"The eigenstate of a Hermitian operator corresponding to different Hermitian operators are orthogonal." Wow. 4 years at MIT and you just explained what all of my professors couldn't.
@ProfessorMdoesScience2 жыл бұрын
Glad you find it useful! Do note however that the precise statement is: "the eigenstates of a Hermitian operator corresponding to different eigenvalues are orthogonal".
@ryanjbuchanan2 жыл бұрын
@@ProfessorMdoesScience Thank you lol. I must have been super tired.
@Dudenobody2 жыл бұрын
Awesome video! Thank you for explaining this, I was struggling to understand it.
@ProfessorMdoesScience2 жыл бұрын
Glad you found this helpful!
@shutupimlearning2 жыл бұрын
very concise explanation, glad i found this channel
@ProfessorMdoesScience2 жыл бұрын
Glad you like the videos! :)
@MrVsoral2 жыл бұрын
I have come across some videos on Quantum simulations where it was mentioned that any Hermitian operator (a Hamiltonian) can be expressed as a linear sum of Identity and Pauli Matrices. For example: H = a₁₂ σ₁ⓧσ₂ + a₁₃ σ₁ⓧσ₃ + a₂₃ σ₂ⓧσ₃ + ... + a₀₁ Iⓧσ₁ + ... where a₁₂ ... a₀₁ .. are scalars. However I am unable to find any details on this; specifically any proof or algorithm to decompose a Hermitian matrix in above form. Do you have a video that explains this, or can you please refer me to an literature that explains this ab-initio? Thanks in advance.
@ProfessorMdoesScience2 жыл бұрын
Great question! We don't yet have a video explaining this, but after we finish with the hydrogen atom series we plan to start with Pauli matrices and we will cover this. But to get you started, it is simplest to first think about a 2-dimensional state space, in which case the Pauli matrices and the 2x2 identity span the full space (and you don't need the tensor product you have in your expression). We'll let you know when we publish the relevant videos!
@nastyavicodin6229 Жыл бұрын
Thank you, very clear explanation!
@ProfessorMdoesScience Жыл бұрын
Glad it was helpful!
@stevenschilizzi41043 жыл бұрын
Outstanding video! Beautiful job.
@ProfessorMdoesScience3 жыл бұрын
Glad you like it!
@floribus55223 жыл бұрын
Sorry when I missed it. Is there a video that proves the existence if eigenvalues/vectors in the first place? Your videos are great, very clear!
@ProfessorMdoesScience3 жыл бұрын
Good insight: we have actually not shown that a Hermitian matrix is diagonalizable, we've simply taken it as a given result. The proof is somewhat more involved than the properties we discuss in this video, and it does not necessarily provide relevant physical insights, and this is the reason we've just taken it as a given.
@floribus55223 жыл бұрын
@@ProfessorMdoesScience I see, thank you very much.
@TheWingEmpire3 жыл бұрын
Thank You very much, it was really clarifying! :)
@ProfessorMdoesScience3 жыл бұрын
Glad you liked it! :)
@tejareddy30763 жыл бұрын
The explanation is very clear and precise. I am unable to catch the speed and watching it at 0.75x. But no problem.
@ProfessorMdoesScience3 жыл бұрын
Glad you like it! And this is the magic of online learning: you can control how you watch it ;)
@kaushikgupta14103 жыл бұрын
for gram Schmidt normalization of more than 2 eigenstates, We have to apply the perpendicular conditions for all eigenstates separately, right?
@ProfessorMdoesScience3 жыл бұрын
This is correct, for each new state that you include, you need to make sure it is orthogonal to all previous states. I hope this helps!
@kaushikgupta14103 жыл бұрын
@@ProfessorMdoesScience thx, it helps :)
@BruinChang2 жыл бұрын
I remember there are s, p, d, f orbitals taught in general chemistry. Does the "degenerate" eigen-values relate to the p, d, f orbitals? Because a single eigen-value maps to multiple eigen-states, and the one-to-many mapping reminds me about diverse orbitals.
@ProfessorMdoesScience2 жыл бұрын
This is a very interesting topic, and a comment is not enough to fully address it. However, let me share a few thoughts. The s, p, d, ... orbitals are first introduced in the context of the hydrogen atom (we have just started a series on hydrogen and will cover this in detail over the next few weeks/months, so stay tuned!). In this context, the s, p, d, ... refer to specific angular momenta associated with the energy eigenfunctions (s is associated with quantum number l=0, p with l=1, d with l=2, and so on). You also need to combine them with the "n" quantum number, and then you get the ground state (1s orbital), the first excited state (2s and 2p orbitals, which are degenerate), and so on. Things get more complicated with relativistic corrections, where, for example, the 2s and 2p orbitals are no longer degenerate. When talking about other elements, it is not known how to find the energy eigenvalues and eigenfunctions analytically, so we need to resort to approximate solutions. Many of these are built on the hydrogen atom solutions above, hence the use of the terms s, p, d, ... orbitals for other systems beyond hydrogen too. I realise this is a very handwavy answer, but we hope to cover these topics in future videos in the detail they deserve, and hopefully this helps you get started in the meantime!
