# Is there actually a 0 probability of finding an electron in an orbital node?

I have recently read that an orbital node in an atom is a region where there is a 0 chance of finding an electron.

However, I have also read that there is an above 0 chance of finding an electron practically anywhere in space, and such is that orbitals merely represent areas where there is a 95% chance of finding an electron for example.

I would just like to know if there truly is a 0 probability that an electron will be within a region defined by the node.

Many thanks.

• – AccidentalFourierTransform Jan 1 '17 at 16:31
• – AccidentalFourierTransform Jan 1 '17 at 16:32
• I do not believe that these are complete duplicates, however thank you for the links, they have been quite interesting. – Benjamin Rogers-Newsome Jan 1 '17 at 16:43
• Orbitals are only models describing behavior of electrons, by no means perfect ones. – Mithoron Jan 1 '17 at 21:16

The probability of finding the electron in some volume $V$ is given by:

$$P = \int_V \psi^*\psi\,dV \tag{1}$$

That is we construct the function called the probability density:

$$F(\mathbf x, t) = \psi^*\psi$$

and integrate it over our volume $V$, where as the notation suggests the probability density is generally a function of position and sometimes also of time.

There are two ways the probability $P$ can turn out to be zero:

1. $F(\mathbf x, t)$ is zero everywhere in the volume $V$ - note that we can't get positive-negative cancellation as $F$ is a square and is everywhere $\ge 0$.

2. we take the volume $V$ to zero i.e. as for the probability of finding the particle at a point

Now back to your question.

The node is a point or a surface (depending on the type of node) so the volume of the region where $\psi = 0$ is zero. That means in our equation (1) we need to put $V=0$ and we get $P=0$ so the probability of finding the electron at the node is zero. But (and I suspect this is the point of your question) this is a trivial result because if $V=0$ we always end up with $P=0$ and there isn't any special physical significance to our result.

Suppose instead we take some small but non-zero volume $V$ centred around a node. Somewhere in our volume the probability density function will inevitably be non-zero because it's only zero at a point or nodal plane, and that means when we integrate we will always get a non-zero result. So the probability of finding the electron near a node is always greater than zero even if we take near to mean a tiny, tiny distance.

So the statement the probability of finding the electron at a node is zero is either vacuous or false depending on whether you interpret it to mean precisely at a node or approximately at a node.

But I suspect most physicists would regard this as a somewhat silly discussion because we would generally mean that the probability of finding the elecron at a node or nodal surface is nebligably small compared to the probability of finding it elsewhere in the atom.

• "The node is a 1D or 2D structure" <-- that probably deserves a reference to the fact the solutions to the wave equation cannot be constant over sets of nonzero volume, else the entire solution is constant. – DanielSank Jan 1 '17 at 18:05
• "The node is a 1D or 2D structure" <-- If I understand the maths right, this should rather say "The node is a 0D, 1D or 2D structure". That is, a point, a curve, or a surface. – Jirka Hanika Jan 2 '17 at 11:06
• @JirkaHanika: yes, thanks, good point. Actually I can't think of any nodal lines in atomic orbitals so the nodes are either points or surfaces. – John Rennie Jan 2 '17 at 11:29
• There is a sense in which we can make the statement of "neglegibly small" (you have a typo there) precise. garyp suggested it to me in a comment to this answer (see the answer for the actual statement). – ACuriousMind Jan 2 '17 at 13:32

You're quite right: the probability of finding the electron at any single point (or on any particular surface) is zero. Nevertheless, the statement makes sense: what it really means is roughly the following.

Consider a box $V$ with width/depth/height $(w,d,h)$. If all of these are sufficiently small that the wavefunction doesn't variate substantially throughout the box, you can approximate $$P = \int_V\!\mathrm{d}V\, |\psi(\mathbf r)|^2 \approx |V| \cdot |\psi(\mathbf r_c)|^2 = w\cdot d\cdot h \cdot |\psi(\mathbf r_c)^2|$$ where $\psi(\mathbf r_c)$ is the wave function evaluated at (say) the centre of the box. Now, if that point happens to lie within a nodal plane then the above approximation yields zero.

Strictly speaking, this is of course just wrong: basically the approximation breaks down since there is no zeroth term in the Taylor expansion, hence the dominant term becomes the linear one even on an arbitrarily small range. However, in that more suitable approximation you'd still get “virtually zero” as the result: say the nodal plane is in xy direction and $h$ measures in z direction. Then the integral becomes \begin{aligned} P \approx&\, w\cdot d\cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,|\psi(\mathbf r_c + z\cdot \mathbf{e}_\mathrm{z})|^2 \\ \approx&\, w\cdot d\cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,\left|z\cdot\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \\ =&\, w\cdot d\cdot \left|\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \cdot \int\limits_{-h/2}^{h/2}\!\!\mathrm{d}z\,(z^2) \\ =&\, w\cdot d\cdot \left|\frac{\partial\psi}{\partial z}\Bigr|_{\mathbf{r}_c}\right|^2 \cdot \frac23 \left(\frac{h}2\right)^3 \propto h^3 \end{aligned} so as you make $h$ small, the probability diminishes not just proportionally as the volume does, but with the third power, making it very small indeed.

• So the probability is not exactly 0, it is just very small? – Benjamin Rogers-Newsome Jan 1 '17 at 23:33
• Well, nothing is really exact in physics, there's always some measurement uncertainty. The probability is zero in the same sense that $(\mathrm{d}x)^2 = 0$ when you're “computing” the derivative of a function like $\frac{\mathrm{d}(x^2)}{\mathrm{d}x} \equiv \frac{(x + \mathrm{d}x)^2 - x^2}{\mathrm{d}x} = \frac{x^2 + 2\cdot x\cdot\mathrm{d}x + (\mathrm{d}x)^2 - x^2}{\mathrm{d}x} \stackrel{!}{=} \frac{2\cdot x\cdot\mathrm{d}x}{\mathrm{d}x} = 2\cdot x$. – leftaroundabout Jan 1 '17 at 23:54
• @leftaroundabout Please notice that the "measurement uncertainty" is something completely different: if you consider it as an experimental uncertainty then it's only an experimental limit but the precise value does exist, by definition; if, instead, you refer to the QM uncertainty then it only holds for non-commuting observables and still doesn't fit the answer. – gented Jan 2 '17 at 14:20
• @GennaroTedesco I consider it as an experimental uncertainty. Whether the exact value of a classical quantity actually exists is a philosophical question; at any rate if a quantity cannot by experiment be determined as $\neq 0$ then it is legitimate to say the quantity is zero. And if $P\propto h^3$ and we assume that the measurement uncertainties of $P$ and $h$ scale by the same factor as we improve our experiment (which is certainly not a given!), then we will at some point conclude that $P=0$ while the volume is still $>0$. – leftaroundabout Jan 2 '17 at 14:29
• @leftaroundabout I would say that you're wrong: "Whether the exact value of a classical quantity actually exists is a philosophical question" is false. In classical mechanics (or any other theory with only commuting observables) the exact values do exist and are precisely defined (as the number of times you have to repeat the unit lenght into the measurement or the like). – gented Jan 2 '17 at 15:16