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Minimum uncertainty states in quantum mechanics 

Professor M does Science
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2 окт 2024

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Комментарии : 26   
@manujsharma1432
@manujsharma1432 Год назад
Very clear explanations, thank you!
@ProfessorMdoesScience
@ProfessorMdoesScience Год назад
Glad you like it!
@weixie4399
@weixie4399 2 года назад
Thanks, Professor Monserrat!
@ProfessorMdoesScience
@ProfessorMdoesScience 2 года назад
Thanks for watching! :)
@drdca8263
@drdca8263 3 года назад
So, if a different choice were made for lambda, it would still be a Gaussian, and it would still have the same expected position and momentum, but the delta x and delta p would be different, right? It seems to me that this would still produce a minimum uncertainty state, but with a different trade-off between the position and momentum uncertainties, but maybe I neglected a term when reaching that conclusion? Would choosing an arbitrary value of lambda make it not a minimum uncertainty state, or just make it not have the particular combination of delta x and delta p? And, the Fourier transform of a Gaussian is also a Gaussian, so in the momentum basis it should look similar (just with the roles reversed, and maybe the hbar in a different position , and maybe a factor of pi or something like that somewhere) right? Thanks again for these videos!
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Great comment! You are absolutely correct: the minimum uncertainty wave function is a Gaussian of arbitrary width and center, there are interesting choices, such as a "balanced" choice where the uncertainty is equivalently spread between Delta x and Delta p. You are also correct about the Fourier transform relation between position and momentum space Gaussians. Edit: Thanks to Oliver Sandberg for pointing out an issue with the original answer.
@ThePianistOS
@ThePianistOS 3 года назад
I'm a bit confused how to interpret the final minimum uncertainty state (the one written in the 'wrap up' section of this video) -- calculation of , and \Delta x both require us to know \Psi, but we cannot specify \Psi without specifying and \Delta x. My best understanding is that these are choices that can be made arbitrarily, i.e., if I set =2, it will appear in the exponential and if I calculate given that particular \Psi, I will get the value 2. Such a choice can be made independently of . It seems like a similar choice can be made for \Delta x, but this will now affect \Delta p, since our state must obey the uncertainty principle. However, in a seperate comment, you say that you have used a particular combination of \Delta x and \Delta p, but this disagrees with the understanding that \Delta x (or \Delta p) can be chosen arbitrarily. I really apprecite the excellent work that you are doing! It is rare to see anything explained so lucidly and carefully, let alone quantum mechanics.
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Thanks for your kind words. Let me see if I understand your points correctly: (i) you can indeed have minimum uncertainty states with different choices of as you describe. The quantum harmonic oscillator provides good examples of this, with the ground state being a minimum uncertainty state with =0, and a coherent state being a minimum uncertainty state which, in general, has =/0 (we will publish a video on the coherent state case in a few weeks). (ii) As to the choice of Delta x and Delta p, one can also choose different values of Delta x (which then fix Delta p). There are various interesting choices one can make, with an example being an uncertatiny which is "balanced" between the two. And I was careless in my response to the other comment, I have now corrected it (with thanks to you).
@urty4395
@urty4395 3 года назад
Hi there ... Lately I've been following your channel ,And I find it pretty intuitive... I have a request .. Can you make a video on density of states... I get confused when people say density of states as no of orbitals per unit energy range.. (I studied solid state by kittel and Charles)
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Glad you like our videos! And thanks for the suggestion, we are actually hoping to start a new series on condensed matter physics (solid state) after we finish with the fundamentals of quantum mechanics, so it's on our list!
@NRUSINGHAPRASADMAHAPATRA
@NRUSINGHAPRASADMAHAPATRA 3 года назад
Thanks sir 😊🙏🙏
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Thanks for watching! :)
@StefanAeschbacher
@StefanAeschbacher 3 года назад
I use your videos to teach myself QM. The content is really clear und understandable and the structure with Background and What Next helps a lot. Could you add more videos in the style of "Try it yourself: the adjoint operator in quantum mechanics"? This would help a lot with cementing the understanding of the content.
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Thanks for the suggestion! We are working an a website to expand our material to also offer problems+solutions, which hopefully would help cement understanding as you ask. It will still take some time, but hopefully it will be worth it! :)
@mohammadg.merdan2493
@mohammadg.merdan2493 3 года назад
Nice lecture. Please note that, you have 1/2i not 1/2 in the generalized Heisenberg uncertainty equation. Check it after 2:40 :).
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Glad you like it! Regarding the point about 1/2i, if I understand you correctly I think our expression is correct. It is true that depending on the derivation of the uncertainty principle you may end up with something like: DeltaA DeltaB >= |(1/2i)| but note that this is inside the absolute value, so we can take the |1/2i| outside to simply become 1/2. We have the mathematical derivation of the principle in this video: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-fsC5Mhd7YUc.html I hope this helps!
@mohammadg.merdan2493
@mohammadg.merdan2493 3 года назад
@@ProfessorMdoesScience Thank you so much for the explanation. Yes you are correct because the absolute value is there. Without the absolute value we can write the equation as Delta A Delta B=1/2i
@jozsefkele7858
@jozsefkele7858 2 года назад
Great video. I think we can avoid introducing the ansatz change of variables by using the same separation of variables technique, completing the square of x^2 + 2x, and absorbing the resulting constant term with ^2 into the integration/normalization constant.
@ProfessorMdoesScience
@ProfessorMdoesScience 2 года назад
Glad you like the video, and thanks for the suggestion!
@TheWingEmpire
@TheWingEmpire 3 года назад
I had one query that is actually from a mathematical point of view....The ket space for position and momentum is the space of complex square integrable functions, and it is required to be complete(so i heard). But when I check the completeness of the space....it is being complete using Lebesgue integration but not complete using Riemann integration. Why is this happening?
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
I am afraid this more mathematical topic is somewhat beyond my area of comfort. However, if you do figure out the answer, I would be keen to hear!
@TheWingEmpire
@TheWingEmpire 3 года назад
@@ProfessorMdoesScience okay
@TheWingEmpire
@TheWingEmpire 3 года назад
Really helpful video..btw please increase the volume level a little
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Thanks for the feedback, we'll do so in upcoming videos!
@prajaktagaikwad531
@prajaktagaikwad531 3 года назад
Very helpful.. Thanku so much sir...🙂
@ProfessorMdoesScience
@ProfessorMdoesScience 3 года назад
Glad you like it, thanks for watching! :)
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