@BruinChang2 жыл бұрын
@@ProfessorMdoesScience Thanks a lot to explain well to me. I am looking forward to seeing the future series!
@sibongumusaws95512 жыл бұрын
where can I learn the bra-ket notation, beginner friendly and advance to more complex stuff like this. Any recommendations
@ProfessorMdoesScience2 жыл бұрын
We have an video where we cover this notation, and use an analogy to more standard vector space notation: kzbin.info/www/bejne/nnvSiIBvn8tjnbc I hope this helps!
@anoopmis.pandey3 жыл бұрын
At time 07:00 and earlier as well, you have mentioned for orthonormal bases the scalar product is delta ij, but shouldn't it be zero for both are different and 1 if both are same like =1 and = 0?
@ProfessorMdoesScience3 жыл бұрын
You are correct, but this exactly what the symbol delta_ij means. This symbol is called the "Kronecker delta" and it is defined as: delta_ij=0 if i is different from j delta_ij=1 if i=j I hope this helps!
@anoopmis.pandey3 жыл бұрын
Ok.. Thank you
@ghanascilab4773 жыл бұрын
great video
@ProfessorMdoesScience3 жыл бұрын
Thanks!
@snjy161911 ай бұрын
@progra_kun43312 жыл бұрын
Hola, no me queda claro, lambda es un real porque, de ser imaginario, la igualdad no podría darse, cierto? Lo digo porque se asume que lambda estrella es de por sí, un complejo conjugado, imagino que es por eso que nos dicen que no trabajaremos con el término imaginario, cierto?
@ProfessorMdoesScience2 жыл бұрын
Exacto. Para hacerlo mas explícito, considera lambda=a+ib, a,b reales. La condición lambda=lambda* implica que: a+ib=a-ib La parte real de esta expresión cumple la igualdad para cualquier a. La parte imaginaria solo cumple la igualdad con b=0. Esto implica que lambda no tiene parte imaginaria, y es real.
@nomanahmadkhan77912 жыл бұрын
Another brilliant video indeed. I am a bit confused about "n" (as in i=1, 2, 3, ..... n,) and "N" (as in N-dimensional space). Are these same/different? Can these be same/different?
@ProfessorMdoesScience2 жыл бұрын
They are not the same in general. "N" refers to the dimension of the vector space. "n" refers to the degeneracy of a given eigenvalue. To give you a simple example, imagine we have a 3-dimensional vector space, so N=3. In this vector space, imagine we have two eigenvalues, one is lambda_1, associated with two eigenstates, and the other is lambda_2, associated with a single eigenstate. This means that the first eigenvalue, lambda_1, is doubly degenerate, so n=2 for it. The second eigenvalue is singly degenerate, so n=1 for it. I hope this helps!
@nomanahmadkhan77912 жыл бұрын
@@ProfessorMdoesScience Oh, definitely got it. Why I was missing it the first time 🤔. Thanks a lot.
@marticircunsiduxans59573 жыл бұрын
Hi professor! I think that there's a mistake in 4:43. The eigenvalue resulting from the operation of the bra phi with the operator A should be a conjugate. Am I right?
@ProfessorMdoesScience3 жыл бұрын
In general you are correct that when going to the dual space, a scalar becomes its conjugate. However, we show in the previous slide to the one you are referring to that the eigenvalues of a Hermitian operator are real numbers, which means that the number is equal to its conjugate, so in this particular case I think there is no mistake. I hope this helps!
@marticircunsiduxans59573 жыл бұрын
@@ProfessorMdoesScience Understood!Thanks professor :)
@detch4082 жыл бұрын
@@ProfessorMdoesScience Hi professor, your answer is clear for the orthogonality section, but I still have doubts about the first part (minute 2:21 when you're showing the proof that a self-adjoint operator always have a real-valued spectrum). In particular, it seems to me that when you apply the bra on both sides of the equation, you are leaving the scalar lambda outside of the brackets just like if it was a real number (which would be circular reasoning, since that's exactly what you're trying to prove btw). But shouldn't that lambda be the coefficient of the second argument of the inner product? And if so, isn't it correct that bringing lambda outside of the brackets would require to conjugate it? What am I missing? Thanks in advance :)
@joeaverage8329 Жыл бұрын
Will you guys make videos on scattering?
@joeaverage8329 Жыл бұрын
Love your videos by the way. Very helpful too!
@ProfessorMdoesScience Жыл бұрын
We do plan to do a series on scattering, but it may take some time before we get there...
@ProfessorMdoesScience Жыл бұрын
Glad you like the videos! :)
@jumpyalexa85262 жыл бұрын
Thank you so much, you may actually be better than Griffiths! Side note, you have a lovely voice!
@ProfessorMdoesScience2 жыл бұрын
Glad you like our approach! :)
@KaranveerSingh-xn4tv3 жыл бұрын
Is every entry of hermitian matrix a real number?
