Quantum Mechanics and Psionics Part II

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

Either one of these two states is an eigenstate of the kinetic energy operator with the same eigenvalue K. A momentum measurement would have told you the exact state of the wavefunction, while a kinetic energy measurement just tells you that it is in one of the two states Y₁,Y₂, or any linear combination of the two. In quantum mechanics lingo these two states are said to be degenerate with respect to kinetic energy, which just means they have the same kinetic energy eigenvalue.

Second, we should reiterate that we are talking only about the wavefunctions of particles in one dimension. In classical mechanics it would take six measurements to determine the state of a particle in three dimensions: one for each component of x and p. Similarly in quantum mechanics it would take three measurements to determine the wavefunction of a particle in three dimensions. Moreover, if a particle has dynamic properties like spin that can't be derived from position and momentum these will require additional measurements to determine.

Finally, we should note that we're talking about idealized measurements here. In practice you can never measure a continuous variable like momentum to infinite accuracy, so you can never get a wavefunction that's an exact eigenstate of momentum. That's good because a pure momentum eigenstate can't be normalized. (Try it and you'll see.) In the real world you measure momentum (or whatever else) to some specified accuracy and you get an answer that tells you that your momentum is somewhere in the vicinity of 7 and your wavefunction is thus made up of a combination of momentum eigenstates. For the most part we're going to continue talking as if perfect measurements were possible, but you should keep this limitation in mind.

Repeated Measurements

Suppose we have a particle in an unknown state, we measure its momentum, and we find the value 23. Whatever the wavefunction of our particle was before, it is now Y=e^23ix/. What if we immediately measure the momentum again? What will we find? You should stop and try to answer this question before reading on. The answer is that we are guaranteed to find the answer 23. A particle that is in an eigenstate of momentum has a definite momentum and we will get that result with probability 1.

Suppose, however, that we decide to measure the position instead. Once again, think about the answer before going on. This time the answer is that we have no idea what we're going to get. The probability distribution for position is given by Y^*Y, which for a momentum eigenstate is just a constant. In other words our particle is equally likely to be in any position as in any other. We've just discovered a special case of the famous Heisenberg uncertainty principle, which states that the better you know the momentum of a particle the less you know about its position, and vice-versa. In this extreme (and unrealistic) case, we know the momentum exactly and therefore the position is completely undetermined. Note that we've followed common practice here in stating the uncertainty principle in terms of knowledge, but this is actually misleading. It's not just that we don't know the position of our particle. A particle in a momentum eigenstate doesn't havea particular position.

Let's play this game once more. We measure the momentum and we find the result 23. Then we measure the momentum again and find the same result. Assuming we're sufficiently determined and/or bored we can measure the momentum 100 times in a row and we'll keep getting the same result each time. Now we measure the position. The result is completely random, but we get some result. In other words we find that the particle is somewhere. Now we go back and measure the momentum again. Do we get 23 yet again? No! When we measure the position we force the wavefunction to become an eigenstate of position, meaning it is no longer an eigenstate of momentum. The position measurement thus erases the value of momentum that the particle had before, and a new measurement of momentum essentially has a clean slate, and any value is equally likely. (We haven't talked about the eigenstates of position. Click here for a discussion of what those look like.)

What if instead of measuring position we had measured kinetic energy, and then went back to measure momentum again? In that case we would still get 23. The reason is that momentum and kinetic energy have the same set of eigenstates, so a measurement of the kinetic energy doesn't erase the momentum information. In general, there are some quantities that can be simultaneously determined, such as momentum and kinetic energy; and some quantities that can not, such as momentum and position. (Pairs of variables that can be determined simultaneously are said to commute, while those that can not are non-commuting variables.)

So now you're ready to solve some interesting quantum mechanics problems. We could give you a situation, iea particle with a known set of forces acting on it, tell you that we had measured some quantity and gotten a particular result, and then ask you for the probability distribution if you subsequently measure any other quantity. Many problems in quantum mechanics are exactly of this type.

What if we don't take the second measurement immediately, however? We can take a measurement and put our particle into a known state, but if we then let it evolve without measuring it for a certain time, we expect that state to change. This idea isn't new. In classical mechanics we know that if we measure the position and momentum of a particle two times in a row we should get the same result both times, but if we wait an hour in between the two measurements we expect to get different answers. Specifically, the way the state of the particle changes is determined by Newton's second law F=ma. In quantum mechanics the situation is very similar. Once we measure something about our particle we know its state, i.e. its wavefunction. If we wait for a while that wavefunction will evolve. Once again that evolution is determined by a differential equation, which in quantum mechanics is called Schrödinger's equation.

