a plane perpendicular to $z$-axis. 1

enter image description here

From the discussion with OP, it seems it must be noted that terms "probability density" and "probability" are defined with respect to a particular random event. Above we analyzed the probability density of finding an electron in an infinitesimal element of volume $\mathrm{d}V$ centered at a point $r$. The radial distribution function arises when considering a probability of a different event, namely, the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$. We start from the probability of finding an electron in infinitesimal volume element $\mathrm{d}V$ at a point $r$ $$ |\Psi|^2 \mathrm{d}V = \underbrace{R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2}_{|\Psi|^2} \underbrace{r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi}_{\mathrm{d}V} \, , $$ and integrate out $\theta$ and $\phi$ dependences $$ \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi} |\Psi|^2 \mathrm{d}V = \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi} R_{nl}^2(r) |Y_l^m(\theta, \phi)|^2 r^2 \mathrm{d}r \sin\theta \mathrm{d}\theta \mathrm{d}\phi \, . $$ And since the spherical harmonics $Y_l^m$ are normalized (to unity) $$ \int\limits_{0}^{2\pi} \int\limits_{0}^{\pi} |Y_l^m(\theta, \phi)|^2 \sin\theta \mathrm{d}\theta \mathrm{d}\phi = 1 \, , $$ for the probability of finding an electron between two concentric spheres of radii $r$ and $r + \mathrm{d}r$ we simply get $$ R_{nl}^2(r) r^2 \mathrm{d}r \, . $$ The corresponding probability density is, of course, $R_{nl}^2(r) r^2$. Such analysis is different from what we did above for the angular $\theta$ dependence: we simply ignored the radial and the other angular dependence instead of integrating them out. This ignorance makes our analysis a bit rough, but not worthless: at least we were able to identify some general trends in the angular $\theta$ dependence.

1) See Wolfram Alpha for details.

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    $\begingroup$ I'm going fine through the mathematics and I agree to your idea. But I doubt the way the question was solved. If the same idea is applied for an electron in 1s orbital, wouldn't we get that sin{\theta} should be maximum for the probability to be maximum in an infinitesimal dV volume element around that point. Now which would imply {\theta}=pi/2. Don't these conclusions seems counter-intuitive or awkward in the sense that in a 1s orbital, the probability of finding an electron in a dV volume element should be spherically symmetric. $\endgroup$ – user104014 Oct 10 '16 at 11:42
  • $\begingroup$ Sorry, there was a mistake: $\sin\theta$ should not have been counted in the probability calculations, since it is a part of $\mathrm{d}V$. $\endgroup$ – Wildcat Oct 10 '16 at 12:14
  • $\begingroup$ This particular mistake was in the question itself in the first place. @user104014, your intuition serves you right: the whole thing was clear as mud. $\endgroup$ – Ivan Neretin Oct 10 '16 at 12:17
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    $\begingroup$ Maximizing probability density, though it has no immediate physical meaning, at least might give you an insight at what the whole 3D density plot looks like. Maximizing probability density times $r^2\sin\theta$ has no meaning whatsoever. As for 1s, that's right, the probability density has a maximum at the origin. What's wrong with that? $\endgroup$ – Ivan Neretin Oct 10 '16 at 12:38
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    $\begingroup$ @user104014 Because, I suppose, they are interested in that radial probability distribution. To obtain it, you take the overall probability density and integrate over all possible values of $\varphi$ and $\theta$, thus making these two go away. Mind you, during integration you will introduce $r^2\sin\theta$. $\endgroup$ – Ivan Neretin Oct 10 '16 at 13:07

Although you ask about hydrogen atoms the same principle is true for any wavefunction. In quantum mechanics we often need to find the probability at some position x. However, x is a continuous variable with an infinite number of values so it does not make sense to ask what is the probability of being at some exact position, say $x=2/3$, as this will be vanishingly small.

Instead we imagine that the probability is that of being at position x to $x+dx$ where dx has some very small value. Suppose there is some general function $f(x)$ from we wish to find the probability density then $f(x)dx $ is probability of being between x and $x+dx$ and $f(x)$ is the probability density which must be a real, non-negative number. It turns out that quantum mechanics postulates that the probability density is given in terms of the wavefunction is $|\psi|^2$ and the $|~|$ means take the absolute value. If the wavefunction is a complex quantity, and it often is, then we write $\psi^*\psi$, where the * indicates the complex conjugate of the wavefunction $\psi$, i.e. replace i with $-i$ . Finding the maximum of the probability density is done in the same way as you would find the maximum/minimum of any function, differentiate, set equal to zero and solve. Check if maximum or minimum.

  • $\begingroup$ I agree to everything you said. But if why are we maximizing probability density? If thats the case, for an electron in 1s orbital, shouldn't it be that the electron is more likely to be found at the center, i.e. the nucleus instead of thr bohr radius? [because for 1s orbital, the wavefunction is an exponentially decreasing function like $ke^(-r)$] $\endgroup$ – user104014 Oct 10 '16 at 12:58
  • $\begingroup$ Notice that for your d-orbital, the wave function is actually something like $r^{2}e^{-r^{2}}$. The maximum value of the norm will occur at some non zero value of $r$. $\endgroup$ – Zhe Oct 10 '16 at 13:15
  • $\begingroup$ The atomic wavefunctions are often defined in spherical polar coordinates and so the volume element over which to integrate is in three dimensions and is $r^2\sin (\theta )dr d\theta d\phi$. The radial part can be separated out and the volume element is therefore $r^2dr$ which means that the probability is not at zero. For the 1s orbital the probability of being at r to $r+dr$ is $(4/a_0^3) r^2\exp(-2r/a_0)dr$ where $a_o$ is the Bohr radius. $\endgroup$ – porphyrin Oct 10 '16 at 22:55
  • $\begingroup$ @porphyrin Exactly, thats where my doubt is! If we know seperate the angular part, we'll be left with $\sin{\theta}d{\theta}d{\phi}$. Now maximizing this gives us ${\theta}=\frac{\pi}{2}$. Whats this? $\endgroup$ – user104014 Oct 11 '16 at 3:07
  • $\begingroup$ You asked specifically about the radial part. I should have mentioned that to calculate the total probability of finding an electron at r to $r+dr$, say in 1s orbital for simplicity, we need to account for the angular parts also, and do this by integrating over them as we seek radial probability, $r^2dr\int_0^{\pi}\sin(\theta)d\theta\int _0^{2\pi}\psi ^*(r,\theta ,\phi)\psi(r,\theta ,\phi) d\phi = (4/a_0^3)r^2\exp(-2r/a_0)dr $ $\endgroup$ – porphyrin Oct 11 '16 at 8:06

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