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Are De Broglie relations only applicable to particles that have zero potential energy?
De Broglie relations are always written as:
$E=h\mathrm{\nu <!--\; \nu \; -->}$
$p=\frac{h}{\mathrm{\lambda <!--\; \lambda \; -->}}$
However, it doesn't make sense when you have waves that are eigenstates of a Hamiltonian operator with a non-zero potential. That is because that would give us a direct relation between momentum and energy: $\frac{E}{p}=v_p$, where $v_p$ is the phase velocity. This doesn't seem to make sense to me. So, would these equations be the same with the electron of the hydrogen atom, for example?

De Broglie wavelength of an electron
If an electron which already possesses some kinetic energy of $X\mathrm{e}\mathrm{V}$ is further accelerated through a potential difference of $X\prime \mathrm{V}$, then is it correct to state that the electron now has a total kinetic energy of $(X+X\prime )\mathrm{e}\mathrm{V}$? Using this information, am I allowed to substitute this value in the de Broglie equation to find the wavelength of the electron?

What is $E$ in below equation? Does it represent total energy or kinetic energy(De-broglie equation for uncharged particle like neutron.)
De-broglie equation for uncharged particle:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{\sqrt{2mE}}$
Where, $\mathrm{\lambda <!--\; \lambda \; -->}$ = wavelength
$h$ = planks constant
$m$ = mass of uncharged particles

Derivation of time-dependent Schrödinger equation from De Broglie hypothesis
In a quantum mechanics script I'm reading, the Schrödinger equation is "derived" (more precisely, motivated) by the De Broglie hypothesis. It starts at
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{2\mathrm{\pi <!--\; \pi \; -->}h}{p}$
$\mathrm{\omega <!--\; \omega \; -->}=\frac{E}{h}$
then takes the partial derivatives of the wave $\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\mathrm{\Psi <!--\; \Psi \; -->}(0,0)e^{\frac{2\mathrm{\pi <!--\; \pi \; -->}ix}{\mathrm{\lambda <!--\; \lambda \; -->}}-<!--\; -\; -->it\mathrm{\omega <!--\; \omega \; -->}}$
$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\frac{ip}{h}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)\end{array}$
$\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=-<!--\; -\; -->i\frac{E}{h}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)$
With the non-relativistic free particle $E=\frac{1}{2m}{p}^2$ one gets
$\begin{array}{}\text{(2)}& ih\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}t}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=E\mathrm{\Psi <!--\; \Psi \; -->}(x,t)=\frac{p^2}{2m}\mathrm{\Psi <!--\; \Psi \; -->}(x,t)\end{array}$
From there, they miraculously get the time-dependent Schrödinger equation. I cannot understand this step. If I insert formula (1) for $p^2$ in (2), I get something with $(\frac{\mathrm{\partial <!--\; \partial \; -->}}{\mathrm{\partial <!--\; \partial \; -->}x}\mathrm{\Psi <!--\; \Psi \; -->}{)}^2$, which is not the second partial derivative of $\mathrm{\Psi <!--\; \Psi \; -->}$ with respect to $x$.
Any hints?

Is the relation c=νλ valid only for Electromagnetic waves?
What is the validity of the relation $c=\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}$? More specifically, is this equation valid only for Electromagnetic waves?
I read this statement in a book, which says:
"de Broglie waves are not electromagnetic in nature, because they do not arise out of accelerated charged particle."
This seems correct, but arises a doubt in my mind.
Suppose I find out the wavelength of a matter wave (or de Broglie wave) using de Broglie's wave equation:
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
Now, can I use $c=\mathrm{\nu <!--\; \nu \; -->}\mathrm{\lambda <!--\; \lambda \; -->}$ to find out the frequency of the wave?

Proof of de Broglie wavelength for electron
According to de Broglie's wave-particle duality, the relation between electron's wavelength and momentum is $\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$
The proof of this is given in my textbook as follows:
1.De Broglie first used Einstein's famous equation relating matter and energy,
$E=m{c}^2,$
where $E=$ energy, $m=$ mass, $c=$ speed of light.
2.Using Planck's theory which states every quantum of a wave has a discrete amount of energy given by Planck's equation,
$E=h\mathrm{\nu <!--\; \nu \; -->},$
where $E=$ energy, $h=$ Plank's constant ($6.62607\times <!--\; \times \; -->{10}^{-<!--\; -\; -->34}\mathrm{J}\mathrm{s}$), $\mathrm{\nu <!--\; \nu \; -->}=$ frequency.
3.Since de Broglie believes particles and wave have the same traits, the two energies would be the same:
$m{c}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
$m{c}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
4.Because real particles do not travel at the speed of light, de Broglie substituted $v$, velocity, for $c$, the speed of light:
$m{v}^2=h\mathrm{\nu <!--\; \nu \; -->}.$
I want a direct proof without substituting $v$ for $c$. Is it possible to prove directly $\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$ without substituting $v$ for $c$ in the equation $\mathrm{\lambda <!--\; \lambda \; -->}=h/mc$?

