Expectation values of operators

[@zacharythatcher7328]

When you integrate by parts, remember that the UV term will go away because you evaluate it at the limits. That held me up for a while.

[@souravsuresh2766]

How does it vanish?

[@AAAAAAAAAAAAAAAAAAAAHH]

@@souravsuresh2766 normalized wave function psi must go to zero at inf and -inf, so the uv| term (evaluated at inf, -inf) is zero-zero=zero u=psi, v=delta function (which came from rightmost integral)

[@schmetterling4477]

​@@AAAAAAAAAAAAAAAAAAAAHH That doesn't mean UV terms are going away. UV terms are caused by very small scales, i.e. hard scattering. IR terms aren't going away, either. A 1/r potential, for instance, decays too slowly. It exerts a measurable influence even at very large distances (they just take a long time to do that). This is no different from classical scattering theory, where all of this is happening just as well. If you are working with Gaussian wave packets you are suppressing both effects because you are choosing an unphysical base with an (implicit) cutoff for finite wave packet size (in both small and large scales). Non-relativistic theory is simply not sensitive to "real" and, admittedly, ugly physics. It's a kindergarten approximation.

[@joel8271]

@@schmetterling4477 you cant be serious

[@schmetterling4477]

@@joel8271 Yes, I am. Imagine a particle beam originating very, very far away. What will happen to it before it ever gets to you?

[@허지원-y3v]

15:14 I was sooooo confused why Prof. Zwiebach substituted the p for a momentum operator with an extra minus sign( - (hbar/i) (d/dx)). But I just realized that he WASN'T using the momentum operator. He was just trying to find a more useful form of the integrand of integral p, with an equivalent result, which just turned out to look similar to the momentum operator in position space. In fact, (as he mentioned) the two integrands above and beneath the underline are exactly the same. Wrote this so no one else gets stuck over nothing for as long as I did ;-)

[@srivishnudasu1694]

or you could say he substituted the operator for the eigen value, which was P ,to take it out of the integral .

[@prasadpawar7027]

That's what heroes do.

[@guilhermecorrer7272]

We can also use the fact that the eigenstates of the operator P in position representation are the exponentials e^(i*hbar*p*x), so he just used the eigenvalue equation for momentum operator of eigenvalue -p (therefore, the minus sign): P*e^(i*hbar*-p*x)= - p*e^(i*hbar*-p*x) (or maybe i'm wrong)

[@KevinS47]

Oh wow, thank you for pointing that out... jesus I also spent like 2 hours trying to figure out that minus sign... I get it now!

[@rougirelarnaud3011]

A great series of lecture ! I am gratefull to the MIT for sharing that. Just a remark : the computations simply much by using Plancherel's Theorem which states that, not only the L^2 norm is conserved, but also the inner product. Then, in particular, the Dirac mass is no more needed.

[@youtubeshortsviral1361]

7:47 pls tell me what that phi stands for?

[@tretolien1195]

@@youtubeshortsviral1361 You probably figured it out by now, but its just the wavefunction in momentum-space as opposed to the normal position space that the wavefunction is normally framed in. We call the generalized form of the wavefunction the state of the quantum system and we can choose which basis to represent it in within the quantum Hilbert space (this is just an infinite dimensional function space of all integrable absolute square functions aka L^2(a,b)) by projecting it down onto an appropriate set of basis vectors. Going from the position space wavefunction to the momentum space wavefunction or opposite is done by the Fourrier transforms you see boxed at 24:07. Hope it helps :)

[@shivamsinghaswal4995]

Really really insightful! Couldn't find any more beautiful explanation elsewhere.

[@paulg444]

Fantastic presentation, great prelude to Wigner's 1932 paper ! Only one quibble, the expectation is really of Q, not Qhat... Qhat is the operator, Q is the variate and it is an expectation of the variate that is of interest. The operator simply gets us there.

[@hyunwoopark131]

My physical chem book doesn't tell me this. This was an amazing lecture.

