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If we solve result into equation 31.19 (0 = A' / @ = AU, p + 2kaBG ) for kaBa and plug the equation 31.18 (0 = A'ut = Aut ~k"B' _ K'Blu + n...

Question

If we solve result into equation 31.19 (0 = A' / @ = AU, p + 2kaBG ) for kaBa and plug the equation 31.18 (0 = A'ut = Aut ~k"B' _ K'Blu + n"'kaBa ), we get 0 = Aut _ k"B' _ kBH _ gn"" A"o (31.20) {Note that I have renamed the index in the last term to avoid confusion with the free LL index) To see that we can teally find values of B" that solve tho equation it helps to take advantage of our freedom to orient our reference frame so

If we solve result into equation 31.19 (0 = A' / @ = AU, p + 2kaBG ) for kaBa and plug the equation 31.18 (0 = A'ut = Aut ~k"B' _ K'Blu + n"'kaBa ), we get 0 = Aut _ k"B' _ kBH _ gn"" A"o (31.20) {Note that I have renamed the index in the last term to avoid confusion with the free LL index) To see that we can teally find values of B" that solve tho equation it helps to take advantage of our freedom to orient our reference frame so that the tzaxis coincides with the direction of the wave s motion. Then ko = [~0,0,0, 0] and we can divide equation 31.20 into individual component equations as follows: 0 = A" _ kB' _ KB' _ {n"A"o = A" [email protected]' + %A? (31.2la) 0 = At _ kB _ kB' _ %ntA"o = At @B* (31.21b) 0 = Aty _ kBY _ kYBt ~ IntA"o = Aty @BY (31.21c) 0 = A" _ KB? _ kB' inA" = Atz @BZ @Bt (31.21d) The first three equations we can solve for B',B*,and BY our solution into the last directly We can then plug equation to solve for B?_ Exercise 31.2.3. Carry out this plan to express the in terms of the original wave components of Bu entirely amplitude matrix Auv



Answers

In Active Figures 18.20a and 18.20b, notice that the amplitude of the component wave for frequency $f$ is large, that for 3$f$ is smaller, and that for 5 f smaller still. How do we know exactly how much amplitude to assign to each frequency component to build a square wave? This problem helps us find the answer to that question. Let the square wave in Active Figure 18.20c have an amplitude $A$ and let $t=0$ be at the extreme left of the figure. So, one period $T$ of the square wave is described by
$$y(t)=\left\{\begin{array}{ll}{A} & {0< t <\frac{T}{2}} \\ {-A} & {\frac{T}{2} < t< T}\end{array}\right.$$
Express Equation 18.13 with angular frequencies:
$$y(t)=\sum_{n}\left(A_{n} \sin n \omega t+B_{n} \cos n \omega t\right)$$
Now proceed as follows. (a) Multiply both sides of Equation 18.13 by sin $m \omega t$ and integrate both sides over one period $T$ . Show that the left-hand side of the resulting equation is equal to 0 if $m$ is even and is equal to 4$A / m \omega$ if $m$ is odd. (b) Using trigonometric identities, show that all terms on the right-hand side involving $B_{n}$ are equal to zero. (c) Using trigonometric identities, show that all terms on the right-hand side involving $A_{n}$ are equal to zero except for the one case of $m=n .$ (d) Show that the entire right-hand side of the equation reduces to $\frac{1}{2} A_{m} T .$ (e) Show that the Fourier series expansion for a square wave is
$$y(t)=\sum_{n} \frac{4 A}{n \pi} \sin n \omega t$$

All right, so in this question, we're dealing with a square wave. Um, and we're kind of being walked through the process of determining the coefficients to before you transform of the square wave. So to start with were given I'm kind of the ideal formula for a square wave, which is wife t, um, is equal to a piece wise function. The first part is an amplitude a, um and that's true. Over a range from zero is less than t, which is less than capital t over to. And then the second part is a negative A, um, and that is the case from, uh, capital t over to its less than lower case T as lesson teeth where capitalism is the period and largest is time. Um, and you know, this is this makes sense, because it will have, um, a constant single amplitude, um, for half the period, and then for the other half, the period will switch to the same magnitude just on the negative side. Um, and then we're also given the kind of general form of before your ah equation, which is he? Some over Len, uh, of a n times sign of n omega T plus bien times co sign of an Omega T. And so this is our starting point. So these two equations will be equal to each other. Um, and in part, a were asked to multiply both sides of more multiplied. Ah, the both sides of this second equation here the y equals T r Y t is equal to the four years, um um, multiplying both sides of that equation by sign of em Omega t. And then we're going to integrate both sides of that equation over the full period of the wave. Um, and then we're supposed to, uh, take a look at the left hand side and consider what happened. And we're supposed to show that when m is even, um, a lot the villa pen signed is equals zero. And then when m is odd, were supposed to show that the left hand side is equal to a constant, which is, um, for a over Evan, Times of makeup. So that's all that we're doing party, which is, you know, a lot of different things. Let's go ahead and get started. This problem is pretty long, so I'm just gonna go ahead and do the different parts in different colors. So I'll do blue for part A. So first thing we're doing is going to multiply both sides by sign of em omega T and put it into an integral over the period of the wave. So for a look inside, we'll end up with the equation, uh, or the expression integral from zero to t of why of tea times Sign of em