Schrödinger's Equation

This equation, like the rules we gave above, is one of the fundamental postulates of quantum mechanics. This postulate can be stated most simply as: "The operator representing energy is i."

It may seem at first that we're contradicting what we said earlier, that the operator for energy is +V(x). In fact, that is true too. So both operators must give the same answer when they act on a wavefunction:

This is it—Schrödinger's equation, the F=maof quantum mechanics. Let's look again at the analogy between these two laws, but mathematically this time.

• Newton's Second Law is a second-order differential equation, F=m. If you know the forces acting on a particle, you can solve this equation to find x. However, your solution will have two arbitrary constants, corresponding to the position and velocity at time t=0. If you plug in those two constants, you get a specific function x(t) (and its derivative v(t)). This is the mathematical equivalent of the physical principle we stated earlier—given the forces and a measurement at a particular time, you can predict the state of the particle in the future.
• Schrödinger's equation is a mixed-order partial differential equation, meaning the function you are solving for (Y) depends on two variables (x and t). If you know the potential energy function V(x), you can solve this equation to find Y. However, your solution will have an arbitrary function, which corresponds to the function Y(x) at t=0. If you plug in that function, you get a specific function Y(x,t). This is the mathematical equivalent of the physical principle we stated earlier—given the potential energy function and a measurement at a particular time, you can predict the state of the particle in the future.

If these mathematical ideas are new to you click here for a brief discussion of partial differential equations.

In general, partial differential equations are hard to solve. Fortunately, this particular equation is easy to solve for one special case: the case where Yhappens to be an eigenstate of energy. By definition, an energy eigenstate satisfies

so Schrödinger's equation becomes

Other than a few constants floating around, this is just dy/dx=y: one of the easiest differential equations to solve. You should be able to quickly convince yourself that the general solution is

(We've used the fact that —i=1/i.) So, when the initial wavefunction happens to be an eigenstate of energy, Schrödinger's equation becomes incredibly easy to solve.

How do we find the time evolution of a wavefunction that is not an energy eigenstate? We rewrite it as a sum of energy eigenstates, and solve for each of them individually. For each energy eigenstate, we can solve Schrödinger's equation simply by multiplying the eigenstate by e^-iEt/ (where E is a constant, the energy of that particular state). In other words, you write a wavefunction at time t=0in the form

Y(x,0)=C₁Y_E1+C₂Y_E2+...

where we are using Y_E1 to mean the energy eigenstate with energy E1. Then the time evolution of this wavefunction is given by

It's worth taking a moment to see what is happening mathematically to these states. For any given energy eigenstate, the time evolution is very simple: the initial state is simply multiplied by a constant. This means it remains an energy eigenstate: if Y_E1 is an eigenstate with energy E1, then 2Y_E1 is an eigenstate with that same energy value; and so is 3Y_E1 and Y_E1. Of course, multiplying by 2 or 3 would disturb normalization, but multiplying by doesn't even do that, since its magnitude is 1.

However, each individual eigenstate is being multiplied by a different constant. So the overall shape of Y, the sum of all these states, may change in all kinds of very complicated ways. We'll see how this works out in an example below.

For completeness, and to map onto the terminology that you may be familiar with, we should mention that most quantum mechanics classes teach that there are two different Schrödinger's Equations.

: The time-dependent Schrödinger's Equation

: The time-independent Schrödinger's Equation

As we have seen, the first one is the real Schrödinger's Equation; the second is just the equation that tells you the energy eigenstates, each with its own constant E. However, the two are interconnected because we solve the first (difficult partial) differential equation by solving the second (easier ordinary) differential equation. Once you have found the energy eigenstates (using the second equation), you can solve Schrödinger's Equation simply by multiplying each eigenstate by e^-iEt/.

So, what does all that tell you? The second equation tells you nothing at all about what the wavefunction of our particle actually is. In general, any wavefunction could be written as a superposition of the energy eigenstates. Knowing what the eigenstates are is useful—but it does not tell you what Yis.

The first equation does seem to say what Y is. It gives us a function Y(x,t) as the answer—and unlike the second equation, it has no E thrown into the mix. Potential energy goes in, Y comes out. But remember the undetermined constants: Y(x,0) can be absolutely anything it wants to. What this equation tells you is how the wavefunction will evolve over time, from a known point.

