Klein-Gordon in QFT: Understanding the KG Equation

[Silviu]

Hello! I am a bit confused about the KG equation in the context of QFT. In QM, the KG equations describes the evolution of a wavefunction, $\phi(x,t)$, in space and time (I will assume we have no potential). This function gives the probability of finding a particle described by this wavefunction at a given position, at a given time. (Please let me know if anything I said or I will say is wrong.) Now in QFT, $\phi(x,t)$ is an operator, which acting on the vacuum state $|0>$ creates a particle at position $x$. I am confused about the meaning of KG equations in this context? What do the space and time evolution of $\phi(x,t)$ mean now? When it acts, it doesn't create a probability distribution, but an actual particle at a point x (it is at x with 100% probability). Also mathematically speaking, if $\phi(x,t)$ is an operator, the KG equation written as $(\partial^2+m^2)\phi(x,t)=0$ doesn't make much sense to me as on the left side you have an operator $(\partial^2+m^2)$ acting on another operator $\phi(x,t)$ while on the right side you have a number, which is 0. How should I understand this equation now? Lastly, once $\phi(x,t)$ acts on the vacuum state and creates a new particle, how does this particle evolve in space and time? I assume it is still the KG equation that dictates this, but I am not sure I understand how to apply it. Any explanation would be greatly appreciated. I am really new to this and I kinda got stuck with understanding this. Thank you!

 

[Mentz114]

Hello! I am a bit confused about the KG equation in the context of QFT. In QM, the KG equations describes the evolution of a wavefunction, $\phi(x,t)$, in space and time (I will assume we have no potential).
[]
Any explanation would be greatly appreciated. I am really new to this and I kinda got stuck with understanding this. Thank you!

Where did you read this ? I ask because I had a similar difficulty until I realized that it is the Hamiltonian operator of the KG equation which is promoted to be an operator on the Fock space. Each application of the creation operator creates a particle which obeys the KG solution (which you are no doubt aware of)
I thought the book was suggesting that $\phi(x,t)$ became an operator which I also find strange.

 

[Silviu]

Where did you read this ? I ask because I had a similar difficulty until I realized that it is the Hamiltonian operator of the KG equation is promoted to be an operator on the Fock space. Each application of the creation operator creates a particle which obeys the KG solution (which you are no doubt aware of)
I thought the book was suggesting that $\phi(x,t)$ became an operator which I also find strange.

I haven't read it specifically, but this is what I understood. To get the Schrodinger equation you use $E=p^2/2m$ to get KG you use relativity dispersion relation. But the transition of E and p to operators is done the same way, so I assumed that the interpretation of the object they are acting on (wavefunction?) is the same. Could you please give me a bit more details about this?

 

[Mentz114]

I can't help much because I'm no further than you in this. But I can quote these snippets,

Define operators $\psi(r)=\sum_k \psi(r)_k b_k$ and $\psi^{*}(r)=\sum_k \psi^{*}(r)_k b^{*}_k$ which satisfy the anticommutation rules.
...
In other words the expectation value of the single particle wave function $\psi(r)$ has been quantized again and is now to be treated as an operator. It is now a many particle operator acting on the abstract Fock space.

I don't understand what that means. It seems to contradict itself.

 

[stevendaryl]

Hello! I am a bit confused about the KG equation in the context of QFT. In QM, the KG equations describes the evolution of a wavefunction, $\phi(x,t)$, in space and time (I will assume we have no potential). This function gives the probability of finding a particle described by this wavefunction at a given position, at a given time. (Please let me know if anything I said or I will say is wrong.) Now in QFT, $\phi(x,t)$ is an operator, which acting on the vacuum state $|0>$ creates a particle at position $x$. I am confused about the meaning of KG equations in this context? What do the space and time evolution of $\phi(x,t)$ mean now? When it acts, it doesn't create a probability distribution, but an actual particle at a point x (it is at x with 100% probability). Also mathematically speaking, if $\phi(x,t)$ is an operator, the KG equation written as $(\partial^2+m^2)\phi(x,t)=0$ doesn't make much sense to me as on the left side you have an operator $(\partial^2+m^2)$ acting on another operator $\phi(x,t)$ while on the right side you have a number, which is 0. How should I understand this equation now? Lastly, once $\phi(x,t)$ acts on the vacuum state and creates a new particle, how does this particle evolve in space and time? I assume it is still the KG equation that dictates this, but I am not sure I understand how to apply it. Any explanation would be greatly appreciated. I am really new to this and I kinda got stuck with understanding this. Thank you!