Omega T DT. So that's the left hand side of the equation. And in the right hand side becomes, ah, much longer things. I'm gonna put it on the bottom or just the next line. Um, and so since those two terms the sign of em Omega t will get distributed to both. And I will go ahead and just do that all in one step so we'll have the integral from zero to t once again. And this will be, uh, the some of a n times sign and omega T times sign m Omega T Plus, that doesn't look like plus plus the n co sign and Omega T times sign. I m omega T d t just barely fed everything in there. So that is a right hand side of the equation. So that's the first part of party now we're going to look exclusively at the left hand side. So the integral of wife t times sign of em omega T d t. And we're supposed to consider. You know what happens when Emma's even and when it m is odd. So the first, the way they were going to do that, is reaction. Going to recall that Wife T is a piece wise function where is equal to a from zero to t over to and where it's equal to negative a, um, from T over two to t. So right, you're going to break the left hand side into two into rules one from zero to t over two and one from t over to to teach, Um, just based on the piece Wise definition. So then left hand side becomes, ah, an integral from zero to t over to. And then for that interval. Why of tea is just equal to capital a. Ah, the amplitude of the wave, and so that will be multiplied by sine of and omega T and D t. And then the second in a role will be negative. A sign of an Omega T e t. But I'm going to go ahead and just bring out the negative sign. Uh, normally, we add the two into girls together, but well, I'll bring up the negative signs so we'll just subtract them. And in this integral will go from T over to to t. And then it will be a sign of em. Omega t going to my parentheses should be over here. So this makes things, uh, ex opinion ice because now we have to intervals that are exactly the same, just with different limits of integration. So now it's a little bit. Now we can start considering what happens when a is even or when m is even and when it's odd. So if M is even if we're integrating a sine function over half its period, if m is even then the first and this the integral for the first in the second half of the period will have the same magnitude and the same sign. So then that value, which, for lack of a better symbol, I'm just gonna go ahead and call Capital E that value e will be subtracted from itself because over the first half of the period it is equal to E. And then the second half has the same magnitude and sign, so we'd have so by evaluating these two into girls, the left hands I would be equal to e minus e, which would be equal to zero. So if m is even the left inside equals zero. But if m is odd, then this first Ah, this first enrolled from zero to t over two is going to have some value. And in the second, integral is going to have the same value but instead of the same magnitude. But instead of having the same sign, it will have opposite signs, so it will be negative. So by evaluating the left hand side, you would have e minus negative e which would instead be equal to true. We. So if m is odd, left hand side is equal to a constant. And now we want to evaluate what that constant is. So now we're going to consider that, uh, M is odd. But instead of evaluating to into girls that air, you know, practically the same, we can go ahead. We've established that they're going to have the same value. So the first integral has the same magnitude as the second in a rule. Um, and that this whole expression is going toe have, ah value of two e. Where is the value of the first one? So we can go ahead and rewrite this expression, saying that the left hand side is gonna be equal to two times the integral from zero to t over to of a sign of em Omega T DT. And this is, you know, actually very easy integral to deal with. So we can go ahead, since as a constant, we can bring it out of the integral, and then we're just integrating Sign of em Omega T, which, due to general will bring out a one over i m omega to the front. And then the sign will become a negative coastline. So then, out of the front of this equation, we're going toe, have a to a over m omega, and then that is going to be multiplied by, uh, evaluating from zero to t over to ah coast negative co sign of M Omega T evaluated from zero t over to. And so we go ahead and plug in zero and t over to those. So we have to a or m omega and then we have that is multiplied by negative co sign of, um, Omega Times t over to plus, because we would have my ass another negative coastline. So we have plus co sign of zero coastline of zero. Of course, it's just one, um and then the in order to evaluate the negative co sign of em omega t over to, we have to remember that omega is defined as being too pie over tea. So, uh, Omega Times t over to is just going to be equal to pi. So then this expression becomes to a over m Omega times, the negative co sign of em pi plus one. And now we need to consider. We need to remember that M is odd. We've made that our condition for pursuing this evaluation. So Emma's odd, which means that the night of coastline is of high or three pie or five pi or negative pie. The point is, it's an odd value of pie, which means that if you take the negative coastline of that, that will always be equal toe one because you're just doing two pi rotations. Um, you know, regardless of whatever value M is, so the negative