The crazy thing is that this is the most premium course on quantum mechanics and it is completely free. Even paid courses aren't this good. Lots of love to you all ❤❤
Thanks for coming out with another great video. This one was a bit easier than some of the others. I would encourage you to use the up and down arrows for the eigenstates which is customary (and which we actually use everyday).
@@ProfessorMdoesScience great. Actually I tend to use the up and down arrow notation for any two state system, they are the easiest and cleanest notation (at least to me, and my professors). 😀
@@YossiSirote One reason we went for the plus/minus notation is because it is somewhat easier to label the eigenvalues and eigenstates collectively (e.g. E_{+/-}) rather than E_{up/down}), but ultimately there isn't much difference. When we discuss spin-1/2 particles, we may well change notation to up/down to make things clearer in that case :)
A small typo at 16:13 in your lecture. When you cleared the board and re-wrote down the expansion of the determinant, you mean to write [(d_o - lamda) + d_3][(d_o -lamba) - d_3] -(d_1-id_2)(d_1 +i d_2) = 0. It is a minor correction and everyone understood your reasoning. But at 16:13, you wrote instead [(d_o - lamda) + d_3][(d_o + lamba) - d_3] -(d_1-id_2)(d_1 +i d_2) = 0. There is one of the special appendix in the Cohen-Tannoudji, et. al, Quantum Mechanics books that has a similar discussion on two-state systems. But your presentation is more detailed. Nice presentation. Really nice. Thank you.
Another great video! just one thing I noticed is at, @ 16:05 you have an expression for the determinate, in the following frame @ 16:14 you flipped the minus to a plus on the (d0 - lambda) term. it was still treated as a minus so all was well.
Thanks for this, you are absolutely correct, there is a typo in the frame where we copy the expression again, but as you also correctly say we still have the correct expression in the following line. Thanks for finding this!
@@ProfessorMdoesScience I was following along and indeed as written in the second line do^2 -(d3 - L)^2 + [...] , rather than (d0 - L)^2 - d3^2 +[...] For a moment I thought I was doing some mistake, until I realize the sign of the lambda was flipped That said: great videos. Very happy I discovered your channel!
I saw the videos about creation- and annihilation operators. There's one discovery, that blows my mind. So, here's my question. If "A" is the matrix of the creation-operator, why does e^A contain the pascal triangle? It becomes more clear, if you square the entries of the matrix B_ij = A_ij² Then, e^B gives exactly the pascal triangle. Has this some special reason? Because the pascal triangle is some pivot point of maths, with connections to fractals, fibonacci, golden ratio, .... That would mean, Q.-Mechanics is "somehow" connected to all these topics.
This is an interesting insight, but I must admit I have not thought deeply about this before. One suggestion may be to consider the Taylor expansion of the exponential, and then look at each term in turn and how it allows you to "navigate" the Pascal triangle. I hope this helps, and please report any progress you make :)
@@ProfessorMdoesScience Sure, I can report my findings. It needs some time to collect and sort them, and I'm actually a bit stressed. Your spamfilter has to allow my mails. But I have to say, I'm not an academic or professional, so it's very likely you're not impressed. I'm just a hobbyist that plays around with maths. Sorry, for my bad english.
I also noticed another case that could be interesting, when the Hamiltonian is traceless. This is the case when H₁₁ = -H₂₂ , and then the eigenvalues will be split into a positive/negative quantity λ = ± √H₁₁² + H₁₂² special cases: traceless, diagonal λ = ±H₁₁ traceless, only nondiagonal λ = ±H₁₂ I think the degenerate part when H₁₁ = H₂₂ acts like a "baseline/reference", and we get energy splitting from the traceless part
Does this mean that if the proportional constant E is equal to one, then the Hamiltonian is identical to the identity matrix? Does this mean that the eigenvalues of the degenerate states of the identity operator can be viewed as quantities of energy?
Yes, although this would not have a deeper meaning than any other value of E. In all cases, it just means that the you have two degenerate energy levels.
I would also like to ask what does a negative energy eigenvalue -E mean? You have probably used it to differentiate between two possible positive energies of quantum states. But could it also be understood that one of the two possible energies is negative?
@@tomaskubalik1952 The absolute values of the energies have no physically important meaning as we could always add a constant energy shift to the Hamiltonian and the physics of the problem would not change. All the matters is the relative energy between the two states. I hope this helps!
The crazy thing is that this is the most premium course on quantum mechanics and it is completely free. Even paid courses aren't this good. Lots of love to you all ❤❤
Thanks for your support! :)
What a coincidence! Prof just today was discussing this. Good to see you after a while
What a coincidence indeed! And stay tuned for the eigenstates video which will (hopefully) come out next week!
Looking forward to the video!@@ProfessorMdoesScience
@@adarshramtel9384 Can now confirmed it is scheduled for Monday :)
After a long time
Thanks for your patience :)
I've been looking for your videos on this topic...glad that you started this.
We have another one coming out next week on the eigenstates, so stay tuned!
Thanks for coming out with another great video. This one was a bit easier than some of the others. I would encourage you to use the up and down arrows for the eigenstates which is customary (and which we actually use everyday).
Yes, definitely costumary in the case of spin at least. Thanks for the suggestion!