The meaning of expressions such as $\phi(x,t)$ changes between single-particle quantum mechanics and quantum field theory. Here's an analogy from ordinary quantum mechanics.

One way of doing quantum mechanics is called the Schrodinger picture. In this approach, the wave function or state evolves in time, while operators such as $p$ and $x$ are time-independent. There is another approach called the Heisenberg picture, in which the wave function, or state, is constant in time, and the operators $p$ and $x$ evolve. In the Heisenberg picture, the equations of motion for $p$ and $x$ are given by:

$\frac{dx}{dt} = \frac{1}{-i \hbar} [H,x]$
$\frac{dp}{dt} = \frac{1}{-i \hbar} [H,p]$

For the usual case of $H = \frac{p^2}{2m} + V(x)$, this implies:

$\frac{dx}{dt} = \frac{1}{m} p$
$\frac{dp}{dt} = - \frac{\partial V}{\partial x}$

So the operator $x$ obeys:

$m \frac{d^2 x}{dt^2} = -\frac{\partial V}{\partial x}$

which is just the classical equation $F = ma$, if you let $F = -\frac{\partial V}{\partial x}$

So the operator $x$ obeys the same equation as the classical variable $x$. Similarly for $p$.

In quantum field theory, a sort of similar thing happens. In going from single-particle quantum mechanics to field theory, the quantity $\phi$ changes from a function to an operator, but obeys the same equations of motion.

 

[Mentz114]

The meaning of expressions such as $\phi(x,t)$ changes between single-particle quantum mechanics and quantum field theory. Here's an analogy from ordinary quantum mechanics.
[..]
In quantum field theory, a sort of similar thing happens. In going from single-particle quantum mechanics to field theory, the quantity $\phi$ changes from a function to an operator, but obeys the same equations of motion.

Thank you. Is the operator a diagonal matrix with $\phi(x,t)$ in position (x,x ) ?

Edit : My guess is wrong. Using the operators $\hat{\psi}(r)=\sum_k \psi_k(r) b_k$ and $\hat{\psi^{*}}(r)=\sum_k \psi_k^{*}(r) b^{*}_k$ makes this right

$H=\int \hat{\psi}^*(r) \nabla^2\psi(r)d^3r + \int \hat{\psi}^*(r) V(r)d^3r = \sum_k E_k \hat{b}^* \hat{b}$

 

[Demystifier]

Hello! I am a bit confused about the KG equation in the context of QFT. In QM, the KG equations describes the evolution of a wavefunction, $\phi(x,t)$, in space and time (I will assume we have no potential). This function gives the probability of finding a particle described by this wavefunction at a given position, at a given time. (Please let me know if anything I said or I will say is wrong.) Now in QFT, $\phi(x,t)$ is an operator, which acting on the vacuum state $|0>$ creates a particle at position $x$. I am confused about the meaning of KG equations in this context? What do the space and time evolution of $\phi(x,t)$ mean now? When it acts, it doesn't create a probability distribution, but an actual particle at a point x (it is at x with 100% probability). Also mathematically speaking, if $\phi(x,t)$ is an operator, the KG equation written as $(\partial^2+m^2)\phi(x,t)=0$ doesn't make much sense to me as on the left side you have an operator $(\partial^2+m^2)$ acting on another operator $\phi(x,t)$ while on the right side you have a number, which is 0. How should I understand this equation now? Lastly, once $\phi(x,t)$ acts on the vacuum state and creates a new particle, how does this particle evolve in space and time? I assume it is still the KG equation that dictates this, but I am not sure I understand how to apply it. Any explanation would be greatly appreciated. I am really new to this and I kinda got stuck with understanding this. Thank you!

See http://lanl.arxiv.org/abs/quant-ph/0609163 Sec. 8.

 

[DrDu]

Just to the meaning of "0". What is meant here, is a function of r and t which is everywhere 0.