coastline of em pie, because I'm Assad has to be equal to one. So inside the parentheses here we just end up with, uh, one plus one. So to a over m Omega Times two. And so the left hand side is equal to four a over m omega. And so I'm gonna box this because this is very important, and we'll come back to it much later. Um, so that's the end of part A in part B, which I'm gonna do in red. Um, we're supposed to use a trig identity to evaluate the B in terms of the right hand side. So what we're dealing with is this expression. So we have the some again, and then we have the integral from zero to t. But then we're only dealing with the B terms, not a terms. S O B N Times CO sign of N omega T times Sign of em Omega T DT. And the goal of this part issues the trigger identity to show that all of the B in terms are equal to zero. So the trick. Identity it up we're going to use is the one where, um we have co sign of Alfa Times. Sign of beta is equal to, um one half times the sign of Alfa plus beta minus the sign of Alfa minus beta. So we'll set Alfa to be co sign of an Omega T and will set our site will set Alfa to be an Omega T and will set baited to be m omega t. So we use this trick identity on the coastline Times sign terms. So then we end up with co sign of an omega T Times sign of em omega T equal to one half times they sign of And I'll do some factoring to make this a little bit cleaner. Uh, n plus m times omega T minus the sign of and minus m times omega t. So then we can plug this back into our integral. So then we have the integral from eso we have the infinite some actually, um and then of the integral from zero to t of bien over to and then that is multiplied by sine of and plus m omega T minus Sign of n minus m omega T do you take? And so that is now our expression for the b terms. Um, and now we have to reason why this means that be terms must be all equal to zero. And the way we do this is we take a look at the fact that you're integrating from zero to t. We're integrating over one period of this Sinus order function. We have to sign terms. So when you integrate over the full period of a sign term, the result will always be zero. And therefore, this means that the B terms must also be equal to zero. So that's actually a full explanation. So the integral of a sign function over it's full period equals zero. So all the n terms equal zero. So that means that this expression that we've just written out ends up producing 20 And therefore, since this is equal to the B n co sign and omega T times, sign em Omega t that we had earlier that in general, that is also equal to zero. And so that's all we're doing for part B in part C. We're doing a very similar thing, but we're dealing with the terms instead of the B turns. And we have to show that all of the terms, um, reduced to zero unless and is equal to em. That is a special case. And we have to, uh, analyze that and figure out, um, and show that what that is equal to as well. So we're gonna use another trick identity, very similar process, just with a slightly different situation. So if we take a look at the expression for just the terms, that is the sum from zero to t um are the some of the integral from zero tt of a n times sign Ah, and Omega T times sign M Omega T duty. So this is our expression for the eight terms, and now we're going to use very similar trick identity to before. But this one is instead that, uh, sign of Alfa times. Sign of beta is equal to one half times the co sign of Alfa minus beta Ah, minus the coast sign of alpha plus beta. So here again, Alfa will be an Omega T beta will be m Omega T. And so we will put use this trick identity on the sign times signed turn. So then the result of that is that sign and omega T times sign m omega T equal to one half times the sign of and minus m Sorry, Not the sign. The co sign Very important detail. The co sign of and minus M Omega T minus the co sign of and plus Omega T And one more time we plug that into our integral And so our new expression is thesis some of the integral from zero to t of a an over to And then we use reporting the rest of the identity so that becomes, uh, co sign of an minus M omega T minus co sign and plus M Oh, my God, See and d t. And so that's a new expression for the A N Terms. Um, so now we look at this again, and we can see that, uh, for very similar reasons to in part B, most of the terms reduced to zero because we're integrating a coastline function over its full period. However we do have the special case of n minus M is equal to zero. So if n is equal to M and my assimilate vehicle to zero. And then this first co sign will be co sign of zero Costa zeros than one, which will give us a constant for the inside. It's in general instead of a coastline function, so we'll have. We will have a non zero value, will have a coincide well, the constant inside the integral instead of coastline function. And so this won't produce 20 if and is equal to so unless and equals. M um, all pay and terms equals zero because integrating a co sign function over it's full period equals zero if m equals. And then we have the some of the integral from zero to t and then because a Z we talked about before Ah, a co sign function integrate over its full period equals zero. We don't have to worry about the right hand term because we have a non zero value for, uh n plus m so that one will reduce to zero because it is a coastline function. But because the first term ends up with coastline zero, that one will be a constant and will not