God Bless you Sir for making this so understandable for me! I struggled in my whole student life to understand logic of physics programming and here I got something! Thank you
In case anyone wants to know, this will work in a platform like Pycharm, and I imagine Spyder also, but the plot opens in your default browser. The barrier in the square-well is glorious......love it. Will be playing with this one for a while. Thank you
Two part question a) how did you find the initial wave function and what exactly is Cn b)while normalizing psi0 we didint take account of i in the second exponent ?
THANK YOU FOR THE AMAZING LECTURE PROFESSOR.I have doubt at time stamp 38:33 why the wavefunction inside the potential barrier is not exponentially decaying?
The eigenfunctions individually (the ones that he solved with the matrix at the beginning) do decay exponentially, but the linear combination of them (which is the “true” solution) don’t. Just as you saw here that the particle has a non-zero probability of going through a wall that has higher energy than it does, similarly, it also has a non-zero probability of reflecting off a dip in potential (a potential well instead of a barrier). Classically, this would be like driving off a cliff and bouncing back when trying to go over the edge (which sounds very weird but can happen in quantum mechanics). So, the particle can also reflect while trying to exit the wall, and the bump you see in the middle that doesn’t fade is because the wave keeps bouncing inside the wall and there is a probability that the particle will be there when you measure it. Different sizes of the wall and momentum of the particle make this effect more or less apparent... if the wall is too big, the wave will just bounce off with a pretty much zero probability of passing through, and if the wall is too short it will just be like if it was transparent.
@@rubencastillo7834 The eigenfunctions we get using finite difference method involves both the terms e^-x and e^x(that we discard when we solve this problem analytically since there is no reflection from the right boundary). So how it decays exponentially inside barrier?
@@shashankpandey2056 for a tldr, read the last part, but i wanted to give a little more in depth explanation of the process If you solve this system analytically, you will find that the eigenfunctions are different on the three regions of space (before, in, and after the barrier). Before and after the barrier the eigenfunctions are imaginary exponentials, and inside the barrier they are real. Before the barrier the eigenfunctions are of the form Ae^(ikx) + Be^(-ikx), inside they are Fe^(kx) + Ge^(-kx), and after the barrier they are Ce^(ikx) + De^(-ikx). Now, when you couple the imaginary exponentials with the time factor ( e^(-iEt/h) ), the positive one is a wave traveling to the right, and the negative is traveling to the left... The A term is a wave coming in from negative infinity, the B term is a wave coming from the barrier and traveling off to negative infinity, the C term is a wave coming from the barrier and traveling off to positive infinity, and the D term is a wave coming in from positive infinity. On experimentation, we send particles from the left, so it is impossible to find a wave coming in from positive infinity, so D must be zero, and the eigenfunction on the third region is only Ce^(ikx), representing a wave leaving the barrier and going off to positive infinity. Now, to solve for the coefficients, since there are no boundary conditions here (meaning that the function doesn't HAVE to be some value somewhere), what we do is impose continuity conditions, that is, the wave function and its derivative must be continuous in all of space; so that way can match all 3 regions of space using these continuity conditions. After setting up these equations, psi1(0)=psi2(0); psi1'(0)=psi2'(0); psi2(a)=psi3(a); psi2'(a)=psi3'(a) (where 0 and a are the points where the barrier begins and ends), we now have 4 equations and 5 unknowns (being the coefficients A, B, F, G, C), which is a problem. What we do here is to look at their relation with A, which is the initial wave we send on experiments, in hopes that we end up with the relative probability of finding the particle in the B, F, G, or C state in comparison to A. Calculating the individual probabilities of each state, you will find that they are |A|^2, |B|^2, |F|^2, and so on, so now we can define the transmission and reflection coefficients as T = |C|^2 / |A|^2, and R = |B|^2 / |A|^2. Which are the relative probabilities of the incident wave to be reflected or transmitted. You can then solve for all of the previous coefficients by dividing the continuity equations by A and solving for that ratio of the coefficients with A, then you can just set A equal to 1, which would kind of mean that 100% of the particles are coming in from the left (which is what we do on experiments). To recap for the answer, the term we eliminated was the D term from the third region, and the transmission and reflection coefficients depend on B and C... The terms inside the barrier are unaffected and are still of the form Fe^(kx) + Ge^(-kx). These are real exponentials, which are not waves, so the eigenfunctions alone do decay exponentially inside the barrier... But by combining all of them on a linear combination (pretty much a Fourier transform), they can create what you saw on the video. The reason why you see a bump inside the barrier is the same, the wave can reflect off of the end of the barrier, so it can stay bouncing inside. If there is confusion with this and the reflection coefficient, I'll add that the reflection coefficient isn't that, it is only the wave that reflects back and actually comes out of the barrier. The bump is its own thing. I made an animation similar to this one but by solving the equations analytically... they don't bounce back on the edges of the screen, and because of the different barrier height and momentum I put in, it might look more similar to what i think you are expecting to see. (The waves inside the 'envelope' are further apart because of the lower momentum)
I don't know how much you know or don't about quantum mechanics, but if you're interested, Introduction to Quantum Mechanics by Griffiths is a great book (the pdf is free on google). It doesn't solve the potential barrier system, but it does solve the very similar finite potential well, though I do recommend reading the book from the beginning until you get there... If you want to go through the calculations of the analytic solution to the potential barrier, Professor Dave Explains made a video about it.
