Shouldn't the electron be everywhere because it is a wave? Or am I wrong, an electron is not a wave? It just has an associated wave function that determines where it might be?
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2$\begingroup$ The wavefunction is a probability density function - in the manner that it is usually defined you can not use it to find the probability of finding a particle at a single point in a continuum, but only the probability of finding it an interval/volume. $\endgroup$– Martin C.Commented Aug 21, 2018 at 14:34
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$\begingroup$ So to rephrase your question, you're asking if the wavelength of a fermion is analogous to the wavefunction, right? Because they aren't: the wavefunction itself is meaningless until you try running your operators to find position probability or momentum or whatever observable. $\endgroup$– user191954Commented Aug 21, 2018 at 16:20
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4 Answers
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Another way of saying "probability of finding an electron is 0.9 at point x" is "If we do an experiment that measures the electron's particle-like properties, then the probability that we find a particle at x is 0.9"
You can make other experiments that measure the electron's wave-like properties, but they would not give the probability at a point.
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An electron is not a wave. It is a quantum object that exhibits both particle-like and wave-like characteristics, but "is" neither. The wavefunction (in position space) determines the probability of finding the electron in any particular part of space during a position measurement, see below.
The probability to detect an electron at a single point is always zero. This is because the square of the wavefunction, $\lvert \psi(x)\rvert^2$, is the probability density to detect the electron at any particular place in space and probability densities necessarily assign zero probability to single points.
When you do a position measurement on an electron, you essentially force it to exhibit particle-like properties, namely to have a sharply defined position, i.e. be localized. Conversely, when one measures the momentum of an electron, one forces it to exhibit wave-like properties, namely to have sharply defined momentum, which by the uncertainty principle means that its has a very ill-defined position, i.e. is delocalized.
answered Aug 21, 2018 at 15:33
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$\begingroup$ If the probability of finding it at a point is zero then how it has a sharply defined position? $\endgroup$ Commented Mar 21, 2022 at 22:28
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$\begingroup$ @AntoniosSarikas It can have a probability close to 1 to be found in some very small interval $[x_0,x_1]$ for $x_0 - x_1$ "small", or, more abstractly, it can have a small position uncertainty. $\endgroup$– ACuriousMind ♦Commented Mar 21, 2022 at 22:30
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$\begingroup$ Would it correct to say that if we make infinite measurements then the number of times we find it in that point divided by the number of experiments will tend to zero? $\endgroup$ Commented Mar 21, 2022 at 22:32
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$\begingroup$ @AntoniosSarikas No, measurements always just determine whether the particle is inside an interval, they do not measure whether it is at a point. $\endgroup$– ACuriousMind ♦Commented Mar 21, 2022 at 22:43
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$\begingroup$ What I can't get is that if we measure the position, it will return a value (e.g. x=0.5). Wouldn't this be counted as finding the electron at a point? $\endgroup$ Commented Mar 21, 2022 at 22:46
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I think your fundamental confusion is coming from the difference between a probability and a probability density. Let's say you have a wavefunction of a particle, $\psi(x)$, and evaluate it at a point $x_0$, and you get $|\psi(x_0)|^2=0.9$. This does not mean the probability of finding the particle at the point $x_0$ is $0.9$. In fact, the probability of finding the particle at exactly $x_0$ is formally zero. What else could it be? So interpreting $|\psi(x_0)|^2$ as a probability is clearly wrong.
The correct thing to say is that $|\psi(x_0)|^2$ is proportional to to the probability of finding the particle in a region around $x_0$. If we have a region of width $dx$, then the probability of finding the particle in that region is given by $|\psi(x_0)|^2dx$. This should make intuitive sense: the probability of finding a particle in a region is larger when the region's width is larger, and it's larger when $|\psi(x_0)|^2$ is larger. A function like this that is proportional to the probability of finding a particle in the nearby region is called a probability density, not a probability.
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Key: An electron possesses properties of both wave and matter. The wave nature of it is described by de Broglie's matter wave. The mathematical representation of this wave is called wavefunction. Its time evolution is described by Schrodinger eqn.
Ans: Assume that you are going to do an experiment wherein you will find the position of an electron. Before doing your experiment you have calculated the probability of finding the electron at x as 0.9. Until you do your experiment the electron is in a linear combination of all possible position eigenstates (x, x', x'',...). And the probability of each eigenstate is different (For x it is 0.9).
Since the electron can be represented as its matter wave (wavefunction), which is nothing but a linear combination of these position eigenstates, you may understand that wavefunction (electron) is extended everywhere.
Now when you do your measurement, this wavefunction collapses to one of its eigenstates (this is a postulate of QM) and you 'see' the electron. When you finish your experiment again the electron goes back to its linear combination state.
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$\begingroup$ Why don't you comment about the reason why you find it unhelpful? That may help me to improve myself. $\endgroup$– DileepCommented Aug 21, 2018 at 16:11
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$\begingroup$ Thank you for your help, please understand this is not easy for a first timer, is the reason that an electron cannot be found at a point (but rather in an interval), because an electron is larger than a point? $\endgroup$– Rob HvCommented Aug 22, 2018 at 14:36
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$\begingroup$ Just because someone didn't understand an answer (because he/she is new to the concept) doesn't make the answer unuseful. Right? Others may find it useful. I am new here and I don't know whether things are different here! :) $\endgroup$– DileepCommented Aug 22, 2018 at 14:45
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$\begingroup$ I think it’s a beautiful thing to witness somebody understand these concepts for the first time....an electron’s wave function collapses when we measure it, and we find it somewhere? My head is still spinning $\endgroup$– Rob HvCommented Aug 22, 2018 at 15:18
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$\begingroup$ I really recommend you to read some basic textbooks on this. The collapse of the wavefunction is a postulate in QM and it works well with the experimental results always. You may start with the book, "Quantum Mechanics Demystified". $\endgroup$– DileepCommented Aug 22, 2018 at 15:23