Hermitian operators in quantum mechanics

[@ryanjbuchanan]

"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.

[@ProfessorMdoesScience]

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".

[@ryanjbuchanan]

@@ProfessorMdoesScience Thank you lol. I must have been super tired.

[@Dudenobody]

Awesome video! Thank you for explaining this, I was struggling to understand it.

[@ProfessorMdoesScience]

Glad you found this helpful!

[@shutupimlearning]

very concise explanation, glad i found this channel

[@ProfessorMdoesScience]

Glad you like the videos! :)

[@MrVsoral]

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.

[@ProfessorMdoesScience]

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]

Thank you, very clear explanation!

[@ProfessorMdoesScience]

Glad it was helpful!

[@stevenschilizzi4104]

Outstanding video! Beautiful job.

[@ProfessorMdoesScience]

Glad you like it!

[@floribus5522]

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!

[@ProfessorMdoesScience]

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.

[@floribus5522]

@@ProfessorMdoesScience I see, thank you very much.

[@TheWingEmpire]

Thank You very much, it was really clarifying! :)

[@ProfessorMdoesScience]

Glad you liked it! :)

[@tejareddy3076]

The explanation is very clear and precise. I am unable to catch the speed and watching it at 0.75x. But no problem.

[@ProfessorMdoesScience]

Glad you like it! And this is the magic of online learning: you can control how you watch it ;)

[@kaushikgupta1410]

for gram Schmidt normalization of more than 2 eigenstates, We have to apply the perpendicular conditions for all eigenstates separately, right?

[@ProfessorMdoesScience]

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!

[@kaushikgupta1410]

@@ProfessorMdoesScience thx, it helps :)

[@BruinChang]

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.

[@ProfessorMdoesScience]

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!

[@BruinChang]

@@ProfessorMdoesScience Thanks a lot to explain well to me. I am looking forward to seeing the future series!

[@sibongumusaws9551]

where can I learn the bra-ket notation, beginner friendly and advance to more complex stuff like this. Any recommendations

[@ProfessorMdoesScience]

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.pandey]

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?

[@ProfessorMdoesScience]

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.pandey]

Ok.. Thank you

[@ghanascilab477]

great video

[@ProfessorMdoesScience]

Thanks!

[@snjy1619]

[@progra_kun4331]

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?

[@ProfessorMdoesScience]

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.

[@nomanahmadkhan7791]

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?

[@ProfessorMdoesScience]

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!

[@nomanahmadkhan7791]

@@ProfessorMdoesScience Oh, definitely got it. Why I was missing it the first time 🤔. Thanks a lot.

[@marticircunsiduxans5957]

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?

[@ProfessorMdoesScience]

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!

[@marticircunsiduxans5957]

@@ProfessorMdoesScience Understood!Thanks professor :)

[@detch408]

@@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]

Will you guys make videos on scattering?

[@joeaverage8329]

Love your videos by the way. Very helpful too!

[@ProfessorMdoesScience]

We do plan to do a series on scattering, but it may take some time before we get there...

[@ProfessorMdoesScience]

Glad you like the videos! :)

[@jumpyalexa8526]

Thank you so much, you may actually be better than Griffiths! Side note, you have a lovely voice!

[@ProfessorMdoesScience]

Glad you like our approach! :)

[@KaranveerSingh-xn4tv]

Is every entry of hermitian matrix a real number?

[@ProfessorMdoesScience]

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-xn4tv]

@@ProfessorMdoesScience Shouldn't the transposed Hermitian matrix be exactly the same as the original one?

[@ProfessorMdoesScience]

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-xn4tv]

Really appreciate it!!🙂🙂

[@stevenschilizzi4104]

@@ProfessorMdoesScience This is a really useful answer, thank you! I was wondering about that myself after watching your (excellent) video.

[@syahrilsabtu1758]

Hi prof. how to Prove that the operators for position x and momentum P are Hermitian. can you make this video.

[@ProfessorMdoesScience]

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!

[@workerpowernow]

awesome!

[@ProfessorMdoesScience]

Glad you like it!

[@wdobni]

what part of a proton is 'observable'? has anybody ever 'observed' an electron?

[@ProfessorMdoesScience]

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-xn4tv]

Is spin operator an hermitian operator?

[@ProfessorMdoesScience]

Yes! For example, the spin 1/2 operators are proportional to the Pauli matrices, which are themselves Hermitian. I hope this helps!

[@mohankurra70]

Hi sir, I have a small doubt Is hermitian and Hamiltonian operators are same are different plz explain sir

[@ProfessorMdoesScience]

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.

[@mohankurra70]

@@ProfessorMdoesScience thank you sir for your valuable reply.

[@mohankurra70]

Hamiltonian is for energy of microscopic objects and hermitian is for all. Am I clear sir

[@ProfessorMdoesScience]

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!

[@ProfessorMdoesScience]

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.

[@fabio_air4230]

Thanks!

[@ProfessorMdoesScience]

Oh wow, thank you!!

[@زينالعابدينماجدمحمد]

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