Hey folks, hope you guys had a fantastic break. It was good getting to meet a bunch of you while I was in town. As for me, I’m back in Europe and I’ve had a couple of weeks to goof around and even get a little work done. More goofing around than work so far, though.
Anyway, back to the blog. Last semester we talked about special relativity and quantum mechanics, so I suppose it’s time to move on. Hopefully you enjoyed the stuff about QM; there’s some fairly crazy nonsense that goes on there. In my opinion, it’s much weirder than special relativity, but it’s also quite a bit more involved (mathematically speaking).
With those two things out of the way, I’d like to move on to something that’s particularly interesting to me: particle physics. You see, those subjects are both very interesting, but no one really “does” special relativity or quantum mechanics (or at least, not very many people). People write textbooks about them all the time, but you’d have to look pretty hard to find a paper published or a seminar given about them, because they’re no longer at the forefront of physics. In fact, they’re usually not viewed as proper theories at all, but rather frameworks, tools people use to do physics. It’s a bit like learning to play chess: knowing how the pieces move is not the same as being able to play the game. It’s certainly a prerequisite; you can’t even begin to play unless you know the rules, but it’s not enough all by itself.
If special relativity and quantum mechanics are the rules, then particle physics is the game. The object of the game is to figure out the fundamental building blocks of the universe, and it has occupied the minds of some of the most magnificent thinkers of the past hundred years. My goal over the next few weeks is to give you a sense of the “lay of the land”; some idea of what we know, and what we think we know, and what we definitely don’t know, about how the universe really works.
Monday, February 11, 2008
Monday, December 17, 2007
Heisenberg's uncertainty principle
Well, it’s getting to be about time to wrap up the semester, and I’ve saved the most important thing for last. Perhaps you’ve gotten the sense in the last few weeks that quantum mechanics is one incredibly bizarre thing after another, and this is certainly true. However, in a very real way, all that unbelievable weirdness actually emanates from a single source: the uncertainty principle. More than that, in fact: everything I’ve told you so far about quantum mechanics can actually be derived from the uncertainty principle.
A friend of mine once referred to uncertainty as “the beating heart of quantum mechanics”. This may be overly poetic, but it is true that the uncertainty principle is one of the two pillars on which all of quantum mechanics is based (the other being Pauli’s exclusion principle). In spite of being so important, it’s usually misquoted, and it’s very often misunderstood.
So what exactly is uncertainty? Actually, when Heisenberg originally wrote his paper in 1925, he didn’t really explain it very well. (I personally think he was hedging his bets. Remember, this was long before the debate about the realist vs. orthodox positions had been resolved.) Basically, what he said was this: the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa.
To be a little more mathy about it, let’s talk about the quantities Δx and Δp. Strictly, these are the standard deviations of the position and momentum, x and p, though people typically just call them the uncertainties. (I’d rather not actually define these, just because I haven’t mentioned a couple of things that go into the definition. However, if you don’t remember what a standard deviation is, it’s safe to think of Δx and Δp as the “spread” in the position and momentum. Of course, Rebecca might be upset with you if you don’t remember what standard deviations are…) Anyway, what Heisenberg said was this:
ΔxΔp ≥ hbar/2
That is, the product of the two uncertainties is greater than some constant. This equation may seem fairly innocuous, but it’s actually an unbelievable result: it doesn’t matter what the actual constant is, the fact that the uncertainties must be greater than zero is the incredible thing. I’m really not sure that there’s anything in human experience which might prepare us for this. What Heisenberg had shown, even though he himself may not have realized it at the time, was the incompatibility of position and momentum.
When a lot of authors (including Heisenberg himself) talked about uncertainty, they made it sound as if it was somehow the experimenter’s fault. For instance, one way to measure a particle’s position is to hit it with a beam of light. If you hit it with low-energy light, you can do your measurement without disturbing the particle too much, so its momentum can be fairly well known. The trade off is that low-energy light isn’t good at resolving the particle, so you don’t know much about where exactly it is. Conversely, you could pummel it with high-energy light, in which case you’d get a great sense of where exactly it is, but the high-energy light would send the particle skittering off to wherever, so we’d have no idea what its momentum is.
This is not only silly, it’s downright misleading. The only conclusion that we would draw from that story is that this particular way of measuring a particle may not be very good. Sheesh, maybe the people doing these experiments just aren’t very smart; it sounds like a clever person would just find a less obtrusive way of measuring the particle’s position. For that matter, maybe the problem is even simpler: perhaps we just need to spend more money to buy a better machine.
But this isn’t it at all! What the uncertainty principle really says is much deeper than this. Recall from the orthodox position that a particle doesn’t always have an exact position or momentum. According to uncertainty, the more definitely a particle has a position, the less definitely it has a momentum, and vice versa. It’s not that an experiment that’s good at finding a particle’s position necessarily has to be bad at finding the particle’s momentum, it’s that when you measure the particle’s position, it doesn’t really have a momentum, so there’s nothing to measure!
At the two extremes, the uncertainty relation is actually even more surprising: if Δx = 0 (that is, the position is known exactly), then Δp → ∞ (that is, the particle has literally every momentum at once). On the other hand, if the particle has an exact momentum (Δp = 0), it exists everywhere simultaneously (Δx → ∞).
