Thank you so much for making these videos. I took quantum chemistry lectures for 2 years now. And i just started the advanced course and im glad these videos are there. They've helped me a looottt :-D Thank you once again
They make for excellent review, it's been a year since I took Quantum chemistry and I'm about to take my next Quantum Mechanics course. It was great seeing everything presented like this!
In the last step, why are you naming fn and fm in the reverse order? if you named them from the beginning, the integrand should became f*m A fn = fm A* fn*
hello^^ let me ask you about taking a complex conjugate of. 1. (exp(a+bi))* = exp(a-bi) ? Or (exp(a+bi))* = (c+di)* = c-di ? 2. when taking a complex conjugate of AΨ=aΨ, I think it should be (AΨ)*=(aΨ)*. and maybe (aΨ)* can be a(Ψ)* or a*Ψ* because 'a' is a real number. but how does it end up being A*Ψ*=aΨ* ?
The complex conjugate of quantity / sum / product / etc. is just the complex conjugates of all of its components. Everywhere you see an i [sqrt(-1)], you substitute in negative i. We can still take the complex conjugate of a real number, it's just going to be the same number. We can also take the complex conjugate of operators, like A*.
You'll see this identity come up in several videos over the rest of the course. Perhaps it will make more sense once you've seen some examples in action.
Especially In chemistry when we measure a property of a system we want it to be something realistic. In that way you can say that yes, the result that we expect from measuring an observable of a system must be real, It must be something that we can work with. Afterall all the QM operators are associated with real physical observables
Good questions, Michael. Expectation values are of the form = , so we don't need to consider the case of m =/= n for expectation values. However, integrals of the form show up frequently as matrix elements in approximation methods, and that's where these properties can be very helpful. As you noted, we show here that = , which implies a_n = a_m = a_m, which means (a_m - a_n) = 0. So unless a_m = a_n, then = 0. This means that *non-degenerate* eigenfunctions must be orthogonal. Two eigenfunctions of the same Hermitian operator can be non-orthogonal, as long as they have the *same* eigenvalue. So energy states with the same energy can overlap. However, we can always choose free parameters within their functions so as to force them to be orthogonal, and give us a completely orthogonal basis set, even in the case of operators with degenerate states.
I saw your video on orthogonality which leveraged the most general definition of a hermitian operator (the one shown here), but I don't see how it follows from the derivation in this video because the eigenfunctions are not distinct.
The extension to the final line comes from more advanced concepts from linear algebra which I'm probably not articulating well, and start to show up in the videos on superposition and in the approximation methods chapter. In general, a quantum state is not a pure eigenstate, but is a linear combination of an arbitrarily large number of eigenstates. In such a case, the expectation value of the energy is still the integral , where each psi is now a linear combination of n eigenfunctions. The sums of psi and psi_star lead to a double sum with n^2 terms, where each integral is of the form of the final light blue line. This is equivalent to representing the wavefunction as a vector of coefficients (C) and the Hamiltonian operator as a matrix (H) whose elements are these integrals. The expectation value then becomes C^T H C, where C^T is the transpose of the vector C. In order for to be guaranteed to be a real number, it turns out that all eigenvalues of matrix H are required to be real. In order for this to be true, H must be a Hermitian matrix. The definition of a Hermitian matrix is a matrix whose elements equal the complex conjugate of those on the opposite side of the diagonal. Since these matrix elements are of the form , our Hermitian requirement imposes the restriction of the final line between any pair of eigenfunctions of the Hermitian operator.
I suppose "continuously differentiable operator" doesn't quite roll of the tongue for a song title in the same way. But smoothness is in fact not a requirement for an operator to be Hermitian.
Correct, I didn't prove why the same thing works for a general function which is not necessarily an eigenfunction of the Hermitian operator in question.
I️ am taking Quantum Chem now in college, I️ am so glad these videos are available. Very helpful.
Thank you so much for making these videos. I took quantum chemistry lectures for 2 years now. And i just started the advanced course and im glad these videos are there. They've helped me a looottt :-D Thank you once again
Glad to help, Vishakha. Wishing you continued success in your studies.
Thank you so much! I'm finally starting to understand what we've been talking about in my Theoretical Chemistry lectures :D
Great video!
Glad to help, Annie.
Thank you so much for taking the time to make all of these videos!
