Klein-Gordon in QFT: Understanding the KG Equation

[Silviu]

OK.That's wrong. A correct equation that you might have in mind is
$$<x|\hat{x}|\psi>=x<x|\psi>$$No it doesn't. To make the state to have a defined position, you must act with the projector $\hat{\pi}(x)=|x><x|$. The relation between the operators $\hat{x}$ and $\hat{\pi}_x$ is
$$\hat{x}=\int dx\, x \hat{\pi}(x)$$

Oh ok ok I see. So now, how can this be expressed for $\phi(x)$?

 

[Demystifier]

Oh ok ok I see. So now, how can this be expressed for $\phi(x)$?

It requires some nitpicking about notation. Let $\phi$ denote the collection of values of $\phi(x)$ at all points $x$. Then $|\phi\rangle$ is a state that has a well defined value of $\phi(x)$ at all points $x$. Let $|\Psi\rangle$ be an arbitrary state. Then we have
$$\langle\phi|\hat{\phi}(x)|\Psi\rangle=\phi(x) \langle\phi|\Psi\rangle$$
$$\hat{\pi}[\phi]=|\phi\rangle\langle\phi|$$
$$\hat{\phi}(x)=\int[d\phi]\phi(x) \hat{\pi}[\phi]$$
where
$$[d\phi]\equiv \prod_{x'} d\phi(x')$$
Introducing the wave functional
$$\Psi[\phi]=\langle\phi|\Psi\rangle$$
the probability (density) of a given configuration $\phi$ is
$$p[\phi]=|\Psi[\phi]|^2$$

 

[Silviu]

It requires some nitpicking about notation. Let $\phi$ denote the collection of values of $\phi(x)$ at all points $x$. Then $|\phi\rangle$ is a state that has a well defined value of $\phi(x)$ at all points $x$. Let $|\Psi\rangle$ be an arbitrary state. Then we have
$$\langle\phi|\hat{\phi}(x)|\Psi\rangle=\phi(x) \langle\phi|\Psi\rangle$$
$$\hat{\pi}[\phi]=|\phi\rangle\langle\phi|$$
$$\hat{\phi}(x)=\int[d\phi]\phi(x) \hat{\pi}[\phi]$$
where
$$[d\phi]\equiv \prod_{x'} d\phi(x')$$
Introducing the wave functional
$$\Psi[\phi]=\langle\phi|\Psi\rangle$$
the probability (density) of a given configuration $\phi$ is
$$p[\phi]=|\Psi[\phi]|^2$$

I am a bit confused about the first equation. If I understand it well that would imply $$\langle\phi|\hat{\phi}(x)=\phi(x) \langle\phi|$$ I can see the resemblance with $\hat{x}$ but if our state is the vacuum i.e. $\langle 0 |$ then $\phi(x)=0$ for all x (the amplitude of the field is 0 for vacuum in the case of a free field, right?). So the equation would be $$\langle 0 |\hat{\phi}(x)=0 \langle 0|=0$$. But acting with $\hat{\phi}$ on the vacuum doesn't return 0, but it returns a field with a particle at the position x, which is not vacuum anymore. What am I missing here?

 

[Demystifier]

I am a bit confused about the first equation. If I understand it well that would imply $$\langle\phi|\hat{\phi}(x)=\phi(x) \langle\phi|$$ I can see the resemblance with $\hat{x}$ but if our state is the vacuum i.e. $\langle 0 |$ then $\phi(x)=0$ for all x (the amplitude of the field is 0 for vacuum in the case of a free field, right?). So the equation would be $$\langle 0 |\hat{\phi}(x)=0 \langle 0|=0$$. But acting with $\hat{\phi}$ on the vacuum doesn't return 0, but it returns a field with a particle at the position x, which is not vacuum anymore. What am I missing here?

Take the analogy with harmonic oscillator in ordinary QM. There
$$\langle 0|\hat{x}\neq \langle 0|0$$
Similarly, in field theory
$$\langle 0|\hat{\phi}(x)\neq \langle 0|0$$
The vacuum is not a state in which position or field is vanishing. The vacuum is just a state with minimal possible energy.

 

[Silviu]

Take the analogy with harmonic oscillator in ordinary QM. There
$$\langle 0|\hat{x}\neq \langle 0|0$$
Similarly, in field theory
$$\langle 0|\hat{\phi}(x)\neq \langle 0|0$$
The vacuum is not a state in which position or field is vanishing. The vacuum is just a state with minimal possible energy.

Yes, but in the case of HO, $\langle 0|$ is an eigenstate of the energy while in this case, based on the way you wrote in the first equation in the previous post, it is an eigenstate of the field operator, isn't it? So isn't it correct to write $\langle 0|\hat{\phi}(x) = \langle 0|0$?

 

[Demystifier]

Yes, but in the case of HO, $\langle 0|$ is an eigenstate of the energy while in this case, based on the way you wrote in the first equation in the previous post, it is an eigenstate of the field operator, isn't it? So isn't it correct to write $\langle 0|\hat{\phi}(x) = \langle 0|0$?

Well, the notation is perhaps somewhat confusing, but no, $\langle 0|$ does not denote an eigenstate of the field operator. Just as in ordinary QM $\langle 0|$ does not denote an eigenstate of the position operator.

 

[stevendaryl]

Yes, but in the case of HO, $\langle 0|$ is an eigenstate of the energy while in this case, based on the way you wrote in the first equation in the previous post, it is an eigenstate of the field operator, isn't it? So isn't it correct to write $\langle 0|\hat{\phi}(x) = \langle 0|0$?

As I sketched, quantum field theory can sort of be thought of as a limit of a lattice theory using many-particle quantum mechanics. It's analogous to lots of harmonic oscillators scattered throughout space connecting particles to each other and to their place in the lattice. In the case of a real scalar field $\phi$, the analogy of $\phi(x)$ is just the amplitude of the harmonic oscillator at point $x$. In my sketch, I let the amplitude be an actual displacement $y$ in a spatial direction perpendicular to $x$. In actual quantum field theory, the amplitude doesn't correspond to a spatial displacement, but the mathematics is the same.

So if $x_j$ is the location of the $j^{th}$ oscillator, then $\phi(x_j) \propto y_j$, the displacement of the oscillator at that position. The "vacuum" in this analogy is the state in which each oscillator is in its ground state. The state in which there is a single particle at position $x_j$ corresponds to the situation in which all oscillators are in their ground state, except the one at point $x_j$, and that oscillator is in its first excited state.

In the quantum mechanics of harmonic oscillators, the way to get from the ground state to the first excited state is to act on it with a "raising operator", $a^\dagger$, which is a linear combination of the displacement $y$ and the corresponding momentum, $p$. Analogously, creating a particle in the corresponding field theory doesn't mean acting on the vacuum with $\phi(x)$, but instead, acting with a linear combination of $\phi(x)$ and $\pi(x)$ (where $\pi(x)$ is the field momentum corresponding to $\phi$).

 

[PeterDonis]

So now the state has a defined position.

Aside from the mathematical issues @Demystifier raised, your logic here is not correct. Acting on a general state $| \psi \rangle$ with the operator $\hat{x}$ does not make the state have a defined position. That is only the case if $| \psi \rangle$ is an eigenstate of position. For such a state, the equation $\hat{x} | \psi \rangle = x | \psi \rangle$ is valid (that equation is called an eigenvalue equation for that reason). But for any other state, it is not.