[@schmetterling4477]

Physics for chemists and engineers is usually atrociously bad.

[@nicktohzyu]

how does he suddenly jump the generalization that just because it works for the momentum operator it works for energy? doesn't this require the wavefunctions fouriered over energy space in terms of coordinates to cancel nicely? or generalise further for any operator?

[@faroukmokhtar6900]

At 14:25 , Prof. Zwiebach used the momentum operator as a derivative , but there is an extra negative sign there (the i is in the denom.) The problem is that: following the derivation (without the negative) , the extra negative we pick from integration by parts will leave me with negative the momentum operator at the end :/

[@gpolix]

at 18.35 (-h/id/dx) is an operator , how can you do integration by parts in that way? Integral uv' = a term that vanishes + u'v ... where is v here ? please can you explain?

[@darkintegrator]

@Farouk Mokhtar That p is not a momentum operator! It's just the momentum itself. The minus sign is actually right. The derivative acts on the second exponencial, cancelling the sign and dropping down the p, as Prof. Zwiebach explains from 14:48 to 15:13.

[@israeltulmo7374]

11:57 I need those blackboards at my University

[@SergeyPopach]

so it only works for k.e. operator. but not for Hamiltonian operator, which includes potential energy?

[@rajinfootonchuriquen]

Interesting that given a particle with the expected value of the momentum = 0, the expected value of the kinetic energy operator is the variance of the momentum

[@schmetterling4477]

A single quantum doesn't have an expectation value. We are always talking about an ensemble of quanta that were generated with the exact same experiment. Technically a quantum with zero momentum is not even measurable directly. There is no energy in that system and without energy we can't perform a measurement. So the only way to measure such a system is by scattering, but then after that measurement the momentum will not be zero any more.

[@Leowlion11]

7:30 Could you help me why we could similarly have(? or find?) the probability density for the momentum space?

[@jacobvandijk6525]

@ 2:24 The expectation value of throwing 1 dice is 3.5. In reality you will NEVER see this outcome!!! Did you 'expect' to measure that???

[@SadatHossain01]

Expected value does not really work that way, if you throw a dice for infinitely many times, you will see the average converges to 3.5. The idea is probably more relevant with the example of a lottery ticket. Suppose, you have a 50-50 chance of winning $10, meaning your expected winning is $5. What does this mean? This means that if you test your luck for many times, your average winning is going to be around $5. Although intuition suggests you are either going to win $10 or 0, after 100 randomized experiments, you are definitely not going to have $1000 in your pocket. Your "average winning" is definitely going to be $500. Hope that helps.

[@jacobvandijk6525]

@@SadatHossain01 In that situation you're right. But what about the expectation value of an electron being in a spin-up or a spin-down state? ;-) It's not what you expect to measure, right?

[@georgeobrien8311]

@@jacobvandijk6525 Right. The term "expectation value" means the amount or value you expect to receive on multiple samplings, not the value "expected" on a single outcome. Dr Zwiebach makes this explicit starting around 5:48. So, in your example, the expected value of the spin is 0, even though that is not a possible measured value. For more on the origin of the term see en.wikipedia.org/wiki/Expected_value

[@jacobvandijk6525]

@@georgeobrien8311 Of course. I just don't like the term. It can confuse people. What's wrong with the mean value? It probably hasn't that scientific flavor, I guess ;-)

[@gpolix]

at 18.35 (-h/id/dx) is an operator , how can you do integration by parts in that way? Integral uv' = a term that vanishes + u'v

[@yyc3491]

That's because the operator works on the delta function. You can see the delta(x-x') function does not change by keeping track of it during the integration by parts.

[@NoName-yp6ow]

The end of the lecture is a bit confusing. It seems that he supposed that u'v' = (uv) '!!!

[@farhanrafid8584]

yes, linear

[@islam13ish]

why didn't use Parseval's identity at 10:30?