@@rubencastillo7834 The eigenfunctions we get as the eigenvectors of the Hamiltonian matrix in finite difference method will not decay exponentially since we cannot make the constant D =0(in this numerical method)which is necessary to make the particle's motion in +x direction.If particles travels in left direction we get wavefunction rises exponentially inside barrier.
Yeah, I really liked that too. I'm only guessing that it is the real and complex pieces he mentions in this video but opts not to include here. I think he said he was going to make another video with those included.
Inside of the envelope are 2 separate waves going up and down, those are the real and imaginary values of Psi, and the envelope is the absolute value of Psi. Depending on the momentum p, is how close or far apart the real and imaginary waves will be... I have a video of it with lower momentum and lower barrier. Also, the actual probability of finding the particle somewhere is the absolute value of Psi squared.
This problem represents a potential barrier in quantum mechnics. In barrier case, the solution should be wave like outside the barrier and frozen(exponantial) inside the barrier. In this animation i have seen that the solution is wave like inside the barrier, which can't be. Why is so ?
Is there a sign error for the Hamiltonian at timestamp of approximately 22 minute mark? The variable H has the wrong sign, I am pretty sure, when compared with the sign used for the lecture slides at timestamp of approximately 6 minutes mark.
I think it's ok - the diagonal elements are negative multiplied by negative - so they are positive. The off diagonals are negative. I wouldn't be surprised if I made a mistake though - it happens.
if anyone is interested watching this, when your cell is markdown, rather than copying and pasting your code into a code cell, you can just change your markdown cell back to a code cell.
It was truly a joy to see a straightforward translation of quantum mechanics into code without any convoluted and non-intuitive steps.
God Bless you Sir for making this so understandable for me! I struggled in my whole student life to understand logic of physics programming and here I got something!
Thank you
In case anyone wants to know, this will work in a platform like Pycharm, and I imagine Spyder also, but the plot opens in your default browser. The barrier in the square-well is glorious......love it. Will be playing with this one for a while. Thank you
Two part question
a) how did you find the initial wave function and what exactly is Cn
b)while normalizing psi0 we didint take account of i in the second exponent ?
How did the normaliztion of psi0 worked for all the psi's
THANK YOU FOR THE AMAZING LECTURE PROFESSOR.I have doubt at time stamp 38:33 why the wavefunction inside the potential barrier is not exponentially decaying?
The eigenfunctions individually (the ones that he solved with the matrix at the beginning) do decay exponentially, but the linear combination of them (which is the “true” solution) don’t. Just as you saw here that the particle has a non-zero probability of going through a wall that has higher energy than it does, similarly, it also has a non-zero probability of reflecting off a dip in potential (a potential well instead of a barrier). Classically, this would be like driving off a cliff and bouncing back when trying to go over the edge (which sounds very weird but can happen in quantum mechanics). So, the particle can also reflect while trying to exit the wall, and the bump you see in the middle that doesn’t fade is because the wave keeps bouncing inside the wall and there is a probability that the particle will be there when you measure it. Different sizes of the wall and momentum of the particle make this effect more or less apparent... if the wall is too big, the wave will just bounce off with a pretty much zero probability of passing through, and if the wall is too short it will just be like if it was transparent.
@@rubencastillo7834
The eigenfunctions we get using finite difference method involves both the terms e^-x and e^x(that we discard when we solve this problem analytically since there is no reflection from the right boundary). So how it decays exponentially inside barrier?
@@shashankpandey2056 for a tldr, read the last part, but i wanted to give a little more in depth explanation of the process
If you solve this system analytically, you will find that the eigenfunctions are different on the three regions of space (before, in, and after the barrier). Before and after the barrier the eigenfunctions are imaginary exponentials, and inside the barrier they are real. Before the barrier the eigenfunctions are of the form Ae^(ikx) + Be^(-ikx), inside they are Fe^(kx) + Ge^(-kx), and after the barrier they are Ce^(ikx) + De^(-ikx). Now, when you couple the imaginary exponentials with the time factor ( e^(-iEt/h) ), the positive one is a wave traveling to the right, and the negative is traveling to the left...