Alright, no doubt about it, that’s weird. But what makes uncertainty so important? What makes it the source of all weirdness in quantum mechanics? Actually, uncertainty is more general than I’ve let on: I started off talking about position and momentum, but the wonderful thing about uncertainty is that it applies to any two physical observables, be they position, momentum, energy, spin, or whatever else we can dream up that we might want to measure. Recall that every physical observable has its own operator in quantum mechanics. What the generalized uncertainty principle says is this: if A and B are any two operators, then
(ΔA)2(ΔB)2 ≥ [A, B]2/4
This second, more general form of the uncertainty principle is the real engine here: we know that physical observables in quantum mechanics are represented by operators, and we know that sometimes two operators don’t commute. What uncertainty does is take this purely mathematical fact and turn it into something physical: because of uncertainty it is now impossible for some observables to have definite values at the same time; it’s why quantum mechanics has wave functions in the first place, and therefore probabilities, and by extension it’s the cause of the realist/orthodox/agnostic debate. So, no fooling, uncertainty really is the motivating force behind everything we’ve talked about in quantum mechanics.
Anyway, now that we know what commutators are (see last week’s blog), we have all the math we need to evaluate this relation. Well, almost. Actually the “absolute value” brackets mean something slightly different when we’re talking about imaginary numbers, but for the problems we’re doing you’ll be fine if you just remember to make sure the square of the commutator is positive. With this, you should be in good shape for this week’s homework.
https://mywebspace.wisc.edu/mweinberg/web/uncertainty.pdf
A friend of mine once referred to uncertainty as “the beating heart of quantum mechanics”. This may be overly poetic, but it is true that the uncertainty principle is one of the two pillars on which all of quantum mechanics is based (the other being Pauli’s exclusion principle). In spite of being so important, it’s usually misquoted, and it’s very often misunderstood.
So what exactly is uncertainty? Actually, when Heisenberg originally wrote his paper in 1925, he didn’t really explain it very well. (I personally think he was hedging his bets. Remember, this was long before the debate about the realist vs. orthodox positions had been resolved.) Basically, what he said was this: the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa.
To be a little more mathy about it, let’s talk about the quantities Δx and Δp. Strictly, these are the standard deviations of the position and momentum, x and p, though people typically just call them the uncertainties. (I’d rather not actually define these, just because I haven’t mentioned a couple of things that go into the definition. However, if you don’t remember what a standard deviation is, it’s safe to think of Δx and Δp as the “spread” in the position and momentum. Of course, Rebecca might be upset with you if you don’t remember what standard deviations are…) Anyway, what Heisenberg said was this:
ΔxΔp ≥ hbar/2
That is, the product of the two uncertainties is greater than some constant. This equation may seem fairly innocuous, but it’s actually an unbelievable result: it doesn’t matter what the actual constant is, the fact that the uncertainties must be greater than zero is the incredible thing. I’m really not sure that there’s anything in human experience which might prepare us for this. What Heisenberg had shown, even though he himself may not have realized it at the time, was the incompatibility of position and momentum.
When a lot of authors (including Heisenberg himself) talked about uncertainty, they made it sound as if it was somehow the experimenter’s fault. For instance, one way to measure a particle’s position is to hit it with a beam of light. If you hit it with low-energy light, you can do your measurement without disturbing the particle too much, so its momentum can be fairly well known. The trade off is that low-energy light isn’t good at resolving the particle, so you don’t know much about where exactly it is. Conversely, you could pummel it with high-energy light, in which case you’d get a great sense of where exactly it is, but the high-energy light would send the particle skittering off to wherever, so we’d have no idea what its momentum is.
This is not only silly, it’s downright misleading. The only conclusion that we would draw from that story is that this particular way of measuring a particle may not be very good. Sheesh, maybe the people doing these experiments just aren’t very smart; it sounds like a clever person would just find a less obtrusive way of measuring the particle’s position. For that matter, maybe the problem is even simpler: perhaps we just need to spend more money to buy a better machine.
But this isn’t it at all! What the uncertainty principle really says is much deeper than this. Recall from the orthodox position that a particle doesn’t always have an exact position or momentum. According to uncertainty, the more definitely a particle has a position, the less definitely it has a momentum, and vice versa. It’s not that an experiment that’s good at finding a particle’s position necessarily has to be bad at finding the particle’s momentum, it’s that when you measure the particle’s position, it doesn’t really have a momentum, so there’s nothing to measure!
At the two extremes, the uncertainty relation is actually even more surprising: if Δx = 0 (that is, the position is known exactly), then Δp → ∞ (that is, the particle has literally every momentum at once). On the other hand, if the particle has an exact momentum (Δp = 0), it exists everywhere simultaneously (Δx → ∞).
Alright, no doubt about it, that’s weird. But what makes uncertainty so important? What makes it the source of all weirdness in quantum mechanics? Actually, uncertainty is more general than I’ve let on: I started off talking about position and momentum, but the wonderful thing about uncertainty is that it applies to any two physical observables, be they position, momentum, energy, spin, or whatever else we can dream up that we might want to measure. Recall that every physical observable has its own operator in quantum mechanics. What the generalized uncertainty principle says is this: if A and B are any two operators, then
(ΔA)2(ΔB)2 ≥ [A, B]2/4
This second, more general form of the uncertainty principle is the real engine here: we know that physical observables in quantum mechanics are represented by operators, and we know that sometimes two operators don’t commute. What uncertainty does is take this purely mathematical fact and turn it into something physical: because of uncertainty it is now impossible for some observables to have definite values at the same time; it’s why quantum mechanics has wave functions in the first place, and therefore probabilities, and by extension it’s the cause of the realist/orthodox/agnostic debate. So, no fooling, uncertainty really is the motivating force behind everything we’ve talked about in quantum mechanics.