[@zphuo]

@18:00, why the minus sign disappeared?

[@akashpremrajan9285]

I don't know if this is relevant to you now: The minus sign disappeared because in integration by parts you get a minus sign when you move the derivative which you can see here(en.wikipedia.org/wiki/Integration_by_parts)

[@abhijithrambo]

physics.stackexchange.com/questions/192029/integration-by-parts-to-derive-d-langle-x-rangle-dt Here's a detailed explanation of the same

[@gpolix]

@@akashpremrajan9285 at 18.35 (-h/id/dx) is an operator , how can you do integration by parts in that way? Integral uv' = a term that vanishes + u'v ... where is v here ? please can you explain?

[@thecritiquer9407]

can I know what language was that and where can I learn it?

[@mitocw]

We are not sure what you asking. He is speaking in English (a very technical English). This is not a computer language (in case that was what you are asking). In order to understand this 'language', you need to know the course prerequisite 18.03 Physics III: Vibrations and Waves: ocw.mit.edu/8-03SCF16. If you don't understand 18.03, you can look at it's prerequisites. All of the courses materials are available on MIT OpenCourseWare for free. You can start where you understand the materials and work your way up. We hope this answers your question!

[@farhanrafid8584]

@@mitocw Relax, he is joking

[@athul_c1375]

@@mitocw He he just trolling He is implying that professor is speaking language of God

[@athul_c1375]

@@mitocw He he just trolling He is implying that professor is speaking language of God

[@shinsuketakasugi692]

@@mitocw lmao it was a joke

[@eskilandersen479]

why did the delta function change from d(x'-x) to d(x-x') at 18:38?

[@chhoutlaychiva4314]

the property of Delta function.

[@hershyfishman2929]

property of Delta function is that d(x'-x) = d(x-x')

[@priyanshibhattacharjee3789]

why we dont operate on psi*??????

[@hershyfishman2929]

We derived the operator hbar/i d/dx in this case by replacing p (the value, not the operator) with -hbar/i d/dx (14:25)(because -hbar/i d/dx acting on exp(-ipx/hbar) happens to be identical with p) and later (20:03) ended up with hbar/i dpsi/dx which happens to be the p operator acting on psi.

[@youtubeshortsviral1361]

7:47 sir pls tell me what that phi stands for?

[@hershyfishman2929]

phi stands for the wavefunction in momentum space that can be converted to psi (the wavefunction in position space) (and vice vera) through a Fourier transform

[@kartickmanna8466]

Can anyone suggest me a good textbook on quantum mechanics, please...

[@mitocw]

You should check out the syllabus: ocw.mit.edu/courses/physics/8-04-quantum-physics-i-spring-2016/syllabus/. It has the required text and a list of reference readings. Best wishes on your studies!

[@kartickmanna8466]

@@mitocw Thank you sir...

[@cafe-tomate]

I think he forgot to write _sub Φ in the expectation value of p

[@souravsuresh2766]

How does the UV term of the integration by parts at 19:25 go to 0?

[@rajinfootonchuriquen]

By definition of the wave function, psi(+- Infinity) = 0. That's a standard definition for taking the adjoint operator of Hilbert space of infinite dimension.

[@schmetterling4477]

@@rajinfootonchuriquen It's also a shortcoming/flaw of non-relativistic quantum mechanics. While it makes treating problems in bound potentials easy, it does not allow us easy access to scattering problems, which happen to be the actual physical case.

[@vibhorbhatnagar7074]

sir will you please tell me where i would find to calculate expectation value of posituon and momentum using raising and lowering operators?

[@hesokaheso855]

What the heck happen here? 12:30 , how the hell can you pull that out of the integral lol.

[@avdeshkumar8500]

That's a property in calculas... Same can happen if instead of derivative there was a limit in its place.

[@sanaymondal6195]

15:14

[@Rebel8MAC]

Oh my gosh he is a good lecturer but does he ALWAYS have allergies?!