I'm learning quantum mechanics in French and I almost gave up on finding anything good on KZbin until I decided to search in English and I'm so glad I did! You, sir, are a life saver. can't thank you enough.
@TMPChem7 жыл бұрын
Thanks, Houda. Good luck with your studies. That's unfortunate about the lack of resources in French. I'm definitely not an expert on what exists for anything besides English. Most of the videos don't have uploaded captions, and the auto-gen captions aren't that good, so if you ever have any issues relating to English comprehension or some weird phrase I use please don't hesitate to ask for further clarification. I'm sure the many other non-native English listeners from many different languages would appreciate having those comments to look at.
@houdazahrane58717 жыл бұрын
Thank you so much that's very generous of you. I'm not having any difficulty so far, everything's clear. sometimes the terms are very different than in French ( like: momentum= quantité de mouvement; Eigenvalue= valeur propre..etc) but nothing a simple google search can't fix. This subject used to scare me but now I'm actually enjoying it thanks to you. I really appreciate it.
@TMPChem7 жыл бұрын
Excellent. I hope things continue to go well for you.
@PunmasterSTP2 жыл бұрын
Hey I know it's been awhile, but I just came across your comment and was curious. How'd the rest of your class go?
@jvanvynck7 жыл бұрын
After struggling all day with my book/class notes, I had pretty much given up (I have an exam on this stuff on tuesday). I watched this and the particle in a box video and I understand more now than I did after a whole day of studying. You da bomb man.
@TMPChem7 жыл бұрын
Thanks John. Good luck on the exam. Hopefully we'll see you again before the next one.
@jvanvynck7 жыл бұрын
TMP Chem no, thank you!
@PunmasterSTP2 жыл бұрын
Hey I know it's been five years, but I just came across your comment and was curious. How'd your exam end up going?
@jvanvynck2 жыл бұрын
@@PunmasterSTP haha better late than never I suppose. I passed, and I’ve had a successful career as analytical chemist since (although I recently just started as a field engineer at an analytical instrument manufacturer) but still overall worth all the pain in the ass studying.
@PunmasterSTP2 жыл бұрын
@@jvanvynck That's really great to hear!
@someguy15766 жыл бұрын
Also TMP Chem. Thank you very much. The teacher I had for quantum mechanics was absolutely useless. We had a mock exam earlier on the year which unfortunately contributed to our final grade but not by an awful lot. A lot of people were able to find the time to teach themselves, which they had to, due to his incompetence, but I wasn’t very lucky and I failed the exam. So I tried again for the real thing. Only this time I used your videos to help me prepare. I had my doubts with this being KZbin but I’m happy to say I got an A in that exam thanks to you. I wanted you to know that. I can’t afford to fund you and I don’t particularly enjoy the area of Chemistry in which you teach but you deserve a lot of support nonetheless.
@TMPChem6 жыл бұрын
Thanks for sharing that story. It's good to hear such anecdotes from time to time, even if they arise due to an unfortunate situation. This kind of resource is probably not going to exceed the quality of a good teacher, but it can certainly exceed the quality of a bad teacher, and be a strong supplement in that case. Also unfortunately more common than would be desired. No need for anything other than making use of the content to progress towards your goals. I do more than ok now thanks to the magic of multinational technology corporations. My primary goal for the channel is student impact.
@jacobvandijk65254 жыл бұрын
@ 1:45 It's a fact that dx = 1/2.dx + 1/2.dx . Therefore, the particle can be found in an position-interval around x like this: x - 1/2.dx < x < x + 1/2.dx . There is no reason why dx always should be to the right of x.
@jessesinger47907 жыл бұрын
im clearly just getting started here, but got to say your videos thus far have been superior to anything else ive found. will be working my way throughout.
@TMPChem7 жыл бұрын
Much appreciated, Jesse. Good luck as you progress. Don't hesitate to ask any questions that come up.
@PunmasterSTP2 жыл бұрын
Oh man, sometimes working through integrals can make me...psi. 😎 But seriously, these videos are solid gold, and I hope you never stop sharing your knowledge!
@GiStormy10 ай бұрын
Thank you so much for this!
@FatieHA98113 жыл бұрын
god bless you , you saved my life thank you so much 👏✨
@TMPChem3 жыл бұрын
Thanks for watching, Fatima.
@Night-Sight8 жыл бұрын
Great, this is what I needed!
@TMPChem8 жыл бұрын
Plenty more where that came from.
@PreciousCollections8 жыл бұрын
@0:35 Sir ok....i understand that when we squared that psi....we got all positive values....but what about the probability below the x axis....!!!unable to connect Sir...pls help....sorry lots of doubts but pls help...me
@TMPChem8 жыл бұрын
A positive number squared is positive. A negative number squared is positive. Zero squared is zero. No matter what number you square, you always end up with a positive number. This is true for all values of all wavefunctions. When we take psi_star * psi, the result is a set of positive real numbers at every position in space. These numbers are the "probability density" that the particle is located at certain positions if we measure its position. A larger number means the particle is more likely to be there. A zero means the particle can't be exactly at that position.
