At the start, you mention that there is for a more rigorous proof in the description. But I've watched all the videos mentioned in the description and I do not remember anything resembing that. Although these statements are really intuitive, I'd be grateful if you could point to some place where one can read the proof, since you mentioned it in the video
@@ProfessorMdoesScience For me, the Python code provides additional insights to the lecture content, especially the part of control buttons. It helps me to build "images" into my mind.
Would you, please, professor dedicate one video upon deriving Fourier transforms for momentum-position spacious change and energy-time evolution (temporal change) using Hilbert space properties? I’ve seen some videos some folks are applying Fourier transforms for “wave packet” model, but it’s still foggy for me.
Is there any video on the channel where you derive the Time indipendent and the time dipendent Schrodinger equation in the position representation? I'm missing something and I can't do the derivation on my own to get the equation you start this video with. Thank you so much :)
Unfortunately we don't currently have a video doing this explicitly. We do hope to make more videos focusing on the position representation, so we'll eventually get there. What have you tried in terms of the derivation?
So for potential well with finite walls the energy eigenvalues won’t be quantized? And particle remains to be “free particle” where k can take any real value (in other words, it’s continuous) It seems like the quantum value k becomes discrete only inside the potential with infinite potential wells. If the potential is a step function, the particle has a probability to penetrate any such “finite wall”… tunneling effect
Not quite: you can still get bound states for a finite well which are quantized, and these would correspond to eigenvalues below the well walls. For higher energies above the finite well walls, you would then a continuum of states that are no longer quantized. We hope to cover the finite square well in the future, but I hope this helps for now!
@@ProfessorMdoesScience please, if you can. I prefer to grip myself to your youtube lectures more than others because seemingly you never distance yourself from bra-ket notation, Hilbert vector space properties (orthogonality and orthonormality, inner and outer products, eigenfunctions and eigenvalues, ext.) I think it’s key to understand how complex-valued wavefunction works in quantum mechanics and how we extract physically meaningful information from it. There are not a lot of great lectures on quantum mechanics out there.
In the beginning of the video, to find the spectrum of the hamiltonian you are using the S.E as an Eigenvalue equation. So the spectrum is just the list of Eigenvalues, meaning the possible measurement outcomes. If we instead act with the Hamiltonian on a general state(in the energy representation), we get a sum over the spectrum weighted by the coefficients of the state. Is there a name for the list of these values? We can pick a value from this list, divide it by the corresponding energy, and get the probability amplitude for that energy. So it seems useful.
Not sure if there is a specific name for this. But note that acting with the Hamiltonian on a state is not the same as measuring the probability amplitude of a given energy eigenstate. You can expand any state (whether you've acted on it with the Hamiltonian or not) in the energy basis, to figure out the probability amplitude of any given energy eigenstate in that particular state. I hope this helps!
@@ProfessorMdoesScience My thought process was like this. I'm trying to understand what acting on a general state does to it and how that is useful. Does the state even live in our state space anymore? I can expand any state in the energy basis. Then the S.E tells me what happens to that state when I act on it with the hamiltonian. If the state is an Eigenstate ui i just get Ei|ui> . But if that state is a more general linear combination Ψ (sum ui>)I get sum ciEi|ui> as the result. (which, because I know the eigenstates of the (time independent)Hamiltonian are eigenstates of the d/dt operator, must be the same as d/dt Ψ up to a constant ihbar) So if I am given this state, i can still figure out the measurement probabilities from the prefactors. But they are not the same as before acting on it with the Hamiltonian, because acting this adds the Ei prefactors. So this would be the collection of the coefficients ui> multiplied by the spectrum of the Hamiltonian?
@@narfwhals7843 A few thoughts: 1. Acting with a general operator A on a state |psi> changes it to another state |phi>=A|psi>. Both states still live in the same state space. 2. I would not think of the Schrödinger equation as an eigenvalue equation. It is the equation of motion of quantum mechanics, which tells you the time dependence of the state. But do not identify ihbar*d/dt with an operator H (otherwise the equation would become a tautology). 3. As to measurement probabilities, they will be different between states |psi> and |phi> after applying any operator A to |psi>. How they differ will depend on both A and the initial state |psi>. I hope this helps!
