Superposition of stationary states

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Brant Carlson

Brant Carlson

Күн бұрын

Пікірлер: 50
@pandogovinda5350
@pandogovinda5350 11 жыл бұрын
All these videos are incredibly helpful. Very well-structured and outstandingly clear at just the right pace. Thank you very much. I would not say no to another lecture series on a different area of physics.
@c_dorado
@c_dorado 4 ай бұрын
For the 18:30 Check your understanding problem I got: ρ(x, t) = c₁²•|X₁(x)|² + c₂²•|X₂(x)|², so, I don't find the time-dependence.
@phillipdavis2290
@phillipdavis2290 4 ай бұрын
I'm really not sure if I'm right, but I think the issue is that we aren't taking the integral of the function. When we don't take the integral, I don't think X1*X2 necessarily equals 0. If I'm right, it means that the time dependencies don't cancel each other out when we multiply out. I'd love if you could give me a real answer though, this is just what I got.
@bol3r0
@bol3r0 3 ай бұрын
@@phillipdavis2290 You are correct. Most people here are confusing the expectation value with the probability density. The question asks for the probability density of the superposition of the two states. This then leads to the integrand at 18:05 (without the energies E1 and E2 as coefficients, since we don't use the hamilton operator for the probability density). So there remains a time-dependent oscillation term in the cross terms.
@JoshuaAalampour
@JoshuaAalampour 4 жыл бұрын
4:13 fo A hehehhe
@Heulerado
@Heulerado 9 жыл бұрын
You saved my homework with this video, thanks! I didn't realize E coult take different values in the exponent (duh...)
@noahshamus4479
@noahshamus4479 Жыл бұрын
In the textbook they make it clear the X(x) is real and not imaginary! This greatly simplifies the math
@rameshraikar4924
@rameshraikar4924 3 жыл бұрын
extremely happy with ur content! u calculated energy using H^ and it is found to be definite, so as we know that for a wave dfunction to have solution of SE it must have definite energy. and hence after calculating energy of superposed wave function it is actually definite. so u proved there that superposed wave function is also solution to SE . am i ryt?..ahh probably yes
@davidhand9721
@davidhand9721 4 жыл бұрын
Thanks so much! I almost abandoned the course due to the green screen!
@Prometheus4096
@Prometheus4096 9 жыл бұрын
The c constants can be complex. They need to be complex conjugated.
@a1ang0r85
@a1ang0r85 8 жыл бұрын
why it can be complex ?
@a1ang0r85
@a1ang0r85 7 жыл бұрын
If it is complex, how should we epxress the psi* function?
@williamolenchenko5772
@williamolenchenko5772 3 жыл бұрын
Yes, this seems to be an error. They need to be complex conjugated.
@Prometheus4096
@Prometheus4096 3 жыл бұрын
@@a1ang0r85 Sorry, I see you replied twice. Once 4 years ago, once more 3 years ago. Sorry, I forgot everything. I also got my degree all for nothing :(
@shubhamkumar-nw1ui
@shubhamkumar-nw1ui 3 жыл бұрын
Can any kind soul explain the solution to check your understanding question ? Please
@hrkalita159
@hrkalita159 3 жыл бұрын
Very nice explanation 🙏♥️😳😎
@a1ang0r85
@a1ang0r85 8 жыл бұрын
for 3:43, the partial derivatives of A and B, should we use product rule?
@a1ang0r85
@a1ang0r85 8 жыл бұрын
@Brant Carlson
@theklaf21
@theklaf21 7 жыл бұрын
no , because they bind together by summation. the product rule is for functions whom being multiply.
@a1ang0r85
@a1ang0r85 7 жыл бұрын
Thanks, I get it
@nopopsicle1984
@nopopsicle1984 8 жыл бұрын
for the pde can they be simplified integrsls
@Lbbettarian
@Lbbettarian 10 жыл бұрын
Several times you mentioned the type of PDE we are dealing with, but I couldn't catch the name: something like "star integrable"? Thank you for the replies. Got it!
@zeenaligog
@zeenaligog 8 жыл бұрын
+Lbbettarian partial differential equation (PDE)
@akashacharyya1030
@akashacharyya1030 8 жыл бұрын
+Lbbettarian Square Integrable
@noesnaidel
@noesnaidel 8 жыл бұрын
+Lbbettarian I was struggling to understand too, but after some googling I found the answer: he says "Sturm-Liouville problem"
@a1ang0r85
@a1ang0r85 7 жыл бұрын
/This is important in quantum mechanics, since the one-dimensional time-independent Schrödinger equation is a special case of a S-L equation./ from wikipedia Still, I cannot read through the equation, the math is difficult for me to comprehend.
@basheer7423
@basheer7423 2 жыл бұрын
Where can I get the solution to the problem @18:30
