quantum mechanics - Recovering QM from QFT

Reading through David Tong lecture notes on QFT.

On pages 43-44, he recovers QM from QFT. See below link:

QFT notes by Tong

First the momentum and position operators are defined in terms of "integrals" and after considering states that are again defined in terms of integrals we see that the ket states are indeed eigen states and the eigen values are therefore position and momentum 3-vectors.

What is not clear to me is the intermediate steps of calculations not shown in the lecture notes, in particular, the computation of integrals involving operators as their integrand, to obtain the desired results.

Answer

OP is asking how to prove $\boldsymbol P|\boldsymbol p\rangle=\boldsymbol p|\boldsymbol p\rangle$ and $\boldsymbol X|\boldsymbol x\rangle=\boldsymbol x|\boldsymbol x\rangle$ where $|\boldsymbol p\rangle$ is a (free) scalar one-particle state, and $\boldsymbol P$ is the momentum operator; $|\boldsymbol x\rangle$ is a "wave packet" centred at $\boldsymbol x$ (defined below) and $\boldsymbol X$ is the "position operator" (also defined below).

PART I

Let

$$\boldsymbol P \equiv\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\, \boldsymbol p\ a_{\boldsymbol p}^\dagger a_{\boldsymbol p}$$

Using $a_{\boldsymbol p}|0\rangle=0$, its easy to see that $\boldsymbol P|0\rangle=0$, which will be useful in a moment.

The CCR are

$$[a_{\boldsymbol p},a^\dagger_{\boldsymbol q}]=(2\pi)^3\delta(\boldsymbol p-\boldsymbol q)$$ (see page 30, eq 2.20)

With this, note that

$$\begin{equation} \begin{aligned} {}[\boldsymbol P,a^\dagger_{\boldsymbol q}]&= \int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\, \boldsymbol p\ [a_{\boldsymbol p}^\dagger a_{\boldsymbol p},a^\dagger_{\boldsymbol q}]= \int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\, \boldsymbol p\ a_{\boldsymbol p}^\dagger[a_{\boldsymbol p},a^\dagger _{\boldsymbol q}]=\\ &=\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\, \boldsymbol p\ (2\pi)^3\delta(\boldsymbol p-\boldsymbol q)a_{\boldsymbol p}^\dagger=\boldsymbol q\,a_{\boldsymbol q}^\dagger \end{aligned}\tag1 \end{equation}$$

Let $|\boldsymbol p\rangle\equiv a^\dagger_{\boldsymbol p}|0\rangle$. Using $(1)$, together with the fact $\boldsymbol P|0\rangle=0$, its easy to see that

$$\boldsymbol P|\boldsymbol p\rangle=\boldsymbol Pa_\boldsymbol p^\dagger|0\rangle=\boldsymbol [\boldsymbol P,a_\boldsymbol p^\dagger]|0\rangle=\boldsymbol p a_{\boldsymbol p}^\dagger|0\rangle\equiv \boldsymbol p|\boldsymbol p\rangle$$ as required.

PART II

Let

$$\psi^\dagger(\boldsymbol x) \equiv\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\,a_{\boldsymbol p}^\dagger \mathrm e^{-i\boldsymbol p\cdot\boldsymbol x}$$

Using $a_{\boldsymbol p}|0\rangle=0$, its easy to see that $\psi(\boldsymbol x)|0\rangle=0$, which will be useful in a moment.

Note that

$$[\psi^\dagger(\boldsymbol x),a_{\boldsymbol q}^\dagger]=0$$ and $$\begin{equation} \begin{aligned} {}[\psi^\dagger(\boldsymbol x),a_{\boldsymbol q}]&=\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\,\mathrm e^{-i\boldsymbol p\cdot\boldsymbol x}[a_{\boldsymbol p}^\dagger ,a_{\boldsymbol q}]=\\ &=-\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\,\mathrm e^{-i\boldsymbol p\cdot\boldsymbol x}(2\pi)^3\delta(\boldsymbol p-\boldsymbol q)\\ &=-\mathrm e^{-i\boldsymbol q\cdot\boldsymbol x} \end{aligned} \end{equation}$$

These relations imply that

$$[\psi^\dagger(\boldsymbol x),\psi^\dagger(\boldsymbol y)]=0$$ and $$\begin{equation} \begin{aligned} {}[\psi^\dagger(\boldsymbol x),\psi(\boldsymbol y)]&=\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\,\mathrm e^{i\boldsymbol p\cdot\boldsymbol y}[\psi^\dagger(\boldsymbol x),a_{\boldsymbol p}]\\ &=-\int \frac{\mathrm d\boldsymbol p}{(2\pi)^3}\,\mathrm e^{i\boldsymbol p\cdot\boldsymbol y}\mathrm e^{-i\boldsymbol p\cdot\boldsymbol x}=-\delta(\boldsymbol x-\boldsymbol y) \end{aligned}\tag2 \end{equation}$$

Let

$$\boldsymbol X=\int\mathrm d\boldsymbol x\ \boldsymbol x\ \psi^\dagger(\boldsymbol x)\psi(\boldsymbol x)$$

First, note that $\boldsymbol X|0\rangle=0$, which is trivial to prove using $\psi(\boldsymbol x)|0\rangle=0$.

Next, using $(2)$, its easy to see that

$$\begin{equation} \begin{aligned} {}[\boldsymbol X,\psi^\dagger(\boldsymbol y)]&=\int\mathrm d\boldsymbol x\ \boldsymbol x\ [\psi^\dagger(\boldsymbol x)\psi(\boldsymbol x),\psi^\dagger(\boldsymbol y)]\\ &=\int\mathrm d\boldsymbol x\ \boldsymbol x\ \psi^\dagger(\boldsymbol x)[\psi(\boldsymbol x),\psi^\dagger(\boldsymbol y)]\\ &=\int\mathrm d\boldsymbol x\ \boldsymbol x\ \psi^\dagger(\boldsymbol x)\delta(\boldsymbol x-\boldsymbol y)=\boldsymbol y\,\psi^\dagger(\boldsymbol y) \end{aligned} \end{equation}$$

Finally, using the relation above, together with $\boldsymbol X|0\rangle=0$, its easy to see that

$$\boldsymbol X|\boldsymbol x\rangle=\boldsymbol X\psi^\dagger(\boldsymbol x)|0\rangle=[\boldsymbol X,\psi^\dagger(\boldsymbol x)]|0\rangle=\boldsymbol x\,\psi^\dagger(\boldsymbol x)|0\rangle\equiv\boldsymbol x|\boldsymbol x\rangle$$ as required.$\tag*{$\square$}$