@@ProfessorMdoesScience Thank you :) My reasoning for this "identification" is that the only functions which satisfy the S.E for eigenstates of the Hamiltionian must be those that are multiplied by a constant when taking the time derivative. So the exponentials. This makes them eigentfunctions of the d/dt operator. I don't think this is actually an identification, though, because not all eigenfunctions of the d/dt operator will satisfy the S.E. Only those with eigenvalues that are a proportional to i/t.
Lets say you write the borders of your potential to be IxI smaller or equal a/2, the rest stays the same. Then the only possible solution would be a cosine, right?
The borders of the potential in the video *are* defined between -a/2 and a/2, see around minute 02:35. If you displaced the potential to, for example, the range (0,a), then your solutions would only be sines.
Dear professor, how do you compute the probability for the state to be in state 1, do you simply just integrate Y1*Yn from 0 to a, and then square the result, I tried this and got a very unconvincing answer, I mean the integral
The state of a quantum system can in principle be anything, and this is not a probabilisitc feature. For example, you could in principle prepare a quantum system to be in the ground state (state 1). Probability comes into quantum mechanics when you perform a measurement. If you had prepared the system in state 1, it is an eigenstate of the energy operator, which means that, if you measured energy, you would get the ground state energy with probability 1. However, you could also prepare the state in a "superposition" of multiple energy eigenstates. In this case, the probability of getting the ground state energy after a measurement would be smaller than 1, and its precise value is given by the measurement postulates in quantum mechanics. We have a few videos covering this, the first one of which is this one: ruclips.net/video/u1R3kRWh1ek/видео.html If you follow the "background" and "what next?" sections in the description of that video, you should find all the relevant information to understand these concepts. Alternatively, we build these ideas from the ground up by following the playlist on the postulates of quantum mechanics: ruclips.net/p/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb I hope this helps!
If I need to consider all dimensions at once, should I have to replace d/dx in the kinetic energy term with the δ/δx term in the laplacian?(I mean second order but it is hard to type that, so I just wrote d/dx)
Yes, in three dimensions the momentum operator is proportional to the gradient, and the kinetic energy to the Laplacian. We very briefly go over it in this video: ruclips.net/video/Yw2YrTLSq5U/видео.html
The wave function in momentum space is the Fourier transform of the wave function in position space, you can find the details of why in this video: ruclips.net/video/2lr3aA4vaBs/видео.html For an infinite square well, if you calculate the Fourier transform of an eigenstate in position space, you will find that the momentum space wave function is made of two sinc functions centered at momenta of +hbar*n*pi/a and -hbar*n*pi/a, corresponding to motion to the right and to the left inside the well.
If we have an operator A, then an eigenstate |psi> is simply a quantum state such that when we act with A on it, we get the same state back (up to a constant lambda, called the eigenvalue): A|psi>=lambda |psi> This notation is the bra-ket notation, but, as you hint in your question, we can represent these states in different ways. Wave functions are simply one of many possible representations, corresponding to the position representation. In this language, the equation above becomes (in 1D): A psi(x) = lambda psi(x), where psi(x) = is the wave function. This equation shows how psi(x) is the eigenstate (also called eigenfunction) of the opreator A. You can learn more about eigenvalues and eigenstates here: ruclips.net/video/p1zg-c1nvwQ/видео.html And about wave functions here: ruclips.net/video/2lr3aA4vaBs/видео.html I hope this helps!
This is my favorite channel for learning QM!!! Keep up the great work...
Glad to hear this! :)
At the start, you mention that there is for a more rigorous proof in the description. But I've watched all the videos mentioned in the description and I do not remember anything resembing that. Although these statements are really intuitive, I'd be grateful if you could point to some place where one can read the proof, since you mentioned it in the video
We haven't yet covered that in a video... hope to do that at some later point!
The Python code is much helpful, thanks a lot!
It is great to hear, we never know whether people take a look at it! :)
@@ProfessorMdoesScience For me, the Python code provides additional insights to the lecture content, especially the part of control buttons. It helps me to build "images" into my mind.
keep going, youre doing us a huge favor professor!
Thanks for the encouragement!
Well done...it fills a gap in available information in the maths part or for those who needs more explanations on the maths.
Glad that you found the video helpful! :)
Thanks Professor nicely explained
Glad you found it useful!
I liked your explanation. It is always good to learn from other points of view.
I agree entirely, what other approach were you familiar with?