 

[A. Neumaier]

Is the operator a diagonal matrix with $\phi(x,t)$ in position (x,x ) ?

No. It is an operator-valued distribution acting on Fock space. Pretending that the distibution is a function (to simplify the explanation), it means that at each point $x$ and time $t$ there is a different operator acting on wave functions with arbitrarily many particles, just as in the Heisenberg picture, $q(t)$ is at each time $t$ a different operator acting on wave functions with one or a few particles.

 

[Mentz114]

No. It is an operator-valued distribution acting on Fock space. Pretending that the distibution is a function (to simplify the explanation), it means that at each point $x$ and time $t$ there is a different operator acting on wave functions with arbitrarily many particles, just as in the Heisenberg picture, $q(t)$ is at each time $t$ a different operator acting on wave functions with one or a few particles.

Thanks. The book I'm using* has some very obscure language and I've found it best to just read the equations which all make sense.
As the author remarks, getting the anti-commutation relations seems a bit abstract !

*A L Ziman, 2002

 

[Silviu]

See http://lanl.arxiv.org/abs/quant-ph/0609163 Sec. 8.

Thank you for this. I am not sure I understand the meaning of $\psi[\phi;t)$. I see that it is used to define a probability distribution of the field itself, but what does this mean? For example in the case of a free scalar field, $\phi(x,t)$ has a defined form (a linear combination of annihilation and creation operators multiplied by incoming and outgoing waves). Isn't this configuration the one that satisfies KG 100% of the time? What is the meaning of this $\psi[\phi;t)$ in this case? Can $\phi(x,t)$ take other form such that $\psi[\phi;t)$ is not trivial (i.e. probability 1 in the above-mentioned case and 0 in other cases)?

 

[Demystifier]

Thank you for this. I am not sure I understand the meaning of $\psi[\phi;t)$. I see that it is used to define a probability distribution of the field itself,

Yes.

but what does this mean? For example in the case of a free scalar field, $\phi(x,t)$ has a defined form (a linear combination of annihilation and creation operators multiplied by incoming and outgoing waves).

This is the field operator (in the Heisenberg picture), which, to avoid the notational confusion, should be denoted as $\hat{\phi}$. Note that $\hat{\phi}$ and $\phi$ are not the same.

This is similar to ordinary QM. There you have the position operator $\hat{x}$ and the position itself $x$ in the wave function $\psi(x,t)$. Here $\hat{x}$ and $x$ are not the same. For harmonic oscillator (in the Heisenberg picture) the position operator can be written as
$$\hat{x}(t)=\hat{a}e^{-i\omega t}+\hat{a}^{\dagger}e^{i\omega t}$$
where $\hat{a}^{\dagger}$ and $\hat{a}$ are the creation and annihilation operators.

Isn't this configuration the one that satisfies KG 100% of the time?

No, the operator is not a configuration.

What is the meaning of this $\psi[\phi;t)$ in this case? Can $\phi(x,t)$ take other form such that $\psi[\phi;t)$ is not trivial (i.e. probability 1 in the above-mentioned case and 0 in other cases)?

See the analogy with ordinary QM explained above.

 

[Silviu]

Yes.This is the field operator (in the Heisenberg picture), which, to avoid the notational confusion, should be denoted as $\hat{\phi}$. Note that $\hat{\phi}$ and $\phi$ are not the same.

This is similar to ordinary QM. There you have the position operator $\hat{x}$ and the position itself $x$ in the wave function $\psi(x,t)$. Here $\hat{x}$ and $x$ are not the same. For harmonic oscillator (in the Heisenberg picture) the position operator can be written as
$$\hat{x}(t)=\hat{a}e^{-i\omega t}+\hat{a}^{\dagger}e^{i\omega t}$$
where $\hat{a}^{\dagger}$ and $\hat{a}$ are the creation and annihilation operators.No, the operator is not a configuration.See the analogy with ordinary QM explained above.