The Bottom Line— How to solve basic quantum mechanics problems

1. You start with a situation where you have a particle with a known potential energy function. In other words, you know all the forces acting on the particle.
2. You solve the "time-independent Schrödinger Equation" for this particular potential energy function. This tells you the energy eigenstates.
3. You take a measurement that tells you the initial state of the particle Y(x,0)and you rewrite this initial state as a sum of energy eigenstates.
4. You see how Y will evolve over time, from its initial state to later states. In principal, you are using the "time-dependent Schrödinger Equation" to do this, but that equation allows you to do it the easy way; for each eigenstate of Y with its own eigenvalue E, just multiply Y by e^-iEt/.
5. To predict the results of the observation of any measurable quantity at any time t, find Y at that time t and rewrite it as a sum of eigenstates for that measurable quantity. The coefficients of these eigenstates will tell you the probabilities of measuring their respective eigenvalues.

Clear as mud? Well, now—at last—it really is time for an example.

Example: the infinite square well

Even if you've already "infinite square welled" to death in class, we encourage you to read this section. We're going to use it to reinforce a lot of the ideas that we've discussed throughout the paper; hopefully, they will come out making more sense than they did before.

Remember that, just as every classical mechanics problem starts with a statement of the forces, every quantum mechanics problem starts with a statement of the potential energy. In the infinite square well, the potential energy is very simple, and has a graph that kind of looks like—well, a big, square, well.

• x<0: V=
• 0<x<L: V=0
• x>L: V=

 

What this potential energy function tells you is that, within the range 0<x<L, the particle is free to move with no forces on it whatsoever. But at x=0 and x=L, there is an absolutely unstoppable force that pushes it back; so the particle just can't go there.For reasons that should hopefully be clear this example is often called a "particle in a box."

So, let's walk through the steps we outlined above. We will start by figuring out the energy eigenstates, using the time-independent Schrödinger equation.

On the left and right sides of the graph, V=. Based on the equation (not based on the physical situation), what does that tell us about Y? Answer: if V=, there is only one way to prevent the middle term from blowing up, and that is to set Y=0. Of course, this mathematical result also tells us what we physicallyexpect: the probability of the position being in this range is zero everywhere. The particle must be in the middle range.

So, what about that range? There, where V=0, the above equation reduces to: =EY_E

This is one of the relatively few differential equations that you can solve just by looking at it right. You start by shoving all the constants over to the right side, for convenience.

In other words, the second derivative of Y_E gives you back minus Y_E, times a constant. It just so happens that both the sin and the cos fill that condition. And since it's a second order differential equation, we only need two linearly independent solutions for full generality, so we can write the answer immediately:

That solves the time-independent Schrödinger equation in the region 0<x<L. However, we still have to apply normalization and boundary conditions. The former we've discussed before and the latter comes about because the wavefunction has to be continuous. (If you like you can think of that as another postulate, but it's really required for the theory to work consistently.) Since we know Y=0 outside this range, we know it must approach 0 at x=0 and x=L. Plugging in Y(0)=0 immediately gives B=0, so one of our terms disappears right away! To get Y(L)=0 we could set A=0, but then we would have Y=0 everywhere, which is a non-physical solution because it can't be normalized. Thus the only way to satisfy our second boundary condition is to set the argument of the sine function to a multiple of p. In other words

What does that tell us? Remember that m and L are fixed parameters of the problem, so the only things that can vary in this equation are E and n. That means that this requirement fixes the possible values for E. Solving, we find that

where n can be any positive integer. It's critical at this point to step back and say: what did we just discover? What we just discovered is that E is not free to be anything it wants to. There is a constant , which is big and ugly but still just a constant. For convenience we're going to call this constant E₀ so we don't have to keep writing it out. E can be equal to E₀, or it can be 4E₀, or 9E₀, or 16E₀, but nothing in between. The energy is quantized:we didn't have to postulate that, the math just spit it out at us.

The final step in finding the energy eigenstates is to normalize this function to find the coefficient A. To do this we calculate

A couple of notes about this calculation: First, we take into account that in principle A could be complex. Second, we chose to rewrite E in terms of n because it's a little simpler, but we could have just left E in explicitly as well. To normalize Y_E we have to set this integral equal to one, so we get |A|²=2/L, which holds true for any A of the form A=eⁱf. The normalized energy eigenstates are thus

with corresponding energy eigenvalues given by E=E₀n²

(We've left out the arbitrary phase in front of Y_n.)