@@ProfessorMdoesScience Basically same method, but for V(x)=0 in 0
@@juanluna2361 For some reason setting the potential to zero in the range 0
good presentation Prof
Glad you liked it!
yes it would be amazing if you share more code in the future. thanks!
Noted! We are hoping to have some more, for example when we work through approximation methods in quantum mechanics.
Would you, please, professor dedicate one video upon deriving Fourier transforms for momentum-position spacious change and energy-time evolution (temporal change) using Hilbert space properties? I’ve seen some videos some folks are applying Fourier transforms for “wave packet” model, but it’s still foggy for me.
We cover some of these topics in our series on wave mechanics, check it out here:
ruclips.net/p/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ
Hope you like it!
Is there any video on the channel where you derive the Time indipendent and the time dipendent Schrodinger equation in the position representation? I'm missing something and I can't do the derivation on my own to get the equation you start this video with. Thank you so much :)
Unfortunately we don't currently have a video doing this explicitly. We do hope to make more videos focusing on the position representation, so we'll eventually get there. What have you tried in terms of the derivation?
loved it , especially the code
Glad you liked it! We are hoping to have more code in some of our future videos :)
So for potential well with finite walls the energy eigenvalues won’t be quantized? And particle remains to be “free particle” where k can take any real value (in other words, it’s continuous) It seems like the quantum value k becomes discrete only inside the potential with infinite potential wells. If the potential is a step function, the particle has a probability to penetrate any such “finite wall”… tunneling effect
Not quite: you can still get bound states for a finite well which are quantized, and these would correspond to eigenvalues below the well walls. For higher energies above the finite well walls, you would then a continuum of states that are no longer quantized. We hope to cover the finite square well in the future, but I hope this helps for now!
@@ProfessorMdoesScience please, if you can. I prefer to grip myself to your youtube lectures more than others because seemingly you never distance yourself from bra-ket notation, Hilbert vector space properties (orthogonality and orthonormality, inner and outer products, eigenfunctions and eigenvalues, ext.) I think it’s key to understand how complex-valued wavefunction works in quantum mechanics and how we extract physically meaningful information from it. There are not a lot of great lectures on quantum mechanics out there.
In the beginning of the video, to find the spectrum of the hamiltonian you are using the S.E as an Eigenvalue equation. So the spectrum is just the list of Eigenvalues, meaning the possible measurement outcomes.
If we instead act with the Hamiltonian on a general state(in the energy representation), we get a sum over the spectrum weighted by the coefficients of the state. Is there a name for the list of these values?
We can pick a value from this list, divide it by the corresponding energy, and get the probability amplitude for that energy. So it seems useful.
Not sure if there is a specific name for this. But note that acting with the Hamiltonian on a state is not the same as measuring the probability amplitude of a given energy eigenstate. You can expand any state (whether you've acted on it with the Hamiltonian or not) in the energy basis, to figure out the probability amplitude of any given energy eigenstate in that particular state. I hope this helps!
@@ProfessorMdoesScience My thought process was like this.
I'm trying to understand what acting on a general state does to it and how that is useful. Does the state even live in our state space anymore?
I can expand any state in the energy basis. Then the S.E tells me what happens to that state when I act on it with the hamiltonian. If the state is an Eigenstate ui i just get Ei|ui> . But if that state is a more general linear combination Ψ (sum ui>)I get sum ciEi|ui> as the result. (which, because I know the eigenstates of the (time independent)Hamiltonian are eigenstates of the d/dt operator, must be the same as d/dt Ψ up to a constant ihbar)
So if I am given this state, i can still figure out the measurement probabilities from the prefactors. But they are not the same as before acting on it with the Hamiltonian, because acting this adds the Ei prefactors.
So this would be the collection of the coefficients ui> multiplied by the spectrum of the Hamiltonian?
@@narfwhals7843 A few thoughts:
1. Acting with a general operator A on a state |psi> changes it to another state |phi>=A|psi>. Both states still live in the same state space.
2. I would not think of the Schrödinger equation as an eigenvalue equation. It is the equation of motion of quantum mechanics, which tells you the time dependence of the state. But do not identify ihbar*d/dt with an operator H (otherwise the equation would become a tautology).
3. As to measurement probabilities, they will be different between states |psi> and |phi> after applying any operator A to |psi>. How they differ will depend on both A and the initial state |psi>.
I hope this helps!
