Quantum Mechanics and Psionics

This article was written by shade000 on 2011-12-05 under Member Pages.

I quote Kenny and Gary Felder:

 

The subject of this paper can be summed up as: "the most basic things you should know after a quantum mechanics class." If you are currently taking or have already taken quantum mechanics, and are having trouble understanding the basic equations and how you're supposed to use them, this paper should give you some solid practical tips. If you haven't had a course in quantum mechanics, this paper will probably be useless and confusing: you might want to check out our paper Quantum Mechanics: The Young Double-Slit Experiment, which gives an overview of the concepts of quantum mechanics without assuming a math or science background.

We should note that there are many important topics in quantum mechanics that we will not touch in this paper, including angular momentum, commutation relations, bra-ket notation, and more. What we are going to discuss is how to use and interpret wavefunctions. Moreover, the way we are going to present the postulates of quantum mechanics is not the most generalizable, but it is in our opinion the simplest form for understanding and solving basic problems.

How Quantum Mechanics Works, Basically

Arguably, all Physics reduces to this question: At some time you have measured a system to be in a certain state (particles with certain positions and momenta, fluids with certain pressures and temperatures, or whatever). At a later time, if you measure the state of the system again, what will you find?

For a classical system made up of particles, you can completely specify the state of the system by giving the position and momentum (or equivalently velocity) of every particle in the system at any particular time. If you consider one particle whose position and momentum is known at this initial time, then if you know all the forces acting on that particle you can write down equations which tell you exactly what its position and momentum will be at any future time. Note that instead of specifying the forces you can specify the potential energy function, which is equivalent: force is the gradient of potential.

In quantum mechanics the situation is a little more complicated. The systems that you study are still made of particles, and the basic procedure is in some ways similar: You measure the state of a particle at some initial time, you specify the forces acting on that particle (or equivalently, the potential energy function describing those forces), and quantum mechanics gives you a set of equations for predicting the results of measurements taken at any later time. There are two key differences between these two theories, however. First of all, the state of a particle in quantum mechanics is not just given by its position and momentum but by something called a "wavefunction." Secondly, knowing the state of a particle (ieits wavefunction) does not enable you to predict the results of measurements with certainty, but rather gives you a set of probabilities for the possible outcomes of any measurement.

The wavefunction which describes the state of a particle is often denoted by the letter Y (spelled "psi" and pronounced "sigh"). It is a complex function which is defined everywhere in space. (Click to jump to a brief description of complex numbers and magnitudes.) No matter what state your particle is in, its wavefunction has some complex value at every point in the universe.

A typical quantum mechanics problem would thus run as follows: You start with a particle in an unknown state subject to a known set of forces, such as an electron in an electromagnetic field. You perform a measurement on that particle that tells you its state, ieits wavefunction. You let it evolve for a certain amount of time and then take another measurement. Quantum mechanics can tell you what result to expect from this second measurement. There are thus three questions we need to address in formulating the basic theory of quantum mechanics.

1. When I measure something about a particle (position, angular momentum, or whatever else), what does that tell me about its wavefunction?
2. Once I have a measurement sufficient to tell me the wavefunction of a particle at some time t=0, how will that wavefunction evolve in time?
3. Assuming I know the wavefunction at some particular time, how can I predict the results of a subsequent measurement?

For reasons that will hopefully become clear as we go along, we are going to answer these questions in the order 3, 1, 2. In other words we will start by talking about how to interpret a wavefunction to predict the results of a measurement, then discuss the opposite question of how to interpret the results of a measurement to tell us the wavefunction, and finally discuss how the wavefunction changes in time. At the end we'll give an example which will tie all of this together.

Position probabilities with a discrete wavefunction

Let's start with the following question: Given a particle with a specified wavefunction Y, how does it tell you where the particle is? Or, to put it more technically: how do you use Y to predict the results of a measurement of the particle's position?

In order to keep this topic as simple as possible, we're going to start by living in a very simple universe. Our universe has only three points: x=1, x=2, and x=3. Our particle must be on exactly one of those points. It cannot be anywhere else, including in between them. Since those three points define the whole universe, the wavefunction itself is defined at all three of those points, and nowhere else. So Yis just three complex numbers, which might look something like this.

