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,

*3x*. If we let

^{4}*D*operate on it, we get

*Df(x) = 12x*

^{3}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) = e*? In that case, we get

^{2x}*Dg = 2e*

^{2x}= 2gDoes 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

*D*? Should you take the derivative of something and then square it? You might think so, but that’s not it. When I say

^{2}*D*, I’m really saying you should apply the operator

^{2}*D*

*twice*. For example, going back to the f

*(x)*from earlier,

*D*

^{2}f = DDf = D(Df) = D(12x^{3}) = 36x^{2}And there you go. To be a bit more explicit, since

*D*is a derivative,

*D*is a second derivative. (If you applied them to a particle’s position,

^{2}*D*would give the velocity, and

*D*would give the acceleration.) Armed with this knowledge, you should be more than a match for this week’s homework.

^{2}https://mywebspace.wisc.edu/mweinberg/web/Operators.pdf?uniq=-9cdw1b

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