This guy is awesome he goes into the history behind the subject matter and has great in-depth explanations. My little pet peeve is that De Broglie is pronounced "De Broy" like boy except with an r in there.
The French are tricky with pronunciation. They often pronounce names and words in a manner that seems lazy. Wiki shows Louis Victor Pierre Raymond, 7th Duc de Broglie's name pronounced several ways. I believe Louis was lazy because of his family's history and importance. They probably pronounced de Broglie as "de Bwa." The guttural "r" would be rolled normally, but here's where the laziness shines and the "r" sounds like a "w" to English speakers. The "...glie" end of his name gets transformed by laziness into the sound of a short "a." You have to love the French. Laid back.
The de Broglie wavelength is a fundamental concept in quantum mechanics that describes the wavelength associated with a particle and is given by the equation: \[ \lambda = \frac{h}{p} \] where \( \lambda \) is the de Broglie wavelength, \( h \) is the Planck constant, and \( p \) is the momentum of the particle. In Different Frames of Reference: Inertial Frame (Rest Frame of the Particle): - In the frame where the particle is at rest, its momentum \( p \) is zero. This leads to an undefined or infinitely large de Broglie wavelength. This emphasizes the wave-particle duality, as a particle at rest can be thought of as having a very long wavelength, suggesting that it is spread out over a much larger distance. Inertial Frame (Moving Observer): - For an observer moving with velocity \( v \) relative to the particle, the momentum of the particle can be found using the relativistic momentum formula: \[ p = \gamma mv \] where \( \gamma \) is the Lorentz factor \( \gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}} \), and \( c \) is the speed of light. Therefore, the de Broglie wavelength in this frame will depend on the relative velocities involved, leading to a shorter wavelength as the speed increases. Frame Moving with Fixed Relative Velocity: - If we consider two observers moving relative to each other, the de Broglie wavelengths measured by each observer will differ due to the relativistic effects. For instance, if one observer moves at a velocity such that the particle's speed reaches a significant fraction of the speed of light, the de Broglie wavelength appears shorter to a stationary observer than to a moving observer. Accelerated Frame: - In an accelerated frame, things become more complex due to the presence of fictitious forces. The de Broglie wavelength can still be computed, but one must also consider how acceleration affects the perceived energy and momentum of the particle, which can lead to different interpretations of its wave nature. The de Broglie wavelength serves as a bridge between classical concepts of particles and wave behavior. Its value and interpretation can change significantly depending on the observer’s frame of reference, demonstrating the fundamental principles of relativity and the wave-particle duality in quantum mechanics. Each frame highlights different aspects of the relationship between momentum, velocity, and wavelength.
At the end he says it’s NOT like the Doppler effect, as in that case the wavelength would not change as a function of velocity, but that is patently false, and indeed exactly what the Doppler effect measures.
Not at all, if you were able to take out a ruler and measure the peaks of sound waves in 2 different inertial frames, two observers would agree on the the length, that's exactly what he's saying here.
DeBroglie formula was very simple but very significant contribution to the quantum physics that ĺed to measuring the diameter of the H nucleus by diffraction.
0:59 Explain__ Why only plane wave instead of spherical or other type of wave? And How to imagine i.e. make thought picture _ a moving point sized particle is also plane wave? I think I may imagine that particle is spherical wave.
A plane wave is a 2d slice of a spherical wave. You should be imagining a wave packet in position space and reciprocal wave packet in momentum space, I.e. Fourier transform.
If you sum up both, with respect to the moving frame the particle would be much far. Imagine that at time t=0s the position of the particle on the stationary frame is x=2m and the particle has a velocity v=2m/s. At time t=1s the particle will be at x=4m. But on on it's frame of reference (the moving frame of reference) the position of the particle is still x'=2m since it was the point where it started. So x'=x-v*t (=) 2=4-2*1.
