# 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.

• Cart before the horse in my opinion, as QFT is based on operators acting on a ground state, and that ground state has to be a solution of a quantum mechanical equation, Dirac or KG. I would rather title it "proving consistency" with underlying quantum dynamical framework Dec 13, 2015 at 6:59
• Comment to the post (v2): It would be good if OP (or somebody else?) could try to make the question formulation self-contained, so one doesn't have to open the link to understand the question. Dec 13, 2015 at 13:04

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{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

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{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}

These relations imply that $$[\psi^\dagger(\boldsymbol x),\psi^\dagger(\boldsymbol y)]=0$$ and \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

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{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}

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$}$