A particle is represented by the following wave function: ...

Key Concepts

Normalization of the Wave Function

In quantum mechanics the wave function must be normalized so that its total probability, when integrated over all space, equals one. This involves ensuring that the integral of the absolute square of the wave function over the entire domain yields unity, and it is crucial for the physical interpretation of the wave function in terms of probability.

Probability Density

The probability density is defined as the square modulus of the wave function, which provides the probability per unit length of finding the particle at a particular location. This concept is used to calculate the probability of the particle being found within a specified interval by integrating (or appropriately approximating) the probability density over that interval.

Expectation Value

The expectation value of a physical quantity is the average outcome of measurements for that quantity on a quantum system, weighted by the probability density. For a particle's position, it is obtained by integrating the product of the position variable and the probability density over all space, providing insight into the probable location of the particle's position.

Root?Mean?Square (RMS) Value

The RMS value is a statistical measure that represents the square root of the average of the square of the position variable. In quantum mechanics, it is used to quantify the spread of the particle's position around its mean value, thereby providing a measure of the uncertainty or dispersion in the particle's location.

Transcript

00:01 Hello, and in this question here, we have a particle defined by a wave function, where the wave function is zero if x is less than l over 2.

00:10 The wave function is given but is c times 2x divided by l plus 1, when x is in between l over 2 and 0.

00:20 The wave function is c times minus 2x divided by l plus 1, when x is between 0 and l over 2, and the wave function is zero when x is greater than l over 2.

00:33 And we're asked to use the normalization condition to find this constant c.

00:39 So, first of all, we define the probability of the particle being located between x1 and x2 as the integral of the absolute value of the wave function with respect to x, and this is the probability of particle being located within x1 and x2.

01:17 So if we take x2 equal to infinity and x1 equals minus infinity, we would expect that the particle has a hundred percent chance of being located in between plus infinity and minus infinity, so therefore, this probability will be equal to 1.

01:41 And this here is our normalization condition.

01:46 So if we calculate this integral, an equation is equal to 1, we should be able to find our constant c.

01:53 To do this, i'm going to first break up the limits of integration of this integral here, and integrate from minus infinity to minus l of the complex, the absolute value of the wave function to be squared plus the integral from minus l over 2 to 0 plus the integral from 0 to l over 2 plus the integral from l over 2 to infinity so now these four terms here equals the red underlined integral up there we've just split the region of integration up into four different regions and there reason why we've done this is that the wave function in this region here is zero and in the final region this region here the way function is equal to zero as well so we had to do this because we've been given our way to function as a piecewise function okay so over different regions the way function has is described by different different functions so solving into this for into this in for the wave function in for the second and third term.

03:14 We get that the integral is equal to integral from minus l over 2 to 0 times c squared times 2x over l plus 1 to be squared, dx plus the integral from 0 to l over 2 times c squared times minus 2x over l plus 1 to be squared dx.

03:44 Okay, now we're going to look at this first integral here and make the substitution u is equal to minus x.

03:54 Okay? we're also going to flip, so we're going to use the fact that's the integral of f of x, dx, from a to b, is equal to integral of f of x, with respect to x from b to a so if you flip the a and the b will introduce a minus sign so flipping minus l over 2 and 0 in this first integral and introducing a minus sign we guess that the integral is equal to minus l over 2 sorry the integral from 0 to l over 2 we have a minus sign because we flipped the the um in the 0 on l over 2, we have c squared times 2x plus l, 2x divided by l plus 1 to be squared dx.

04:55 We now make this substitution that we said above of u is equal to minus x, so du is equal to minus dx.

05:06 Okay? so this gives, so we have our original minus.

05:10 We now have our when x is equal to minus l over two, u is equal to l over two.

05:19 So this is why this is our new limits of integration.

05:24 C squared.

05:25 We now introduce it's going to be two minus two u over l plus one d x to be squared.

05:36 And we also have minus d u.

05:47 Because d x is equal to minus d u so this minus sign here we'll cancel with this minus sign here and using the fact that u is just going to be a dummy variable and we could replace you by w or by a or any letter we're going to replace u by x again so we're going to write this as the integral from 0 to l over 2 times c squared minus 2x over l plus 1 dx.

06:19 Okay? so we've shown that we can write this first integral here as this integral underlying of blue as this quantity here.

06:30 However, we now notice that this red underlined integral is the exact same as this integral here.

06:38 So therefore, the total integral of the wave function from minus infinity to plus infinity is equal to twice the integral from 0 to l over 2 times c squared minus 2x over l plus 1 dx.

06:57 Oh, that should be to be squared, sorry.

07:00 So now, computing this integral, well, we could just square the brackets and then it just becomes the integration of a polynomial and quoting the final answer because it's quite a long question we're going to get that this is equal to c squared times l over 3 just by computing this integral here and this proper normalization condition is equal to 1 so hence we can find c is equal to the square root of 3 over l now for the second part of the question we want the probability that we find the particle within a small interval around x is equal to 0 .225 l okay so the integral the probability to find x um to find the particle between x1 and x2 is given by this integral here and in this case we're letting x2 is equal to x1 plus delta where delta is is very small, okay? and the question here says to do no integration.

08:24 So we're going to make a reasonable approximation that because delta is very small, the region in which we're integrating is a very small region.

08:33 So we assume that the wave function is approximately constant over the region because it is so small.

08:39 So this means the integral.

08:42 If the wave function is approximately constantly constant, we can take the wave function outside the integral.

08:47 However, this is an approximation, but it is a pretty valid approximation.

08:54 Dx.

08:55 So evaluating this, evaluating this integral now, gives the absolute value of the wave function to be squared times x, evaluated x1 plus delta, and it x1.

09:11 So this gives that the probability, if the interval is small, is equal to the absolute value of the wave function to be squared times the width of the interval.

09:24 Now, what value of the wave function will be most accurate? so if we evaluate this at the midpoint of the interval, we're going to get the most accurate answer.

09:36 So we evaluate at midpoint for most accurate approximation.

09:44 And we're told on the question, the midpoint that we're to evaluate is x and subscript m for midpoint is equal to 0 .25l and delta the width of this small interval is 0 .1l.

10:01 Okay, so what part of the wave function should we use? so we go up here and see that our x is equal to 0 .225l is in this region here.

10:13 So we should be using this description for the wave function...

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