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Make use of the matrix expression (13.18) for $x$ in a representation defined by the harmonic-oscillator wave functions to obtain a similar matrix for $x^{2}$, usin...

Question

Make use of the matrix expression (13.18) for $x$ in a representation defined by the harmonic-oscillator wave functions to obtain a similar matrix for $x^{2}$, using purely matrix methods. Compare with the answer to Prob. 3, Chap. IV.

Make use of the matrix expression (13.18) for $x$ in a representation defined by the harmonic-oscillator wave functions to obtain a similar matrix for $x^{2}$, using purely matrix methods. Compare with the answer to Prob. 3, Chap. IV.



Answers

(a) Show by direct substitution in the Schr$\ddot{o}$dinger equation for the one-dimensional harmonic oscillator that the wave function $\psi_1(x) = A_1xe^{-a^2x^2/2}$, where $\alpha^2 = m\omega/\hslash$, is a solution with energy corresponding to $n$ = 1 in Eq. (40.46). (b) Find the normalization constant A1. (c) Show that the probability density has a minimum at $x$ = 0 and maxima at $x = \pm1/\alpha$, corresponding to the classical turning points for the ground state $n$ = 0.

Okay. Good day, ladies and gentlemen, Um, today we're considering, uh, problem number 47 from section 4.7. And it's, ah, two part problem and be both, which are very similar. You know, all that is involving is okay, so we have this parameter, which is how, um, and in both cases, if we very l here, we're gonna work where we have two distinct um, solutions. Um or, ah, you know, linearly end up 2 to 1 solution, um, to the equation. And we have to find a second, uh, linearly independent solution using reduction of order. Um, okay, let's let's just charge straight. And, um, first off, the reduction of order formula is, um maybe familiar. By now, it looks like this. And, um, the way I always do this kind of thing is I first calculate the inside integral here, and then I look at the outside integral, um, here and then finally multiply. But in this case, we're not going to calculate the outside integral. We're really just plugging in pretty much the the numbers. That's about all we're really doing our plugging in the the functions. And that's really all we're doing. So It's actually very straightforward. In the case we have, Ellie was four. They say that why one of T equals to one minus duty squared is a the first solution. So, um, by looking at the formula here ah, the differential equation. You can see that r p p f t is negative to t um so when we do the the inner girl, the inside integral, we just get, um, t squared s. So then all we do is we get Why two of tea, Of course. And dots, um, this is the form of it and plugging in the numbers. This is what we get. Uh, sorry, de ti umm Yeah. It's very straightforward, actually. Um, and this is the answer. Bass players. I don't know that you can actually do this inter girl very easily, but, um, that's That's the That's the answer. So, for the 2nd 1 again, we're given a, um, were given Al Eagle six. And why one of tea is that, um, and we use the same formula in this case, um, plugging in what we already know. We already know this guy here. Um, and then just the why one squared. Really comes directly from that. So, yeah. So there's nothing really else about it. That's both a M. B. Um, it's just a question of getting the, um, you know, right things in the right spots. Pretty much so, Uh, thank you very much. Have a good day.

So we know if we want to find the position as a function of time of a simple harmonic oscillator. But we must use the law of conservation of energy. So we can say that we can start off by saying velocity is gonna be equal to plus or minus the square root of the spring constant divided by the mass times a squared minus X squared. So that would simply be the of formula for the velocity. Now, this is gonna equal d x over g t. This is simply the definition of philosophy, a change in position over with respect to time. Therefore, D x divided by a squared minus X squared to the positive 1/2 power is going to be equal to one plus or minus the square root of K over m times DT. And at this point, we can actually integrate and say that, um, the integral from acts not that's with this The interviewer from X not to x of D. X over a squared minus X square to the 1/2 power. Uh, this would this is gonna equal plus or minus, um, this square root of the spring constant divided by the mass times the integral from zero to t of DT. And so we can say that this rather let's, uh we can actually evaluate at this point and say that negative co sign negative our coastline rather of X over a, um, evaluated at X not an ex is gonna be equal to plus or minus the square root of K over m times t and we find that weaken. Ah, express this. We can expand this rather and say negative are co sign of X over a plus Our co sign of ex not over a will be equal to plus or minus t times the square of the spring constant divided by the mass We know that the angular velocity is equal to the square root of the spring constant divided by the mass. And we know that the co sign of our coastline rather of ex not divided by X divided by the amplitude, would be equal to five. The face constant. Therefore, we can say that negative co sign of a negative are co sign of X over a plus five. The face constant is gonna be equal to plus or minus the angular velocity times the time and rearranging and isolating X X is going to be equal a co sign of plus or minus omega T. So the angular velocity times time plus the face constant. And this is the general formula for a simple harmonic oscillator. That is the end of the solution. Thank you for watching.