@@ProfessorMdoesScience great. Actually I tend to use the up and down arrow notation for any two state system, they are the easiest and cleanest notation (at least to me, and my professors). 😀
@@YossiSirote One reason we went for the plus/minus notation is because it is somewhat easier to label the eigenvalues and eigenstates collectively (e.g. E_{+/-}) rather than E_{up/down}), but ultimately there isn't much difference. When we discuss spin-1/2 particles, we may well change notation to up/down to make things clearer in that case :)
Indeed this is a wonderful lecture professor, also explain the eigen values of 3*3 and diagonalization of matrixes .
Thanks
Thanks for the suggestion!
welcome back❤❤❤
Thanks for watching!
Clearly explained. Greatly appreciated.
Glad you like it!
It is always so very exciting to see you 😊
Thanks for watching! :)
A small typo at 16:13 in your lecture. When you cleared the board and re-wrote down the expansion of the determinant, you mean to write
[(d_o - lamda) + d_3][(d_o -lamba) - d_3] -(d_1-id_2)(d_1 +i d_2) = 0. It is a minor correction and everyone understood your reasoning. But at 16:13, you wrote instead [(d_o - lamda) + d_3][(d_o + lamba) - d_3] -(d_1-id_2)(d_1 +i d_2) = 0.
There is one of the special appendix in the Cohen-Tannoudji, et. al, Quantum Mechanics books that has a similar discussion on two-state systems. But your presentation is more detailed.
Nice presentation. Really nice. Thank you.
Thanks for catching the typo, and glad you like our presentations!
Another great video! just one thing I noticed is at, @ 16:05 you have an expression for the determinate, in the following frame @ 16:14 you flipped the minus to a plus on the (d0 - lambda) term. it was still treated as a minus so all was well.
Thanks for this, you are absolutely correct, there is a typo in the frame where we copy the expression again, but as you also correctly say we still have the correct expression in the following line. Thanks for finding this!
@@ProfessorMdoesScience I was following along and indeed
as written in the second line do^2 -(d3 - L)^2 + [...] , rather than (d0 - L)^2 - d3^2 +[...]
For a moment I thought I was doing some mistake, until I realize the sign of the lambda was flipped
That said: great videos. Very happy I discovered your channel!
@@Entropy3ko Yes, sorry about the typo when copying the expression from one slide to another... hopefully it's not causing too much trouble!
wow. that's what i am looking for. thank you very much for such nice presentation.
Thanks for watching!
welcome back!
Thanks for watching!
Thanks a lot for this great presentation!
Glad you like it!
Great video! When are you going to upload the next one of the series?
Hopefully soon, but our day job is currently very demanding so struggling to find the time...
Please consider a series on statistical mechanics
Thanks for the suggestion!
I saw the videos about creation- and annihilation operators.
There's one discovery, that blows my mind. So, here's my question.
If "A" is the matrix of the creation-operator, why does e^A contain the pascal triangle?
It becomes more clear, if you square the entries of the matrix B_ij = A_ij²
Then, e^B gives exactly the pascal triangle. Has this some special reason?
Because the pascal triangle is some pivot point of maths, with connections to fractals, fibonacci, golden ratio, ....
That would mean, Q.-Mechanics is "somehow" connected to all these topics.
This is an interesting insight, but I must admit I have not thought deeply about this before. One suggestion may be to consider the Taylor expansion of the exponential, and then look at each term in turn and how it allows you to "navigate" the Pascal triangle. I hope this helps, and please report any progress you make :)
@@ProfessorMdoesScience
Sure, I can report my findings. It needs some time to collect and sort them, and I'm actually a bit stressed. Your spamfilter has to allow my mails.
But I have to say, I'm not an academic or professional, so it's very likely you're not impressed. I'm just a hobbyist that plays around with maths.
Sorry, for my bad english.
Great video! When are you going to release the one about eigenstates of a two state quantum system?
Coming up on Monday!
Excellent. Thank you.
Thanks for watching!
I also noticed another case that could be interesting, when the Hamiltonian is traceless.
This is the case when H₁₁ = -H₂₂ , and then the eigenvalues will be split into a positive/negative quantity
λ = ± √H₁₁² + H₁₂²
special cases:
traceless, diagonal
λ = ±H₁₁
traceless, only nondiagonal
λ = ±H₁₂
I think the degenerate part when H₁₁ = H₂₂ acts like a "baseline/reference", and we get energy splitting from the traceless part
Good insights! We'll look at plenty of examples in future application videos :)
Does this mean that if the proportional constant E is equal to one, then the Hamiltonian is identical to the identity matrix? Does this mean that the eigenvalues of the degenerate states of the identity operator can be viewed as quantities of energy?
Yes, although this would not have a deeper meaning than any other value of E. In all cases, it just means that the you have two degenerate energy levels.
Does every complete set of commuting operators contain the identity operator? Thanks a lot@@ProfessorMdoesScience
I would also like to ask what does a negative energy eigenvalue -E mean? You have probably used it to differentiate between two possible positive energies of quantum states. But could it also be understood that one of the two possible energies is negative?
@@tomaskubalik1952 The absolute values of the energies have no physically important meaning as we could always add a constant energy shift to the Hamiltonian and the physics of the problem would not change. All the matters is the relative energy between the two states. I hope this helps!
@@tomaskubalik1952 Not sure I fully understand your question. But just in case: the identity operator does commute with every other operator.