Anyway, now that we know what commutators are (see last week’s blog), we have all the math we need to evaluate this relation. Well, almost. Actually the “absolute value” brackets mean something slightly different when we’re talking about imaginary numbers, but for the problems we’re doing you’ll be fine if you just remember to make sure the square of the commutator is positive. With this, you should be in good shape for this week’s homework.
https://mywebspace.wisc.edu/mweinberg/web/uncertainty.pdf
Monday, December 3, 2007
Compatibility in quantum mechanics
Last week I claimed (and in the Bell’s theorem blog I proved) that particles generally don’t have physical properties until those properties are specifically measured. This week I’d like to explain a little more about how measurement actually works. We already have some sense of how to “do” quantum mechanics: we know what we’re looking for (the wave function Ψ) and, at least in principle, how to get it (solve the Schrodinger equation).
But what do we do once we have Ψ? Well, we start applying operators to it, and see what happens. To review, if we let an operator A act on Ψ, there are generally two possible outcomes:
(1) We get back the original wave function times a constant: AΨ = aΨ, where a is just some number. In this case the particle already had a definite value for that property. For example, take the momentum operator p. If we apply it to a particle’s wave function and find that pΨ = 4Ψ, then the particle was already in a state with a definite momentum (of 4, though we’d need to know the units before this would mean much).
(2) We get something else. Then the particle does not have a definite value for this property, and the wave function can only tell us the probability of getting a particular value if we make a measurement.
So what’s the big deal? We find Ψ and then start applying operators to it. If we get outcome (1), fantastic, we get a value for that physical property without doing any more work. If we get outcome (2), well, we’ll just go measure that property! Then it will have a definite value, and we can move on to the next operator. When we get done with all the operators we care about, we’ll have a long list of values representing the properties the particle has.
Well, this is tempting, but unfortunately there’s a teensy little problem. Recall from last week that a measurement actually changes the wave function to something else. In many cases, measuring one property will destroy any information you had about another property, because these two properties cannot exist simultaneously. If this is true, the two measurements are said to be “incompatible”. Incompatibility is yet another unbelievably weird feature of quantum mechanics.
As an example, let’s consider the spin of an electron. I can measure the electron’s spin in the x-direction or the y-direction, but I can never know both simultaneously. If I measure the spin in the x-direction I will always find it pointing exactly left or exactly right. Say I measure it pointing left. If I measure it again in the x-direction then I will definitely see it pointing left again. If I measure it in the y-direction, however, then I will have 50/50 odds of seeing it point either way. Once I do that measurement, however, I will have fixed it to be pointing either up or down, and a left/right measurement will once again yield 50/50 odds. Conclusion: the electron cannot have a definite spin in more than one direction at a time.
If you happen to have a more experimental turn of mind, you might be starting to wonder how we can ever get any work done at all. If making a measurement can cause you to lose the information you got from your previous measurement, why bother? Fortunately, not all measurements are incompatible. The trick, as it turns out, is figuring out which measurements can be made while still keeping the information from previous measurements.
So how do we determine which observables (physical properties) are compatible? We use an operator called a “commutator”. If A and B are any two operators, then the commutator of A and B is
[A, B] = AB – BA
Uh-oh. AB minus BA? Isn’t this always going to be equal to zero? Actually, no, not always. It would be, of course, if A and B were numbers, but they’re not; they’re operators. In fact, if you’re in EM this semester, you’ve probably already run into an example of something like this: the cross product. If instead we have two vectors a and b, then when you take their cross product the order matters quite a bit, since a x b is not the same as b x a.
Anyway, In the homework this week we’ll look at examples of two operators that have a nonvanishing commutator, but for now you’ll just have to take my word for it that the commutator is not always zero. (By the way, to use one more bit of jargon, if the commutator of A and B does happen to be zero, then A and B are said to “commute”.)
Once we compute the commutator of two operators, we have a mathematical way of determining whether their observables are compatible (that is, can be measured at the same time): if the two operators do commute, then their observables are compatible, and vice versa. (If they don’t commute, then their observables aren’t compatible.)
As an example, the energy and momentum of a free particle are represented by the operators H and p, respectively. As it turns out, these two operators commute ([H, p] = 0), so it’s possible to measure a particle’s energy without damaging what you know about that particle’s momentum, because the two quantities are compatible.
What about an example of operators that don’t commute? I’m glad you asked, because that brings us to this week’s homework. (As always, if you get stuck, feel free to ask me for help.)
https://mywebspace.wisc.edu/mweinberg/web/commutation.pdf
But what do we do once we have Ψ? Well, we start applying operators to it, and see what happens. To review, if we let an operator A act on Ψ, there are generally two possible outcomes:
(1) We get back the original wave function times a constant: AΨ = aΨ, where a is just some number. In this case the particle already had a definite value for that property. For example, take the momentum operator p. If we apply it to a particle’s wave function and find that pΨ = 4Ψ, then the particle was already in a state with a definite momentum (of 4, though we’d need to know the units before this would mean much).
(2) We get something else. Then the particle does not have a definite value for this property, and the wave function can only tell us the probability of getting a particular value if we make a measurement.
So what’s the big deal? We find Ψ and then start applying operators to it. If we get outcome (1), fantastic, we get a value for that physical property without doing any more work. If we get outcome (2), well, we’ll just go measure that property! Then it will have a definite value, and we can move on to the next operator. When we get done with all the operators we care about, we’ll have a long list of values representing the properties the particle has.
Well, this is tempting, but unfortunately there’s a teensy little problem. Recall from last week that a measurement actually changes the wave function to something else. In many cases, measuring one property will destroy any information you had about another property, because these two properties cannot exist simultaneously. If this is true, the two measurements are said to be “incompatible”. Incompatibility is yet another unbelievably weird feature of quantum mechanics.
