Commutator: Hamiltonian and position

[@lseinjr1]

(for *mobile* users) 00:15 Introduction 00:34 Hamiltonian and position operators in QM 01:06 Explicit forms of kinetic energy operator T and position operator x 01:30 Use of property of commutator to write commutator as sum of two commutators 02:04 Since position commutes with potential energy function V, second commutator equals 0 02:16 Definition of commutator 02:35 Addition of dummy argument Ψ 02:59 Grouping of operators 03:15 Replacement of "first" (rightmost) operator by explicit form 04:05 Replacement of "second" (leftmost) operator by explicit form 04:40 Pull constants in front of derivatives 05:53 Use fact that the second derivative is the first derivative of the first derivative 06:05 Use product rule to find first derivative 07:22 Differentiate second time 08:01 Combine like terms 08:15 Cancel additive inverses 08:38 Cancel factor of 2 08:52 Use fact that 𝘪 × 𝘪 = -1 09:30 Use explicit form of momentum operator in QM Color-coded, step-by-step calculation of the commutator of the Hamiltonian operator (in one dimension) and position. Don't forget to like, comment, share, and subscribe!

[@brittanyh1228]

Nice video! Great job!

[@lseinjr1]

Thank you very much!

[@TheOmega1102]

Thank you very much for this Video! Great and easy explanation.

[@lseinjr1]

Glad it was helpful!

[@nurra4515]

Thank you so much!! This really helps me a lot 😊

[@lseinjr1]

You're so welcome!

[@Jekku1987]

Thank you! Excellent video.

[@lseinjr1]

Glad it was helpful!

[@SMc-1235]

Thanks. I was wandering why commutation between position 'x' & potential 'V' is zero. But now I've got the answer.

[@arnabghosh2575]

Thank you so much sir. It really helps a lot💖

[@lseinjr1]

You're welcome 😊

[@muzammilsaeed321]

thank you soo much sir

[@Boooommerang]

Thanks a lot!

[@hmena9934]

What would you consider is the interpretation of being able to obtain the momentum operator from the commutator of the Hamiltonian and position operator?

[@lseinjr1]

We have a number of interpretations of the value of the commutator in pure math. In quantum chemistry, there is one important interpretation of the value of the commutator: if the commutator is 0, the two operations commute; if the commutator is NOT 0, then the two operations DON'T commute. If two operations commute, it means (in quantum chemistry) that we could know both, simultaneously, to perfect precision. On the other, if the two operations DON'T commute, then the Heisenberg Uncertainty Principle applies. Momentum (p) and position (x) do not commute, so that Δ x Δ p ≥ ℏ/2, whereas if they DID commute, Δ x Δ p ≥ 0.

[@tomstalley3179]

thanks!

[@lseinjr1]

Welcome!

[@Ither1023]

thanks :}

[@rflorezablan1]

Thanks!

[@a-factoracademy]

Thank you so much ... Actually how can find the commutator of Hamiltonian and a function of coordinate, q, and p

[@lseinjr1]

The commutator is defined as a binary operator. Why are you trying to take the commutator of three functions simultaneously?

[@a-factoracademy]

@@lseinjr1 I saw a question like that and I was worried too.. I don't know if I can share the question with you!

[@lseinjr1]

If the function of position q and momentum p is a LINEAR function, then you might be able to use one of the commutator identities derived here: kzbin.info/www/bejne/mWO0pmqoqpt_a9E

[@مفكال]

So they don’t commute

[@lseinjr1]

- and so, by the Heisenberg Uncertainty Principle, there are limits to how well we can know the values of each at the same time.