Quantum Chemistry 3.9 - Average Position

[@woojinchoi5239]

For those wondering about the sign change on right side (from orange line to blue line), there should be a *negative* sign in front of the 1/4k^2 in the general trig formula at the top most line on the right side: www.colby.edu/chemistry/PChem/notes/PrtinboxHeisenbergn.pdf

[@gurinderathwal7224]

quick question, when plugging in out k value into the equation we put in npil/l but it was k^2 @4:13 is what im referring to

[@TMPChem]

Hi Gurinder. I believe you're correct. I guess I wasn't too careful in the substitutions since those terms end up going to zero, and the value I substituted in appears to be k rather than k^2. I believe this is due to the previous version where I had 1/8k and corrected that error to 1/8k^2, but didn't carry the correction down to the next line. Looks like I'll need to correct that for the next version. Thanks.

[@gurinderathwal7224]

no worries! Im trying to learn it just like everyone else so its awesome i can help. Also can you help me understand how the K value is made? its not a constant is it?

[@TMPChem]

In this case, I'm trying to find an integral from a table which closely corresponds to the integral that I have to solve. I won't find x sin^2 (n pi x / L) exactly in a table (unless I use Wolfram Alpha, Maple, etc.), but I will find x sin^2 (kx). The value of that indefinite integral is given in the purple line. I then find what value of k makes this integral equal to mine, and that's true for k = n pi / L. The value of pi is constant, and n and L are parameters, which are constant once we have made a choice for their values. Any dependence on x must be explicitly included in the integral I'm searching for, otherwise the table value won't end up being correct.

[@saidileepkumarseera9099]

While integrating x ^2* sin^2(x) dx you had made a mistake in writing the answer it will be --(1/4*k^2)x*cos(2kx) instead you had written +(1/4*k^2)x*cos(2kx), but you corrected while plug in the limit values hope you'll look into this

[@nouraalfakie]

I’m not sure how the cosine term reaches zero. Shouldn’t cosine be “1” at 2npi?

[@andrewminczeski8528]

I had the same thought. But it becomes 1 at l and 1 at 0. 1 - 1 = 0 after the integration from 0 to l.

[@Ganoub2010]

why you are telling that the particle is on average in the middle of the box even if it came from calculations? however it should be most likely at one of the two tops ?

[@300pianist]

The average value is not the point of highest probability. There are two maximums in the probability distribution, one to the right and one to the left of L/2. In fact, the probability distribution is symmetric about x=L/2, so the probability of it being to the right and left of it cancel, leading to the average of x=L/2. So this simply means that it is equally likely to be on each side of the box, yet it does not say anything about where on each of those sides it is most likely to be found.

[@panyakorntaweechat8586]

I have tried calculating < n | x | m > and < n | x^2 | m >, when m=/=n. They are still 0, is it due to orthogonal effect? But these case are not < m | n >.

[@jacobvandijk6525]

@ 1:50 A-hat . psi = a . psi. Therefore, psi* . a . psi = a . psi* . psi. If psi is normalized psi* . psi is just a ratio (like 1/6 in the case of throwing a dice). So this integral is just another version of the usual calculation of a mean value. @ 6:12 But the particle will NEVER be in this average position!!! The question remains: "Why are we interested in expectation values in QM?" Doing calculations without a purpose is useless.

[@nouraalfakie]

When evaluating the integral for , shouldn’t k be squared? It is not when substituting

[@lukaskostal9844]

You are correct but it doesn't matter in the final answer since the terms in which he made the mistake cancel out.

[@גלעדפישר-ח3צ]

Hello! Thank you very much for the videos. I would like to ask a question: calculating in the yellow formula on the right side of the screen when you put the value l the sin and cos become zero and thus you remain with (X^3/6). When you put the value 0: (X^3/6) = 0 sin = 0 cos = 1 X = 0 so when you insert the value 0 the hole formula should be reduced to 0 but in your lecture you didnt insert the value X=0. What am I missing here? Hope I was clear when asking. Thank you very much, Gilad, Hebrew University - Israel

[@TMPChem]

Hi Gilad. The first value in parenthesis is the entire value at X = L, and the second parenthesis is the entire value at X = 0. As noted, both terms go to zero at X = 0. Both terms are non-zero at X = L, and both contribute to the final answer. Perhaps it was unclear which term was which inside the parentheses.