I have a question: in the end he says that the wavelength shouldn't change as he gave an analogy with dopler shift. However, if the matter waves are associated with the moving particle i.e. we can take the particle as the source of the wave, then we know that if the source is also moving then wavelengths would change. For example, if the source is moving towards the observer then the wavelength will shrink and if the source is moving away from the observer then the wavelength will increase, which is exactly the case here. But, I couldn't understand why he said it is so unusual in this case. I hope the uploader or anyone will give an answer. Thanks in Advance
As far I understand he compared two cases : 1 - matter waves and 2 - ordinary waves. In the case of matter waves lambda changes, as he showed, however, in the case of classical ordinary waves, the wavelength doesn’t change. So the hypothesis is “if the de Broigle waves are classical, then the wavelength shouldn’t change ”. He derived the opposite result, and therefore de broigle waves are not classical waves
There is no source of waves; in addition, you're not able to precisely measure the electron's position if the momentum is well defined (Heisenberg's principle); the matter wave just represents a probability amplitude that relates with the trajectory of the electron.
Yes sure but OP is correct in that the professor incorrectly suggests classical waves do not change wavelength by relative velocity, which they do, that is exactly what the Doppler effect measures. Also the source of the matter wave in position space is the source in momentum space, its source is therefore constant in phase space.
Zwiebah is my favorite quantum mechanic's professor. I am not sure why he doesn't use the total relativistic energy as De Broglie used since that is how De Broglie came up with the idea. I think Zwiebah knows that if he used that energy, the results would then be inertial frame invariant. It is interesting though that it doesn't make a difference in the end though as the group velocity still works out regardless.
I think non relativistic means that frame 1 velocity is small compared to c (speed of light). What he does is called the Galilean transform (as opposed to the Lorentz transform, which is relativistic)
since the velocity is small compared to c, it is nonrelativstic so we do not need to use lorentz transform and instead can just use the classical, nonrelativistic galilean transform
No, because that's the formula for photons. It need not apply to all particles. Here's a fun thing though - if you take the sum of a particle's kinetic and mass energy (using the equation from relativity) and take mass equal to 0, you do get E=hf, because a photon has zero mass. Not applicable to particles with mass though.
@@green0563 You're totally wrong. E=hf=`hw which is de Broglie hypothesis and valid for all matter particles, not only photons. It is one of the fundamental postulates of Quantum Mechanics that de Broglie came up with. So, E=hf is applicable to all matter particles as well as photons.
Yes OP is correct. A typical wave function for a free electron is cos(kx - ωt), in fact this is any classical plane wave. You can re-write k as p/ℏ and ω as E/ℏ using the de Broglie and Plank relations respectively. Momentum is the generator of space symmetry and energy is the generator of time symmetry in Lagrangian dynamics, which are indeed Lorentz invariant.
three ( 3 ) formulas is ok seven ( 7 ) formulas is missing drei ( 3 ) formula sind vertig sieben formula sind fehlen üç ( 3 ) formül tamam yedi ( 7 ) formül eksik
Professor seems as astonished as deBroglie's advisor who almost threw him out of the PhD program when deBroglie proposed matter was a wave for his one-page dissertation. When the advisor couldn't disprove deBroglie, he asked Einstein who told him to give deBroglie his PhD. deBroglie's dissertation was expanded by committee to six-pages and to this very day, physics fails to comprehend deBroglie's brilliance. Because you see, there are no particles, there are only waves. And just as a photon is a sinusoidal wave of finite temporal dimension, so too are what we incorrectly refer to as "particles."
they are particles AND waves, not "only waves". It depends on the level of interaction between matter: Most matter at our macroscopic level behave like particle, but at subatomic level they behave like wave. Like ocean waves (but inversely, since ocean behave like wave from afar, and water particle from nearer).
@@jackmaxwell3134 - At a macroscopic level, the sugar-cube disappears as if ants were a stream of water washing it away; therefore ants are a continuous stream. Upon closer inspection, ants are individual organisms, each taking a tiny portion of the sugar-cube. Therefore, ants are both discrete organisms and continuous streams? Preposterous.
@@tomnoyb8301 I've thought about, you're right. Probably every experiment proving the particle theory can be explained by the wave theory. But the inverse is false: You can't explain the double split experiment with particle theory.
It's so good to see that there are such brilliant minds as You, who can singlehandedly outsmart all the professors teaching quantum physics! Of course these old fashioned teachers can not comprehend the meanings behind de Broglie's words, but thankfully here you are, taking time out of your day to explain us mortals the secrets of the universe. Or maybe not? Maybe these scientists actually know what they are talking about?
