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I have been told that an electron beam has a wave function equivalent to a plane wave $\psi(x) = Ae^{ikx}$, however I would like to know why? Also, if an electron beam can be shown to have a wave function $\psi(x) = Ae^{ikx}$ how do we reconcile this with the fact that this function can not be normalized?

I would like to stress this is not a homework problem, I just genuinely would like to understand why this is the case.

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  • $\begingroup$ A freely propagating electron in the x direction would be described in what other fashion? $\endgroup$
    – Jon Custer
    Commented Dec 12, 2020 at 16:58
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    $\begingroup$ Related post by OP: physics.stackexchange.com/q/599845 $\endgroup$ Commented Dec 12, 2020 at 17:01
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    $\begingroup$ In what context were you told this? $\endgroup$
    – nasu
    Commented Dec 12, 2020 at 17:01
  • $\begingroup$ The wave function describes a single electron, however this can not be normalized. To get around this issue, I have been told that this wave function can be viewed as a beam. I would like to understand why the function $\psi(x) = Ae^{ikx}$ can be viewed as an electron beam and if it can why would it be allowed given the function doesn't look like it can be normalized? $\endgroup$
    – Hello123
    Commented Dec 12, 2020 at 17:03
  • $\begingroup$ @nasu I am studying an introductory course in QM, my college professor briefly mentioned that the issue of normalizing can be overcome by viewing it as a beam of electrons instead of a single electron in a powerpoint slide. $\endgroup$
    – Hello123
    Commented Dec 12, 2020 at 17:09

2 Answers 2

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an electron beam has a wave function equivalent to a plane wave $\psi(x)=Ae^{ikx}$, however I would like to know why?

In an electrom beam the electron is supposed to have a well-defined momentum $p_x$, i.e. measuring the momentum will with 100% probabilty result in a certain value $p_x$.

Or saying it mathematically: the wave function $\psi(x)$ must be an eigenfunction of the momentum operator $\frac{\hbar}{i}\frac{\partial}{\partial x}$ with an eigenvalue $p_x$.

$$\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)=p_x\psi(x)$$

The solutions of this equation are $$\psi(x)=Ae^{ikx},\quad \text{ for eigenvalue }p_x=\hbar k \tag{1}$$ with any parameter $k$.


how do we reconcile this with the fact that this function can not be normalized?

You correctly noticed that the eigenfunctions (1) are not normalizable. That means that such a state is not physically possible.

To overcome this issue, we can modify the solution (1) and make a wave-function which is confined to a large but finite region of space. For example, a function like $$\psi(x)=\begin{cases} \frac{1}{\sqrt{2L}}e^{ikx} &\text{, if } -L\le x \le +L \\ 0 & \text{, else } \end{cases} \tag{2} $$ with some large length $L$ would be normalized and hence be physically possible. But of course it would not exactly be an eigenfunction of momentum.

Nevertheless the functions (1) are handy as a mathematical idealization. Therefore we can (with some care) use them to approximate physical reality. This is usually much easier than using the functions (2) instead, calculating the physical results, and at the end doing $\lim_{L\to\infty}$.

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The plane wave is indeed an idealized approximation. As it was correctly addressed by Thomas Fritsch, in many experiments the electrons in the beam should have a well defined momentum. He also answered the question on normalizablity. The only thing I want to add is a response to the first question on why a plane wave: Even if you start with a relatively localized electron wave packets, as they accelerate through the electric field in a cathode ray tube, the wave packets spread very quickly so that within a very short time you will have a widely dispersed wave packet, which can be approximated as a plane wave.

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