We're done with the hard part. Now, remember that at t=0, Y can be pretty much anything it wants (subject to the normalization and boundary conditions)—you can't figure out Y(0) from the potential energy, you have to be told Y(0) as an initial condition of the problem. So, just suppose someone tells you that at time t=0the wavefunction for your particle is:

You can confirm that this is properly normalized. We're now going to ask and answer a number of questions about this wavefunction and the particle it describes. We strongly encourage you to try to answer them before looking at our answers.

Question 1: If you were to measure the energy of this particle at time t=0what would you find?

Answer: The correct answer is that there is a 1/4 chance that you would find E=E₀ and a 3/4 chance that you would find E=25E₀. Note that after you did this measurement the wavefunction would change completely. Depending on what you got it would either become Y₁ or Y₅. In all the questions that follow we're going to assume you haven't done this or any other measurement before the one asked about in the question.

Question 2: If you were to measure the position of the particle at time t=0what would you find?

Answer: The odds of your different results would be given, as always, by the probability distribution |Y|². We've plotted this function below.

Question 3: What will the wavefunction be at some later time t?

Answer: Each energy eigenstate will be multiplied by a phase e^-iE/, so the wavefunction will be

Question 4: If you measure the energy at time twhat will you find?

Answer: Exactly the same thing you found at time t=0. Multiplying the coefficients by a phase doesn't change their magnitudes, so the odds of the two results are still 1/4 and 3/4.

Question 5: If you measure the position at time twhat will you find?

Answer: Once again the probability distribution for position is given by |Y|², but now that's a different function. Taking t=/5E₀, for example, the probability distribution will look like

You should be able to check this yourself by plugging in t=/5E₀into the equation above and calculating the squared magnitude. Phases do matter!

Summary and Conclusions

With the information given above, you can take a wavefunction and analyze it to find the probability of position falling into any given range; the expectation value of position; the probability of momentum falling into any given range; and the expectation value of momentum. For any other observable quantity made up out of position and momentum you can figure out the appropriate operator and use it to find the expectation value of that observable, and with a bit more work you can find the basis functions for that observable and use it to find the probability of that observable falling into any given range (if it's continuous) or having specific values (if it's discrete).

It all boils down to the following rules.

• |Y|²gives you the probability density function for position.
• The operator for momentum is —i, the operator for position is x (meaning "multiply by x"), and other operators are built from these because other observables are built from position and momentum.
• The eigenfunctions for any operator are the basis states for its observable, and the eigenvalues are the values associated with those functions. For instance, e^ipx/ is a basis state for momentum with momentum=p, because —i(e^ipx/)=p(e^ipx/).
• When you write Y as a sum of eigenfunctions for any particular observables, the magnitude squared of the coefficients of the functions give you the probabilities of their eigenvalues. For continuous variables this sum becomes an integral and these probabilities become a probability density.

Applying these rules can lead to many of the strange consequences of quantum mechanics. We've already mentioned one—Heisenberg's uncertainty principle. Another one is quantization. For example, in our infinite square well, we saw that the energy could take certain discrete values, but nothing in between. It is the existence of such quantized systems that gave quantum mechanics its name.

Purists may be annoyed by the fact that we have one general rule for finding every observable except position, and a completely different (albeit simpler) rule for position. Every other observable uses the operator to find the basis states, and work from there. Is there a basis state for position? The answer is, yes there is—you can treat position exactly the same way as any other observable, basis states and all. Click hereif you want to see how.

We should note for the sake of completeness that not all observable quantities can be built out of position and momentum. For classical particles that's all there is, but quantum mechanical particles have an entirely new dynamic property called spinthat's unrelated to position and momentum. The basic rules for spin are the same. There is a spin operator and there are spin basis states and so forth, but a full discussion of this would be beyond the scope of this paper.

Finally, we should note once more that once you have measured a quantity for a particular particle, that particle's wavefunction changes drastically. Whatever the probability distribution might have been for the quantity you are measuring, once you've measured it you know it has exactly the value you measured. For example, suppose a particle has the wavefunction

Y = 3/5 Y₁ + 4/5 Y₂

where Y₁ and Y₂ are eigenstates of some observable O with eigenvalues 2 and 7 respectively. The squared coefficients 9/25 and 16/25 tell you the probabilities that a measurement of O will give the results 2 or 7 respectively. If you make the measurement and find the result 2, however, then at that point the particle's wavefunction has become Y₁. A subsequent measurement performed immediately afterwards is guaranteed to give you the result 2 again. This change that occurs in the wavefunction as a result of measurement is called the "collapse of the wavefunction," and it forms one of the principal differences between quantum and classical physics.

