Klein-Gordon in QFT: Understanding the KG Equation

[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.