• x=1: Y=1+i
• x=2: Y=2-2i
• x=3: Y=2+2i

Now, here's the rule for position: the magnitude of Y squared gives the probability of finding the particle at a particular position. (This value is written |Y|². Its definition is described in the discussion of complex numbers and magnitudes.) For instance, the probability of finding our particle at position x=1 is |1+i|² which is 2. The probability of x=2 is 8. So, for any given measurement, we are four times as likely to find the particle at position 2 as at position 1.

Now, if we sum up the probabilities of all three locations, we get the odds that the particle will be in any one of those three states. Since we've already said that the particle must be at one of those states, we should get a probability of 1. But at the moment, we find a total probability of 18. That means this wavefunction is unnormalized: it can tell us the relative probabilities of two positions, but not the absolute probabilities of any of them. To normalize it, we multiply all the values by a constant, to make the total probability equal to 1. In this case, we have to multiply every value of |Y|² by 1/18, which means we multiply every value of Y by the square root of 1/18. This will keep all the relative probabilities the same, but ensure that the total probability of finding the particle somewhereis 1.

So now, at our three places, we have the following.

• , probability=2/18
• , probability=8/18
• , probability=8/18

That is a properly normalized description of the particle's position. You can't say for sure what any given measurement will find, but you know exactly what the odds are.

Now, by applying some basic probability theory to those numbers, we can answer another very important question: if I had a lot of particles in exactly this state, and I measured all their positions, what would the average result be? This is called the "expectation value" of position and is often written as <x>. It is calculated as: one times (probability of x=1), plus two times (probability of x=2), plus three times (probability of x=3). (Click to jump to a brief description of expectation values and this calculation.) You should verify for yourself that this gives 42/18=7/3. So if we had a whole bunch of these particles and measured them all, and averaged all the positions, the average position would be at 2; even though, of course, none of the particles is actually at that position.

I want to make it clear what we've introduced so far. We've said that "the magnitude of the wavefunction squared gives the probability of finding the particle at a particular position." This is a totally out-of-the-air rule; or, to put it another way, a fundamental postulate of quantum mechanics, that we will not make any attempt to justify.

Based on that, we applied some basic probability math (no new Physics or crazy ideas) to do two things. First, we normalized Y so that the total probability sum was one. Second, we calculated the expectation valueof position. Hopefully, there was nothing too surprising in any of that.

We're going now to repeat that whole exercise, but for a slightly more realistic case, where the particle can be anywhere on the x axis (instead of only in three possible places). Of course, the math will get a bit hairier, and our sums will become integrals, but all the basic ideas will be the same.

Position probabilities with a continuous wavefunction

Now we're playing in a slightly more complicated universe. Instead of three discrete points, our particle has the whole x-axis to play on. So instead of specifying the wavefunction as three complex numbers, we have to specify it as a different complex number for every point on the x-axis. In other words, Y is now going to be a complex function of x.

(Of course, we could make the universe even morecomplicated by making it three-dimensional, such as, say, the real world. But we'll stick entirely with one dimension in this paper. That means that the answers we find for position and momentum will be numbers, not vectors.)

As before, let's start by assuming we've already calculated Y. And just for fun, let's say it wound up looking like exactly one wavelength of a sin wave.

• x<0: Y=0
• 0<x<2p: Y=sin(x)
• x>2p: Y=0

Note that we've chosen a purely real function (no imaginary part); that makes things a bit easier to picture, but doesn't really change what we're going to do.

At any given point, Y has some value: for instance, Y(p/2)=1. By analogy to what we did before, you might suppose that the square magnitude of this value gives the probability of finding the particle there: so the probability (unnormalized) of finding the particle at x=p/2 is 1.