Thank you! Now it makes a lot more sense. However there is still something I am confused about. In QM, $\hat{x}$ is a position operator while $x$ is a parameter (like time). Now when you calculate $|\psi(x,t)|^2$ you see the probability of finding a particle around position x at time t. In QFT, $\hat{\phi(x,t)}$ is an operator that creates a particle at position x (in the free theory at least), while $\phi(x,t)$ is a parameter of this new wave function $\psi[\phi;t)$. What is the object whose probability we get when we square this? In QM, it was the probability of finding a particle at position x. Now, it is the probability of finding what, in the field configuration $\phi(x,t)$? I guess my confusion is that in QM i knew what x was, but I am not sure what $\phi(x,t)$ is in QFT (i.e. its physical meaning).

 

[A. Neumaier]

What is the object whose probability we get when we square this?

The absolute square of the wave function $\psi(\phi,t)$ would be the probability density for finding the field at time $t$ to be $\phi$ - if it were possible to measure this. But measurements can be done only locally, so one cannot give $|\psi(\phi,t)|^2$ an operational meaning.

 

[stevendaryl]

A way to sort of see the connection between quantum field theory and quantum mechanics is (and @Demystifier has a lot to say about this) to look at how a QFT can arise from many-particle quantum mechanics.

Here's a toy problem that is similar to real solid state physics problems. Imagine that you have a lattice of atoms that are fixed in place in the x-direction, but are free to move in the y-direction. I'm going to model the forces on the atoms using harmonic oscillators. Each atom is connected to its base with a spring, that will tend to keep it in place vertically. In addition, there is a second spring connecting each atom to its nearest neighbor. For simplicity of the calculation, let me assume that the equilibrium length of the horizontal springs is negligible.

We can write down a classical (nonquantum, nonrelativistic) lagrangian for the collection of atoms:

$L = \sum_j \frac{1}{2} m (\dot{y_j})^2 - \frac{k_1}{2} (y_j)^2 - \frac{k_2}{2} (y_{j+1} - y_j)^2$

where $m$ is the mass of an atom, and $y_j$ is the vertical position of atom number $j$ and $k_1$ and $k_2$ are two spring constants, and where $\dot{A}$ means $\frac{dA}{dt}$.

If the number of atoms is very large, and the energies of the atoms are smallish, then we can approximate the discrete lattice by a continuous function $\phi(x)$:

$y_j = K \phi(j \Delta x)$ where $\Delta x$ is the horizontal distance between atoms. ($K$ is a scaling factor to be determined later)
$\dot{y_j} = K \dot{\phi}$
$y_{j+1} - y_j \approx K \Delta x \phi'$ (where $'$ means a derivative with respect to $x$).

In terms of $\phi$, the lagrangian becomes:

$L = \sum_j \Delta x [\frac{mK^2}{2 \Delta x} (\dot{\phi})^2 - \frac{k_1 K^2}{2 \Delta x} \phi^2 - \frac{k_2 K^2 \Delta x}{2} (\phi')^2]$ (I've factored out $\Delta x$ in anticipation of approximating the sum by an integral).

Now, let's choose our scaling factor $K$ so that $\frac{mK^2}{ \Delta x} = 1$. Then let's define new constants:
$\mu^2 = \frac{k_1 K^2}{\Delta x}$
$\bar{c}^2 = k_2 K^2 \Delta x$

In terms of these constants,

$L = \sum_j \Delta x [\frac{1}{2} (\dot{\phi})^2 - \frac{\mu^2}{2} \phi^2 - \bar{c}^2 (\phi')^2]$
$\approx \int dx [\frac{1}{2} (\dot{\phi})^2 - \frac{\mu^2}{2} \phi^2 - \bar{c}^2 (\phi')^2]$

This is exactly the lagrangian for a relativistic neutral spin-zero quantum field in one spatial dimension, (if you interpret $\bar{c}$ as the speed of light).

So the classical (nonrelativistic, nonquantum) theory of the toy lattice problem transforms, in the continuum limit to what looks like a relativistic (nonquantum) field theory.