reduce to zero. So we end up with a N over to times co sign zero times d t. And then since em is equal to end then we have some. This is the only his that it works when m and N are equal to each other. So that takes care of the infinite some. And then we just change the sub script on a N two b m. So then we can say that this is equal to the integral from zero to t of a m over to Times co sign zero, which, of course, is just one. And so we just have be into Groll from zero to t of a m over two d t. And so that is equal to just, uh, actually one half AM times capital t the period. And so this is our result for the eight terms. Um, all the eight terms refused to zero as we talked about, unless em is equal toe end, in which case it is equal to this. So this is S O. This is the result of the infinite sums. So all the A terms reduced to one half a m. T. So that is the end of part C. Now we move on the part D here we are basically showing that the entire right hand side of the equation reduces toe one half a m t. So let's go ahead and review what we had for the right hand side in the beginning. Ah, before we started analyzing at the way that we didn't parts being see. So there we had that the right hand side was equal to Ah, the in a row from zero to t of He's some of a and times sign omega r n Omega t times sign Mm Omega T plus the n Times co sign and Omega T times Sign M Omega T beauty. So that was the right hand side in the very beginning or not very beginning, but import a So from port be We saw that this reduces 20 and from part C, we saw that this reduces to one half a m t. So the result of this is then that the right hand side is equal to one half a m t plus zero. And so the right hand side is equal to one half a and t. So the very simple step we're just showing that you know, we can split up the integral and some to be, you know, on its individually on the terms and the terms. And so we can just simply add what they reduce, what those infinite sums and inter crawls, uh, reduced to individually to give us with the one half a m. T. That's all we have to do for Part D. Um, And then in part E, we want to, uh, just show that the ah for your expansion for this square wave is equal to are is why of t is equal to the sum of four a over n pi times sign of an omega T. So in part E, what we need to do is we need to set are reduced values or reduce expressions for the left hand side and the right hand side against each other. So remember, from part A, we said that the left hand side was equal to for a over m o May go and the right hand side, which we just, you know solve for is one half a m. T. So now we want to do is want to solve for a m simply we just multiply both sides by two and divide by T So am equal to eight a over m omega t But then we need to recall again that Omega is by definition two pi divided by t. So if we multiply by t, we're gonna have a a new too pie in the denominator. So omega T is equal to pi. So now our expression becomes eight a over to em high eight to buy by two is four and so a M is equal to four a over him Hi. So then we can plug this into the, uh, a m where we had that in the four year equation of beginning. And they will give us the four year spent for your series expansion. So Ah, we also need to remember that the terms on Lee survived on left hand side when m was odd and they only survived on the right hand side when end was equal. Tow em. So that means that why of tea is equal to see some whole end for of for a divided by n pie times the sign of n omega t and then we're all done. There you go

In this question we have given a uniform A string of land 20 m and it is suspended from a rigid support. A short way purses introduced at its lowest end and it starts moving up the string we have to find the time taken to reach it to the support. So let this is this thing we have given here and the length of the string is X. Now I'm assuming a small portion, this is the X at the lower stand and the existence of X. So if lambda is mass per unit land, then we considered muscle X length. Yes, that is equal to lambda into X. So we can say that rate of X land that is equal to this is Mx. So this is M X in two G. Which is equal to lambda X in two G. So we can say that this equals to detention in this string. So we can say that attention at X portion That is T. is equal to this is λ X in two G. Now we know that velocity of pulses young but it is under a lot of do you buy lambda here? So we can see this will come out to be under a lot of blame. The X. J upon this is linda and we know that we've written that this is the X by DT. So from here we can see that This is the X by DT and this will consult with the disease and the root of GS 10 into this road tax. Now we can simplify this equation. So we can say disease the X Upon route X is equal to disease route turn into DT I'm going to differentiate on both sides. So at time T is he goes to zero the policies at lowest and which is europe. And with time T the policies at the Support which is 20 m. So we can see that from here. After integration we get this is No integration of one. Biotechs is equal to disease. Go into root tex and the limited from 0 to 20 and this is route 10 into the and the limit is from 0 to 80 year. Now put the apparent lower limit so you can see this is to into route 20 and this is equal to route 10 into T. From here we get the value of T and this all comes out to be 2/2 seconds. So this is the answer of argumentation for that option. B is the correct choice. Thank you.