But that isn't quite true. x=p/2 is some teeny tiny little point, one of infinitely many such points in the range 0<x<2p. The probability of being at exactly that point, or any other specific position, is zero. So Y does not exactly give a probability but a "probability density," and our golden rule is: the integral of |Y|² between any two points, gives the probability of the particle appearing between those two points. For instance, |Y|²dx gives the probability of finding the particle somewhere between x=0 and x=1.

If you haven't worked with probability densities before, this idea takes a bit of getting used to, but if you play with it, it becomes very natural. You find that in areas where Y is uniformly zero, the probability is zero. In areas where Y is high, the probabilities tend to be high. You find that the probability of being at any exact point is zero (as we said earlier), since the integral from any number to itself is always zero. The probabilities sum in a natural way: the odds of being between 0 and 1, plus the odds of being between 1 and 2, give you the odds of being between 0 and 2. All this corresponds to what you would expect. (Of course, remember that we're working with |Y|² and not just Y. So you neverhave a negative probability. In our example, the right half of the wave has exactly the same probability as the left half.)

And of course, we have a normalization condition, similar to the one we had in the discrete case. The integral

gives the probability of finding the particle anywhere,and should be 1.

Question:The wavefunction given in the example above is unnormalized. What do we have to do to it (what do we have to multiply it by) in order to normalize it? We highly recommend you try to figure this out for yourself before reading any further—these little exercises are an important way to make sure you're still following along, and haven't missed anything.

Answer: You have to multiply Y by 1/. To confirm that answer, you have to calculate |Y|²dx for the original function given above. The result of this integral works out to be p. So once we divide Y by then |Y|²dxwill give us 1, which is what we want.

So we now have a properly normalized Y:

• x<0: Y=0
• 
• x>2p: Y=0

We can use this to find the probability of the particle being in any given range, by the magic formula of integrating |Y|² between the two bounding points. For instance, if you integrate from 0 to p, you find that the particle is 50% likely to be in that range (as we would expect). What is the probability of finding the particle between p and 2p? Between 0 and 2p? Between 2p and 4p? Calculate all these answers, and make sure the answers make sense to you.

What about the expectation value? In our discrete-position universe, we calculated the expectation value by doing a sum: the x-value of a point, times the normalized probability of the particle appearing at that point, summed over all the points. As you might expect, we will calculate expectation value by the same basic principal, but using an integral: x|Y|²dx. If we had a million particles with this particular Y, and we measured all their positions and averaged the results, this formula would give us the average.

So, this is another good point for a do-it-yourself exercise. First, looking at the drawing of Y above, what would you intuitively think the expectation value of position for the particle would be? (In other words, on average, where would you expect to find the particle?) Now, actually do the integral. Did you get the result you expected? (Hint: integrating x cos(x)requires integration by parts.)

We're done with position. So before we go on, let's once again summarize where we are.

We have essentially learned only one rule, which is that |Y|²gives you the probability of finding a particle at a given position. But we've layered a considerable amount of math onto that rule.

• We've talked about normalizingthe probabilities (meaning, making sure that the total probability over all space is 1).
• We've talked about finding the expectation value of position by summing up (x-times-probability-of-x) over all possible positions.
• We've talked about the fact that, when x values are continuous instead of discrete, |Y|² is actually a probability density instead of a probability, and you have to integrate over some finite range of x-values to get a non-zero probability.

If any of the math was new or difficult for you, you definitely need to bone up on it. You can spend a whole class feeling like you're always confused and always one step behind, and it turns out in the end it was all because you were never quite comfortable with some aspect of integrals, or something like that. Now is the time to make sure that your prerequisite math isn't getting in your way.

But if you're on top of all the math, then we've just learned one little rule so far, which is all you need for position.

Momentum probabilities

So, someone tells us that a particle is in a particular state, represented by a particular wavefunction Y. We know how to analyze that state to yield all the possible positions that the particle might be in, and their respective probabilities. How do we analyze the exact same wavefunction to find the momentum probabilities? How do we determine something like "this particle has a 1/4 probability of momentum 23, and a 3/4 probability of momentum 42"?

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=e^5ix/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=e^7ix/ then the momentum is 7. In general, if Y=e^ipx/ for any constant p, then the momentum of the particle is exactly p.

