How exactly does one go about being a handicapper for pay?
To answer this question, we're going to start with a special case that turns out to be quite simple. If a particle is in a state described by Y=e5ix/ then its momentum is 5. No ambiguity, no probabilities, the momentum is just 5. (Don't worry about the normalization of this wavefunction for the moment; we'll return to that point later.)
Now, as you might guess, there is nothing particularly special about the number 5. If Y=e7ix/ then the momentum is 7. In general, if Y=eipx/ for any constant p, then the momentum of the particle is exactly p.
Functions of this form (Y=eipx/) are known as the basis states of momentum, which means they represent particles that have exactly specified momentum. They may seem like such a special case that it isn't very interesting: how often is Y going to happen to be in just exactly that form? But in fact, the basis states are the key to the whole process. The general strategy is to represent Y as a sum of different basis states, and once Y is represented in that way, you can easily get all the information you need about momentum.
We can illustrate the key "momentum strategy" by moving to a slightly more complicated example.
This is not a basis state, so the momentum is not exactly specified. But this wavefunction is a sum of two different basis states, e23ix/ and e42ix/. We can therefore say with confidence that the momentum of the particle must be either 23 or 42. To find the relative probabilities, we look at the coefficients?the numbers multiplied by these basis states?and we square them. (2)2=4, |2+2i|2=8, so the particle is twice as likely to have momentum 42 as it is to have momentum 23.
In summary, there are two key rules you need to know about momentum.
If Y=eipx/ for any constant p, then the particle's momentum is exactly p. Functions of this form are known as the basis states of momentum.
When the wavefunction is a sum of different momentum basis states, the coefficient squared of each of those basis states gives the probability of measuring the particle's momentum to have that value.
Now, as we did with position, let's proceed up from these very simple and limited examples to apply the same rules to more general cases. In our last example, there were only two possible values of the momentum. What if momentum can be any positive integer? Then we need a formula that looks something like:
where |f(3)|2 gives the relative probability of finding the particle with momentum 3; and, in general, |f(p)|2 gives the relative probability of finding the particle with momentum p, for any p. Since there are infinitely many possibilities, we aren't looking for a bunch of individual numbers: we're looking for one general function, f(p), that will give us all our coefficients and therefore all our probabilities.
Finally, moving (as we did with position) from a discrete world to a continuous world where p can be any real number, what we really want to write is:
The numerical coefficient in front of the integral is just a convention that makes certain equations simpler. Aside from that, this formula simply says "We are representing Y as a combination of different momentum basis states eipx/, each with its own coefficient f(p)." Any function Y(x) can be written in this form, and once you do, you can find the probability of the particle being in any particular momentum range. Some of you may recognize this as the formula for a Fourier Transform. Click for a brief introduction to Fourier transforms, including how to invert this formula to find f(p) for a given wavefunction Y(x).
Of course, moving from a discrete world to a continuous world forces us to view things a little differently, exactly as it did in the case of position. The squared magnitude |f(p)|2 does not represent the probability of finding the particle with momentum p exactly?that probability will always be zero. Instead, you integrate |f(p)|2 between any two numbers to find the probability of the momentum falling into that range: for instance, |f(p)|2 dp gives the probability that the particle will have a momentum between 0 and 1. Analogously to the position case, p|f(p)|2 dp gives you the expectation value of p.
Dealing with a continuous world allows us to finally address the question of normalization that we've been ducking so far in this section. A wavefunction with exact momentum such as Y=e5ix/ cannot be normalized. (Try it!) This means that you can't have a nonzero probability of having your momentum exactly equal to 5, just as the probability of the position being at exactly any point was zero. A wavefunction of the form
can be normalized, however. Recall that to normalize Y you set |Y(x)|2 dx=1, meaning the total probability of finding the particle somewhere is equal to one. It turns out that if you do that you will necessarily find that |f(p)|2 dp=1, meaning that the total probability of finding the particle with some momentum is equal to one. This fact, which follows directly from the properties of Fourier Transforms, is one of those cases where the math seems to almost magically do what it has to in order to give you the right answer.
Since the function f(p) contains all the information that Y(x) contains, but represented in a different form, the two functions Y(x) and f(p) are sometimes referred to as the "position representation" and the "momentum representation" of the wavefunction, respectively. This is pretty elegant because the way you treat f(p) to find momentum probabilities is exactly the way you treat Y(x) to get position probabilities.
So let's summarize what we've learned about momentum.
Y=eipx/ is a "basis state" of momentum, which represents a particle with momentum p.
Any wavefunction can be written as an integral over momentum basis states Y(x)= f(p)eipx/ dp. The function |f(p)|2 then gives you the probability density for momentum.
From there, the rest of the math?normalization, expectation values, and probabilities within specific ranges?is exactly like the math that you do for position, as discussed in the previous section.
All that is great: hopefully it makes sense, and if someone gave you a wavefunction you would be ready to extract all the information on the particle's position and momentum. So that's the good news.
But the bad news is, we used one set of arbitrary looking rules for position, and a completely different set of arbitrary looking rules for momentum. The wavefunction is supposed to tell us everything we would ever want to know about a particle, including its kinetic energy, angular momentum, hair color, cup size, and comparable worth. Are we going to have to learn a whole new set of rules for each observable quantity?
The answer is no. In fact, there is one absolutely general rule that enables you to analyze a wavefunction to find the probabilities of any measurable quantity about a particle. Unfortunately, it's a pretty ugly looking sort of rule, which is why we've been putting it off. Now we're going to get there, and the way we're going to get there is by talking some more about momentum.
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