Okay and this problem in party we have to normalize the function. So we have the will continue its say X. Zero equal to. Yeah. So you want thanks. Which is the first component of the extraordinary state plus. Uh huh. Say two X. Second component of the exhibition estate. So to normalize ah the weekend right? It test sized eric exactly zero into say into X. Yeah D. X. Which is equal to one wow. So this is again yeah. So we can write it as mode of a scared into seven X. Ex blessed side do thanks staring into say one X. Yes. Side two X. Dx. Therefore a more into this become seven X. Scared plus side two eggs scare plus side to stick X. Seven X. Yeah place say one steak Eggs into side two weeks A. And E. X. So this is equal to one. This is again it went to one these two were functional cabinets. So that's why they are able to zero. So we can write the test more of a scare into one plus one plus zero plus zero. So we can write more of a scare is equal to one out of are too so therefore equal to one out of one route to which is the believable of. So we can write the will function function say X. It will do one apart and due to in to say one eggs plus say two X. So for but be. Uh huh. A real function is given us say eggs T. Equals do able to submission. I equal to want to end I want to end CN. Which is the constant into the wave function by the end stationary state into E. X. Into your race to the bottom minus. I had see E. X. E minus actor E. D. O. D. It get. So if 40s it will do zero. We function looks like this. This is a X. Equal to submission. I got to want to end into C. N. C. A. N. X. So dear for this is equal to one upon and route to into seven eggs blast. Say do X. X. Equal to submission. I equal to want to end and to see and say in X. So comparing the coefficient we get is C. one equals 2 one apart. And due to and she do it will do again one upon and do it two c. 3 and c. four are equal to zero. So this become the way one can say into eggs the equal to one upon. Andrew too say one X. E. Rest to the ball minus at the E. D. Out of H. Good plus as I do X. Into areas to the power minus after a a duty. This is even out of it's good. So as we know that the Energy he and is equal to four end state is equal to end scare by its good scared old off uh two. And I'm scared. So this becomes and scared oh my God this time is equal to oh my God Story into one out of age cut. So this whole town is equal to omega. So this is becoming scared Omega. So therefore the wave function as I into XD equal to one I would have. And due to into Andrew two out of a sign into and by out of a X. E. I don't only that the Plus and do two out of a a sign into to buy out of X into. He raised to recover minus iota for omega T. So therefore this can be written as on the side XT equal to one upon and rode A into sign into by by eggs. He raised to the power to only 30 plus sign into to buy. I would have X. He raised to level minus out for omega D. We also need to find out the probability identity. So to find out the probability density weekend. All right, the function is yes. Say hysteric XT into say into XT in will too one out of a into sign into by by X E I O to omega T. A blessed sign into to buy out of X into uh I oughta for omega T into sign. Sign into bye bye. X. E minus omega T. Plus sign into to buy out of a X E minus I would have four Omega T. So this E and e arrested the computer. Oh my God, they will can sell this and this will get cancelled. So we have, so we get run out of there signs scared into by by X. Plus sign scare Into two by by X. Plus sign into by by X. Into sign into too. Bye bye X. I want to apply by he I had uh three or May 30 plus E. O. Dot minus three or may that D. Yeah. So this becomes This is equal to three. Of course three only the D. So therefore we can write it as one out of a sign scared into pie by X. Plus sign scared into two by by a X. Plus two. Sign into by by a eggs into sign into to buy buy a X. Of course Do three Omega T. Yeah. So to find out the expected expectation value X. We know that the expectation value X. He is equal to Integral from 0 to a eggs into sigh into X. A. D scared D. X. As we have calculated definitive. So we will put in this integral. So this integral becomes zero to a eggs. Sign scared into pie by X. D. X. Into zero to a eggs signs scared to buy buy a eggs into 0 to AX. Sign into bye bye eggs sign into to buy buy a X. And dx. So to solve such a land integral. It is a difficult process so we break it and dissolve it. So there's let care equal to bye bye. Yeah. And then Above integrate reduced to 0 to a. And these eggs into sight. Sign into Yeah ex. D. Ex. So therefore we can. All right, this is integral science care. Okay X. D. X. As we know that. Mhm. This can be written as integral one minus cause Into two K. X. Out