Thanks for a nice clear explanation of the textbook description of a free particle (its 30years since I learned this a college). However, I struggle with the normalization of a wave packet in more than one dimension. I can't see how you can have a wave packet that doesn't expand over space. In th 1D situation the wave packet propagates in one direction. In two dimensions why does it propagate in one direction, say along x, but remains bounded along the y direction? If you throw a stone into water you get ripples i.e propagation in x and y. If you solve a 2D wave equation, after separating the variables, you get two equivalent expressions that are equal to a common constant. Even 1D solutions have two wave propagating in opposite directions. We conveniently select the direction we want. This is practical but glosses over a full description of the situation. Wave packets appear free in the direction of propagation but bounded in the orthogonal directions. Are there solutions to the wave equation that replicate this behaviour?
@@Valeria-ib6gj that's true cuz mathematicians work with pure statements and sentences. When my mom saw me working on math problems she was like: "Is this your English literature hw?"
So to confirm, the superposition wave packet won't be a solution to the Schrodinger equation itself, only the wave functions that it is composed of will? All solutions of the Schrodinger equation over zero potential were e^ikx waves so all solutions were unnormalisable. Therefore the superposition equation is not a solution to the Schrodinger equation since the wave packets could be normalisable?
So for the check your understanding parts: Higher energy means higher velocity (I reasoned based on E=mc²) Linking K with wave number 2pi/wavelength and solving for wavelength gives wavelength = 2pi*hbar/sqrt(2mE) so higher energy results in shorter wavelength For frequency, knowing that wavelength × freq = c (assuming vacuum and c constant), as wavelength goes down, freq increases to maintain speed of light For the final check understanding: 1. Wave is moving towards +x so right 2. I think uncertainties in momentum and position should be very high (infinity) but am not quite sure. I tried to solve using variance, but I did go a bit fast over integrals and may have made an error, but i think infinite uncertainties. 3. Knowing de Broglie equation p= hbar/wavelength and using knowing K and wavenumber are equal, subbing p instead of wavelength gets sqrt(2mE)/hbar = 2pi×p/hbar resulting in p = sqrt (2mE)/2pi or p = K×hbar/2pi
For the uncertainties I am not sure as it kind of contradicts my understanding that if high uncertaintity in position, results in greater accuracy in momentum, so if anyone or you professor could explain, I would be grateful.
c is constant, its not E=mv^2 but E=mc^2, therefore your argument for higher energy means higher velocity is invalid, this equation is not even meant to relate the velocity of a particle with its mass, but more of equvalence of mass and energy for the particle
It's not only dependent on x. The wave function evolves with time. It's $\psi$(x, t). He doesn't show it, but to solve the schrodinger equation, you have to use separation of variables.
Since the electron (particle) is not bounded by any boundaries unlike in particle in a box in whose case the potential is infinite beyond the boundaries [V(x)=inf ; xa] and zero inside the box [V(x)=O ; x>0 and x
Good Morning. I'd like to sample the first seconds of this video for the intro of an ambient electronic music track. Is it a problem for you ? :o) Thanks in advance...
Remember that i = √−1 (the square root of negative one) First, (i * k)^2 = (i * k) * (i * k) = (√−1 * k) * (√−1 * k) = (√−1 * √−1 ) * k * k = -1 * k = -k^2 (Because square root of negative one times the square root of negative one equals negative one) Second, (-i * k)^2 = (-i * k) * (-i * k) = (-i * -i) * (k * k) = (i * i) * ( k * k) = (-1 * k * k) = -k^2 Hope that helps, sorry for having to type it out, would prefer to write it on paper for a better explanation.
how exactly does a 14 year old understand differential equations, partial derivatives, complex numbers and Euler's formula, or even oscillations and wave functions? How do you even know about the energy of particles or De Broglie matter waves?
You explained in 15 minutes what my prof couldn't in 90
Tiny error in 5:25 with factoring. Correction: First term should have factored out -ik and second term should have factored out ik.
Literal godsend, all your quantum videos. thank you!
Much appreciate it. Greetings from the Dominican Republic.
Thank you so much for making this lecture. Your explanation makes the subject so intuitive.
are there answers to the "check your understanding" section?
I am also searching for that!
Thanks for a nice clear explanation of the textbook description of a free particle (its 30years since I learned this a college). However, I struggle with the normalization of a wave packet in more than one dimension. I can't see how you can have a wave packet that doesn't expand over space. In th 1D situation the wave packet propagates in one direction. In two dimensions why does it propagate in one direction, say along x, but remains bounded along the y direction? If you throw a stone into water you get ripples i.e propagation in x and y. If you solve a 2D wave equation, after separating the variables, you get two equivalent expressions that are equal to a common constant. Even 1D solutions have two wave propagating in opposite directions. We conveniently select the direction we want. This is practical but glosses over a full description of the situation. Wave packets appear free in the direction of propagation but bounded in the orthogonal directions. Are there solutions to the wave equation that replicate this behaviour?
Welcome to Quantum Physics where you're mostly working with symbols and not numbers :p
as a pure mathematician I must say u guys work with numbers :)
@@Valeria-ib6gj that's true cuz mathematicians work with pure statements and sentences. When my mom saw me working on math problems she was like: "Is this your English literature hw?"
Great explenation! Making it soo easy to understand. Thank you!!
