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
Monday, December 17, 2007
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
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