of two D. X. So this becomes accepted to minus sign Into two K. X. Out of two uh four K. Plus C. Therefore abo interior will become going to which is 0 to a exit sign into K. X. D. X. Equal to equal to eggs Into eggs out of two sign two K. X. Divide by. For okay As the limit from uh 20 minus zero to a. into eggs out of two minus sign two K X. Divide by four K. & two d. x. So substituting the value we get it is A scared of the four minus a sign to K. Out of four K minus Ex scared out of four. Bless cost to K X. Out of eight K Scare. And they limit from 0 to a. So this can get written as Is scared out of two a sign took Out of four K. Okay into scared out of four plus cause to get out of Out of eight K Scare. Mine is one of the eight K scare. So this can be written as Hey scared out of four minus uh signed to a out of for game minus goes took a ow and gas care -1 out of eight. Yes. Yeah. So put put okay able to buy, buy a in a bold vision we get it is as the interior, I got too scared of stuff afford and put in case you will do by by air we get I equal to uh scared by four. So both the integral same. So now we have to find out the integral. The other integral, which is he went to, so we have another integral which is zero to a eggs. Say into okay, eggs sign into okay X into sign into two X A D X. So as we know that we have a formula which is it were to sign uh Sign be able to one out of two cause into a men S b minus goes into a place. So we use this formula in both integral. So this becomes zero to a eggs into one out of two because K x minus because Into three KX A Dx. So therefore this weekend equal to April 2, 1 out of two into eggs into sign K X. Out of okay minus saying into three K eggs out of tricky. And the interior from 0 to air limit from zero train and then integral from 0 to into sign the exhale dot okay minus sign into three eggs out of three. Okay, into dX. Yeah. And this interior become, And this interior, is it weird to one out of two into a sign. Get elder okay minus Signed three K out of three K. He came minus the integral. So putting, he's ignorant too. Bye bye in above the vision we get and this is equal to zero. This whole trial music to zero and 0 Manistee integral. So no we calculate the interior which is equal to zero to a sign into three game eggs out of three game minus sign K. X. Out of K. Into dx needs. So this becomes equal to minus of course to get a X Out of nine case here of place because K. X. Out of K scare and the limit is from and two zero so far case equal to bye bye bye bye. This is equal to minus because three. Okay, Divide by nine K Scare. Bless cause get hold of Okay scared- into one out of nine. Okay scared Plus one out of get scared so this is equal to one out of nine K scare -1 out of K Scare plus one out of nine K. Scared minus one out of K. Scare. So this is equal to Into -8 out of nine nine P scare. So finally knee Expectation really accessible to one out of there. They're scared of the four Plus is scared of the four plus To into -8 a scare Out of nine Byte Scare. So this expectation value is equal to A out of two into 1 -32 out of nine by scare goes into three or more that day.

In this question. We have a pocket in the infinite square brown. Uh, with this initial wave function, Sigh X zero is good to a times Taiwan X plus side to act. Okay, so, um, for Infinite Square. Wow. Yeah, And that's where our problem. Okay, so the diagram looks like this. He from zero to a, uh, and the potential is infinity. Okay. And then so the re function Sai wan x is 12 over a sorry pi X or a and then side two X. Yes, Uh, ritual a sorry two pi X or a Okay, we have the one because two hi square HPR square over two and a square. And then you have me to he goes to, uh, for high square H Bar square by to m a square. Okay, so, um, first, there are five parts in the question. Okay? So in part, they want to find a normalization constant. Okay, so we'll use the normalization condition. Yeah, which states that, uh, the integral from negative infinity to infinity model. Big Sigh, x zero square, E X. A secret one. Okay, so So what you are going to have is, uh, a square well, Taiwan side too. Star times. I want side too. Okay. And the X And then this is a good one. And then you're going to have what's so one square plus one say to a square. Hi. Um, Taiwan star, like two plus trying to start Taiwan. Okay. And the X goes to one. Since all the functions are real. So, um, and then we have our uh huh energy. I can function. Normalize. So this is because to one, and we have, uh, one side too Square yanks. It's also it was one and then someone inside to, ah, diagonal to each other. So we had Sai Wan side, too. Yeah. Thanks. Is it good to zero? Okay. And similarly inside to Taiwan. Okay, so the above, uh, the line above, we'll get