[@גלעדפישר-ח3צ]

Thank you for your quick answer. Yet I would like to ask a following qustion to what you have wrote: When X=L (X^3/6) = (L^3/6) cos function = cos(2npi) = 0 sin function = sin(2npi) = 0 I do not understand where does the L/4n^2pi^2 comes from. Gilad.

[@TMPChem]

Note: cos(2 n pi) = 1. Cosine equals 1 at zero, and repeats every 2pi thereafter.

[@גלעדפישר-ח3צ]

Thank you.

[@minkway522]

Thank you so much for your dedication!

[@danejordan221]

You probably dont give a shit but if you're bored like me atm then you can stream all the latest movies and series on instaflixxer. Been binge watching with my girlfriend these days =)

[@kinglamkwong2929]

Is there any mathematical / physical meaning of putting the A-hat between the psi* and psi inside the integral of calculating expected value?

[@pok5532]

Tks it is very useful for me

[@himanshusonone7870]

in right side , when switching from 3rd to 4th line u changed signs , i dont know what was that?

[@christyzhang9371]

I was wondering for that last two lines of computing should it be a plus sign in the parentheses because in the line above you have addition but then in the final answer you have a minus sign.

[@TMPChem]

Hi Christy. I think you're right that there is a typo somewhere. I know the final expression is correct, so it's just a matter of what is messed up. I suspect it might be in the translation from the general solution to the integral to substituting to the specific case of this example.

[@PunmasterSTP]

Average position? More like "Amazing videos, and I'm on a mission"...to watch them all!

[@pitantichalowa6477]

Hi Thank you very much. Would you please suggest references i.e. Books.

[@TMPChem]

kzbin.info/www/bejne/fqqTh4Gka8qqi7c

[@abbiravindhran8424]

How would you calculate the average position if you had a linear combination as your wavefunction?

[@TMPChem]

For that example I highly recommend viewing videos 4.10-4.12 on the superposition principle, where we do exactly that.

[@abbiravindhran8424]

why does a wave move in a sine motion and not in another type of motion?

[@TMPChem]

Well in the classical wave equation we have a second derivative of position equaling a second derivative in time. In the classical wave equation video we show how through separation of variables, this means that a classical wave must be a function whose second derivative is equal to itself times a negative constant. The general solution for such a case is a (co)sine wave. But that doesn't mean the wave itself has to look like a sine wave, just that the basis set we use to represent it as a linear combination is a sine function. So long story short, sine functions have a lot of the properties of waves, and any wave can be expressed as a sum of sine functions.

[@solsticetwo3476]

Abbi Ravindhran in

[@jacobvandijk6525]

The example shows (once again) that an average (or expectation value) on its own does not tell you much about a system. In this case, the particle will NEVER be at the "expected" position! It's like throwing a dice. The average outcome is 3.5, but in reality it's an outcome that can't exist. SO BEWARE WITH AVERAGES!!!

[@300pianist]

That's only true for even n values. Odd n values have non-zero probability at x=L/2. The drawing of the graph is just one n value (n=2). But really, what =L/2 tells us is that the probability of the particle being above this value is the same as it being below this value.

[@jeffpowell860]

did anyone mention deviation being misspelled ;) "devition" != deviation love your videos. Definitely a great resource!

[@TMPChem]

Interesting. In the copy of this slide on my local hard drive, it's spelled correctly, as it is if you click on the "Notes Slide" Imgur link in the description. Yet here it is in this video, plain as day, misspelled as "devition". Perhaps we have a whodunnit mystery on our hands. Perhaps I just noticed it immediately after filming and forgot that I changed it. Perhaps we'll never know. #spooky

[@jeffpowell860]

whoa! it is spelled right in the thumbnail!

[@TMPChem]

I'm less surprised by that. I make the thumbnails after the video is filmed by placing a text on top of the image. I must have fixed it immediately after filming.

[@abbiravindhran8424]

damn a sine function that just became so complicated. Lol if particle in a box is this complicated ...

[@TMPChem]

Hi Abbi. The integrals often become somewhat involved, but the concept should become clear with practice: The average value of a physical observable (like position) is the integral over all space (0 to L, here) of psi_star times the operator that represents the property (multiplying by x) acting on psi integrated with respect to all position variables (in this case, only x).