# The operation of scalar field $\phi (\vec x)$ on vacuum state

I am now learning the Quantum Field Theory by reading the lecture notes by David Tong.

I have some question about the mode expansion about the real scalar field that is canonically quantized by promoting the classical Klein Gordon field to a quantum field.

The mode expansion of the field is given by

$$\phi (\vec x) = \int \frac{d^{3}p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_{\vec p}}} (a_{\vec p} e^{i {\vec p} \cdot {\vec x}} + a^{\dagger}_{\vec p} e^{-i {\vec p} \cdot {\vec x}})$$

where $\omega_{\vec p} = \sqrt{p^{2} + m^2}$ and $a^{\dagger}_{\vec p}$ will creat a spin $0$ particle in the momentum state $\left| \vec p \right\rangle$, namely $a^{\dagger}_{\vec p} \left| {0} \right\rangle = \left| {\vec p} \right\rangle$.

The question I am now curious about is about what will I get if I operate the quantum field on the vacuum state, that is $\phi (\vec x) \left| {0} \right\rangle = ?$

It seems that in the lecture note $\phi (\vec x) \left| {0} \right\rangle = \left| {\vec x} \right\rangle$, where $\left| {\vec x} \right\rangle$ is the spin $0$ particle in position state at $\vec x$.

(eq 2.52 in http://www.damtp.cam.ac.uk/user/tong/qft/two.pdf)

However this seems nontrivial to me so I carried out the following derivation.

$$\phi (\vec x) \left| {0} \right\rangle = \int \frac{d^{3}p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_{\vec p}}} (a_{\vec p} e^{i {\vec p} \cdot {\vec x}} + a^{\dagger}_{\vec p} e^{-i {\vec p} \cdot {\vec x}})\left| {0} \right\rangle$$ $$= \int \frac{d^{3}p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_{\vec p}}} (\left| {\vec p} \right\rangle e^{-i {\vec p} \cdot {\vec x}})$$

However, I have no idea how to prove that $$\int \frac{d^{3}p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_{\vec p}}} (\left| {\vec p} \right\rangle e^{-i {\vec p} \cdot {\vec x}}) = \left| {\vec x} \right\rangle$$

I know that in elementary quantum mechanics we have $$\left| {\vec x} \right\rangle = \int d^{3}p \left| {\vec p} \right\rangle \left\langle {\vec p} \right| \cdot \left| {\vec x} \right\rangle$$ However this doesn't resemble what I want to prove.

It seems a stupid question but I was just stuck on it.

I would be grateful for any suggestion!

• It looks like you have basically shown it, you just need to recognise that this is a fourier transform from $x$ to $p$ space Apr 23, 2018 at 13:14
• related: Recovering QM from QFT. Apr 23, 2018 at 13:34
• Thanks, but there is a $\omega_{\vec p}$ in the integral, and I am not sure how to eliminate it to actually get $\left| {\vec x} \right\rangle$. Apr 23, 2018 at 13:47
• The question is properly answered below, but I would just say you can see equation 2.62 onwards for Tong's normalizations. Apr 23, 2018 at 15:22

## 2 Answers

You almost did everything except the relativistic considerations, which are not obvious of course.

The factor $\sqrt{\omega}$ is the relativistic normalization for the eigenstates, and also you need to pay attention to the measure, $d^3 p$. This kind of normalization comes from the fact that the eigenstates and the measure should be both Lorentz invariant individually, eventhough the whole integral is invariant.

Therefore, you can check that $\frac{d^3 p}{2 \omega}$ and $\sqrt{\omega} | p \rangle$ indeed will be Lorentz invariants. Since the normalization of the position eigenstate is $\frac{1}{\sqrt{\omega}} | x\rangle$, there remains only a $\sqrt{\omega}$ in the denominator.

Sometimes there is a convention where the normalization done on the creation and annihilation operators, instead of the eigenstates. It is better to check your textbook which convention it follows, or your own notation, in order not to confuse the definitions and get some weird results in your calculations.

you can proove that if you use the fourier transform from x to impulse space.

• Thanks, but there is a $\omega_{\vec p}$ in the integral, and I am not sure how to eliminate it to actually get $\left| {\vec x} \right\rangle$. Apr 23, 2018 at 13:59