@@ProfessorMdoesScience Thank you :)
My reasoning for this "identification" is that the only functions which satisfy the S.E for eigenstates of the Hamiltionian must be those that are multiplied by a constant when taking the time derivative. So the exponentials. This makes them eigentfunctions of the d/dt operator.
I don't think this is actually an identification, though, because not all eigenfunctions of the d/dt operator will satisfy the S.E. Only those with eigenvalues that are a proportional to i/t.
This is the best derivation
Glad you like it! :)
Thank you sir!!
Most welcome!
Is there any videos about tunneling effect in quantum mechanics?
We don't yet have any videos on tunneling, but hoping to publish some in the future!
@@ProfessorMdoesScience Thanks, please keep us update.
Lets say you write the borders of your potential to be IxI smaller or equal a/2, the rest stays the same. Then the only possible solution would be a cosine, right?
The borders of the potential in the video *are* defined between -a/2 and a/2, see around minute 02:35.
If you displaced the potential to, for example, the range (0,a), then your solutions would only be sines.
yep code definitely helps!
Thanks for the feedback!
Dear professor, how do you compute the probability for the state to be in state 1, do you simply just integrate Y1*Yn from 0 to a, and then square the result, I tried this and got a very unconvincing answer, I mean the integral
The state of a quantum system can in principle be anything, and this is not a probabilisitc feature. For example, you could in principle prepare a quantum system to be in the ground state (state 1). Probability comes into quantum mechanics when you perform a measurement. If you had prepared the system in state 1, it is an eigenstate of the energy operator, which means that, if you measured energy, you would get the ground state energy with probability 1. However, you could also prepare the state in a "superposition" of multiple energy eigenstates. In this case, the probability of getting the ground state energy after a measurement would be smaller than 1, and its precise value is given by the measurement postulates in quantum mechanics. We have a few videos covering this, the first one of which is this one: ruclips.net/video/u1R3kRWh1ek/видео.html
If you follow the "background" and "what next?" sections in the description of that video, you should find all the relevant information to understand these concepts. Alternatively, we build these ideas from the ground up by following the playlist on the postulates of quantum mechanics: ruclips.net/p/PL8W2boV7eVfmMcKF-ljTvAJQ2z-vILSxb
I hope this helps!
Yes, more code samples.
Thanks for the feedback!
If I need to consider all dimensions at once, should I have to replace d/dx in the kinetic energy term with the δ/δx term in the laplacian?(I mean second order but it is hard to type that, so I just wrote d/dx)
Yes, in three dimensions the momentum operator is proportional to the gradient, and the kinetic energy to the Laplacian. We very briefly go over it in this video:
ruclips.net/video/Yw2YrTLSq5U/видео.html
@@ProfessorMdoesScience thank you very much
How do you represent the wave function in a momentum space
The wave function in momentum space is the Fourier transform of the wave function in position space, you can find the details of why in this video: ruclips.net/video/2lr3aA4vaBs/видео.html
For an infinite square well, if you calculate the Fourier transform of an eigenstate in position space, you will find that the momentum space wave function is made of two sinc functions centered at momenta of +hbar*n*pi/a and -hbar*n*pi/a, corresponding to motion to the right and to the left inside the well.
yes please code in future videos
Thanks for the suggestion!
I LOVE YOU
Hi, why is the eigenstate also a wave function? isn't wave function a kind of representation? thanks!
If we have an operator A, then an eigenstate |psi> is simply a quantum state such that when we act with A on it, we get the same state back (up to a constant lambda, called the eigenvalue):
A|psi>=lambda |psi>
This notation is the bra-ket notation, but, as you hint in your question, we can represent these states in different ways. Wave functions are simply one of many possible representations, corresponding to the position representation. In this language, the equation above becomes (in 1D):
A psi(x) = lambda psi(x),
where psi(x) = is the wave function. This equation shows how psi(x) is the eigenstate (also called eigenfunction) of the opreator A.
You can learn more about eigenvalues and eigenstates here: ruclips.net/video/p1zg-c1nvwQ/видео.html
And about wave functions here: ruclips.net/video/2lr3aA4vaBs/видео.html
I hope this helps!
@@ProfessorMdoesScienceThanks！ it is clear!
❤
Hi professor, can I get u r mail id . iam from india , masters in physics from IIT
Thanks, you can find our email for business enquiries in the "About" page of the channel.