[BLANK_AUDIO]. Okay, so in the presentation on the Born interpretation of the wave function, we said that Born interpreted the wave functions as corresponding to probability amplitude waves. So if we think of a system, a too-far simple system that we've been concentrating on so far, where we have our particle in this one dimensional line, going from zero to L, particle-in-a-box problem. What is the probability, the probability on finding that part between zero and L? It must be somewhere between zero and L. So what that means is that the probability, total probability of finding the particle between zero and L is equal to one. So, what we must do is, to achieve this, is we need to, what's called normalize or its normalization, it's normalize our way functions. So, the way we, we do that is as follows, so Born said going back to the Born lecture, said that the psi squared of x times dx. That's the probability of finding the particle in the region x plus dx. Now if we go along this line here, and we add up all these regions, what we should know because the particle is definitely between zero and L, that if we add these all up together, then we should get, get one. Now, another way of saying that for a continuous function, is that if we integrate between zero and L, and here we have our psi functions, now we labeled them with the, the quantum numbers n, sin x dx, what we're saying is that, that's equal to, equal to one, sorry. The wave function squared is equal, equal to one. So what we can do now is that we can solve to normalize our wave functions. We can solve this function. And we can get normalize, what we called normalized wave functions from it, using this, using this condition. So lets apply this to the, to the system. Again the simple system, we're talking about the, particle, in the, in the box. So we have the integer from zero to L, sin squared x, again if you go back, that was the constant A squared sin squared, and here we had pi nx all over L. Dx, and we'll say, the condition we have to satisfy is that, that total integral is equal to one. So let's let's do that pretty quickly. So you can bring, this is a constant of integration. A squared comes, comes outside of integral. So now we have from zero to L. Sin squared n pi x all over L, dx and that's equal to, equal to one. Now, what we're trying to do here really is we're trying to solve for the value of A, because A is a con, we, we wish it to serve as our constant, but that actually is what we call our normalization constant. And by solving the above equation, we can solve for that, and then we can express our wave functions as normalized wave functions. Okay, so to solve these integral, this integral here, it's handy to use the trigonometric identity, so where we can express the sin squared at, in pi x all over L. [BLANK_AUDIO]. You can use the following transformation, but that's equal to a half minus a half, cosine 2n pi x over L,. [NOISE] So now you would plug that, this here, into the equation above, so you'd get A squared outside the integral sin zero to L, and now you have a half minus a half cosine 2n pi x all over L dx is equal is equal, that's all equal to zero. So now you want to solve for A, so let's just do a part of that integration, so you have A squared, and what you'd have is you would have x over 2 minus 1 over 2, and now you've integrate this bit inside. And if you do that you get L over 2n pi, sin of 2n pi x all over L, and you're integrating from zero to L, and that's all equal to, equal to one. So, if you substitute in for L and zero, what you should come out with is that A squared L all over 2 is equal to one. Or you should show that A is equal to the square root of 2 all over L. Or another way of expressing that is 2 over L to the power of one-half. So what you show therefore, is that your normalized wave function is. Let's try general form psi n of x is equal to, so A is equal to square root of 2 over L. So it's square root of 2 over L sin n pi x all over L. For this here, is your normalization constant. [BLANK_AUDIO]