### Video Transcript

A wave function of a particle with mass ๐ is given by ฮจ as a function of ๐ฅ equals cosine ๐ผ๐ฅ if ๐ฅ is greater than or equal to negative ๐ over 2๐ผ and less than or equal to positive ๐ over two ๐ผ and zero otherwise, where ๐ผ is 1.00 times 10 to the 10th inverse meters. Find the probability the particle can be found in the interval 0.00 to 0.50 times 10 to the negative 10th meters.

So given this wave functions ฮจ of ๐ฅ and this constant value for ๐ผ, we want to find the probability of a particle being located within particular distance boundaries; weโll call this probability capital ๐. Given this wave function, the value for ๐ผ, and these boundaries, the first thing weโd like to do is figure out how the given boundary conditions match up with the boundaries in our piecewise function.

When we know that, weโll have figured out what our wave functions ฮจ actually looks like over the interval weโre interested in. So our first question is to look at the lower bounds of our piecewise function and the lower bounds of the given range in our problem statement. We would like to know is negative ๐ over two ๐ผ less than or equal to 0.00.

Since ๐ผ is a positive number, that means overall negative ๐ over two ๐ผ is negative. So indeed negative ๐ over two ๐ผ is less than 0.00. Next we want to figure out is the upper bound, 0.50 times 10 to the negative 10th meters, is that less than or equal to positive ๐ over two ๐ผ.

If we plug in the given value for ๐ผ, we see that the right side of this expression is ๐ over two times 10 to the negative 10th meters, whereas the left side is one-half times 10 to the negative 10th meters. ๐ over two is roughly 1.5, so that means this inequality also is true. Taken together, taken together, these two conditions being satisfied means that ฮจ of ๐ฅ over the range that weโre interested in exploring it is always equal to the cosine of ๐ผ๐ฅ, never zero.

Since ฮจ of ๐ฅ is a wave function, one condition it might satisfy as all wave functions do is the condition of normalization. All that means is that if we integrate the square of the function ฮจ of ๐ฅ for over all space, from negative infinity to positive infinity, then that integral must be equal to one. This is the mathematical statement of saying that if we search everywhere over all space for this particle, then the chance of our finding it is one.

In other words, it exists somewhere. Now when we go to normalize our function ฮจ of ๐ฅ, instead of using boundaries of negative infinity to positive infinity, we use boundaries of negative ๐ over two ๐ผ to positive ๐ over two ๐ผ. The reason we can do that is because ฮจ is zero elsewhere, so thereโs no point in integrating over a place where the particle canโt exist. Now hereโs the wave function ฮจ of ๐ฅ weโre going to use in this integration equation.

Weโre going to use the equation ๐ถ, a constant, times the cosine of ๐ผ๐ฅ. Now you may say why weโre using ๐ถ; there was no ๐ถ that appeared in the original piecewise function, so why weโre introducing that? And the reason weโre using ๐ถ is because we donโt know if this originally stated ฮจ of ๐ฅ is normalized. It may be; that is, it maybe we find a ๐ถ is equal to one, but it may not be. It may be that ๐ถ is a different value, in which case weโll be properly normalizing this function by solving for ๐ถ.

So plugging in for ฮจ of ๐ฅ with ๐ถ cosine ๐ผ๐ฅ, our expression simplifies to ๐ถ squared cosine squared of ๐ผ๐ฅ ๐๐ฅ. And remember, by the normalization condition, all of this equals one. So we want to integrate this expression and find out what capital ๐ถ is equal to. Since ๐ถ is a constant, we can remove it from inside the integral. To integrate this cosine squared function, we can split this expression up into two different simpler integrals.

We can do that by recalling the half-angle identity that the cosine squared of a variable ๐ฅ equals one plus the cosine of two ๐ฅ all divided by two. With this substitution in place, we can now make two integrals out of this one single expression. The single integral splits up into two integrals: the first one, simply, of the constant one-half and the second one of cosine of two ๐ผ๐ฅ. Notice that both integrals still have the same upper and lower boundary conditions and that we havenโt forgotten about the factor of one-half in the second integral.

When we evaluate the first integral in this expression, we find that itโs equal to ๐ถ squared over two times ๐ฅ evaluated at the lower boundary condition and then to the upper boundary condition. When we enter these values in for ๐ฅ, we find that this is ๐ over two ๐ผ minus minus ๐ over two ๐ผ or ๐ divided by ๐ผ. So thatโs the first integral. For the second integral, we can recall that the integral of cosine of ๐ฅ is sin of ๐ฅ, and weโll also remember to divide our answer for the integral of cosine of two ๐ผ๐ฅ by the derivative of the argument of the cosine of two ๐ผ๐ฅ, to be consistent with the chain rule in differentiation.

So our integral is the sine of two ๐ผ๐ฅ divided by two ๐ผ evaluated from ๐ over two ๐ผ down to negative ๐ over two ๐ผ. Now as we insert our upper and our lower boundaries into this function, notice what happens in each case. Two ๐ผ cancels in the numerator and denominator, leaving us with, in the first case, sine of ๐ and, in the second case, sine of negative ๐. As we consider the sine function though, we recall that the sine of ๐ is equal to zero and the sine of negative ๐ is also zero.

So the entire second integral evaluates to zero over the interval weโre considering. All that means weโre left with an expression thatโs simpler than we might have expected: ๐ถ squared over two times ๐ over ๐ผ is equal to one. When we rearrange this expression to solve for ๐ถ, we find itโs equal to the square root of two ๐ผ over ๐, so ๐ถ is not one. But the good news is now our wave function is normalized, so our final normalized wave function is ฮจ of ๐ฅ equals the square root of two ๐ผ over ๐ times the cosine of ๐ผ๐ฅ.

Now weโre ready to solve for the probability ๐ that a particle described by this wave function appears in the bounds that weโve been given in the problem statement. That probability is equal to our wave function ฮจ of ๐ฅ, the magnitude squared, integrated from our lower bound to our upper bound. Weโll solve this integral in a similar way as the one we solved to find the normalization constant ๐ถ. First, we factor out the constant, then we use the half-angle identity to rewrite our integral into more manageable separate integrals.

Before splitting them up, we can simplify by canceling out the factor of two in numerator and denominator. Then with the integrals split into two separate integrals with the same upper and lower bounds, we start by evaluating the first one. The integral of one with the respect to ๐ฅ is ๐ฅ. And evaluating this from the upper to the lower bounds, since the lower bound is zero, the first integral overall simply equals ๐ผ over ๐ times the upper bound.

Looking at the second integral, once again weโll remember that the integral of the cosine of ๐ฅ is the sine of ๐ฅ, and weโll also remember to divide by the derivative, with respect to ๐ฅ, of the argument of this trigonometric function. This integral, sine of two ๐ผ ๐ฅ over two ๐ผ, means that the ๐ผ in the denominator and the one multiplied in the constant cancel out. Factoring out the one- half from beneath the sine function, now weโre ready to evaluate this trigonometric function at the given boundary conditions.

If we look at the second of these two terms, notice that the argument of the sin function comes down to zero, and the sine of zero is itself zero. So the second integral reduces to the sine of two ๐ผ times the upper bound. Weโre now ready to take our given value of ๐ผ and plug it in where it appears in this expression. With ๐ผ plugged in, when we look at our first term, it simplifies to one over 2.00๐.

And our second term, when we evaluate it using a calculator, evaluates to approximately 0.841 over two ๐. And when we add these two numbers together, we find that, to three significant figures, ๐ is 0.293 or 29.3 percent. Thatโs the probability of our finding a particle between the lower and upper bounds given.