@@billiesherman7 - I began making these points more than twenty years ago; the majority always clings to their mistakes. Physics began to turn around on this and other related issues a few years ago. Soon, you will be in the minority, because truth always prevails. Schrodinger's wave equation always holds, particle-models are situation-dependent, they only sometimes work. "Duality" is already disappearing at universities.
Writing with chalk and talking to the board instead of the students is ridiculous. It robs the students of time, the time it takes for him to write on the board. MIT needs to invest in new, modern infrastructure to come into the contemporary world. Even Prof Zwiebach's facial expressions can influence the students' learning experience. Support your professors, MIT. Stop being cheap.
that's the problem with people. They are just extremely ungrateful. Be grateful that he is teaching you Quantum Mechanics for FREE and that too is on a superior level. There are very few people on the planet who can give such great lectures on QM.
lol they figured out a solution to their problem and still decided to complain about it -- go up to 2x speed if you want fam, no one's stopping you lol
@@1eV Its not like I am not grateful. I passed this subject only because of Prof Barton. He has one of the best course lectures I have seen anywhere including my own college.
I just wish my university lecturers had been as clear, concise & engaging as this.
You haven't taken the tests. These lectures barely cover the depths that the recitations, texts and homeworks reach.
@Anubhav Mahapatra
Yes, whether you pass the tests does depend on your “level”. Very good.
😆
This guy is awesome he goes into the history behind the subject matter and has great in-depth explanations. My little pet peeve is that De Broglie is pronounced "De Broy" like boy except with an r in there.
me too
daily8150 0
It's fine. Just remember that even Feynman didn't know how to pronounce his name ;)
@@stauffap when he was little?
The French are tricky with pronunciation. They often pronounce names and words in a manner that seems lazy. Wiki shows Louis Victor Pierre Raymond, 7th Duc de Broglie's name pronounced several ways. I believe Louis was lazy because of his family's history and importance. They probably pronounced de Broglie as "de Bwa." The guttural "r" would be rolled normally, but here's where the laziness shines and the "r" sounds like a "w" to English speakers. The "...glie" end of his name gets transformed by laziness into the sound of a short "a." You have to love the French. Laid back.
The de Broglie wavelength is a fundamental concept in quantum mechanics that describes the wavelength associated with a particle and is given by the equation:
\[
\lambda = \frac{h}{p}
\]
where \( \lambda \) is the de Broglie wavelength, \( h \) is the Planck constant, and \( p \) is the momentum of the particle.
In Different Frames of Reference:
Inertial Frame (Rest Frame of the Particle):
- In the frame where the particle is at rest, its momentum \( p \) is zero. This leads to an undefined or infinitely large de Broglie wavelength. This emphasizes the wave-particle duality, as a particle at rest can be thought of as having a very long wavelength, suggesting that it is spread out over a much larger distance.
Inertial Frame (Moving Observer):
- For an observer moving with velocity \( v \) relative to the particle, the momentum of the particle can be found using the relativistic momentum formula:
\[
p = \gamma mv
\]
where \( \gamma \) is the Lorentz factor \( \gamma = \frac{1}{\sqrt{1 - (v^2/c^2)}} \), and \( c \) is the speed of light. Therefore, the de Broglie wavelength in this frame will depend on the relative velocities involved, leading to a shorter wavelength as the speed increases.
Frame Moving with Fixed Relative Velocity:
- If we consider two observers moving relative to each other, the de Broglie wavelengths measured by each observer will differ due to the relativistic effects. For instance, if one observer moves at a velocity such that the particle's speed reaches a significant fraction of the speed of light, the de Broglie wavelength appears shorter to a stationary observer than to a moving observer.
Accelerated Frame:
- In an accelerated frame, things become more complex due to the presence of fictitious forces. The de Broglie wavelength can still be computed, but one must also consider how acceleration affects the perceived energy and momentum of the particle, which can lead to different interpretations of its wave nature.