This is the story all about how....my life got flip turned upside down!! Sorry, hahaha. This got me reminiscing on Fresh Prince of Bel-Air!!
I literally thought of the same thing!
Compared to low energy particle, high energy particle have faster velocity, but how we derive the change of wavelength and frequency?
Try using the wavenumber k, dependent on E, to find the wavelength. Then you should be able to find the frequency aswell
So to confirm, the superposition wave packet won't be a solution to the Schrodinger equation itself, only the wave functions that it is composed of will? All solutions of the Schrodinger equation over zero potential were e^ikx waves so all solutions were unnormalisable. Therefore the superposition equation is not a solution to the Schrodinger equation since the wave packets could be normalisable?
Great explanation
Amazing all respect to you :)
superb stuff thanks
So for the check your understanding parts:
Higher energy means higher velocity (I reasoned based on E=mc²)
Linking K with wave number 2pi/wavelength and solving for wavelength gives wavelength = 2pi*hbar/sqrt(2mE) so higher energy results in shorter wavelength
For frequency, knowing that wavelength × freq = c (assuming vacuum and c constant), as wavelength goes down, freq increases to maintain speed of light
For the final check understanding:
1. Wave is moving towards +x so right
2. I think uncertainties in momentum and position should be very high (infinity) but am not quite sure. I tried to solve using variance, but I did go a bit fast over integrals and may have made an error, but i think infinite uncertainties.
3. Knowing de Broglie equation p= hbar/wavelength and using knowing K and wavenumber are equal, subbing p instead of wavelength gets sqrt(2mE)/hbar = 2pi×p/hbar resulting in p = sqrt (2mE)/2pi or p = K×hbar/2pi
For the uncertainties I am not sure as it kind of contradicts my understanding that if high uncertaintity in position, results in greater accuracy in momentum, so if anyone or you professor could explain, I would be grateful.
c is constant, its not E=mv^2 but E=mc^2, therefore your argument for higher energy means higher velocity is invalid, this equation is not even meant to relate the velocity of a particle with its mass, but more of equvalence of mass and energy for the particle
Nice video, but since $\Psi$ is only dependent on _x_, the usage of the partial symbol is redundant, or even wrong.
It's not only dependent on x. The wave function evolves with time. It's $\psi$(x, t). He doesn't show it, but to solve the schrodinger equation, you have to use separation of variables.
This video is literally about the time _independent_ Schrödinger eqution, TISE, so time is not a factor, hence i commented it.
Cazo it’s still a function of time and position. It doesn’t change the properties of the WF
Cazo that doesn’t change the fact the wave function depends on time, too.
@@CazoDKit is redunant yes, wrong idk. Maybe if we set psi(x,t=0) then everything should be fine
I can barely pass algebra 2
Thank you
why electron's potential energy is assumed 0 at every where?? and what does if V is not 0 and has time dependent characteristic?
Since the electron (particle) is not bounded by any boundaries unlike in particle in a box in whose case the potential is infinite beyond the boundaries [V(x)=inf ; xa] and zero inside the box [V(x)=O ; x>0 and x
Good Morning.
I'd like to sample the first seconds of this video for the intro of an ambient electronic music track. Is it a problem for you ? :o)
Thanks in advance...
can we check the track?
What is the lower case e supposed to stand for
Euler's number, e ≈ 2.7181. It's a really important number in math and physics, or the e^x function is I should say.
@@crosisbh1451 Thank you for replying.
What brought you here? If your not even sure what e is
@@laelfoo2285 I am 6th grade and I only learned theory until now also I am from serbia not sure if that is relevant
@@laelfoo2285 Currently in mathematics I am learning rational numbers.But in my spare time I study complex analysis.
i don't get why ik^2 and -ik^2 are the same
Remember that i = √−1 (the square root of negative one)
First, (i * k)^2 = (i * k) * (i * k) = (√−1 * k) * (√−1 * k) = (√−1 * √−1 ) * k * k = -1 * k = -k^2 (Because square root of negative one times the square root of negative one equals negative one)
Second, (-i * k)^2 = (-i * k) * (-i * k) = (-i * -i) * (k * k) = (i * i) * ( k * k) = (-1 * k * k) = -k^2
Hope that helps, sorry for having to type it out, would prefer to write it on paper for a better explanation.
when can we take the solution as sinusoidal answer me please
mekkiih h when you have the cosine of something, and the frequency is the coefficient of t
fresh prince brought me here
Good Luck :)
@@DeltaSigma16
I dont even remamber what was the joke
@@yohaijohn holy shit dude 3 years
@@yohaijohn Do you remember it now?
Thanks bro
Is this the free particle solution for one dimensional problem?
WE WILL REBUILD
What is this???? Is it about space???
sir u was saving wrong -i squre=-1
No. -i^2 = 1
i^2 = -1, so -(-1) = 1
This will blown my mind lol
Ah yes i still have no fucking clue
im still 14 and there's no time to say it's to early
how exactly does a 14 year old understand differential equations, partial derivatives, complex numbers and Euler's formula, or even oscillations and wave functions? How do you even know about the energy of particles or De Broglie matter waves?
Exactly
@@GeneralPet im 21 and i still dont know all that lol T_T
GeneralPet I’m 14 and I understand those. In fact anyone could understand the things you listed as long as it’s taught in a way you can understand.
fresh prince brought me here