---

Footnotes

Complex Numbers and Magnitudes

If you take any real number and square it, you get a positive number (or zero). There is no number that, when you square it, gives you a negative number. So at some point, someone just made one up, and designated it by the letter i (which stands for "imaginary"): i²=–1, by definition. You can readily see that (–i)²=–1, that (2i)²=–4, that i³=–i, and so on.

A real number is defined as one that has no i in it. An imaginary number is a real number times i (such as 4i). A complex number has both real and imaginary parts. Any complex number z can be written in the form x+iy where x and y are real numbers. x is the "real part" and y is the "imaginary part." For instance, 2+3iis a complex number.

The complex conjugate of a complex number, represented by a star next to the number, is the same real part but the negative imaginary part. So if z=2+3i then z^*=2–3i. Note that the reverse is also true: if z=2–3i then z^*=2+3i. So you always have two numbers that are complex conjugates of each other.

Real numbers can be graphed on a line (the number line). Complex numbers, since they have two separate components, are graphed on a plane called the "complex plane" where the real part is mapped to the x-axis, the imaginary part the y-axis. Every complex number exists at exactly one point on the complex plane: a few examples are given below.

Every complex number has a magnitude which gives its distance from the origin (0) on the complex plane. So 1 and i both have magnitude 1. The magnitude is written with absolute value signs. So we can say that if z=2+3i then |z|²=2²+3².

If you multiply any number by its own complex conjugate, you get the magnitude of the number squared (you should be able to convince yourself of this pretty easily). So we can write that, for any z, z*z=|z|².

Expectation Values

Suppose you roll a 4-sided die. What number do you expect to get? Answer: you have no idea. Any number is as likely as any other, presumably.

OK, but on averagewhat number will you get? For instance, if you rolled the die a million times, and averaged all the results (divided the sum by a million), what would the answer probably be? This is called the "expectation value." You can probably see, intuitively, that the answer in this case is 2½. Never mind that an individual die roll can never come out 2½: on the average, we will land right in the middle of the 1-to-4 range, and that will be 2½.

But you can't guess the answer quite so easily if the odds are uneven. For instance, suppose the die is weighted so that we have a 1/2 chance of rolling a 1, a 1/4 chance of rolling a 2, a 1/8 chance of rolling a 3, and a 1/8 chance of rolling a 4? Your intuition should tell you that our average roll will now be a heck of a lot lower than 2½, since the lower numbers are much more likely. But exactly what will it be?

Well, let's roll a million dice, add them, and then divide by a million. How many 1s will we get? About 1/2 million. How many 2s? About 1/4 million. And we will get about 1/8 million each of 3s and 4s. So the total, when we add, will be (1/2 million)*1+(1/4 million)*2+(1/8 million)*3+(1/8 million)*4. And after we add—this is the kicker—we will divide the total by a million, so all those "millions" go away, and leave us with (1/2)1+(1/4)2+(1/8)3+(1/8)4=1 7/8.

The moral of the story is: to get the expectation value, you do a sum of each possible result multiplied by the probabilityof that result. This tells you what result you will find on average.

Fourier Transforms

Before explaining Fourier transforms, we will start by quickly reviewing a related topic you may be more familiar with, namely Fourier series. If you have any function f(x) that is periodic with period L you can write it as a sum of sine and cosine terms:

For each sine or cosine wave in this series the wavelength is given by L/m or L/n respectively. In other words the indices m and n indicate the wave number(one over wavelength) of each wave. Using the formula

e^ix=cos x + i sin x

you can rewrite the Fourier series in the equivalent form

The function e^2pipx/L is called a plane wave. Like a sine or cosine wave it has a value that oscillates periodically, only with plane waves this value is complex. That means that even if f(x) is a real function the coefficients c_p will in general be complex. Either way you represent it, however, a Fourier series is simply a way of rewriting a periodic function f(x)as an infinite series of simple periodic functions with numerical coefficients.

A Fourier transform is the exact same thing, only f(x) doesn't have to be periodic. This means that the index pcan now take on any real value instead of just integer values, and that the expansion is now an integral rather than a sum.