So essentially here will be taking the derivative of the amplitude with respect to the angular velocity. And we're gonna set this equal to zero and then solve for omega Weren't the angular velocity So we can say that the amplitude is going to be equal. Thio uh, the force divided by the mass times the rather omega squared minus omega not squared plus the damping constant squared times omega squared, divided by the mass squared. Um, this is gonna be equal to the force divided by the mass ah times. And this is to the one halftime of my apologies. So this would be Omega squared minus omega, not squared, plus B squared, omega squared, divided by I'm squared and says to the negative 1/2 power. So at this point, we can then, um, finding derivative and say that the derivative of the amplitude with respect to omega will equal zero. And this legal negative force divided by two m and then this would be multiplied by Omega squared minus omega not squared, uh, squared plus B squared, omega squared daughter by m squared. And I think I I think yeah, my apologies. This is squared and this is square. This would b to the negative three over to power, and then this would be multiplied by two times Omega squared minus omega, not squared times to omega plus two times B squared omega over I'm squared. And after this, we can simply say that if we're going to really all this and this entire term equal zero Ah, we can just say that two times Omega squared minus omega, not squared times to omega plus two B squared omega over M squared. This will equal zero because all other terms will actually cancel out, given that there's a zero on the other side. So zero times to em would be a 00 divided by force would be zero Ah, and then this entire term is could a negative three half's power. So this is actually one over this entire term to the positive three house power. Um, so this being a denominator term, essentially, this would equal zero as well. Uh, moving it over would essentially eliminate the term completely. And the only term that is positive and, um is a sum. This term is a sum of two terms. Therefore, you can't cancel out uh, completely. Oh, and we know that this is gonna equal zero because this is again this would be to the first power. Essentially. So at that point, we can say that will make a squared equals omega, not squared, minus the damping Constant squared, divided by two m squared. And this and solving For this we have the Omega equals Omega not squared, minus the damping. Constant squared, divided by two m squared all to the 1/2 power. And this is your final solution. That is the end of the solution. Thank you for watching.

So in this problem we share to obey functions. Why wanted by two on? So this problem we need Teoh, understand that everyone function no matter what the form is, must satisfy the general wave equation which is given here. So in order to actually get is two solutions on show that their linear combination is also where you function We need to address a certain property, solve the differentiation or functions So when we show, for example, if off eggs this will hold true for the partial derivatives off multi variable functions this is for the single variable function. For example, he could NGO facts plus a low off eggs than the partial derivative when the function half off eggs is going to be able to the linear combination off these two personal religious for these two functions that our own day on the side here So this property allows us to the date is a standing wave function, for example, or generally, when we have ah, they hear way functions linear combination of two way functions you can call it a standing wave function but anyway, this combination or two we'll do away functions and say the combination equals y one. Plus, why do now from the property here we can say that further. We want to partnering with you or ext. It's going to be the derivative. The first functional Rex also edition secondary with no brakes. Now what happens when we try to make A to create the second only the differentiation on this side. So this will be the second order partial derivative over X which will be equal. And now we will apply standard those of depreciation. So this is Barschel derivatives. Why line over X and so for the white too. And now we will apply these derivative here on the first and on second tearing. So the Sequels second burst derivative Rx, or the white one function plus the second partial derivative or the white to function over the horizontal position coordinates. So this is very important because this relationship you could see here shows that the first term of the general way function, which is the turning off the differentiation over the X coordinate. So this is the turn will be leaner if the wise addition off two functions So the whole the river there will be a linear combination off the relatives of the two functions or the spatial coordinate. No need to brew sending for the dive. And also we don't really to prove because this since this holds true for the X coordinate, which is one of the variables in the function, why axity the same the whole 40. So both the second partial derivative over X and the second part of the river do over t behold the same the same property off linearity. So in that case, we need to actually see what is the final conclusion of this. Well, we can now substitute into the baby question directed, so start from partial derivative of their Y function. Second brush of the remedy over X is one over the lost squared far from the religion for the first one over tea. In addition part for the relative why do over tea? Then we can see that this term right here from the Santa disproven before is is equal to one over been lost two squared and then we will will move the differentiation outside on the brackets of the linear combination. So it will be the secondary with you over tea and inside little camp. Why one plus y two and notice one thing since linear combination. Basically, this this deserve here is also it will do over axle. Why want plus who I do. And this means that that the common in since this is a solution, this is the solution is equal to this. So it is the solution for the wave equation. Therefore, this proves that linear combination. Why one plus one? Why? To really be the actual solution as well to the wave equation, the proper solution wave equation. And they're using the same property in the teeter. The second order derivative over time do they in this form? Why? Well, because now we can just say these equals do why and we finally obtain it right? Partial derivative off the wife function, which is the function it is obtained by addition of two functions is in fact equal to we lost two squared and again second. Why should they really be over time off? Do I function so on both the time and the horizontal position aspect this fashion Why one plus one? That's something by additional y one and y two indeed satisfies the veil equation and therefore it is A. It is a proper solution. Fourth of eight equation. After this, Siri's off Bruce and substitution. So this is the end of the problem. I hope you find this helpful, and they hope to see you in another lesson.


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