Functions of operators in quantum mechanics

📚 Many important phenomena in quantum mechanics are described by functions of operators. For example, spatial translations are given by an exponential function of the momentum operator, and time evolution is given by an exponential function of the Hamiltonian. In this video we explain how to construct the function of an operator, and explore what their properties are.

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If you are interested in the mathematics behind the Baker-Campbell-Hausdorff formula, further details can be found in the following links:

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Director and writer: BM
Producer and designer: MC

[fernandojimenezmotte]

Very clear, didactical explanation of Operators in Quantum Mechanics Professor M with beautiful edited math and very interesting stuff. I just came up with your channel 2 days ago and I will definitively follow You.

[vladimirkolovrat]

As always, your explanations are just brilliant. This is quite an involved topic, but it's redeemed by having a friendly face explain it in a clear way. Thank you for posting this, and for your hard work.

[rodrigoappendino]

I have test of quantum mechanics this week, and we are already learning about spinors and approximation methods. I use your videos a lot, because the subjects I was supposed to know already are not entirely clear in my mind. So thanks a lot.

[gufo__]

Oh wow, the timing, that's what I needed today

[tariklahcen]

Amazing explanation and great English pronunciation. Many thanks

[armalify]

Your videos are always great, I can say that [ ALWAYS, GREAT ] commute ;-)

[vipinnegi]

thnx i am following Zettili and your videos are helping me a lot

[armalify]

Can we say that If A and B commute then [A^n, B^n] and [ B^n, A^n] are always commute ? Thanks !

[attaurrahmanphysicist]

Thank you sir. I was searching for such a video. Please sir tell me from which book you are explaining this topic.

[gergodenes]

I am unsure whether we're actually using the fact that A commutes with [A, B] in the proof that [A, F(B)] = [A, B] F'(B).
It seems that we're only using that B commutes with [A, B] at 15:14, so the relation should still hold even if A does not commute with [A, B].
Obviously both commutative relations are needed for the example of x and p, but I figure this somewhat more general result should be noted.

[clemcaramel]

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!

[xiaoyu]

Thanks a lot for the video. It seems to me the addition of operators is not defined? Or maybe I am missing something? Is (A1 + A2)(f(x)) defined as A1(f(x)) + A2(f(x))? The definition of addition is needed for the expansion? Oh, perhaps it is just a given? I can sort of understand it if I map it to matrices and vectors.

Oh, I see from the later equations that it is the question then....Thanks for the video.

[cunningham.s_law]

do we have some constraints on these functions?
if they are not bijective then we can't invert them
so if our measure is related to an non invertible function of an operator, how can we knoow which of the base state of the operator we are in?

[_stellar_saha]

CAN YOU PLEASE PROVIDE THE PDF FORM OF THE NOTE IN DESCRIPTION?

[abhishekshrivastava]

why we write function of an operator in a power series

[Upgradezz]

What is the operator sqrt{ x^2 + y^2 + z^2 } ?

[MessedUpSystem]

The last property is just the chain rule in disguise since the commutator is just a Lie derivative :p