They make for excellent review, it's been a year since I took Quantum chemistry and I'm about to take my next Quantum Mechanics course. It was great seeing everything presented like this!
Thanks for watching. Glad you found them useful. Good luck in the next course. Feel free to ask any questions in the comments.
In the last step, why are you naming fn and fm in the reverse order? if you named them from the beginning, the integrand should became f*m A fn = fm A* fn*
hello^^ let me ask you about taking a complex conjugate of.
1. (exp(a+bi))* = exp(a-bi) ? Or (exp(a+bi))* = (c+di)* = c-di ?
2. when taking a complex conjugate of AΨ=aΨ,
I think it should be (AΨ)*=(aΨ)*. and maybe (aΨ)* can be a(Ψ)* or a*Ψ* because 'a' is a real number. but how does it end up being A*Ψ*=aΨ* ?
The complex conjugate of quantity / sum / product / etc. is just the complex conjugates of all of its components. Everywhere you see an i [sqrt(-1)], you substitute in negative i. We can still take the complex conjugate of a real number, it's just going to be the same number. We can also take the complex conjugate of operators, like A*.
Thank you^^ though i still wonder why, I think i have to just accept it for this time.
You'll see this identity come up in several videos over the rest of the course. Perhaps it will make more sense once you've seen some examples in action.
Quick question : Measured properties are not necessarily real, in that case, unlikely to be Hermitian operator?
Especially In chemistry when we measure a property of a system we want it to be something realistic. In that way you can say that yes, the result that we expect from measuring an observable of a system must be real, It must be something that we can work with. Afterall all the QM operators are associated with real physical observables
If = (a_n) = a_n, and ()* = (a_m)* = a_m, does a_n need to equal a_m for = * ? And if |m> =/= |n>, does still equal 1?
EDIT: = *
Good questions, Michael. Expectation values are of the form = , so we don't need to consider the case of m =/= n for expectation values. However, integrals of the form show up frequently as matrix elements in approximation methods, and that's where these properties can be very helpful. As you noted, we show here that = , which implies a_n = a_m = a_m, which means (a_m - a_n) = 0. So unless a_m = a_n, then = 0. This means that *non-degenerate* eigenfunctions must be orthogonal. Two eigenfunctions of the same Hermitian operator can be non-orthogonal, as long as they have the *same* eigenvalue. So energy states with the same energy can overlap. However, we can always choose free parameters within their functions so as to force them to be orthogonal, and give us a completely orthogonal basis set, even in the case of operators with degenerate states.
I saw your video on orthogonality which leveraged the most general definition of a hermitian operator (the one shown here), but I don't see how it follows from the derivation in this video because the eigenfunctions are not distinct.
The extension to the final line comes from more advanced concepts from linear algebra which I'm probably not articulating well, and start to show up in the videos on superposition and in the approximation methods chapter. In general, a quantum state is not a pure eigenstate, but is a linear combination of an arbitrarily large number of eigenstates. In such a case, the expectation value of the energy is still the integral , where each psi is now a linear combination of n eigenfunctions. The sums of psi and psi_star lead to a double sum with n^2 terms, where each integral is of the form of the final light blue line. This is equivalent to representing the wavefunction as a vector of coefficients (C) and the Hamiltonian operator as a matrix (H) whose elements are these integrals. The expectation value then becomes C^T H C, where C^T is the transpose of the vector C. In order for to be guaranteed to be a real number, it turns out that all eigenvalues of matrix H are required to be real. In order for this to be true, H must be a Hermitian matrix. The definition of a Hermitian matrix is a matrix whose elements equal the complex conjugate of those on the opposite side of the diagonal. Since these matrix elements are of the form , our Hermitian requirement imposes the restriction of the final line between any pair of eigenfunctions of the Hermitian operator.
...but is it a smooth operator?
I suppose "continuously differentiable operator" doesn't quite roll of the tongue for a song title in the same way. But smoothness is in fact not a requirement for an operator to be Hermitian.
in this video you didn't cover why Hermitain operator also works for general functions, right?
Correct, I didn't prove why the same thing works for a general function which is not necessarily an eigenfunction of the Hermitian operator in question.
is the same as * ? thank you
Not sure, but I think that's likely the case.
if u integrate it just like in the video, the answer is yes
no idea whats going on