Why $c$ needs to be added in numerator and denominator while solving a de Broglie equation word problem?
Calculate the de Broglie wavleength of a 0.05 eV ("theremal") neutron.Making a nonrelativistic calculation,
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}=\frac{h}{\sqrt{2m_{0}K}}=\frac{hc}{\sqrt{2\left(m_0{c}^2\right)K}}=\frac{12.4\times <!--\; \times \; -->{10}^3\mathrm{e}\mathrm{V}\cdot <!--\; \cdot \; -->\text{Å<!-- Å -->}}{\sqrt{2(940\times <!--\; \times \; -->{10}^6\mathrm{e}\mathrm{V})\left(0.05\mathrm{e}\mathrm{V}\right)}}=1.28\text{Å<!-- Å -->}$
I am confused as to why it is necessary to add $c$ in the numerator and denominator (red equation) while solving a word problem like this?

Blackbody radiation, de Broglie equation, and lightwaves to be shifted left
I'm having a hard time figuring this out.
1.Say we heated a lead ball to 1,000 Kelvin. Not all of the particles are at the exact same temperature--some parts are a little hotter, some are a little cooler. But for now, let’s assume that it follows a normal distribution that is centered at 1,000K.
2.The heat (or movement of the molecules) causes it to emit light.
3.Because the light photons have to be discrete (you can't have a 1/2 photon), this causes the observed light wavelengths to be shifted left.
4.This means we observe more red light than we might otherwise expect.
5.This is a long wind up to my specific question--does a particle vibrating at a specific frequency emit light at the same frequency? (i.e. a particle vibrating at 4.3 MhZ emits light at 4.3 Mhz). Because it seems like the whole thing hinges on that.
I mention this because I asked a physics teacher this, and he said, “No, the particles emit light following the de Broglie equation.” This would mean that the light emitted ignores the frequency and instead is based solely on its momentum. But, if this were true, then I would assume it would emit light in a standard distribution of frequencies as opposed to the left-skewed distribution that is actually observed.
Any help on this would be greatly appreciated!

De-Broglie equation for other particles except photon
If E=hf is applicable for electron and other particles, the De Broglie wavelength should be λ=hv/pc. Because, mc^2=hf which implies mc^2=hc/λ which implies m=h/λc and thus λ=hv/pc. But I have found in my text book that λ=h/p is applicable not only for photon but also for all particle. But how can λ=h/p=h/mv be applicable for all particle?

Wavelength and relativity
From de Broglie equation λ=h/p. But p=mv and velocity is a relativistic quantity so also wavelength is relative ? In other words does wavelength depends on the reference frame ?

Special Relativity, Louis de Broglie Equation Dilemma
While learning atomic structure I stumbled upon a very unusual doubt.
As we know that the energy of a wave is given by the equation: $E=\frac{hc}{\lambda }$ and Louis de broglie wave equation is given by the equation ${\lambda }_B=\frac{h}{p}$. My doubt is that, that is ${\lambda }_B=\lambda$. Do the ${\lambda }_B,\lambda$ represent the same thing $?$
My teacher equated $E=\frac{hc}{\lambda }$ and $E=mc²$ to form $\frac{hc}{\lambda }=mc²$ and rearranged to form $\lambda =\frac{h}{mc}$ and then replaced $\lambda$ by ${\lambda }_{{\rm B}}$ and $c$ by $v$ for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of $E=mc²$ because the $E=mc²$ is for particles while the former is for waves.
Is my understanding correct$?$ Please help and thanks in advance$!$

Which formula for the de Broglie wavelength of an electron is correct?
So, I have my exams in physics in a week, and upon reviewing I was confused by the explanation of de Broglie wavelength of electrons in my book. Firstly, they stated that the equation was: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$, where $h$ is the Planck constant, and $p$ is the momentum of the particle. Later, however, when talking about electron diffraction and finding the angles of the minima, the author gave the formula equivalent to that for light: $\mathrm{\lambda <!--\; \lambda \; -->}=\frac{hc}{E}$. Now, what I don't understand is if it is simply a mistake made by the author, or whether a different formula have to be used for electron diffraction, as the two formulae are very clearly not equivalent. In the latter case, I don't understand why the formula would be different. I greatly appreciate the help, as the exams are really close, and I would like to make sure I get this right!
Edit: I was told that pictures of text are taking away from the readability of the posts, and thus they were removed. Essentially, the difference between the two cases are that in the first case, the proton did not have any significantly large kinetic energy, while in the second example, the kinetic energy was $400MeV$

Does the wavelenth of matter waves depend upon the kinetic energy of the particle or object?
Do the properties of the waves (wavelength,frequency) emitted by a particle or object depend upon the velocity, or as to say its kinetic energy? Is the De Broglie equation $E=h\mathrm{\nu <!--\; \nu \; -->}$ applicable to matter waves as well?