Functions of this form (Y=e^ipx/) 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 Yis 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, e^23ix/ and e^42ix/. 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.

1. If Y=e^ipx/ for any constant p, then the particle's momentum is exactly p. Functions of this form are known as the basis states of momentum.
2. 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 wherepcan 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 e^ipx/, 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=e^5ix/ cannot be normalized. (Try it!) This means that you can't have a nonzero probability of having your momentum exactlyequal 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 momentumis 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=e^ipx/ 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)e^ipx/ dp. The function |f(p)|²then 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 anymeasurable 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.

More about momentum, and other measurable quantities

The first thing we said about momentum is that it has certain "basis states," which are wavefunctions where the momentum is certain and determined. A general wavefunction for a particle with no determined momentum can be written as an integral over different momentum basis states, with the magnitude squared of the coefficients of those basis states giving you the probability density for momentum.

All of that applies to any other measurable quantity. So for instance, there are certain functions that represent basis states of energy.If the wavefunction is in a basis state of energy, then the energy is determined; if it isn't, you can rewrite it as a sum of different basis states of energy, and use their coefficients to find the probabilities of different energy values.

But different observables have different basis states. The function e^ipx/ happens to be a basis state for momentum, but it isn't a basis state for all quantities. In fact, the Heisenberg Uncertainty Principle guarantees that you can never be in a basis state for everything at once: if some things are precisely determined, other things are vague and probabilistic! (The rules we've given so far are actually enough to derive the Heisenberg Principle for position and momentum. Think about what we've said so far and try to convince yourself that a particle couldn't be in a state of exact momentum and position. Click for an intuitive derivation of the uncertainty principle.)

What we need in order to generalize to all the different things we want to measure is to know their basis states. How do you find the basis state for a particular quantity? In order to answer that, we first have to introduce the mathematical concept of an operator.This is pretty simple, as concepts go: an operator is something you do to a function, and you get a different function. For instance, an operator might be "add seven" or "take the second derivative and then take the sin" or something like that.

In quantum mechanics, every measurable quantity is associated with an operator. For instance, the position operator happens to be "x" meaning "any function you give me, I will multiply by x." The momentum operator is "—i" meaning "any function you throw at me, I will take the first derivative of it, and then multiply by —i." So if you act with the position operator on the function x⁵ you get x⁶; if you act with the momentum operator on x⁵ you get —5ix⁴. Nothing too fancy-looking there, right? Doesn't look particularly useful, but at least it isn't hard.

Now, what happens when you act on Y=e^5ix/ with the momentum operator? (Try this yourself to make sure you get the same answer we do!) You get 5e^5ix/. Note that you got back the exact same function you started with, times 5. It is exactly this fact that makes e^5ix/a basis state with momentum 5: when you act on it with the momentum operator, you get the same function back, times 5.

This brings us to the general rule for finding basis states, and it's a really ugly and arbitrary-looking rule, but it is a key postulate of quantum mechanics, so it's worth stating on its own:

When the operator for a particular quantity, acting on a wavefunction, produces that same wavefunction times a constant, then that wavefunction is a basis state for that quantity, and the constant is the value for that quantity.

This is not only ugly, it's hard to say. We can't make it less ugly, but we can make it easier to say by introducing some terminology. Any function that has this property (the operator returns the same function times a constant) is called an eigenfunction of the operator, and the constant is called its eigenvalue.We can now state our general rule more succinctly:

The eigenfunctions of an operator represent its basis states, and their eigenvalues are the corresponding values of the measurable quantity.

We can get even more concise by writing it mathematically. Let "A" be an operator, so "AY" means "the operator A acting on Y." And let "a" be a constant. Then we can define "Y" as an eigenfunction (or basis state) of A and "a" as an eigenvalue, all by writing:

AY = aY (—>"A operating on Y produces a times Y")

Since this is so abstract, it's worth coming back to the simple example we started with, because the example really says it all. The momentum operator is —i. It just so happens that if Y=e^i5x/, then when you act on Y with the momentum operator you get 5Y. This tells us that Y=e^5ix/is a basis state (or eigenfunction) that represents a particle with a definite momentum of 5 (the eigenvalue).