@karm00n29
@karm00n29 Жыл бұрын
i would like to know too :(
@supern0is349
@supern0is349 4 жыл бұрын
amazing video
@spasojekukuric7192
@spasojekukuric7192 4 жыл бұрын
Can someone tell me the answer to the last problem to check myself?Thanks in advance!
@shubhamkumar-nw1ui
@shubhamkumar-nw1ui 3 жыл бұрын
Did you get the answer brother ?
@a1ang0r85
@a1ang0r85 8 жыл бұрын
I do not know what it is asking for "non-trivial time dependence" in 19:00
@II-op5vv
@II-op5vv 8 жыл бұрын
having some variables (or terms) not equal to zero or identity
@a1ang0r85
@a1ang0r85 7 жыл бұрын
I get c1square + c2square + c1c2X1starX2exponetial +c1c2X1X2starexponetial, which is not equal to zero
@stopwatcher8930
@stopwatcher8930 Жыл бұрын
@@a1ang0r85 That can’t be correct from my point of understanding since X1*X2 and X2*X1 when integrated equal zero, so your time depending Term vanishes.
@a1ang0r85
@a1ang0r85 Жыл бұрын
@@stopwatcher8930 Okay, now I understand, the problem here is to deal with time-dependence superposition that a particle can be in a superposition of states with different energies. Since there are different value of energies involved, the two terms will have different frequencies and the oscillation can be out of phase. The the complex terms to be cancelled out, the frequencies need to be the same to have stationary state. When we expand the absolute term, we'll obtain the expectation value of the Hamiltonian + the complex terms. I did not understand at that time because I was too focus on the math part of the whole video, but now it is more clear to me.
@stopwatcher8930
@stopwatcher8930 Жыл бұрын
@@a1ang0r85 Is it like this then? so the cancellation only works when the stationary states have both the same energy, but when this isn’t the case there is no stationary state to which the Hamiltonian can be applied to, since the operator acting on a physical state only has one eigenvalue, but two stationary states would have to energies and therefore two eigenvalues.
@debasishraychawdhuri
@debasishraychawdhuri 3 жыл бұрын
Is it called the 'Stermal Uvul' equation? How do you spell this?
@tszchunlau223
@tszchunlau223 3 жыл бұрын
sturm liouville
@esorse
@esorse Жыл бұрын
"Energy has to live for an infinite amount of time", implying it existed before our universe, consequently "t = 0" being technically impossible, "has to live" connoting the imperative verb 'lives' and some number x required to define , - , meaning negative on it *, concommitantly defining it's opposite, +x and implying that -x and +x are opposite forms of the same entity, written, -x+, breaking the law of non-contradiction : nothing is it's opposite **, may exclude a physical interpretation of your work. * The English concept for "-x" is obtained by concatenating negative and x, while even though the logical opposite of not x, ¬ x, is x, ¬¬ x, the numeric opposite of -x when noun x is the numeral for number 'ex', is adjective positive x, +x, because the noun plus, + and minus, - , operators are opposites. ** Without the law of non-contradiction, the opposite to any argument is as plausible, so a reasonable conclusion cannot be drawn and Hawking's title "God Created The Integers" focuses attention upon an issue this raises.
@FlintPet
@FlintPet 5 ай бұрын
Do you know the Solution to the checkup question though?
@fiftysevenforce
@fiftysevenforce 5 ай бұрын
And that's a really ugly Psi Must fix.
@siddarthjawahar5452
@siddarthjawahar5452 8 жыл бұрын
Answers to those try yourselves anyone?
@a1ang0r85
@a1ang0r85 7 жыл бұрын
I get c1square + c2square + c1c2X*X2exponetial +c1c2X1X2*exponetial, which is not equal to zero
@Evultz
@Evultz 7 жыл бұрын
Alan Leung why does that make it non trivial?
@aprodriguesoficial
@aprodriguesoficial 6 жыл бұрын
Because the exponents don't vanish. This exponents contain the time arguments. So, the distribution r(x,t) is under the influence of time! Being under the influence of time means non-triviality as to time dependence.
@THEBESTOFLEO96
@THEBESTOFLEO96 4 жыл бұрын
So...becouse X1 X2* and X1* X2 they are not integrated they are not equal to 0,so i dont have and identity and a non-trivial solution, right?
@나지태-j2y
@나지태-j2y 2 жыл бұрын
@@a1ang0r85 but the integral of X1X2* is zero, which makes me left with only c1 square + c2 square. Since I don't have any time arguments in my final answer doesn't it mean that this is independent of time?
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