As an example, let’s consider the spin of an electron. I can measure the electron’s spin in the x-direction or the y-direction, but I can never know both simultaneously. If I measure the spin in the x-direction I will always find it pointing exactly left or exactly right. Say I measure it pointing left. If I measure it again in the x-direction then I will definitely see it pointing left again. If I measure it in the y-direction, however, then I will have 50/50 odds of seeing it point either way. Once I do that measurement, however, I will have fixed it to be pointing either up or down, and a left/right measurement will once again yield 50/50 odds. Conclusion: the electron cannot have a definite spin in more than one direction at a time.
If you happen to have a more experimental turn of mind, you might be starting to wonder how we can ever get any work done at all. If making a measurement can cause you to lose the information you got from your previous measurement, why bother? Fortunately, not all measurements are incompatible. The trick, as it turns out, is figuring out which measurements can be made while still keeping the information from previous measurements.
So how do we determine which observables (physical properties) are compatible? We use an operator called a “commutator”. If A and B are any two operators, then the commutator of A and B is
[A, B] = AB – BA
Uh-oh. AB minus BA? Isn’t this always going to be equal to zero? Actually, no, not always. It would be, of course, if A and B were numbers, but they’re not; they’re operators. In fact, if you’re in EM this semester, you’ve probably already run into an example of something like this: the cross product. If instead we have two vectors a and b, then when you take their cross product the order matters quite a bit, since a x b is not the same as b x a.
Anyway, In the homework this week we’ll look at examples of two operators that have a nonvanishing commutator, but for now you’ll just have to take my word for it that the commutator is not always zero. (By the way, to use one more bit of jargon, if the commutator of A and B does happen to be zero, then A and B are said to “commute”.)
Once we compute the commutator of two operators, we have a mathematical way of determining whether their observables are compatible (that is, can be measured at the same time): if the two operators do commute, then their observables are compatible, and vice versa. (If they don’t commute, then their observables aren’t compatible.)
As an example, the energy and momentum of a free particle are represented by the operators H and p, respectively. As it turns out, these two operators commute ([H, p] = 0), so it’s possible to measure a particle’s energy without damaging what you know about that particle’s momentum, because the two quantities are compatible.
What about an example of operators that don’t commute? I’m glad you asked, because that brings us to this week’s homework. (As always, if you get stuck, feel free to ask me for help.)
https://mywebspace.wisc.edu/mweinberg/web/commutation.pdf
Thursday, November 29, 2007
Bell's theorem
Hey folks,
For those of you who are interested, I went ahead and wrote up the proof for the "orthodox" position. Let me know what you think.
https://mywebspace.wisc.edu/mweinberg/web/BellTheorem.pdf
For those of you who are interested, I went ahead and wrote up the proof for the "orthodox" position. Let me know what you think.
https://mywebspace.wisc.edu/mweinberg/web/BellTheorem.pdf
Monday, November 26, 2007
The Schrodinger equation
I just know you’ve been in suspense for two weeks now, so I think it’s time I told you which of the three viewpoints (realist, orthodox, or agnostic) is correct. It’s worth mentioning that only recently did physicists discover there even was a right answer to this question. (Well, not that recently; the news is over forty years old by now.) Before the 60s you could basically pick whichever viewpoint appealed most to you, and each one had a large group of fans.
In 1964 though, an Irish physicist named John Bell demonstrated that it makes a real, observable difference whether the particle had a definite (though unknown) position before the measurement. As Griffiths puts it: “Bell’s discovery effectively eliminated agnosticism as a viable option, and made it an experimental question whether the realist or orthodox position is the correct choice.” So if any of you picked agnostic thinking it was the safest route, you might be surprised to learn it was the first one to bite the dust.
I’ve struggled a lot with the question of whether to show you Bell’s proof, but I finally decided against it. I realize this probably means that Taylor will never believe me, but unfortunately the proof requires a fair bit of background information, as well as being somewhat complicated in its own right.
But let’s say for the sake of argument that you believe me, and agree that there is a right answer. So what is it? Alright, fine, I’ll tell you: it turns out that the correct position is… the orthodox one. Believe it or not, the particle did not have a position before you measured it. This is why quantum mechanics is so insanely weird. Before the measurement, that particle wasn’t entirely anywhere; however, it was somewhat everywhere. It’s a safe bet that no human on earth can really “comprehend” this, and it’s difficult to even communicate this fact through language, but there it is anyway: particles just don’t have positions.
And this isn’t just true of positions! Get down small enough, and you’ll find that particles don’t have definite momenta either, or angular momenta, or energy, or anything else that can be measured. The act of measurement gives these things definite values. If you thought relativity was weird, get a load of that.
In the comments last week, Taylor made an extremely good point. The complaint was this: suppose you measure a particle’s position and find it was at point P (from the graph in the “wave function” blog post). Then, immediately after that, you make a second measurement. Who’s to say you won’t find it far away in, say, Atlanta? If you make the measurement fast enough, wouldn’t the particle have had to travel faster than the speed of light?
That’s a legitimate objection, and if quantum mechanics allowed things like that to happen, we’d have a real problem. But wait! We’re now asking a very different question! We’re now asking what happens when we perform a second measurement on the same particle. The trick is, the first measurement changes the particle’s wave function, so now it looks like this:
In 1964 though, an Irish physicist named John Bell demonstrated that it makes a real, observable difference whether the particle had a definite (though unknown) position before the measurement. As Griffiths puts it: “Bell’s discovery effectively eliminated agnosticism as a viable option, and made it an experimental question whether the realist or orthodox position is the correct choice.” So if any of you picked agnostic thinking it was the safest route, you might be surprised to learn it was the first one to bite the dust.
I’ve struggled a lot with the question of whether to show you Bell’s proof, but I finally decided against it. I realize this probably means that Taylor will never believe me, but unfortunately the proof requires a fair bit of background information, as well as being somewhat complicated in its own right.