The A term is a wave coming in from negative infinity, the B term is a wave coming from the barrier and traveling off to negative infinity, the C term is a wave coming from the barrier and traveling off to positive infinity, and the D term is a wave coming in from positive infinity. On experimentation, we send particles from the left, so it is impossible to find a wave coming in from positive infinity, so D must be zero, and the eigenfunction on the third region is only Ce^(ikx), representing a wave leaving the barrier and going off to positive infinity.
Now, to solve for the coefficients, since there are no boundary conditions here (meaning that the function doesn't HAVE to be some value somewhere), what we do is impose continuity conditions, that is, the wave function and its derivative must be continuous in all of space; so that way can match all 3 regions of space using these continuity conditions. After setting up these equations, psi1(0)=psi2(0); psi1'(0)=psi2'(0); psi2(a)=psi3(a); psi2'(a)=psi3'(a) (where 0 and a are the points where the barrier begins and ends), we now have 4 equations and 5 unknowns (being the coefficients A, B, F, G, C), which is a problem. What we do here is to look at their relation with A, which is the initial wave we send on experiments, in hopes that we end up with the relative probability of finding the particle in the B, F, G, or C state in comparison to A. Calculating the individual probabilities of each state, you will find that they are |A|^2, |B|^2, |F|^2, and so on, so now we can define the transmission and reflection coefficients as T = |C|^2 / |A|^2, and R = |B|^2 / |A|^2. Which are the relative probabilities of the incident wave to be reflected or transmitted. You can then solve for all of the previous coefficients by dividing the continuity equations by A and solving for that ratio of the coefficients with A, then you can just set A equal to 1, which would kind of mean that 100% of the particles are coming in from the left (which is what we do on experiments).
To recap for the answer, the term we eliminated was the D term from the third region, and the transmission and reflection coefficients depend on B and C... The terms inside the barrier are unaffected and are still of the form Fe^(kx) + Ge^(-kx). These are real exponentials, which are not waves, so the eigenfunctions alone do decay exponentially inside the barrier... But by combining all of them on a linear combination (pretty much a Fourier transform), they can create what you saw on the video. The reason why you see a bump inside the barrier is the same, the wave can reflect off of the end of the barrier, so it can stay bouncing inside. If there is confusion with this and the reflection coefficient, I'll add that the reflection coefficient isn't that, it is only the wave that reflects back and actually comes out of the barrier. The bump is its own thing.
I made an animation similar to this one but by solving the equations analytically... they don't bounce back on the edges of the screen, and because of the different barrier height and momentum I put in, it might look more similar to what i think you are expecting to see. (The waves inside the 'envelope' are further apart because of the lower momentum)
I don't know how much you know or don't about quantum mechanics, but if you're interested, Introduction to Quantum Mechanics by Griffiths is a great book (the pdf is free on google). It doesn't solve the potential barrier system, but it does solve the very similar finite potential well, though I do recommend reading the book from the beginning until you get there... If you want to go through the calculations of the analytic solution to the potential barrier, Professor Dave Explains made a video about it.
@@rubencastillo7834
The eigenfunctions we get as the eigenvectors of the Hamiltonian matrix in finite difference method will not decay exponentially since we cannot make the constant D =0(in this numerical method)which is necessary to make the particle's motion in +x direction.If particles travels in left direction we get wavefunction rises exponentially inside barrier.
How did you program the envelope of the wave in the teaser animation at the beginning of this video?
Yeah, I really liked that too.
I'm only guessing that it is the real and complex pieces he mentions in this video but opts not to include here. I think he said he was going to make another video with those included.
Inside of the envelope are 2 separate waves going up and down, those are the real and imaginary values of Psi, and the envelope is the absolute value of Psi. Depending on the momentum p, is how close or far apart the real and imaginary waves will be... I have a video of it with lower momentum and lower barrier. Also, the actual probability of finding the particle somewhere is the absolute value of Psi squared.
This problem represents a potential barrier in quantum mechnics. In barrier case, the solution should be wave like outside the barrier and frozen(exponantial) inside the barrier. In this animation i have seen that the solution is wave like inside the barrier, which can't be. Why is so ?
Is there a sign error for the Hamiltonian at timestamp of approximately 22 minute mark? The variable H has the wrong sign, I am pretty sure, when compared with the sign used for the lecture slides at timestamp of approximately 6 minutes mark.
I think it's ok - the diagonal elements are negative multiplied by negative - so they are positive. The off diagonals are negative.
I wouldn't be surprised if I made a mistake though - it happens.
Yes, it seems that near the 6 minute mark, there is a missing minus sign on the factor that the entire matrix is multiplied by.
I noticed it too.
if anyone is interested watching this, when your cell is markdown, rather than copying and pasting your code into a code cell, you can just change your markdown cell back to a code cell.