The de Broglie wavelength serves as a bridge between classical concepts of particles and wave behavior. Its value and interpretation can change significantly depending on the observer’s frame of reference, demonstrating the fundamental principles of relativity and the wave-particle duality in quantum mechanics. Each frame highlights different aspects of the relationship between momentum, velocity, and wavelength.
At the end he says it’s NOT like the Doppler effect, as in that case the wavelength would not change as a function of velocity, but that is patently false, and indeed exactly what the Doppler effect measures.
Not at all, if you were able to take out a ruler and measure the peaks of sound waves in 2 different inertial frames, two observers would agree on the the length, that's exactly what he's saying here.
I am keeping watching. Great
DeBroglie formula was very simple but very significant contribution to the quantum physics that ĺed to measuring the diameter of the H nucleus by diffraction.
Great professor! very good historical overview and connection amongst topics. Everything is connected together very well and explained even better.
0:59 Explain__ Why only plane wave instead of spherical or other type of wave? And How to imagine i.e. make thought picture _ a moving point sized particle is also plane wave? I think I may imagine that particle is spherical wave.
A plane wave is a 2d slice of a spherical wave. You should be imagining a wave packet in position space and reciprocal wave packet in momentum space, I.e. Fourier transform.
shouldn't x'= x+vt because the position of frame on the left will become that on the right as soon as distance "vt" is covered
If you sum up both, with respect to the moving frame the particle would be much far. Imagine that at time t=0s the position of the particle on the stationary frame is x=2m and the particle has a velocity v=2m/s. At time t=1s the particle will be at x=4m. But on on it's frame of reference (the moving frame of reference) the position of the particle is still x'=2m since it was the point where it started. So x'=x-v*t (=) 2=4-2*1.
no
x' is the position of the particle in the S' frame, not the position of the particle in the S frame.
I have a question: in the end he says that the wavelength shouldn't change as he gave an analogy with dopler shift. However, if the matter waves are associated with the moving particle i.e. we can take the particle as the source of the wave, then we know that if the source is also moving then wavelengths would change. For example, if the source is moving towards the observer then the wavelength will shrink and if the source is moving away from the observer then the wavelength will increase, which is exactly the case here. But, I couldn't understand why he said it is so unusual in this case. I hope the uploader or anyone will give an answer. Thanks in Advance
As far I understand he compared two cases : 1 - matter waves and 2 - ordinary waves.
In the case of matter waves lambda changes, as he showed, however, in the case of classical ordinary waves, the wavelength doesn’t change.
So the hypothesis is “if the de Broigle waves are classical, then the wavelength shouldn’t change ”. He derived the opposite result, and therefore de broigle waves are not classical waves
There is no source of waves; in addition, you're not able to precisely measure the electron's position if the momentum is well defined (Heisenberg's principle); the matter wave just represents a probability amplitude that relates with the trajectory of the electron.
Yes sure but OP is correct in that the professor incorrectly suggests classical waves do not change wavelength by relative velocity, which they do, that is exactly what the Doppler effect measures. Also the source of the matter wave in position space is the source in momentum space, its source is therefore constant in phase space.
Zwiebah is my favorite quantum mechanic's professor. I am not sure why he doesn't use the total relativistic energy as De Broglie used since that is how De Broglie came up with the idea. I think Zwiebah knows that if he used that energy, the results would then be inertial frame invariant. It is interesting though that it doesn't make a difference in the end though as the group velocity still works out regardless.
I think non relativistic means that frame 1 velocity is small compared to c (speed of light). What he does is called the Galilean transform (as opposed to the Lorentz transform, which is relativistic)
since the velocity is small compared to c, it is nonrelativstic so we do not need to use lorentz transform and instead can just use the classical, nonrelativistic galilean transform
can we use E=hf for an electron moving with velocity v ? If yes ,why ? If no,why too ?
Thanks for the help :))
No, because that's the formula for photons. It need not apply to all particles. Here's a fun thing though - if you take the sum of a particle's kinetic and mass energy (using the equation from relativity) and take mass equal to 0, you do get E=hf, because a photon has zero mass. Not applicable to particles with mass though.
@@green0563 You're totally wrong. E=hf=`hw which is de Broglie hypothesis and valid for all matter particles, not only photons. It is one of the fundamental postulates of Quantum Mechanics that de Broglie came up with. So, E=hf is applicable to all matter particles as well as photons.