How to interpret the velocity in de Broglie's equation?
Just wondering if anyone can help me understand the basic principle of quantum theory.
De Broglie's equation allows one calculate the wave length of the physical object, following the fundamental wave-particle duality of quantum theory.
$\mathrm{\lambda <!--\; \lambda \; -->}=h/mv$
Since velocity $v$ is always relative to the reference frame of observer, does it imply that the wave property is not inherent but displays itself differently to different observers?

Importance of Schrodinger equation
Louis de Broglie suggested that for microparticles like electrons, wave-like properties can be applied in order to explain some phenomena. Schrodinger wrote down an equation, a wave equation, describing these waves. What I do not understand is why is Schrodinger's contribution so important; if the concept of wave-like property of an electron was known already, then why writing a mere wave equation was an important step?

De Broglie Wavelength interpretation
I've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:
In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.
I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.
In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.
y = ${\displaystyle \frac{m\mathrm{\lambda <!--\; \lambda \; -->}L}{d}}$ (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)
If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.
Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).
The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.
Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

Origin of the de Broglie Equation
I was curious about the famous $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ equation. In high school I think you are just exposed to this equation with the explanation of "something something matter waves." But early in a undergraduate QM course you solve the time-independent Schrodinger's equation for a free particle in 1D and get the following solution:
$\frac{-<!--\; -\; -->{\mathrm{\hslash <!--\; \hslash \; -->}}^2}{2m}\frac{{\mathrm{\partial <!--\; \partial \; -->}}^2}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}\mathrm{\psi <!--\; \psi \; -->}=E\mathrm{\psi <!--\; \psi \; -->}⟹<!--\; ⟹\; -->\mathrm{\psi <!--\; \psi \; -->}=e^{\pm <!--\; \pm \; -->ikx}$ and $E=\frac{{\mathrm{\hslash <!--\; \hslash \; -->}}^2{k}^2}{2m}$
where I believe $k$ is just defined by the $E(k)$ equation. And then do we just realize that the $E(k)$ is the classical $p^2/2m$ equation if we set $p=\mathrm{\hslash <!--\; \hslash \; -->}k$, and thus we have "derived" $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ or do we "know" $p=\mathrm{\hslash <!--\; \hslash \; -->}k$ beforehand and this just confirms it?

Uncertainty Principle in 3 dimensions
I'm trying to understand how to write Heisenberg uncertainty principle in 3 dimensions. What I mean by that is to prove something of the form $f(\mathrm{\Delta <!--\; \Delta \; -->}{p}_x,\mathrm{\Delta <!--\; \Delta \; -->}{p}_y,\mathrm{\Delta <!--\; \Delta \; -->}{p}_z,\mathrm{\Delta <!--\; \Delta \; -->}x,\mathrm{\Delta <!--\; \Delta \; -->}y,\mathrm{\Delta <!--\; \Delta \; -->}z)\ge <!--\; \ge \; -->A$
This is what I got: The unknown volume that a single particle can be in is $\mathrm{\Delta <!--\; \Delta \; -->}V=\mathrm{\Delta <!--\; \Delta \; -->}x\mathrm{\Delta <!--\; \Delta \; -->}y\mathrm{\Delta <!--\; \Delta \; -->}z$. The uncertainty in the size of the momentum is $\mathrm{\Delta <!--\; \Delta \; -->}p=\sqrt{\mathrm{\Delta <!--\; \Delta \; -->}{p}_x^2+\mathrm{\Delta <!--\; \Delta \; -->}{p}_y^2+\mathrm{\Delta <!--\; \Delta \; -->}{p}_z^2}$
Now this is where I get stuck. In my textbook, for the 1d case, they used De-Broglie equation for connecting the uncertainty of the particle wavelength and its momentum along the $x$-axis. But does De-Broglie equation is correct per axis or for the size of the vectors?
Thanks for you help

Is there a significant error in using De Broglie's equation for an electron at really high speed?
I was wondering if using the De Broglie equation
$\mathrm{\lambda <!--\; \lambda \; -->}=\frac{h}{p}$
for object traveling at really high speeds would result in a significant error. For example if an object travelled at $0.02c$ would the error be negligible? How can I calculate the uncertainty in the result?

Direction of momentum given by the de Broglie relation
$p=mv$
where $m$ is the mass of an electron, and $v$ is its velocity. In this case, since $v$ is a vector, it's clear that the momentum will be also a vector.
However if the momentum is a vector quantity (and it is), what is the direction of the electron's momentum given by the de Broglie relation
$$\begin{gathered} p=h/\mathrm{\lambda <!--\; \lambda \; -->} \\ p=\mathrm{\hslash <!--\; \hslash \; -->}k \end{gathered}$$
if the Planck constant $h$ is scalar and the wavelength $\mathrm{\lambda <!--\; \lambda \; -->}$ is also scalar. Similarly the reduced Planck constant $\mathrm{\hslash <!--\; \hslash \; -->}$ is scalar and the wavenumber $k=2\mathrm{\pi <!--\; \pi \; -->}/\mathrm{\lambda <!--\; \lambda \; -->}$ is also scalar.