How to derive the de Broglie Relation

In summary: So k = 2π/λ is just the distance from the origin to the point at which the wave amplitude is half its maximum. k is also related to the wavelength of the wave in terms of its distance from the origin. I see that now, thanks! If you're still lost after reading that, please feel free to ask for clarification.In summary, Schrodinger's Equation and Wave Functions can be derived from k = 2π/λ.
  • #1
jigsaw21
20
0
Homework Statement
I'm trying to figure out how to derive the appropriate de Broglie Relation: k = p / h-bar
Relevant Equations
E = mc^2

E = hf

lambda = h / p

h-bar = h / 2*pi

Kinetic Energy = 1/2m*v^2
I began by taking E = mc^2 and E = hf , where h is Planck's constant, and then rewrote E as 1/2mv^2.

I rewrote f as c / λ, which made hf become h*c / λ. I then set this expression equal to the Kinetic Energy equation 1/2mv^2, which gave me:

1/2mv^2 = h*c / λ

I then replaced c on the right side with v, because although that equation initially represented a photon (from E = hf), it can apply to the energy of any particle, which means we can use v for that speed (is that correct??)

So then one of the v's on the left will cancel out with the v term on the right and simplify to:

1/2 mv = h / λ

I then solved for λ and got λ = 2h / mv. I then replaced mv with p (momentum) and got λ = 2h / p

From here, I may have done some redundant steps that were unnecessary, but I was trying to do this from scratch. So I multiplied both sides by (h/2π), and got: hλ / 2π = 2h^2 /p*2π

On the right side, the 2's canceled out. On the left, I recalled from a prior lesson that a value of k was k = 2π / λ. So since I had its reciprocal on the left side next to the, I rewrote the left side as just h / k, which would be equal to h^2 / p*π

From here, I was a bit lost, and decided to multiply both sides by k. That would give me h = h^2 k / p*π

I then multiplied both sides by pπ and got h*p*π = h^2*k

I then canceled out the h from the left with one of the h's on the right, and got pπ = hk

I then divided both sides by h which gave pπ/h = k. At this point I thought I was close, but not sure. I decided to multiply both sides by 2 since I knew that ħ = h / 2π. So after that step, I got p*(2π / h) = 2k.

I then replaced (2π/h) with 1/ħ , since that's the reciprocal, and that gave me p / ħ = 2k, which was really close to the answer I should've gotten which should be k = p / ħ. I have 2k instead of k, and I"m not sure how I got that, or even if this would still be correct since k is a constant. I'm not sure.

Can someone please just check and verify that my steps are correct, and let me know if there's another equation I may be missing, or if I made any mistakes with my math.

Thanks for any help, and apologize for the lengthiness of this.
 
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  • #2
It's all in what you have posted except you couldn't just pull it together. You know that ##\lambda=h/p## and you also know that ##k=2\pi/\lambda##. What happens if you take ##\lambda## from the first equation and put it in the second?
 
  • #3
kuruman said:
It's all in what you have posted except you couldn't just pull it together. You know that ##\lambda=h/p## and you also know that ##k=2\pi/\lambda##. What happens if you take ##\lambda## from the first equation and put it in the second?

I see that now, thanks!

I guess I don't even need the Energy equations at all to derive that k = p / ħ

I then guess one of my other curiosities was what is k = 2π/λ derived from, and how does it relate to the topic of Schrodinger's Equation and Wave Functions? Because that's the topic that I'm currently going over in my studies. I'm at the point where I'm motivated enough to know where these things come from instead of just accepting/memorizing them. I've seen it being referred to as "Circular Wavenumber", but I'm not totally sure what that really means.

From my years of doing math, I've always seen k in most equations referred to as a constant, but it looks like it isn't in this situation since its value is dependent on p. Is that right?
 
  • #4
Correct me if I'm wrong, but What I've gathered is the general form of the wave equation is given as ψ = e^i(kx - ωt). And since this represents a wave, I related it to my past history of mathematics with sin and cos graphs when there were transformations, in the form of for example: y = Acos(Bx + C), where the B value is akin to the k value of the wave function, ψ. Are these values in fact related in terms of how they affect the graphs by shortening the wavelengths the larger those values are ??

Thanks again for your reply, and/or anyone else that wouldn't mind chiming into help broaden my understanding.
 
  • #5
You cannot derive ##k=2\pi/\lambda##; it's a definition of the wave vector. It's a measure of how many wavelengths you can fit in ##2\pi## worth of phase of the sinusoidals. It is true that ##e^{i(kx-\omega t)}## is a solution of the Schrodinger equation with ##E=\hbar \omega## and ##p=\hbar k##. As you probably know, you can write ##\psi(x,t)=e^{i(kx-\omega t)}=\cos(kx-\omega t)+i~\sin(kx-\omega t)##, that is the wavefunction ##\psi(x,t)## has a real and an imaginary part.
 

Related to How to derive the de Broglie Relation

What is the de Broglie relation?

The de Broglie relation is a fundamental law in quantum mechanics that describes the wave-like behavior of particles. It states that every particle with a momentum p has an associated wavelength λ, given by λ = h/p, where h is Planck's constant.

Who first derived the de Broglie relation?

The de Broglie relation was first proposed by French physicist Louis de Broglie in his 1924 doctoral thesis. He was inspired by the wave-particle duality concept proposed by Einstein, which states that particles can exhibit both wave-like and particle-like behavior.

What is the significance of the de Broglie relation?

The de Broglie relation is significant because it provides a theoretical basis for the wave-like behavior of particles, which was previously only described by experimental observations. It also helped bridge the gap between classical mechanics and quantum mechanics, leading to a deeper understanding of the nature of matter.

How is the de Broglie relation derived?

The de Broglie relation can be derived using the principles of classical mechanics and special relativity. By equating the momentum of a particle with its wavelength, de Broglie was able to show that the kinetic energy of a particle can be expressed in terms of its frequency, leading to the de Broglie relation.

What are the applications of the de Broglie relation?

The de Broglie relation has many practical applications in various fields of science and technology. It is used in particle physics, quantum mechanics, and material science to understand the behavior of particles and waves. It also plays a crucial role in the development of technologies like electron microscopy, scanning tunneling microscopy, and particle accelerators.

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