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Background: I'm currently doing self-study of quantum physics by following university level quantum physics lectures on YouTube (e.g. the YouTube MIT 8.04 Quantum Physics I, Spring 2013 or Stanford Quantum Lectures by Leonhard Susskind, plus supplemental YouTube courses like DrPhysicsA, with the occasional popular level YouTube on specific subjects). My classic math and physics is somewhat robust (high-school) but rusty (about 30 years ago).

In one of the MIT lectures I came about the de Broglie formulas:

de Broglie.

I did google and kind of understand how the de Broglie idea for wavelength lambda is determined for a particle with a certain mass (somewhat analog to the Einstein formula for photons) p= h/lambda and then lambda= h/p and then going classic on p via mass*velocity.

However, in the lecture the professor is using $k$ and refers to $k$ as as the wave-number (in conjunction with the reduced Planck constant).

$p = \hbar k $ with

$ k = \frac {2 \pi} \lambda $

So I guess, in intuitive terms (maybe this question is due to my rusty knowledge of classic physics), what's best way to think of $k$? Or maybe someone can just point me to a lecture/explanation of $k$ in terms of classic waves.

Terminality
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3 Answers3

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It's simply a measure of wavelength. Wavelength ($\lambda$) should be reasonably intuitive. The reason we use $k$ is simply that formulae involving $\lambda$ tend to be cluttered with factors of $2\pi$. Using $k$ leads to less clutter. But every such formula may be written in $\lambda$ terms.

John Doty
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$k$ is the spatial frequency of the periodic wavefunction for the particle represented in angular as opposed to cyclic units. In cyclic units the spatial frequency is $\nu = k/2\pi = 1/\lambda$.

Jagerber48
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I wanted to add something to Jagerber's answer as $k$ is very useful even in classical waves. $\vec{k}$ also tells the direction of propagation in the waves.

First consider a monochromatic wave traveling in 1-D. Intuitively we should use a periodic function to describe the amplitude of a periodic wave wave! The simplest function is

$$\psi = Asin(bx)\tag{1}$$

where $\psi$ is the amplitude of the wave. Note we can deduce the value of b by realising that the wave should be periodic in its wavelength, i.e $\psi(x) = \psi(x+\lambda)$. This just means that after traveling one wavelength, we should come to the same point in the wave. Using equation (1), we have

$$ sin(bx) = sin(bx+ b\lambda)$$

Using the fact that $sin(\theta) = \sin(\theta +2 \pi)$, we deduce that

$$b \equiv k = \frac{2\pi}{\lambda}$$

Therefore, we have a definition for the $\textbf{wavevector}$ $k$. Now let's move on to $3$-D to see what it corresponds to then.

Consider again a monochromatic wave emanating from a single source at some point in space. The amplitude is given by

$$ \psi \sim \frac{ e^{i \vec{k} \cdot \vec{x} }}{r}$$

where the $1/r$ power fall-off ensures energy conservation. In $3$-D, we can talk about contours of constant amplitude. Consider sitting at a point $\vec{x}$ and displacing it slightly by some vector $\delta\vec{x}$. If we are to have the same amplitude, we must have $\vec{k} \cdot \delta\vec{x} = 0$. This means that vector $\vec{k}$ tells you the direction in which the magnitude increases. Another way to see this is to compute $\nabla \psi$ and show that it is proportional to $\vec{k}$. Therefore, $\vec{k}$ also tells you the direction of propagation of the wave!

Hope this helps.

$\textbf{EDIT:}$

I forgot to add the connection with QM. The reason why $\vec{p} = \hbar \vec{k}$ makes sense is the last paragraph. The momentum of the wavepacket is in the direction of the propagation!

emir sezik
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