But let’s say for the sake of argument that you believe me, and agree that there is a right answer. So what is it? Alright, fine, I’ll tell you: it turns out that the correct position is… the orthodox one. Believe it or not, the particle did not have a position before you measured it. This is why quantum mechanics is so insanely weird. Before the measurement, that particle wasn’t entirely anywhere; however, it was somewhat everywhere. It’s a safe bet that no human on earth can really “comprehend” this, and it’s difficult to even communicate this fact through language, but there it is anyway: particles just don’t have positions.
And this isn’t just true of positions! Get down small enough, and you’ll find that particles don’t have definite momenta either, or angular momenta, or energy, or anything else that can be measured. The act of measurement gives these things definite values. If you thought relativity was weird, get a load of that.
In the comments last week, Taylor made an extremely good point. The complaint was this: suppose you measure a particle’s position and find it was at point P (from the graph in the “wave function” blog post). Then, immediately after that, you make a second measurement. Who’s to say you won’t find it far away in, say, Atlanta? If you make the measurement fast enough, wouldn’t the particle have had to travel faster than the speed of light?
That’s a legitimate objection, and if quantum mechanics allowed things like that to happen, we’d have a real problem. But wait! We’re now asking a very different question! We’re now asking what happens when we perform a second measurement on the same particle. The trick is, the first measurement changes the particle’s wave function, so now it looks like this:
As you can see, there’s now zero chance of finding the particle anywhere except at point P. So if you make a second measurement (and third, forth, etc.) you will continue to get the same value back. That is, before the first measurement there was no telling where it would be, but once you’ve made that measurement, any subsequent measurement will give the same value. (To throw around a bit of jargon, the first measurement is often said to “collapse” the wave function into the function above, which is called a “Dirac delta function”.)
Alright, time to get back to work. In the “wave function” blog post I did a sort of comparison between classical and quantum mechanics. In both cases, we consider a particle of mass m, and for simplicity, we only let it move in the x-direction. Then we subject it to some kind of force, F(x, t), and we’re ready to roll. In classical mechanics, the goal is to figure out the particle’s position at any particular time, x(t), and in quantum mechanics (where particles often don’t have positions), the goal is to figure out the particle’s wave function at any particular time and place, Ψ(x, t).
But how do you get these in the first place? In classical mechanics, the answer is easy: that’s what Newton’s second law is for! Once you know that F = ma, your worries are pretty much over. Sometimes it’s a teeny bit more complicated, because you’ll be given a potential energy instead of a force. But that’s no big deal: you know from the definition of work that F = -dU/dx (in this case, I’m using U to mean potential energy). If we put this in, Newton’s second law reads: md2x/dt2 = -dU/dx. Plug stuff in with the right initial conditions and you’re done.
So that’s how you get x(t) in classical mechanics, but how do you get Ψ(x, t) in quantum mechanics? As it turns out, we get it by solving something called the Schrodinger equation. So the Schrodinger equation does the exact same thing in quantum mechanics that Newton’s second law does in classical mechanics. This may not seem like a big point, but I went through an entire year of quantum mechanics before anyone bothered to tell me this. So in case you ever take a course in QM, maybe this will save you some grief.
Anyway, maybe you’re wondering why I haven’t actually shown you what the Schrodinger equation looks like yet. That’s because it’s your job to tell me, which brings us to this week’s homework.
Alright, time to get back to work. In the “wave function” blog post I did a sort of comparison between classical and quantum mechanics. In both cases, we consider a particle of mass m, and for simplicity, we only let it move in the x-direction. Then we subject it to some kind of force, F(x, t), and we’re ready to roll. In classical mechanics, the goal is to figure out the particle’s position at any particular time, x(t), and in quantum mechanics (where particles often don’t have positions), the goal is to figure out the particle’s wave function at any particular time and place, Ψ(x, t).
But how do you get these in the first place? In classical mechanics, the answer is easy: that’s what Newton’s second law is for! Once you know that F = ma, your worries are pretty much over. Sometimes it’s a teeny bit more complicated, because you’ll be given a potential energy instead of a force. But that’s no big deal: you know from the definition of work that F = -dU/dx (in this case, I’m using U to mean potential energy). If we put this in, Newton’s second law reads: md2x/dt2 = -dU/dx. Plug stuff in with the right initial conditions and you’re done.
So that’s how you get x(t) in classical mechanics, but how do you get Ψ(x, t) in quantum mechanics? As it turns out, we get it by solving something called the Schrodinger equation. So the Schrodinger equation does the exact same thing in quantum mechanics that Newton’s second law does in classical mechanics. This may not seem like a big point, but I went through an entire year of quantum mechanics before anyone bothered to tell me this. So in case you ever take a course in QM, maybe this will save you some grief.
Anyway, maybe you’re wondering why I haven’t actually shown you what the Schrodinger equation looks like yet. That’s because it’s your job to tell me, which brings us to this week’s homework.
Thursday, November 8, 2007
The wave function
Okay, now I’m going to sum up in one paragraph almost every problem you can get in your mechanics class. Let’s say we find a particle of mass m and we let it move back and forth in the x-direction. While we’re at it, I’m going to apply a force to the particle, F(x, t). Basically the whole idea of classical mechanics is to figure out, given that force, what the particle’s position will be at any particular time, x(t). Pretty much everything else follows from that. Yours for the asking are the velocity, dx/dt; the momentum, mv; the kinetic energy, (1/2)mv2; or whatever else Rebecca decides to ask you for.
The point of this is simply to give you something to compare quantum mechanics to, to help you get oriented. So let’s compare: suppose we have the exact same setup (a particle of mass m moving in the x-direction) in quantum mechanics. The problem is that at the microscopic level, particles don’t necessarily have positions, so it doesn’t make sense to ask about x(t). Instead, what we’d like to find out is something called the wave function of the particle: Ψ(x, t). This is the interesting thing, and once we’ve got it we can figure out anything else we want.
