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. enter image description here

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    – Philipp
    Sep 10, 2014 at 6:10
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    – Philipp
    Sep 10, 2014 at 6:11

1 Answer 1


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 $\Omega$ $$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$

  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 $$\left\langle x\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{x}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)x\psi\left(x\right)dx$$ $$\left\langle p\right\rangle ={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\hat{p}\psi\left(x\right)dx={\displaystyle \int}_{\Omega}\psi^{\star}\left(x\right)\left[-i\hbar\dfrac{d\psi\left(x\right)}{dx}\right]dx$$


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