Yes OP is correct. A typical wave function for a free electron is cos(kx - ωt), in fact this is any classical plane wave. You can re-write k as p/ℏ and ω as E/ℏ using the de Broglie and Plank relations respectively. Momentum is the generator of space symmetry and energy is the generator of time symmetry in Lagrangian dynamics, which are indeed Lorentz invariant.
4:39 to 4:52 Why is this case for ordinary waves is strange?
Because these are wave functions, probability densities.
Dont stop this plzz
Thanks ❤️🤍
Please increase the volume of the uploads. Thanks.
time for hearing aids
Please make frequency f.
Do the problem solutions happen to be freely available somewhere?
We don't have solutions available on our site but hopefully someone else does.
@@mitocw Alright, good to know
So the slower a massive object moves the longer its wavelength?
Yes
Sometimes 2pi can be very useful....yes indeed it can.
three ( 3 ) formulas is ok seven ( 7 ) formulas is missing
drei ( 3 ) formula sind vertig sieben formula sind fehlen
üç ( 3 ) formül tamam yedi ( 7 ) formül eksik
This is helpful ❤️🤍
Acerto a lambida todas as vezes
Ui! Cuidado com a localidade da sua bolinha, eu estou com o meu taco e não erro.
Professor seems as astonished as deBroglie's advisor who almost threw him out of the PhD program when deBroglie proposed matter was a wave for his one-page dissertation. When the advisor couldn't disprove deBroglie, he asked Einstein who told him to give deBroglie his PhD. deBroglie's dissertation was expanded by committee to six-pages and to this very day, physics fails to comprehend deBroglie's brilliance. Because you see, there are no particles, there are only waves. And just as a photon is a sinusoidal wave of finite temporal dimension, so too are what we incorrectly refer to as "particles."
they are particles AND waves, not "only waves". It depends on the level of interaction between matter: Most matter at our macroscopic level behave like particle, but at subatomic level they behave like wave. Like ocean waves (but inversely, since ocean behave like wave from afar, and water particle from nearer).
@@jackmaxwell3134 - At a macroscopic level, the sugar-cube disappears as if ants were a stream of water washing it away; therefore ants are a continuous stream. Upon closer inspection, ants are individual organisms, each taking a tiny portion of the sugar-cube. Therefore, ants are both discrete organisms and continuous streams? Preposterous.
@@tomnoyb8301 I've thought about, you're right. Probably every experiment proving the particle theory can be explained by the wave theory. But the inverse is false: You can't explain the double split experiment with particle theory.
It's so good to see that there are such brilliant minds as You, who can singlehandedly outsmart all the professors teaching quantum physics! Of course these old fashioned teachers can not comprehend the meanings behind de Broglie's words, but thankfully here you are, taking time out of your day to explain us mortals the secrets of the universe. Or maybe not? Maybe these scientists actually know what they are talking about?
@@billiesherman7 - I began making these points more than twenty years ago; the majority always clings to their mistakes. Physics began to turn around on this and other related issues a few years ago. Soon, you will be in the minority, because truth always prevails. Schrodinger's wave equation always holds, particle-models are situation-dependent, they only sometimes work. "Duality" is already disappearing at universities.
Writing with chalk and talking to the board instead of the students is ridiculous. It robs the students of time, the time it takes for him to write on the board. MIT needs to invest in new, modern infrastructure to come into the contemporary world. Even Prof Zwiebach's facial expressions can influence the students' learning experience. Support your professors, MIT. Stop being cheap.
The explanation is almost like in Samara State University
WHy is he so damn slo. I have to watch the video in 1.75x speed damn
that's the problem with people. They are just extremely ungrateful. Be grateful that he is teaching you Quantum Mechanics for FREE and that too is on a superior level. There are very few people on the planet who can give such great lectures on QM.
lol they figured out a solution to their problem and still decided to complain about it -- go up to 2x speed if you want fam, no one's stopping you lol
@@1eV Its not like I am not grateful. I passed this subject only because of Prof Barton. He has one of the best course lectures I have seen anywhere including my own college.