So the wave function, Ψ, does much the same thing for you in quantum mechanics that the position, x, does for you in classical mechanics. It’s the big thing, from which pretty much anything else can be figured out. You say you want to know about the particle’s momentum? Simply apply the momentum operator to Ψ and see what happens. The kinetic energy? We’ve got an operator for that, too. Potential energy? Oh, jeez, we’ve got dozens of operators for that. (Of course, you can’t actually apply any of these operators yet, because I haven’t actually told you what they are.)
Alright, so that’s what the wave function does, but what exactly is it? The best answer was provided by a German dude named Max Born, who provided the “statistical interpretation”:
If Ψ(x, t) is the wave function of a particle, then Ψ2 is the probability of finding the particle at point x, at time t.
To be a teeny bit more mathy about it:
∫ab Ψ(x, t)2 dx = {probability of finding the particle between a and b, at time t}
That is, if you’ve got the wave function of a particle Ψ, and you’d like to know the probability of finding it between two particular points, just take the area under the curve of Ψ2 from a to b. For example, the picture is a graph of Ψ2, so we would be much more likely to find the particle near the middle than out at the edges.
If you’re thinking all of this is a little bit hokey, I can hardly blame you. What kind of a theory is this? We get the wave function, the most important thing we can learn about this particle, and the best we can do with it is find the probability that a particle will be in a certain place? Why can’t we say beforehand exactly where it’s going to be when we measure it? Is there just something wrong with quantum mechanics?
Actually, all this can be summed up in one question. Let’s say a particle has the wave function in the picture, and I actually do measure the particle’s position. It so happens that I discover the particle was somewhere right around point P. The question is: where was the particle the instant before we measured it?
There are three general opinions you might have on this (courtesy of David Griffiths):
1. The realist position: “Well, obviously, the particle had to have been at point P. That’s where we found it after all. If you find your socks under your bed, clearly they were under you bed right before you found them, you just didn’t know it.” It certainly seems hard to argue with this logic, and Einstein himself took this position. If it’s true, the theory of quantum mechanics is a bit of a letdown, because it must be incomplete. That is, the particle had a position before we measured it, and quantum mechanics just couldn’t tell us what it was.
2. The orthodox position: “Believe it or not, the particle wasn’t really anywhere until you measured it. The measurement itself caused the particle to have a position.” So how about that? In addition to being really, really weird, this claim requires us to believe some pretty farfetched things. In particular, it means there really is such a thing as a “random” event: it was not possible, even in principle, to know where the particle was going to be before you measured it. Even if you knew every true thing in the entire universe, you still could not have said where that particle would be.
If this surprises you, maybe you can relate to this: when I was a kid, I thought everything in the universe was completely determined. A coin flip, for instance, only seems random; if you knew exactly how it was flipped, and exactly how the air moved around it, you could figure out whether it would come up heads or tails every time. A computer claims to generate “random” numbers, but it really just takes a seed number from its internal clock and performs calculations on it. Even humans would just be machines, following deterministic paths (though they might think they had free will). In short, I figured if you knew all the starting conditions of the universe, you could figure out everything that would happen, from a coin flip to what a person would say at a particular moment. However, if the orthodox position is right, even with a perfect knowledge of the universe, there are some things that can’t possibly be known beforehand.
3. The agnostic position: “I refuse to answer the question. I don’t want to, and you can’t make me.” Actually, this isn’t as ridiculous as it sounds. The point is, it makes no sense to ask about where the particle was before a measurement: the only way to know that would be to make a measurement! In that case, it wouldn’t be before the measurement anymore, would it? It’s just goofy philosophy to ask about something that can’t, by definition, be known. It’s like asking “what if time suddenly started to run at half-speed?” There’d be no way to check. It may sound like you’re describing two different things, but you’re actually just describing the same thing in two different ways.
This question bugged me for years, but I always thought it was basically a matter of opinion. Believe it or not, though, there actually is a correct answer! Incredibly, it makes an observable, testable difference which position is right. I’ll give it away next week, but I’d like to know: what do you think? Position 1, 2, or 3? Or do you have a different position altogether? Write a comment and let me know.
In the meantime, here's the problem for this week. Feel free to write me if you get stuck, too.
https://mywebspace.wisc.edu/mweinberg/web/waveFunction.pdf
The point of this is simply to give you something to compare quantum mechanics to, to help you get oriented. So let’s compare: suppose we have the exact same setup (a particle of mass m moving in the x-direction) in quantum mechanics. The problem is that at the microscopic level, particles don’t necessarily have positions, so it doesn’t make sense to ask about x(t). Instead, what we’d like to find out is something called the wave function of the particle: Ψ(x, t). This is the interesting thing, and once we’ve got it we can figure out anything else we want.
So the wave function, Ψ, does much the same thing for you in quantum mechanics that the position, x, does for you in classical mechanics. It’s the big thing, from which pretty much anything else can be figured out. You say you want to know about the particle’s momentum? Simply apply the momentum operator to Ψ and see what happens. The kinetic energy? We’ve got an operator for that, too. Potential energy? Oh, jeez, we’ve got dozens of operators for that. (Of course, you can’t actually apply any of these operators yet, because I haven’t actually told you what they are.)
Alright, so that’s what the wave function does, but what exactly is it? The best answer was provided by a German dude named Max Born, who provided the “statistical interpretation”:
If Ψ(x, t) is the wave function of a particle, then Ψ2 is the probability of finding the particle at point x, at time t.