@peytoncleary27516 жыл бұрын
I really love your videos! I'm a little confused how we got from sigh squared x to to Bsin(npix/l) though... isn't that the same as sigh(x)? Thank you!
@PunmasterSTP2 жыл бұрын
There is a distinction between the wavefunction psi(x) and the sine function sin(x). Is it possible you thought that one of them referred to the other?
@mariorpg117 жыл бұрын
In the last step you take B to be the +sqrt(2/l). Why can't it be -sqrt(2/l)?
@TMPChem7 жыл бұрын
It's an arbitrary decision to choose the + or - root, and neither choice affects any measurable physical property. When we compute observable physical properties (position, momentum, energy, etc.) the formulas always contain a multiplication of psi_star and psi, thus if there is a factor of (+-1), it always gets squared and ends up as a 1. So ultimately the sign of a wavefunction doesn't matter, as long as the entire wavefunction is multiplied by a factor of +1 or -1.
@mariorpg117 жыл бұрын
Oh yeah, that makes sense. Thx for the fast reply
@andreipaulau32627 жыл бұрын
Why can we assume that the factor B is a real number, so not a complex number?
@TMPChem7 жыл бұрын
Because we take the complex conjugate of the wavefunction times itself the integrand is guaranteed to be a real function. Integrating a real function over a set of real numbers is guaranteed to produce a real number for the integral. 1 is a real number as well, and the quotient of two real numbers is guaranteed to be a real number. Additionally, the integrand is required to be non-negative at all x (because psi_star psi is the square magnitude of a function), so the integral is also guaranteed to be non-negative, meaning its square root is also guaranteed to be real.
@andreipaulau32627 жыл бұрын
I'm still a little bit confused. to specify, I was talking about the transition from equation 1 to equation 2 on the right hand side. I get, that the whole integrand is definitely real, as of course are the sine functions alone. But both of those statements do not tell me anything about B. You can still have a real integrand in the first equation if you just keep B* times B (and then pull that out of the integral like which would probably be more annoying for the calculation). And though it might become clear later that it had had to be real in that first integral, but i still miss the instant justification for the transition from the first equation to the second. Could you please give some more clues for that?
@TMPChem7 жыл бұрын
Ah. Yes, I could keep that as B* B and not assume a real number. In that case, B* B = re(B)^2 + im(B)^2. Since the result of the integral is real, when you solve for the real and imaginary parts of B you'll find that im(B) = 0. This is usually not done because the integrand is required to be a positive real number, thus im(B) = 0 for all wavefunctions.
@andreipaulau32627 жыл бұрын
thank you very much! this made it much clearer
@lhyere97305 жыл бұрын
@@TMPChem I know this is an old post and I apologize. but I'm not really following the logic here. Your final result, had you not assumed a real number would have been 2/L = re(B)^2 + im(B)^2. It is clear that 2/L is real... but that does not help because so are im(B)^2 and re(B)^2 - thus I see a problem of one equation and two unknowns here. Wouldn't it have been equally valid to set B=i(2/L)^(1/2)? You did mention in a separate post that the choice positive and negative B did not affect any measurements due to the fact that the multiplication of psi* and psi would always eliminate the relevance, couldn't the same be said for the choice of B as effectively any complex number of magnitude (2/L)^(1/2) of which there are infinite possible choices, but (2/L)^(1/2) is more convenient to work with?
@thinklearn51306 жыл бұрын
Sir what is the significance of normalization ??
@TMPChem6 жыл бұрын
Hi Duljit. Normalization is a concept from probability theory which ensures that the probability of all possible outcomes adds up to 1, i.e. 100%, as it must. The odds that something happens must be 100%, and this is a constraint on the overall total probability density, which is the wavefunction squared. This means that the integral of the wavefunction squared (i.e. the probability density) over all space must be 1, i.e. a 100% chance of finding the particle somewhere.
@basedoppenheimer14972 жыл бұрын
@@TMPChem It's also just for the mere fact of convenience. You can theoretically calculate quantum mechanical problems without normalization but it is very very hard and complicated. Normalizing the wavefunction just makes things easier as the probability density is always proportional to 100%, it just simplifies problems.
@anshulkapila75727 жыл бұрын
What does the normalization constant mean ????????
@TMPChem7 жыл бұрын
The normalization constant enforces the restriction that the probability of finding the particle somewhere in space is equal to 1 (i.e. 100%, or a guarantee that the particle will be found *somewhere*; that the particle exists). There isn't a physical interpretation to the value of the normalization constant, as the wavefunction is not an observable physical property. Only expectation values of operators acting on the wavefunction have any physical meaning (see videos on average position and average momentum).
@someguy15766 жыл бұрын
Is anybody watching in 2018?
@Lukinaification6 жыл бұрын
Ja, reviewing for a test covering chapters 1-4 that was covered in three weeks. I've learned more from these than in lecture by faaaar.