 Okay, so we've seen for the one-dimensional particle in a box, this is one-dimensional particle in a box problem, that the solutions to the Schrodinger equation have this form. Some constant times the sign of some n pi x over a where n has to be an integer and x has to be confined to this box we've defined somewhere between zero and a is what we've defined to be the boundaries of our one-dimensional box. So that's the particle in a box problem. We've arrived at the solution by thinking about what values are allowed inside of the constants inside of the sine function in order to make the function be continuous but we haven't talked at all about this value of a. So let's see if we can understand what these wave functions look like. So here's a box where x ranges from zero to a and I've drawn walls of the box here that the wave function is going to be zero only inside non-zero only inside this box representing the fact that we've confined the particle inside this box. So if I take let's say the n equals one solution, sine of one pi x over a goes up comes down when n is one and when I plug in x the value of x is equal to the edge of the box a. So x over a cancels and if n is one I just have sine of pi and so that's the first time this sine function has come back down and hit zero. So that would be a graph of size of one. If I plug in n equals two I'll get a different function it's still going to be a sine function that's just going to oscillate twice as fast so it's going to go down and hit zero halfway through the box and then it's going to oscillate back upwards and hit zero at the far edge of the box. So each of these constants inside the sine function are engineered so that the function always comes back in at zero maybe after one half an oscillation maybe after a full oscillation maybe after one and a half or two halves or four halves and so on. So each of these wave functions is just a different number of oscillations fitting inside the box. That's the meaning of this constant n. The meaning of the constant a that's just a constant that multiplies the entire function. So let's say for this size of one function with a particular value of a we get this function if I were to choose a different value of a I can make it taller or even taller still or even shorter. If I choose different values of a so some value of a or a different value of a or a different value of a different values of this constant a just change the height of this function. So which of these functions is right which ones are wrong which values of a make the function better or worse. Remember as always when you're confused about what the wave function means let's go back to our requirement that the wave function squared has to represent the probability of finding a particle somewhere. So again let's say for this top function if we have this wave function I've got a high probability of finding the particle in the middle of the box zero probability of finding the particle at the edges of the box. With a different a that just changes the size of the probability but we know I have a hundred percent probability of finding the particle somewhere somewhere between negative infinity infinity if I add up all the probabilities of finding the particle at all of those places inside the box I have to get a hundred percent. So since probability is equal to the square of the wave function and since I don't have any complex numbers anywhere I don't have to worry about the complex conjugate in the square squaring the wave function if I integrate that everywhere I have to get one so that's going to end up being how we determine the value of a. So let's see what that looks like I want to require that one is equal to the integral of wave function squared from negative infinity to infinity. The wave function looks like this in the place we're interested in inside the box outside the box the wave function is just zero so this wave function that I'm plotting is zero outside the box so I can break the integral up into three pieces I can integrate psi squared from negative infinity to zero so I integrate this portion of the curve the area under this green line when I square it is just going to be zero because the wave function is zero so wave function zero so this whole piece goes away this whole piece goes away because the wave function is zero outside the box so the only part of the integral I care about is the part inside the box and what psi squared looks like inside the box if psi is this function psi squared is going to be a squared sine squared not k but n pi x over a and I have to integrate with respect to dx so that's the integral I want to do I want to do let's move that I guess over here I want to make sure that when I integrate sine squared n pi x over a between zero and a I want to make sure I get one there's only going to be one value of a that allows this integral to come out and have that value so how do we do that integral let's integral of sine squared doesn't look too tough but to get rid of this messy stuff inside the trig function let's use use substitution so if I let u be n pi x over a so the du is n pi over a times dx and dx is du I'll move the a and the n pi to the other side that lets me rewrite my integral so I want one to be equal to the integral of a squared sine squared n pi x over a I chose my u so that that's n pi x over a would simplify and just become a u dx has now become a little more complicated dx is a du multiplied by some constants a over n pi and because I have a definite integral I have to remember and transform the limits as well so the integral used to go from x equals zero to x equals a now my integral needs to go from u equals something to you equals something when x equals zero if I plug into this expression if I put a zero in here the value of u is n pi times zero over a so that's still zero when x equals a n pi times a over a the a's cancel and I'm on the upper side upper limit just integrating up to n pi so let's pull all the constants out of this integral so I've got a squared a over n pi and now I'm integrating from zero to n pi the quantity sine squared u du all right so what do I do with that probably realistically probably what you do that is ask your calculator what it is or look it up in an integral table and that's fine that will give you the right answer we can in fact let me leave a little room and tell you that in fact the answer if I integrate sine squared from zero to n pi what I'm going to get is n pi over two and there's a nice reason that's true so I can draw a picture that will show you exactly why that's true if I so the quantity I'm integrating is sine squared so here's a graph of sine squared so sine squared looks like a sine except instead of dropping below the axis it bounces back above the axis because I've squared the sine wave so here's u the graph I've drawn there is sine squared of you I want to know the area under that curve as I go out to n pi that's what sine squared look like looks like it starts at zero and oscillates from that point on if I draw and let me say that's sine squared if I were to draw cosine squared that looks very very similar it just starts high and oscillates exactly the same but starts at one drops to zero goes back to one drops back down to zero and so on so that the second drop graph I've drawn there that's the graph of cosine squared and we know if I add those cosine squared and one squared together I get one right that's one of the tree identities almost nobody can forget so if I add the area under the green curve the sine squared curve if I add that to the area under the orange curve then what I'm getting is the area under one the area under this flat line one so the area under one if I go from zero out to n pi that's just a rectangle with height one and length n pi so that tells me that what I get from integrating just the sine squared part from just one of these two equal pieces is the the rectangle n pi divided by 2 so that's a way to remember if you want to remember how to do this definite integral of sine squared that's remember a way to remember how you get it but again the answer your calculator gives you is certainly just fine there's some cancellation that happens n pi over 2 the n pi in the numerator cancels this n pi in the denominator so we find that what we want to be true is that the one on the left side is equal to a squared times a little a over 2 we are solving for the value of a remember we're looking for the specific value of capital a that allows this integral to be equal to one so when I rearrange and solve for capital a a squared is 2 over a a is equal to the square root of 2 over a right so what that means is that the value of a that we've solved for in this wave function our actual particle in a box wave function we know the value of a now it's square root of 2 over a times sine of n pi x over a with the same requirement that this is only true inside the box and that n has to be some positive integer value so the process that we've just gone through taking the wave function squaring it integrating it everywhere and finding the value of this constant out front that guarantees that I have a hundred percent probability of finding the particle at some location that's a pretty common process that will go through any time we want to find the constant out in front of an integral but often we don't need to bother to do that notice that we got this far without having to discuss the value of a because this value of a didn't change whether or not this wave function solved the shorting equation whether I use this value of a or some different value of a it's still a solution to the shorting equation it still reaches zero at the edge of the box no matter how large the function is the only thing it affects is whether the probability integrates to one the way it's supposed to supposed to so if I truly want the full wave function including the value of this constant out front I need to go through this process which is often a little bit tedious to find this constant that we call a normalization constant this process that we just went through is called normalizing the wave function to find the normalization constant so it's give that a name and once I've normalized the wave function that's a good thing but if all I care about is is whether the wave function solves shorting there's equation or not or whether it has the right value at the boundaries of the box then I don't need to bother with the wave function so sometimes we'll solve quantum mechanical problems without bothering to think too hard about the normalization constant so the next thing we'll talk about normalization this this idea of normalization it turns out to be very closely related to another idea called orthogonalization and that's what we'll talk about in the next lecture