# How do electrons travel through nodes

I understand this is a basic question, but I'm having such a hard time wrapping my head around it. I'm trying to avoid thinking about it as an actual "particle" but as a wave, but that confuses me too. Wouldn't a wave be the entire area the electron travels ? Or does the associated wave represent the path and not the electron? If so, where is the electron? Here's a picture to refer to in your explanation if it helps. • Welcome to Chemistry.SE! Please have a look at this tutorial to acquaint yourself with the way math and chemical formulae can be nicely formatted on this site. Furthermore, when you include formulae or pictures in your questions, please make sure that you define the appearing symbols/variables and declare what is actually shown. Sep 10, 2014 at 6:10
• Have a look here and here. Does this answer your question? Sep 10, 2014 at 6:11

From the image above and from what you're asking, I presume you are referring to a particle confined in a certain space region. You can find everything you need on the internet but I wanna clarify a few things about QM:

1. There is no concept in QM for "path of a particle": position and momentum are incompatible observables, which means that they cannot be measured simultaneously each with zero standard deviation: in fact the product of standard deviations for position and momentum must be greater than $\dfrac{\hbar}{2}$. $$\sigma_{x}\sigma_{p}\geq\dfrac{\hbar}{2}$$ So it's meaningless in QM to talk about trajectories.

2. Meaning of the wave function: Let's assume that your system consists of a single particle in a one dimensional space. The wave function $\Psi\left(x,t\right)$ is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared $\left|\Psi\left(x,t\right)\right|^{2}$ , which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx , gives you the probability for finding the particle between $x$ and $x+dx$ $$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$ Of course you can integrate the previous eq. to find the probability over a certain region of space $\Omega$ $$P_{\Omega}=\int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}d$$

3. Stationary state: If the potential is time independent (in your case it's zero), the time dependence of the wave function can be factored out like $$\Psi\left(x,t\right)=\mathrm{e}^{-i\omega t}\psi\left(x\right)$$ In this case it's easy to prove that the probability density is time independent.

4. The quantization: For this simple problem, you must solve the eigenvalue equation for the energy $$-\dfrac{\hbar^{2}}{2m}\dfrac{d\psi\left(x\right)}{dx}=E\psi\left(x\right)$$ By solving it and by imposing that the particle cannot exists outside a certain region (outside the region $\left[0,3\right]$ in your picture), you find a natural quantization condition for the energy: the energy can assume only certain discrete values, which are called eigenvalues, $E_{n}$ . Of course, this means that for every eigenvalue there is an associated wave function, which is called eigenfunction $\psi_{n}\left(x\right)$ . The general solution to the eigenvalue equation is in fact a linear combination of eigenfunctions $$\psi\left(x\right)=\sum_{n}c_{n}\psi_{n}\left(x\right)$$

5. In the above picture, you can see the first three $\left|\psi_{n}\left(x\right)\right|^{2}$ . They of course represent probability density functions: note that where $\left|\psi_{n}\left(x\right)\right|^{2}$ is zero the probability of finding the particle is of course zero (for example in 0 and 3 , like we requested before).

6. So where is the electron? Like said before, in QM locating a particle it's a non-sense. All you can do is to compute the average position and momentum, which for a system in the state $\psi\left(x\right)$ are given by $$\left\langle x\right\rangle =\int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx=\int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)d$$ $$\left\langle p\right\rangle =\int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx=\int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]d$$