Superposition of stationary states

[@pandogovinda5350]

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.

[@davidhand9721]

Thanks so much! I almost abandoned the course due to the green screen!

[@rameshraikar4924]

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

[@shubhamkumar-nw1ui]

Can any kind soul explain the solution to check your understanding question ? Please

[@noahshamus4479]

In the textbook they make it clear the X(x) is real and not imaginary! This greatly simplifies the math

[@Heulerado]

You saved my homework with this video, thanks! I didn't realize E coult take different values in the exponent (duh...)

[@nopopsicle1984]

for the pde can they be simplified integrsls

[@Prometheus4096]

The c constants can be complex. They need to be complex conjugated.

[@a1ang0r85]

why it can be complex ?

[@a1ang0r85]

If it is complex, how should we epxress the psi* function?

[@williamolenchenko5772]

Yes, this seems to be an error. They need to be complex conjugated.

[@Prometheus4096]

@@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 :(

[@Lbbettarian]

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]

+Lbbettarian partial differential equation (PDE)

[@akashacharyya1030]

+Lbbettarian Square Integrable

[@noesnaidel]

+Lbbettarian I was struggling to understand too, but after some googling I found the answer: he says "Sturm-Liouville problem"

[@a1ang0r85]

/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.

[@c_dorado]

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]

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]

@@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.

[@hrkalita159]

Very nice explanation 🙏♥️😳😎

[@spasojekukuric7192]

Can someone tell me the answer to the last problem to check myself?Thanks in advance!

[@shubhamkumar-nw1ui]

Did you get the answer brother ?

[@supern0is349]

amazing video

[@a1ang0r85]

for 3:43, the partial derivatives of A and B, should we use product rule?

[@a1ang0r85]

@Brant Carlson

[@theklaf21]

no , because they bind together by summation. the product rule is for functions whom being multiply.

[@a1ang0r85]

Thanks, I get it

[@basheer7423]

Where can I get the solution to the problem @18:30

[@karm00n29]

i would like to know too :(

[@a1ang0r85]

I do not know what it is asking for "non-trivial time dependence" in 19:00

[@II-op5vv]

having some variables (or terms) not equal to zero or identity

[@a1ang0r85]

I get c1square + c2square + c1c2X1starX2exponetial +c1c2X1X2starexponetial, which is not equal to zero

[@stopwatcher8930]

@@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]

@@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]

@@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]

Is it called the 'Stermal Uvul' equation? How do you spell this?

[@tszchunlau223]

sturm liouville

[@JoshuaAalampour]

4:13 fo A hehehhe

[@esorse]

"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]

Do you know the Solution to the checkup question though?

[@siddarthjawahar5452]

Answers to those try yourselves anyone?

[@a1ang0r85]

I get c1square + c2square + c1c2X*X2exponetial +c1c2X1X2*exponetial, which is not equal to zero

[@Evultz]

Alan Leung why does that make it non trivial?

[@aprodriguesoficial]

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]

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]

@@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?

[@fiftysevenforce]

And that's a really ugly Psi Must fix.