two times more a square. Is he going to one? Which means that on a is equal to 1/2. Okay, so this is our normalization Constant. Probably came to be, um, we want to find, uh uh huh. The re function in time. And also the probability density at Time T. Okay. So, um uh huh. So we are given this. It's the would you show a function? Okay. And we want to find, uh, we function at time T. Okay. What we need to do is just to attach the face sector into the negative i e one t or h five and then to each, uh, come Okay, foresight to will just be negative. I e to t over h by. Okay, then we define omega s, uh, e one over h bar. So which is hi square each bar over to M E Square so we can simplify the txt to become Taiwan ex into the minus. I Omega Key has. And then e two is four times of e one, so the Omega would be four times minus I for Omega T. Yeah. So this is, uh, re function at time key. Okay, then if you want to find a probability density, okay, which is, uh, Mark's Square. Mm. So, uh huh. Yeah. So just record. Just remember, that's I want X is, uh, to over a, um sorry. Hi. X over a inside two. X is to a Sorry, um, two packs of a Okay. Yeah. So I'm not putting the exact conscience here yet, but if you want to substitute. Yeah, you can just substitute. Okay, So you do the more square, then you have half. What? Uh, so yeah, so just okay. Complete. Okay, so the more square will be Sorry, star. Sorry. Right? Yes. Uh, 1/2 will become half mhm. And then you have a Sidewinder X. Yeah, because space the spatial wave function is real. So I'm not going to put a star. Yeah, but, uh, time based factor from minus I becomes I he and then decide to x into the eye for Megawati. Okay, so this is the side star, and then the sigh Uh uh, we got the star will be you can just copy. Okay. And then, um then you just multiply So you have Sai wan square. Okay, so the this term multiplied this times the exponential term cancer out. And then, um, you also have the site to square as well. If these two times world, there's no time dependence. And then you have, um, Taiwan Side two. Me to the minus. I trio Megawati and then you also have the side one side, two e to the I tree omega t mhm. He didn't continue to algebra. If so, here you can pull out the Sai Wan inside, too. You have each of the I three omega T plus into the minus. I tree omega T. Okay, this can be, uh, we return there. So to Hussein, using the oil a formula Jose tree. Omega T. Yeah. And you want to substitute ah, functions inside here. Um, so you have house, um, times. So there's a two over a mhm I'm saying. And then, uh, science Square. Hi. Excellent. A plus science square to my ex or a class, um, to sign. Hi. Absolutely sorry. Two packs over a Jose tree. Omega T. Okay. And then you can simplify further. Okay, so this is, uh, the function. The probability density? Yeah. After we function at time T. Okay. Okay. So that's your baby. Then we go to Passy. Passy, you want to find the expectation value of position? Okay, So expectation value opposition is according to the definition. Spicer X Sorry. Um, the X And since X doesn't operate, X can be moved everywhere. So you actually get this x Times months? I sexy star square. What square? So, uh, then you just need to copy what you have in the previous question. Okay to here. Okay, Go from zero to a, um X Times one over a a square high X over a plus. Thanks. Where two packs of a thus to sign. Hi. X over a sorry two pi x over a cause. I try Omega T and then the X. Okay, So here there are three integral that we need to do. Okay. I'm just going to do them one by one. It makes it easier. Okay. So, uh, zero to a x Zion square. Hi X over a, uh, the X Mhm. This gives us a square over to Okay. This is also the same call. Um, x Times Square to buy X over a the x Mhm. Yeah. And then the integral Jeez. Zero to a X. Yeah, to hear, um, sign Hi. X over a sorry two packs over a Hussein trio Omega t the X, Then if you use so that the cosine omega T doesn't participate in the integration is a constant in this case. And then So if you use a wolfram Okay. All right. And you would get, um we are going to get negative It's a square or like high square. There's a two in friends. And that's because I try Omega T. Okay, So here use. Uh huh. Yeah. Okay. You sure? The answer. Okay. See you sometime. Mhm. And so, uh, then we just put everything together. Okay, So the expectation value of X, we have won over a in front, and then we have a square over to us. Uh, the square over to uh huh. Mm. You know, I think this one is a square before. Yeah, okay. And then plus, uh, minus. Okay. Minus 16. A square over nine Pi square. Of course. I trio McGaughey. Okay, then you can cancel the A. So you have me over to minus 16. 