The factor of is a matter of convention and is set differently by different authors. Likewise some people add a factor (usually ±2p) in the exponential. The basic properties of the Fourier transform are unchanged by these differences. Note that this formula has nothing to do with position and momentum. The variable x could be position, or time, or anything else, and the variable p tells you the wave number (one over wavelength) of each plane wave in the expansion. (If x is time then the wave number is the same thing as the frequency of the wave.) To invert this formula, first multiply it by e^—ip'x (where p' is just an arbitrary number) and then integrate with respect to x.

It is possible to evaluate the integrals on the right hand side of this equation, but doing so requires a familiarity with Dirac delta functions. If you are not familiar with these functions you can just skip the next paragraph and jump straight to the result. (We discuss Dirac delta functionsin a later footnote, but if you're not already somewhat familiar with them trying to follow the argument below may be an exercise in frustration.)

The only x dependence in this integral comes in the exponential, which is an oscillatory function. If p-p' is anything other than zero then this exponential will have equal positive and negative contributions and the integral will come out to zero. (Actually e^ipx is complex so the equal contributions come from all directions in the complex plane. If you can't picture this just take our word for it.) If p-p' is zero, however, then the exponential is simply one everywhere and the x integral is infinite. Thus the x integral is a function of p-p' that is infinite at p-p', and zero everywhere else, which is to say a delta function. Of course this isn't a rigorous proof. It can be proven rigorously that the integral of a plane wave is a delta function, however, and when you do the math carefully you find that the integral produces a delta function with a coefficient of 2p. So

where the integration over p uses the defining characteristic of a delta function. Replacing the dummy variable name p' with pwe finally arrive at the formula

We should emphasize once more that mathematically x is just the argument to the function f(x) and p is just the wave number of the plane waves in the expansion. The fact that the relationship between Y(x) and f(p) in quantum mechanics takes the form of a Fourier transform means that for a given wavefunction Y(x)the momentum happens to be inversely related to the wavelength. (Once again, though, a realistic wavefunction will be made of many waves with different wavelengths.)

We said before that the numerical coefficients in the Fourier transform definition were arbitrary. In quantum mechanics these coefficients pick up an extra factor of , so the full relation between Y(x) and f(p)is

Finally, we should note one convenient theorem about Fourier transforms. Consider the normalization of the wavefunction, which is determined by the integral of its squared magnitude. Writing that in terms of the Fourier expansion we find

This result can be proven by taking the complex conjugate of the expression for Y(x) above, multiplying it by Y(x), and integrating over x using the same delta function trick we used above. We won't bother to show this proof here. Recall that f(p) is the coefficient of the momentum basis state with momentum p, so what this result says is that if the wavefunction is properly normalized so that the total probability of the position being anywhere is one, then it will automatically be normalized so that the total probability of the momentum having any value is one.

The Uncertainty Principle

In classical physics a particle at any given moment has a position and a momentum. We may not know what they are, but they exist with some exact values. In quantum mechanics a particle has a wavefunction which gives it some probability distributions for position and momentum. The uncertainty principle says that the more definite position is the less definite the momentum, and vice-versa.

We can see how this comes out of the rules we've given so far by considering the basis states of momentum, which are of the form e^ipx/, which is equal to cos(px/)+i sin(px/). In other words the basis states of momentum are waves whose wavelength is determined by the value p. (Specifically, the wavelength is 2p/p.) When you expand a wavefunction in momentum basis states you are writing Y(x)as a superposition of these waves.

Those familiar with Fourier expansions know that a localized wave is made up of components with many different wavelengths. At the other extreme a function made up of a single wave isn't localized at all. If a particle is in a momentum eigenstates Y=e^ipx/ then its position probabilities are equally spread out over all space. The bigger the range of wavelengths contained in the superposition the more localized the function Y(x) can be. From this we can see that a particle highly localized in momentum (a small range of wavelengths for Y) will have a very uncertain position (Y(x) spread out over a large area), and a particle highly localized in position (Y(x) concentrated near a point) will have a very uncertain momentum (many wavelengths contributing to the Fourier Transform of Y). You can certainly cook up a wavefunction with a large uncertainty in both position and momentum, or one that's highly localized in one but uncertain in the other, but you can't make one that's highly localized in both.

It's possible to make a mathematical definition of what you mean by the <

Register or Login to comment!

15 Comments:

MyTK Shoutbox:

Hello Newbie. If you want to enjoy the full MyTK experience, sign up!