*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

**and**

*a***, then when you take their cross product the order matters quite a bit, since**

*b***x**

*a***is not the same as**

*b***x**

*b***.**

*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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