Hello, science enthusiast! This is Professor M does Science, where university level quantum mechanics is made simple(r). One M is a faculty member at the University of Cambridge in the UK, researching the quantum mechanical properties of matter. The other M is another Cambridge academic with a particle physics background. We have taught quantum mechanics to undergraduate and graduate students for years, and this channel aims to cover all the most important concepts and questions in a mathematically rigorous, concise, and transparent way. If you have any feedback or suggestions, please comment in our videos!
At 11:17 you added the terms for ni=0, instead you could have subtracted those two terms and would have obtained the result that the commutator is 1 instead of anticommutator [because (ci dagger ci) gives 0, so it doesn't matter if we add or subtract we get the same RHS but LHS would be different giving both commutator and anticommutator equals to 1]. Same for ni=1 case. Can you please answer this query.
I don't get how one could represent things like position and momentum here. Those are continuous basis. Energy is fine, just a really long unit vector of coefficients for all posible energies. Do we just work in discrete basis like in matrix mechanics? If so, whats the form the hamiltonian will take in matrix form in the energy basis? What about the position operator in the energy basis? (To calculate useful properties like the expectations value)
What is the recommended viewing sequence of the 95 videos. Is it suggested by viewing them from oldest to newest? Each video I have viewed refers to other videos. I find these very helpful in my learning quest!!!
May i clarify my understanding for this lecture here? So what u mean is that different values of alpha and beta can represent a same physical state, since exchanging identical particles should not change anything. Right? And we should expect that the probability of measuring |up, up> remains the same, or rather be independent of alpha and beta. Am i right? But then, upon calculation, results completely contradicted this idea and thus we need to dismiss exchange degeneracy? Just like that?
Hello, for the proof that [A,B^n]=n[A,B]B^(n-1) if [A,[A,B]]=0 and [B,[A,B]]=0 ( 15:39 ), I don’t understand why we need [A,[A,B]]=0 ??? It seems we do not use this condition, so does it works if we only suppose [B,[A,B]]=0 ? Thanks for yours videos, those are great!
Thanks - I do have an issue though with the notation towards the end; how can g_{ijkl} be dependent on a specific pair (q,q') when we have eliminated the sum over (space) indices? Shouldn't the expression for $g_{ijkl}$ be free of the q,q' just like the $f_{ik}$ and $h_{jl}$ at the top of the page?
Interesting question, we are currently looking at non-relativistic quantum mechanics, but we do hope to cover relativistic QM in the future. We are in fact building towards that with our series on quantum field operators :)
Hallo sir ! Unfortunately, in general, neither symmetric nor ant-isymmetric wavefunctions can be said to be eigenfunctions of the Hamiltonian. The wave function for an electron in a hydrogen-like atom with atomic number Z in the ground state is RZ(r)=2(Z/a0)^(3/2)*exp(-Zr/a0). RZ(r) is an eigenfunction of HZ=1/(2m)*p^2-Ze^2/(4πε0r). But RZ(r) is not an eigenfunction of HZ'=1/(2m)*p^2-Z'e^2/(4πε0r), Z'≠Z. Let us consider the case where a hydrogen-type atom with atomic number Z and a hydrogen-type atom with atomic number Z' are sufficiently separated from each other. And each electron in each atom is in the ground state. The anti-symmetric wave function Ψ={RZ(r1)RZ'(r2)-RZ(r2)RZ'(r1)}/2^(1/2) is not an eigenfunction of the Hamiltonian H=1/(2m)*p1^2-Ze^2/(4πε0r1)+1/(2m)*p2^2-Z'e^2/(4πε0r2). It should be an ironclad rule of quantum mechanics that the wave function is an eigenfunction of the Hamiltonian.
Thank You professor M does Science for your wonderful series course in Quantum Mechanics. I am a scientist , that comes from Electrical Enginering course work/career and my final metamorphosis dream is to fully mutate to a mix of Teoretical Physics + Computational Mathematics scientist
The kinetic energy is associated with the curvature of the wave function. You can qualitatively understand this by considering that, in the position representation, the kinetic energy operator is the second derivative with respect to position. In this context, the more "wiggly" the wave function, the higher the kinetic energy. If you look at the higher excited states of hydrogen, you will see that they do become increasingly "wiggly" (the number of nodes increases, and therefore so does the curvature). As a result, the kinetic energy grows with the excited state. I hope this helps!
The eigenfunctions are the position representation of the eigenstates, so they do describe the same thing. You can check out our videos on wave functions to explore the relationship between wave functions and more abstract states here: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2lr3aA4vaBs.html I hope this helps!
What about energy and time? We know that time and energy can be also related by Fourier transform. So, as the momentum is related to k value, p=hk, energy related to angular frequency, E=hw. Can we relate the energy representation to time representation in similar manner or that's not the case simply because time quantity is not an operator?
Your last sentence is the key point: time is not an operator, it is a parameter, so we cannot do as you ask. However, one can still study the relationship between energy and time, for example we explore it in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-auLZ2WOKiqE.html I hope this helps!
I tried Nielsen & chuang ,but it was your video that cleared so many things about density operator formalism for me ,I hope this channel grows more and more,thank you.
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: ru-vid.com/group/PL8W2boV7eVfnHHCwSB7Y0jtvyWkN49UaZ Hope you like it!
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.
I was just wondering that how were you able to understand these conepts without animation ? Because in old days, there were no animations, how people were able to imagine such complicated topics ?
Dear Professor, I would request you to make videos about Classical Mechanics as well. Like not the Newtonian one, but the Lagrangian, Hamiltonian and Hamilton-Jacobi theories. From one of your well-wishers. God bless!