To be a teeny bit more mathy about it:
∫ab Ψ(x, t)2 dx = {probability of finding the particle between a and b, at time t}
That is, if you’ve got the wave function of a particle Ψ, and you’d like to know the probability of finding it between two particular points, just take the area under the curve of Ψ2 from a to b. For example, the picture is a graph of Ψ2, so we would be much more likely to find the particle near the middle than out at the edges.
If you’re thinking all of this is a little bit hokey, I can hardly blame you. What kind of a theory is this? We get the wave function, the most important thing we can learn about this particle, and the best we can do with it is find the probability that a particle will be in a certain place? Why can’t we say beforehand exactly where it’s going to be when we measure it? Is there just something wrong with quantum mechanics?
Actually, all this can be summed up in one question. Let’s say a particle has the wave function in the picture, and I actually do measure the particle’s position. It so happens that I discover the particle was somewhere right around point P. The question is: where was the particle the instant before we measured it?
There are three general opinions you might have on this (courtesy of David Griffiths):
1. The realist position: “Well, obviously, the particle had to have been at point P. That’s where we found it after all. If you find your socks under your bed, clearly they were under you bed right before you found them, you just didn’t know it.” It certainly seems hard to argue with this logic, and Einstein himself took this position. If it’s true, the theory of quantum mechanics is a bit of a letdown, because it must be incomplete. That is, the particle had a position before we measured it, and quantum mechanics just couldn’t tell us what it was.
2. The orthodox position: “Believe it or not, the particle wasn’t really anywhere until you measured it. The measurement itself caused the particle to have a position.” So how about that? In addition to being really, really weird, this claim requires us to believe some pretty farfetched things. In particular, it means there really is such a thing as a “random” event: it was not possible, even in principle, to know where the particle was going to be before you measured it. Even if you knew every true thing in the entire universe, you still could not have said where that particle would be.
If this surprises you, maybe you can relate to this: when I was a kid, I thought everything in the universe was completely determined. A coin flip, for instance, only seems random; if you knew exactly how it was flipped, and exactly how the air moved around it, you could figure out whether it would come up heads or tails every time. A computer claims to generate “random” numbers, but it really just takes a seed number from its internal clock and performs calculations on it. Even humans would just be machines, following deterministic paths (though they might think they had free will). In short, I figured if you knew all the starting conditions of the universe, you could figure out everything that would happen, from a coin flip to what a person would say at a particular moment. However, if the orthodox position is right, even with a perfect knowledge of the universe, there are some things that can’t possibly be known beforehand.
3. The agnostic position: “I refuse to answer the question. I don’t want to, and you can’t make me.” Actually, this isn’t as ridiculous as it sounds. The point is, it makes no sense to ask about where the particle was before a measurement: the only way to know that would be to make a measurement! In that case, it wouldn’t be before the measurement anymore, would it? It’s just goofy philosophy to ask about something that can’t, by definition, be known. It’s like asking “what if time suddenly started to run at half-speed?” There’d be no way to check. It may sound like you’re describing two different things, but you’re actually just describing the same thing in two different ways.
This question bugged me for years, but I always thought it was basically a matter of opinion. Believe it or not, though, there actually is a correct answer! Incredibly, it makes an observable, testable difference which position is right. I’ll give it away next week, but I’d like to know: what do you think? Position 1, 2, or 3? Or do you have a different position altogether? Write a comment and let me know.
In the meantime, here's the problem for this week. Feel free to write me if you get stuck, too.
https://mywebspace.wisc.edu/mweinberg/web/waveFunction.pdf
Friday, October 26, 2007
Quantum mechanics
Offhand, I can’t think of any physics theory that was discovered just by one person. Often, though, there’s one go-to guy who wrapped the whole thing up in a paper somewhere, and whenever we think of that theory we think of them. Newton’s mechanics, for example, or Maxwell’s electrodynamics, or even Einstein’s relativity. But quantum mechanics isn’t like that; it was put together by a collection of extremely clever folks over the span of maybe twenty years or so. Many of them didn’t even like each other. Heisenberg, for example, couldn’t stand Schrodinger’s work, even though it turned out they were doing the same thing.
As a result, even today no two physicists completely agree on what quantum mechanics is. No one knows what its fundamental principles are, or how it should be taught. (Although a friend of mine likes to say that “quantum mechanics is just Heisenberg’s uncertainty plus Pauli’s exclusion.” I’m not sure what that means, but I like it because it’s short and snappy.) I heard a story once about a kid who misses the very first day of kindergarten, when the teacher explains what “numbers” are. He gets there the second day and everyone already seems to know, so he’s too embarrassed to ask. He doesn’t want anyone to catch on that he’s got no idea what a number is, so he works incredibly hard. He learns arithmetic and algebra, and later trigonometry and calculus. Eventually he becomes a famous mathematician and wins the Fields Medal. But inside he always feels like a fraud, because the truth is he still doesn’t know what a number is, just how it works. That’s sort of how I feel when I “do” quantum mechanics.
Over the next few weeks, I’d like to crack the QM door and let you peek in. Unfortunately, a peek is really all that’s possible, because there’s a fair amount of math that goes along with the subject. Things like spherical harmonics, Fourier transforms, and Hilbert spaces, along with a host of others. Probably you’ve never heard of any of that stuff before, but you’ll run into some of it if you’re interested in technical careers like engineering or the sciences.
Okay, so in my opinion, the very first thing to know about quantum mechanics is the idea of operators. In physics there are all kinds of observable quantities that might interest us; things like position, velocity, momentum, angular momentum, potential energy, kinetic energy, and so forth. In classical mechanics (what you’ve been studying) these quantities are just variables, which hopefully you’re familiar with by now. In quantum mechanics, each of these quantities gets an upgrade—they’re turned into operators.