@ProfessorMdoesScience3 жыл бұрын
Good question! The answer is in general "no". This is because in matrix form, A=A^{dagger} becomes A_ij=A*_ji, which means that the diagonal elements must be real numbers, but the only constraint on the off-diagonal elements on either side of the main diagonal is that they are complex conjugates of each other. An example of a Hermitian matrix that we often use in quantum mechanics and that has entries that are not real numbers is the second Pauli matrix, a 2x2 matrix: 0 -i i 0 Having said this, the precise entries of a matrix in quantum mechanics always depend on the basis in which you are writing the states and operators. As Hermitian operators have real eigenvalues, then writing the Hermitian operator in the basis of its own eigenstates will always lead to a diagonal matrix whose entries will all be real. I hope this helps!
@KaranveerSingh-xn4tv3 жыл бұрын
@@ProfessorMdoesScience Shouldn't the transposed Hermitian matrix be exactly the same as the original one?
@ProfessorMdoesScience3 жыл бұрын
It is the transposed conjugate matrix that should be exactly the same as the original one. This means you need to do two things: (1) exchange rows and columns, and (2) take complex conjugates of all entries. If we use the second Pauli matrix as an example, the starting point is: 0 -i i 0 If we now apply step (1), we exchange rows and columns to get: 0 i -i 0 We next need to apply step (2), so we calculate the complex conjugate of each entry: 0 -i i 0 This gives back the original matrix, so this matrix is Hermitian. I hope this helps!
@KaranveerSingh-xn4tv3 жыл бұрын
Really appreciate it!!🙂🙂
@stevenschilizzi41043 жыл бұрын
@@ProfessorMdoesScience This is a really useful answer, thank you! I was wondering about that myself after watching your (excellent) video.
@syahrilsabtu17583 жыл бұрын
Hi prof. how to Prove that the operators for position x and momentum P are Hermitian. can you make this video.
@ProfessorMdoesScience3 жыл бұрын
Thanks for the suggestion! In the meantime, we have a video in which we explain how to write the position and momentum operators in the position basis (how they act on wave functions): kzbin.info/www/bejne/j6iViqWKgbika7c From these expressions, you can then show that these operators are Hermitian by showing that \int dx phi*(x psi) = \int dx psi (A^{dagger} phi)* and an analogous expression for the momentum operator. I hope this helps!
@workerpowernow3 жыл бұрын
awesome!
@ProfessorMdoesScience3 жыл бұрын
Glad you like it!
@wdobni Жыл бұрын
what part of a proton is 'observable'? has anybody ever 'observed' an electron?
@ProfessorMdoesScience Жыл бұрын
Interesting question, which fundamentally addresses what we mean by "observed". When you observe an everyday object with your eyes or a camera, you are looking at the photons scattered by that object. In a similar fashion, an electron can be observed, for example, in a bubble chamber or in a silicon detector by its interaction with surrounding matter. I hope this helps!
@KaranveerSingh-xn4tv3 жыл бұрын
Is spin operator an hermitian operator?
@ProfessorMdoesScience3 жыл бұрын
Yes! For example, the spin 1/2 operators are proportional to the Pauli matrices, which are themselves Hermitian. I hope this helps!
@mohankurra703 жыл бұрын
Hi sir, I have a small doubt Is hermitian and Hamiltonian operators are same are different plz explain sir
@ProfessorMdoesScience3 жыл бұрын
This is an important question. They are not the same. A Hermitian operator A is an operator that is equal to its ajoint, A^dagger=A. The Hamiltonian is an example of a Hermitian operator, but there are other Hermitian operators beyond the Hamiltonian, for example the position operator, the momentum operator, or the angular momentum operator.
@mohankurra703 жыл бұрын
@@ProfessorMdoesScience thank you sir for your valuable reply.
@mohankurra703 жыл бұрын
Hamiltonian is for energy of microscopic objects and hermitian is for all. Am I clear sir
@ProfessorMdoesScience3 жыл бұрын
A Hermitian operator is an operator that is equal to its ajoint. This is a purely mathematical statement without physics content. What happens is that operators that are Hermitian (so that they obey the mathematical property above) are then used in quantum mechanics to represent physical observables. The Hamiltonian is the operator associated with the energy of the system, and is an example of a Hermitian operator because it obeys the mathematical definition of Hermitian operators. But there are other operators used in quantum mechanics that are also Hermitian, for example the operator for the position of a particle, or the operator for the momentum of a particle. If we focus on the Hamiltonian, it is the operator representing the total energy of any system: microscopic and macroscopic. For example, you can write down the Hamiltonian of a material by including the kinetic and potential energy of all electrons and ions in it. I hope this helps!
@ProfessorMdoesScience3 жыл бұрын
Following up on your question, we now have a video on the Schrödinger equation: kzbin.info/www/bejne/eXzTqWyeoLZmfq8 In this video we see why the Hamiltonian is such an important operator: it is the operator that drives the time dependence of quantum states.