So—for those of you keeping track at home—here's what we've done so far. We have introduced two extremely arbitrary-looking and ugly postulates. One is "the basis states are the wavefunctions with the peculiar property that, when you operate on them, you get them back times a constant." The other is "the operator for momentum is —i". Using these two rules, we've been able to easily prove the rule that seemed totally arbitrary in the last section, "the basis states of momentum are of the form Y=e^ipx/." We can now see that that formula wasn't arbitrary; it was a special case of a general principal. Still, this hardly seems like an improvement, does it?

But the reason we went through all that is because, once we have generalized in this fashion, we can very easily get at all the other observable quantities. The reason is that virtually all observable quantities are built from position and momentum, and the way you build them in quantum mechanics is through the operators. (The main exception is spin.)

For instance, consider kinetic energy. Classically, we know that kinetic energy is p²/2m. (This is another way of writing the more familiar "½mv².") In quantum mechanics, the same rule applies, except that instead of numbers we're working with operators. The momentum operator is —i. We find kinetic energy by "squaring" that operator (ie doing it twice) and then dividing by 2m, so the kinetic energy operator is . We can use this to find the basis states for kinetic energy, and all the math works the same from there. (In this particular example, the basis states for kinetic energy turn out to be the same as the basis states for momentum, since if p is determined then p²is too!)

(Click if you're bothered by the way we just happily squared an operator, for a brief justification of why it makes some sense.)

What about expectation values? If you have an observable quantity represented by an operator O, you can always write the wavefunction Yas a sum of basis states

Y=C₁Y₁+C₂Y₂+...

where Y₁, Y₂,... are the eigenfunctions of O with some set of eigenvalues v₁, v₂,... If we measure this observable quantity we will find the result v₁ with probability |C₁|² and v₂ with probability |C₂|², and so on. So the expectation value is given by v₁|C₁|²+v₂|C₂|²+... If Orepresents a continuous variable you do the same thing with integrals instead of sums.

Although this expression gives you a way to calculate expectation values in principle, it turns out in practice that there is an easier way that doesn't require calculating basis states or eigenvalues. If we use the same letter Ofor both the observable quantity and the operator representing it, we can write

In other words to find the expectation value of O you act with the operator O on the wavefunction Y, multiply the result by the complex conjugate Y^*, and integrate the whole thing over the x-axis. For a somewhat hand-waving explanation of why it works, click for our footnote on the general formula for expectation values.

The Energy Operator

This section will not introduce any new concepts at all. It will, however, showcase a specific example of what we've been talking about—and it will also lay some important groundwork for a later section, in which we introduce Schrödinger's Equation.

We said earlier that the operator for kinetic energy was . Remember that this was not a new postulate. It came from the momentum operator, and from the fact that kinetic energy is p²/2m—in quantum or classical mechanics. What about total energy? This does not need a new postulate either. In quantum mechanics, just as in classical mechanics, total energy is kinetic energy plus potential energy.

OK, what's potential energy? It is the energy a particle has based on what forces are acting on it and where it is in space. For instance, consider an electron orbiting around a proton. We can say that the electron experiences a force of F=C/r² in the direction of the proton, where r is the distance from the proton and C is a constant. Or, equivalently, we can say that the electron has a potential energy given by V=—C/r. Particles tend to move from regions where they have high potential energy to regions where they have low potential energy—hence, the electron will move toward the proton. The laws of classical mechanics can be expressed either in terms of forces or in terms of potential: since F=—dV/dx, the two are different mathematical ways of arriving at the same results. In quantum mechanics, it is more convenient to work with V than with F.

Total energy is thus kinetic plus potential energy, and the operator for total energy is +V(x). What does that mean? It means that when that operator acts on the wavefunction of a particle which is in an eigenstate of energy, it gives back the same eigenstate times the energy:

Let's make sure you're comfortable with where we are. Earlier we said, as a fundamental postulate of the system, that the momentum operator was —i. This means that if you act with that operator on an eigenstate of momentum, you get back the same eigenstate times the momentum: —i=pY. And we solved that differential equation to find that Y=e^ipx/ are the eigenstates of momentum, with momentum p.

