### Responsibility

Many years ago, when I was an assistant professor at Princeton, there was a cocktail party at Curt Callan’s house to mark the beginning of the semester. There, I found myself in the kitchen, chatting with Sacha Polyakov. I asked him what he was going to be teaching that semester, and he replied that he was very nervous because — for the first time in his life — he would be teaching an undergraduate course. After my initial surprise that he had gotten this far in life without ever having taught an undergraduate course, I asked which course it was. He said it was the advanced undergraduate Mechanics course (chaos, etc.) and we agreed that would be a fun subject to teach. We chatted some more, and then he said that, on reflection, he probably shouldn’t be quite so worried. After all, it wasn’t as if he was going to teach Quantum Field Theory, “That’s a subject I’d feel *responsible* for.”

This remark stuck with me, but it never seemed quite so poignant until this semester, when I find myself teaching the undergraduate particle physics course.

The textbooks (and I mean *all* of them) start off by “explaining” that relativistic quantum mechanics (e.g. replacing the Schrödinger equation with Klein-Gordon) make no sense (negative probabilities and all that …). And they then proceed to use it anyway (supplemented by some Feynman rules pulled out of thin air).

This drives me up the #@%^ing wall. It is *precisely* wrong.

There is a *perfectly* consistent quantum mechanical theory of free particles. The *problem* arises when you want to introduce interactions. In Special Relativity, there is no interaction-at-a-distance; all forces are necessarily mediated by fields. Those fields fluctuate and, when you want to study the quantum theory, you end up having to quantize them.

But the free particle is just fine. Of course it has to be: free field theory is just the theory of an (indefinite number of) free particles. So it better be true that the quantum theory of a single relativistic free particle makes sense.

So what is that theory?

- It has a Hilbert space, $\mathcal{H}$, of states. To make the action of Lorentz transformations as simple as possible, it behoves us to use a Lorentz-invariant inner product on that Hilbert space. This is most easily done in the momentum representation $\langle\chi|\phi\rangle = \int \frac{d^3\vec{k}}{{(2\pi)}^3 2\sqrt{\vec{k}^2+m^2}}\, \chi(\vec{k})^* \phi(\vec{k})$
- As usual, the time-evolution is given by a Schrödinger equation

where $H_0 = \sqrt{\vec{p}^2+m^2}$. Now, you might object that it is hard to make sense of a pseudo-differential operator like $H_0$. Perhaps. But it’s not *any* harder than making sense of $U(t)= e^{-i \vec{p}^2 t/2m}$, which we routinely pretend to do in elementary quantum. In both cases, we use the fact that, in the momentum representation, the operator $\vec{p}$ is represented as multiplication by $\vec{k}$.

I could go on, but let me leave the rest of the development of the theory as a series of questions.

- The self-adjoint operator, $\vec{x}$, satisfies $[x^i,p_j] = i \delta^{i}_j$ Thus it can be written in the form $x^i = i\left(\frac{\partial}{\partial k_i} + f_i(\vec{k})\right)$ for some real function $f_i$. What is $f_i(\vec{k})$?
- Define $J^0(\vec{r})$ to be the probability density. That is, when the particle is in state $|\phi\rangle$, the probability for finding it in some Borel subset $S\subset\mathbb{R}^3$ is given by $\text{Prob}(S) = \int_S d^3\vec{r} J^0(\vec{r})$ Obviously, $J^0(\vec{r})$ must take the form $J^0(\vec{r}) = \int\frac{d^3\vec{k}d^3\vec{k}'}{{(2\pi)}^6 4\sqrt{\vec{k}^2+m^2}\sqrt{{\vec{k}'}^2+m^2}} g(\vec{k},\vec{k}') e^{i(\vec{k}-\vec{k'})\cdot\vec{r}}\phi(\vec{k})\phi(\vec{k}')^*$ Find $g(\vec{k},\vec{k}')$. (Hint: you need to diagonalize the operator $\vec{x}$ that you found in problem 1.)
- The conservation of probability says $0=\partial_t J^0 + \partial_i J^i$ Use the Schrödinger equation (1) to find $J^i(\vec{r})$.
- Under Lorentz transformations, $H_0$ and $\vec{p}$ transform as the components of a 4-vector. For a boost in the $z$-direction, of rapidity $\lambda$, we should have
$\begin{split}
U_\lambda \sqrt{\vec{p}^2+m^2} U_\lambda^{-1} &= \cosh(\lambda) \sqrt{\vec{p}^2+m^2} + \sinh(\lambda) p_3\\
U_\lambda p_1 U_\lambda^{-1} &= p_1\\
U_\lambda p_2 U_\lambda^{-1} &= p_3\\
U_\lambda p_3 U_\lambda^{-1} &= \sinh(\lambda) \sqrt{\vec{p}^2+m^2} + \cosh(\lambda) p_3
\end{split}$
and we should be able to write $U_\lambda = e^{i\lambda B}$ for some self-adjoint operator, $B$. What is $B$? (N.B.: by contrast the $x^i$, introduced above, do
*not*transform in a simple way under Lorentz transformations.)

The Hilbert space of a free scalar field is now $\bigoplus_{n=0}^\infty \text{Sym}^n\mathcal{H}$. That’s perhaps not the easiest way to get there. But it is a way …

#### Update:

Yike! Well, that went south pretty fast. For the first time (ever, I think) I’m closing comments on this one, and calling it a day. To summarize, for those who still care,

- There is a decomposition of the Hilbert space of a Free Scalar field as $\mathcal{H}_\phi = \bigoplus_{n=0}^\infty \mathcal{H}_n$ where $\mathcal{H}_n = \text{Sym}^n \mathcal{H}$ and $\mathcal{H}$ is 1-particle Hilbert space described above (also known as the spin-$0$, mass-$m$, irreducible unitary representation of Poincaré).
- The Hamiltonian of the Free Scalar field is the direct sum of the induced Hamiltonia on $\mathcal{H}_n$, induced from the Hamiltonian, $H=\sqrt{\vec{p}^2+m^2}$, on $\mathcal{H}$. In particular, it (along with the other Poincaré generators) is block-diagonal with respect to this decomposition.
- There are other interesting observables which are also block-diagonal, with respect to this decomposition (i.e., don’t change the particle number) and hence we can discuss their restriction to $\mathcal{H}_n$.

Gotta keep reminding myself why I decided to foreswear blogging…

## Re: Responsibility

What book would you recommend for undergraduate particle physics and similar level/beginner QFT?