8/9. Pi square for sign. She Omega T, and this is equal to able to one minus 32 a over nine point square. Course. I tree Omega T. Mm. Right. So this is the expectation value of the position. Then we need to determine the empty too. Okay. The M P two is, uh, 32 a or nine pi square A. That should be It is not here. Sorry. Okay. After you find out a So there is not there. 8 30 to over 95 square times a day or two. You can use your calculator. You get 0.360 times are over to Yeah, and then the frequency. Hey, the frequency. Is he good to tree omega? Okay, which is three pi square, each bar the right by two m a square. Okay. Next, we want to find, uh, expectation, value of the momentum, the expectation, value or momentum. By definition, it is, um, size scar, and then the momentum The coordinate representation of momentum is, uh, Bob, uh, I I show up at your ex. Uh huh. And this is a star sign. XT Yes. Okay, So this thing we need to do it, uh, carefully and step by step. Okay. So, uh, yeah, we need to do the differentiation first. So So let's just write down what our sexy is. Society is one or a sorry. Hi x or a me to the minus. I only got t Uh huh. Sorry. Two Pi x or a you to the minus. I call my body Mhm. Passion. Passion X Sorry. XT. You go to one over a um hi. Over a cause I Hi X over a. You do the minus. I only got 80 plus to buy over a co sign two pi x over a e to the minus I for Omega T. Yeah. And then, um you can simplify by pulling out. Uh, pyro a. Okay, I'm going to ignore the H Power I for the moment. It just put it back three. No. Okay. We are just going to focus on multiplying functions. Okay, so do this, Um, before that, Yeah, it's just right down the complex contributors. Wow. Well, okay, so this is the complex congregate, and then we are going to multiply. That's I start with the G d x of the sign. So, uh, we have pie over a square. What? Sign packs over a call. Sign packs over a. Okay, so the we first write down the terms that exponential terms cancer out. Oh, okay. And then the next times Yeah, I'm not. Wow. Yeah. Okay. Okay. And then we are going to integrate this thing. Okay, Then you can use, uh, from again. You realize that, uh, the science Jose with the same arguments. They just cancer out. They just give you zero, okay? They don't cancel, but they give you zero okay? Yeah. Yeah, because they are Darwin. Okay. Yeah. So basically, this is equal to zero after integration. Okay? After integration. Okay, not in this, uh, when you multiply the two functions, but when you integrate and then using them again, Okay, so if you use your friend or the other two terms too. Sorry. Hi, Axl. A sorry. Two packs of a the X, and this gives you minus for a trip. I and then you do the other 10 to a I'm sorry. Two pi x over a was saying I actually a Yes. This gives you, um for a or tree pie T. So, um, so when it comes to the integral Okay, the integral of Mm hmm. So I x c the x Uh huh. I I star 60 d d x I x T. Yes. Can you have mhm um, hi. Over a square. And then you have, um, negative for a trip. I into the minus. I tree omega t plus for a or tree pine. Eat the eye tree. Omega T. Okay. Okay. So the pie cancels the A also cancer. Okay, so you have 43 A e tree omega T minus E minus. I treat omega t mhm so far over three a, uh, times, too. I sign tree omega heat. Mhm. And so the expectation value of P Yes. You go to thus far over. I did you grow off. Sorry, star XT the V X I XT the x. Okay, so this is equal to H bar Over I for three. A two. I sign tree omega T. Okay, so the eye cancels and you get meet each power P a Sorry. Three omega t no. Okay, so this is the expectation value of the momentum. Okay, so just want to highlight you can do from how far Here. Okay. Yeah. Okay. In part five, you're in part E. Um, You want to find, uh, probability of getting each energy? Okay, So probability of e equals two e one. His house? Yeah, and the probability of getting equals e to It's also hot. Okay. It's basically because we wrote our sy x zero s in the form of, um see, I, uh yes, I I right. And so the probability of getting any egos to ei, it's just mon see, I square. Okay. And each c I, in this case is one of the two. So when you square it, you get half okay. Mhm. Then you need to find the expectation value of the Hamiltonian. Hey, expectation value of the Hamiltonian is is equal to the probability time stuff, uh, individual energy and sum them up. So it's half e one plus half too. Yeah, so half b one C two. And you know that E two is for everyone. So we have five over to E one. Yes, Yes, 5/2. Hi. Square H bar square over to M E Square. Okay. Yeah. So, um, the expectation value of the Hamiltonian is, uh, in between you on and you too. Okay, So this is the expectation value of Hamiltonian, and then this is how it compares to, uh, you want to need to, and that's all for this question.


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