The general Schrödinger equation. x, p commutator

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7 жыл бұрын

MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-04S16
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
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Пікірлер: 47

@teddychibuye 11 ай бұрын

Brilliant physics professor

@nelsonmichaelvillegasjuro4362 4 жыл бұрын

Barton is the best! Now I can see SE more easy!

@byronwatkins2565 4 жыл бұрын

There is a physical reason why we must multiply V(x, t) by Psi. Psi(x, t) indicates where the particle is likely to be observed (its magnitude squared) so the system's potential energy is the overlap between V and Psi. Classically, V(x, t) is the potential energy of the particle IF IT IS LOCATED AT (x, t); but quantum particles have no location in this respect. The product of V and Psi is large only where and when both V and Psi are large such that potential energy is large at (x, t) and the particle is likely to be observed at (x, t).

@prasadpawar7027 4 жыл бұрын

Also not multiplying it would be dimensionally incorrect.

@byronwatkins2565 4 жыл бұрын

@@prasadpawar7027 Psi has no units, so no it wouldn't.

@prasadpawar7027 4 жыл бұрын

@@byronwatkins2565 Psi has units of 1/L^.5 in one dimension.

@byronwatkins2565 4 жыл бұрын

@@prasadpawar7027 True. Two physical reasons (and counting).

@jacobvandijk6525 4 жыл бұрын

Psi only gives information about position when the position-operator is acting on the wave function. Psi does not do that in general.

@paulg444 2 жыл бұрын

The challenge is in interpreting the -ih d/dt as a total energy operator, it makes perfect sense as a kinetic energy operator from first principles for a free particle but once you add the potential the meaning of it as a total energy operator is very subtle. This guy is as good as they get as an instructor !

@lepidoptera9337 4 ай бұрын

It is also problematic because most expressions that you get this way are neither energy nor momentum conserving, i.e. the theory we get through this path is not even physical. We teach it that way, anyway, because it's simple.

@kemalm9383 4 жыл бұрын

Awesome work B.

@iwonakozlowska6134 4 жыл бұрын

Why can't we also enter the time operator?

@whatitmeans 4 жыл бұрын

WHY if the conmutator of 2 linear operators is not zero it means they cannnot be measure at the same time??

@kaushaljain5999 4 жыл бұрын

Why did physicist not made time operator formally as position operator?

@kaushaljain5999 4 жыл бұрын

7.21 to 7.35 What about if x^ is operated on a function which have not x as argument.

@jacobvandijk6525 4 жыл бұрын

You yourself always has a position in spacetime, right? It's the same thing with a non-relativistic particle: IF it exists, it must be somewhere in the universe ;-)

@kaushaljain5999 4 жыл бұрын

4:22 Now SE is non linear eq but why did you say in L1.1 that it is linear eq i.e. Hamiltonian is linear operator?

@mikhailmikhailov8781 4 жыл бұрын

It is linear. (1.1) Suppose psi_1 and psi_2 solve ih*del_t*psi=H*psi, where del_1 is the partial derivative operator and H is the Hamiltonian, then (1.2) ih*del_t*psi_1=H*psi_1 and (1.3)ih*del_t*psi_2=H*psi+2. Obviously alpha*psi also solves the SE and you can add 1.2 and 1.3 to and factor out the operators and see that the SE is linear. In general linearity should be viewed as something multiplicative, so any time you see anything written down in a multiplicative way - it is linear. In fact in QM you never even think of non-linear operators, all operators have to be linear, so their solutions form vector spaces, so that you can talk about superposition of states.

@surendrakverma555 2 жыл бұрын

Excellent lecture Sir. Thanks 🙏🙏🙏🙏🙏🙏 🙏🙏🙏🙏🙏🙏🙏🙏

@kaushaljain5999 4 жыл бұрын

9:31 to 9:37 But you said on this a little bit. So how are these things analogous in general way?

@jacobvandijk6525 4 жыл бұрын

Through linear algebra.

@michaellewis7861 3 жыл бұрын

Why not product rule when differentiating the x*Φ

@qcmesil12 2 жыл бұрын

At the 14 min mark? He did apply the product rule, he just didn't state it

@kemalm9383 4 жыл бұрын

I hope all is well with this guy.

@GBY13 3 жыл бұрын

You would get confused in learning Quantum Mechanics if you do the following things: 1. Start learning QM from wave function Φ(x, t) instead of state vector |Φ(t)> 2. Start learning QM from Schrödinger equation to solve Φ(x, t) 3. Keep learning QM without knowing the difference between an electron and photon; a photon does not obey Schrödinger equation.

@abhinandanmehra7765 2 жыл бұрын

Why photons has quanta as well then why it doesn't satisfy SE?

@matrixate 4 жыл бұрын

14:13 - Something from a teaching point of view...don't assume experience when teaching something new. Always remember that. This goes for all you future teachers and soon to be teachers. He applied the product rule. Trivial step but still worth the 1 second to mention that.

@gauravagarwal8528 4 жыл бұрын

If someone is taking the 8.04, it is understood that they have taken calculus previously.

@gkollias14 4 жыл бұрын

the product rule is something learnt in high-school. Studying physics at MIT, especially by the time they are doing quantum physics 1, they would have had tons of practice with much more complicated rules than the product rule.

@No_Tutorial 3 жыл бұрын

I majored in math and I was totally thrown off in this moment. Thank you for this comment, totally forgot Phi is a function of x and t lol

@kemalm9383 4 жыл бұрын

I wonder what country is this guy orginally from? He has an accent. (X)

@jacobvandijk6525 4 жыл бұрын

He is from Peru and has German parents.

@kemalm9383 4 жыл бұрын

What's (h not )? This dude, fellas, homie, B, homestead,buddy, dog, cat, son mentioned hydrogen. What about Oxygen ? Is H2O included in it?

@kemalm9383 4 жыл бұрын

This fella needs a lab.