Now we have a similar equation for energy—same logic, same definitions, just a different operator, based on the fact that in quantum mechanics, just as in classical mechanics, E=p²/2m+V. Can we do the same thing, solve it to find the eigenstates of energy? Yes, we can, but only when we know the function V(x). The eigenstates of energy—unlike the eigenstates for momentum, position, or kinetic energy, for instance—depend onthe potential energy function. In other words, they depend on the forces acting on the particle. When we know the potential energy function, we can find the eigenstates of energy by solving the differential equation given above. We will show you how to do this in our example toward the end of this paper.

Measurements: How to Find Y

Now that we've talked about how to use the wavefunction to find measurable quantities, we can turn the question around and ask how I use a measurement to determine the wavefunction. Suppose we have a particle in a known force field, such as an electron in a hydrogen atom. We know the eigenstates of position, momentum, kinetic energy, etc. Moreover, since we know the potential energy function (V=—C/r) we can also figure out the eigenstates of potential energy, total energy, and anything else we might want to measure. What we don't know, however, is what state our particle is actually in. The wavefunction of this electron could in principle be any normalized complex function. It might be tightly concentrated near the nucleus, or floating somewhere around Alpha Centauri, or spread out with equal probabilities everywhere in between. The only way to know the state of the particle is to measure it.

But what measurement should we use? If we want to know the wavefunction of our particle, should we measure its position, its momentum, its kinetic energy, or what? Classically the answer is that to know the state of a particle you need to make two independent measurements to determine its position and momentum. Quantum mechanically, the state of a particle is much more complicated, so you might think you would need many more measurements to determine the full wavefunction of a particle.

Surprisingly, the answer is that you need only one! If we measure the position, the momentum, the kinetic energy, the angular momentum, or any other function of position and/or momentum, we know our particle's wavefunction. Specifically, we know that our particle must be in an eigenstate of whatever we measured, and the particular eigenstate must be the one whose eigenvalue corresponds to the result we got from our measurement. For example, if we measure the particle's momentum and we find the result p=7 (in some units), we know that our particle's wavefunction (up to a normalization constant) is Y=e^7ix/.

How can this be? We said that before we took our measurement the wavefunction could have been anything. Now we take a momentum measurement, and we're claiming that no matter what result we get the wavefunction will be in the form Y=e^ipx/, where p is some number. What if our wavefunction beforehand was a polynomial, or a combination of hyperbolic trig functions, or a graph that vaguely resembled Mt. Rushmore on a cloudy day? The answer is that we have no idea what the wavefunction was before we took the measurement. Measuring the momentum of the particle causesits wavefunction to become an eigenstate of momentum.

This point is crucial and bears repeating. At each moment in time a given particle has a wavefunction. We may or may not know what it is, but we presume that it exists just like the position and momentum of a classical particle exist whether we know them or not. When we measure anything about our particle, that wavefunction determines the probabilities of the different results we might get. As soon as we make such a measurement, the wavefunction instantly changes into an eigenstate of whatever we chose to measure. So we might never know what the wavefunction of our particle was before we measured it, but we know with certainty what it is once the measurement has been done.

This process of changing the state of a particle by measuring it is usually called collapsing the wavefunction.Many physicists and philosophers have debated at length exactly what constitutes a measurement and how to interpret this strange result. Such questions are, in our opinion, some of the most interesting issues in quantum mechanics or in science in general, but we're not going to say anything more about them here.

Before we move on, we need to make several caveats to what we've said in this section:

First, we said that it takes only one measurement to determine the wavefunction of a particle, but technically that's not always true because the quantity you measure could have more than one eigenstate with the same eigenvalues. For example if you measure the kinetic energy and get the result K, this tells you that the magnitude of the momentum must be given by p=. However your particle could still be in either of the two wavefunctions Y₁=e^ipx/ or Y₂