Thank you. Elegantly presented! I do not know why it is that grappling with the underlying math helps so much in understanding the physics -- but it does!
@@ProfessorMdoesScience LOL, I was just kidding. It was kinda heavy math tho, but heavily enjoyable at the same time. I wish my professors taught in this way.
@@sandippaul468 I mean you are chemistry background but I think you will get used to it. As a physics background, it's the exceptions and theory that fries my brain more.
At around 19:09, shouldn't there be a minus sign inside the exp-bracket of the third term of the RHS of the last equation (in blue)? I mean, if we take the operator A = sqrt(m w/ 2h_bar)lambda x and the operator B = - i / sqrt(2 m h w) lambda p, as we can actually read from the first two term in the RHS of the last expression, then when we compute 1/2 [A,B] and take out all the coefficients, aka "sqrt(m w/ 2h_bar)lambda" for A and " - i / sqrt(2 m h w) lambda" for B, shouldn't we end up with exp(- i lambda^2 / 4 h_bar [x,p])? Or am I missing something?
Note that the formula is e^Ae^B = e^{A+B+(1/2)[A,B]}. In our case, [A,B] is just a scalar, so we can write this as: e^Ae^B = e^{A+B}e^{1/2[A,B]}. This then becomes: e^Ae^Be^{-1/2[A,B]} = e^{A+B}. I hope this helps!
Not completely sure about which precise step you are referring to. But in general, if we act with the operator \hat{x} on one of its eigenstates |x>, we get: \hat{x}|x> = x|x>, where the "x" in the right hand side is now a scalar eigenvalue. I hope this helps!
@@ProfessorMdoesScience yeah my issue got solved by itself. It's just an eigenvalue. The eigenstates is still a vector. It had nothing to do with exoectation value
Very enjoyable video. I do wonder who first came up with this approach. I can imagine MANY sheets of paper in the waste basket on the way to the solution.
Clear explanation, as usual. Thank you. I feel a little sorry for the lonely 'hermit' polynomials, and prefer to retain the French pronunciation of Hermite. You feel free to use the Baker-Campbell-Hausdorff theorem. I hope your intended audience does too. You only need the special case that the commutator itself is a c-number that commutes with everything. I wonder if there is a 'simple' proof of BCH for this particular case. Are you aware of such proof? (I know the elementary proof of of BCH by Martin Eichler, which could of course be adapted to the special case, but I'm hoping for something even simpler than that.)
Thanks for your feedback! Haven't tried to look for a "simple" proof, but could be interesting. For the video itself, this would have taken us somewhat off-topic, and we always have to balance length and clarity with detail... Thanks again!
If I understand your point correctly, I guess one could put it the other way around: the expression for psi that we have is normalized, so when we compare it with the expression with h the equality will have the correct prefactors to ensure normalization. I hope this helps!
Ok, this is a bit much for me. I'm a floor layer. I must have took a wrong turn somewhere. I typed n=5 in Maple and got this nonsense -e^(z^2)*(-32*e^(-z^2)*z^5*ln(e)^5+160*e^(-z^2)*z^3*ln(e)^4-120*e^(-z^2)*z*ln(e)^3) and simplified to 32*ln(e)^3*z*(ln(e)^2*z^4-5*z^2*ln(e)+15/4). I think I would rather go with the recurrence relationship. At least I know those. From what I know about recurrence relationships, there is a polynomial associated with it and that polynomial is reflected in the recurrence relationship. A cubic recurrence has four terms (a,b,c,d). a*z^3+b*z^2+c*z+d becomes A[n]=-(b/a)*A[n-1]-(c/a)*A[n-2]-(d/a)*A[n-3]. It uses an identity matrix for the A[0] and A[1] terms. My favorite cubic atm is λ^3-z*λ^2+λ*z-1. Hermite Polynomials are different. H[n]=2*z*H[n-1]-H'[n-1]. So the n-1 was repeated. I don't quite know how to deal with that just yet. But from this, I would make the leap that H[n]=2*z*H[n-1]-H'[n-1]+2*z*H[n-2]-H'[n-2] might mean something. Or maybe H[n]=2*z*H[n-1]-H'[n-1]+2*z*H[n-2]-H''[n-2]. In much the same way a quadratic is to a cubic. Or maybe it's even easier (H[n]=2*z^2*H[n-1]-z*H'[n-1]-H'[n-2]). These patterns are never clear, I can be here all day guessing. The characteristic equations of the rotation matracies are absolute chaos. I never came close to guessing that pattern (rotation matracies) until I figured it out. I'll spend the next few days looking at the zeros of the polynomials to see what they tell me. There are a couple things that stick out like a sore thumb. e is a product in derivative recurrence, where e is a sum in polynomial recurrence. The graphs in polynomial recurrence are typically bound in the y axis, if within a particular domain. The graphs of derivative recurrence in the y axis look bound by a factor of e for the domain. That definitely looks like the growth I've seen in a dozen other places.
It is always quite a lot of fun to play with the maths associated with polynomials such as these, and there are always very intriguing relationships to discover!
There is a more direct way of getting to the Hermite polynomials via using the Rodrigues formula. The nth power of the raising operator, when written in coordinate space, can be converted, via a Hadamard lemma transformation, into the Rodrigues formula, and from there one can read off the Hermite polynomials multiplied by the Gaussians. Just pointing this out in case you are not familiar with it.
If I understand your question (there are a few exponentials around!) then what happens is that we are "moving" the exponential from one side of the equation to the other, hence the change in sign in the exponent. You can think of this as multiplying the whole equation (both left and right) by e^(z2). I hope this helps!
@@ProfessorMdoesScienceWouldn't this also be applicable to (-1)^(n-1)? Shouldn't it be (-1)^(1-n)? I don't understand why there wasn't this exchange. Thanks! (4:34)
Lol i dont think she did that type of research bro, hermite polynomials are in many quantum mechanics textbooks. The entire point is to create a countable "polynomial basis" that can approximate any continuous polynomial function in some interval (Stone-Weierstrass theorem) which helps in solving ODE/PDE problems. Similarly Fourier Decomposition approximates almost any periodic function. You are going to have to read a book or find a video/article for solutions that show higgs boson exist