So what is an operator? Well, I suppose it doesn’t help to say they operate on something, so maybe an example is the way to go here. Let’s say I invent an operator called D. What D does is take the derivative of something with respect to x. That is,
D = d/dx
Sitting out there, all by itself, D doesn’t mean very much. It’s a derivative, sure, but of what? To be useful, D has to operate on something, say a function f(x). Suppose f(x) is something simple, like, oh, 3x4. If we let D operate on it, we get
Df(x) = 12x3
I just made up the operator D, but I should actually mention that it’s not the first operator you’ve run into. In trigonometry, things like sine and cosine are operators. Obviously a “cos” just sitting around doesn’t mean anything, but everyone’s seen cos θ. It doesn’t mean you should multiply θ by cos, rather the cosine operates on θ.
Groovy? Hopefully you’re happy with this, but if not you can always write me with questions. While we’re at it, I should also mention a couple of pitfalls so you’ll know to avoid them. Let’s go back to my made up operator D again. What if this time it operates on the function g(x) = e2x? In that case, we get
Dg = 2e2x = 2g
Does this mean we can cancel out the g’s and conclude that D = 2? Of course not! Remember, we’re not multiplying by g here, we’re operating on g. In this case, g was an exponential, so we got it back when we used D on it. If you’re still not convinced, consider this equation:
cos θ = sin θ
Is it okay to cancel out the θ’s and say cos = sin? Nope.
Here’s another thing you should be aware of. What does it mean if I have D2? Should you take the derivative of something and then square it? You might think so, but that’s not it. When I say D2, I’m really saying you should apply the operator D twice. For example, going back to the f(x) from earlier,
D2f = DDf = D(Df) = D(12x3) = 36x2
And there you go. To be a bit more explicit, since D is a derivative, D2 is a second derivative. (If you applied them to a particle’s position, D would give the velocity, and D2 would give the acceleration.) Armed with this knowledge, you should be more than a match for this week’s homework.
https://mywebspace.wisc.edu/mweinberg/web/Operators.pdf?uniq=-9cdw1b
As a result, even today no two physicists completely agree on what quantum mechanics is. No one knows what its fundamental principles are, or how it should be taught. (Although a friend of mine likes to say that “quantum mechanics is just Heisenberg’s uncertainty plus Pauli’s exclusion.” I’m not sure what that means, but I like it because it’s short and snappy.) I heard a story once about a kid who misses the very first day of kindergarten, when the teacher explains what “numbers” are. He gets there the second day and everyone already seems to know, so he’s too embarrassed to ask. He doesn’t want anyone to catch on that he’s got no idea what a number is, so he works incredibly hard. He learns arithmetic and algebra, and later trigonometry and calculus. Eventually he becomes a famous mathematician and wins the Fields Medal. But inside he always feels like a fraud, because the truth is he still doesn’t know what a number is, just how it works. That’s sort of how I feel when I “do” quantum mechanics.
Over the next few weeks, I’d like to crack the QM door and let you peek in. Unfortunately, a peek is really all that’s possible, because there’s a fair amount of math that goes along with the subject. Things like spherical harmonics, Fourier transforms, and Hilbert spaces, along with a host of others. Probably you’ve never heard of any of that stuff before, but you’ll run into some of it if you’re interested in technical careers like engineering or the sciences.
Okay, so in my opinion, the very first thing to know about quantum mechanics is the idea of operators. In physics there are all kinds of observable quantities that might interest us; things like position, velocity, momentum, angular momentum, potential energy, kinetic energy, and so forth. In classical mechanics (what you’ve been studying) these quantities are just variables, which hopefully you’re familiar with by now. In quantum mechanics, each of these quantities gets an upgrade—they’re turned into operators.
So what is an operator? Well, I suppose it doesn’t help to say they operate on something, so maybe an example is the way to go here. Let’s say I invent an operator called D. What D does is take the derivative of something with respect to x. That is,
D = d/dx
Sitting out there, all by itself, D doesn’t mean very much. It’s a derivative, sure, but of what? To be useful, D has to operate on something, say a function f(x). Suppose f(x) is something simple, like, oh, 3x4. If we let D operate on it, we get
Df(x) = 12x3
I just made up the operator D, but I should actually mention that it’s not the first operator you’ve run into. In trigonometry, things like sine and cosine are operators. Obviously a “cos” just sitting around doesn’t mean anything, but everyone’s seen cos θ. It doesn’t mean you should multiply θ by cos, rather the cosine operates on θ.
Groovy? Hopefully you’re happy with this, but if not you can always write me with questions. While we’re at it, I should also mention a couple of pitfalls so you’ll know to avoid them. Let’s go back to my made up operator D again. What if this time it operates on the function g(x) = e2x? In that case, we get
Dg = 2e2x = 2g
Does this mean we can cancel out the g’s and conclude that D = 2? Of course not! Remember, we’re not multiplying by g here, we’re operating on g. In this case, g was an exponential, so we got it back when we used D on it. If you’re still not convinced, consider this equation:
cos θ = sin θ
Is it okay to cancel out the θ’s and say cos = sin? Nope.
Here’s another thing you should be aware of. What does it mean if I have D2? Should you take the derivative of something and then square it? You might think so, but that’s not it. When I say D2, I’m really saying you should apply the operator D twice. For example, going back to the f(x) from earlier,
D2f = DDf = D(Df) = D(12x3) = 36x2
And there you go. To be a bit more explicit, since D is a derivative, D2 is a second derivative. (If you applied them to a particle’s position, D would give the velocity, and D2 would give the acceleration.) Armed with this knowledge, you should be more than a match for this week’s homework.
https://mywebspace.wisc.edu/mweinberg/web/Operators.pdf?uniq=-9cdw1b
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