Now, to make the description quantum-mechanical, we can go back to the original problem, and interpret the $y_j$ and $\dot{y_j}$ as operators, instead of numbers. We can transform from the lagrangian form to the hamiltonian form by introducing a momentum $p_j$ defined by:

$p_j = \frac{\partial \mathcal{L}}{\partial \dot{y_j}} = m \dot{y_j}$

Then we introduce the commutation relation: $[p_j, y_k] = -i \delta_{jk}$. This gives rise to the equivalent commutation relation for $\phi$:

$[\dot{\phi}(x), \phi(x')] = -i \delta(x-x')$

So this reinterpretation of the many-particle quantum mechanical problem as a quantum field-theory problem illustrates some facts about $\phi$. It is not a wave function. Going back to the original problem, you can see that $\phi$ is the amplitude for the classical displacement of the atoms in the lattice. The quantum mechanics does not come in by interpreting $\phi^2$ as a probability. It isn't a probability. Instead, for each atom, there would be a corresponding wave function $\psi_j(y)$ giving the probability amplitude that atom number j has vertical displacement $y$. When you shift to the field-theoretic interpretation, there would correspondingly be a wave function giving the probability amplitude that the classical wave $\phi$ has a particular value at a particular point.

 

[Demystifier]

Now, it is the probability of finding what, in the field configuration $\phi(x,t)$?

Almost. The probability of finding the configuration $\phi({\bf x})$ at time $t$.

 

[A. Neumaier]

Almost. The probability of finding the configuration $\phi({\bf x})$ at time $t$.

No, the probability of finding simultaneously at every $x$ the configuration $\phi({\bf x})$ at time $t$.

 

[Demystifier]

No, the probability of finding simultaneously at every $x$ the configuration $\phi({\bf x})$ at time $t$.

How is that different from my statement?

 

[A. Neumaier]

How is that different from my statement?

Well, in your statement there is an $x$ without an explanation what it means and how it enters the probability. Quantifying it outside the sentence (which is the standard default) gives a wrong meaning.

 

[Silviu]

Almost. The probability of finding the configuration $\phi({\bf x})$ at time $t$.

But what is the meaning of this $\phi(x)$? In QM its meaning would be that by squaring it, you get the probability of finding a particle at position x (leaving aside the problems with negative probability and Feynman interpretation of antiparticles associated with KG equation). But in QFT, what does it represent. You don't have a single particle anymore, so I assume you can't have the same interpretation.

 

[Mentz114]

A way to sort of see the connection between quantum field theory and quantum mechanics is (and @Demystifier has a lot to say about this) to look at how a QFT can arise from many-particle quantum mechanics.
[..]
It isn't a probability. Instead, for each atom, there would be a corresponding wave function $\psi_j(y)$ giving the probability amplitude that atom number j has vertical displacement $y$. When you shift to the field-theoretic interpretation, there would correspondingly be a wave function giving the probability amplitude that the classical wave $\phi$ has a particular value at a particular point.

This is like the derivation I just went through and your comments help to make sense Ziman's reference to the 'expected value of $\phi$' being quantised. I think what he means is that there is a function which gives the amplitude of the classical field $\phi$ being in a certain range. The exact moment of this 'gear-change' is a bit obscure but it does not bother me because the end result seems consistent and useful ( Ziman calls it an 'elegant and powerful' theory !)

 

[Demystifier]

But what is the meaning of this $\phi(x)$? In QM its meaning would be that by squaring it, you get the probability of finding a particle at position x (leaving aside the problems with negative probability and Feynman interpretation of antiparticles associated with KG equation). But in QFT, what does it represent. You don't have a single particle anymore, so I assume you can't have the same interpretation.

In this context you have to think of $\phi$ as a field similar to electromagnetic field in classical electrodynamics. It means that the field itself is a physical object distributed around.

 

[Silviu]

In this context you have to think of $\phi$ as a field similar to electromagnetic field in classical electrodynamics. It means that the field itself is a physical object distributed around.

Ok... but in classical electrodynamics the EM field can be observed (measured) i.e. you can see for example the effect of E and B on a test charge. But I still don't see what $\phi(x)$ is. How can you measure it and what are (some of) its observables (being a physical objects, I guess these questions should have an answer, right?)?

 

[Demystifier]

Ok... but in classical electrodynamics the EM field can be observed (measured) i.e. you can see for example the effect of E and B on a test charge. But I still don't see what $\phi(x)$ is. How can you measure it and what are (some of) its observables (being a physical objects, I guess these questions should have an answer, right?)?

You are right, the electromagnetic field is measured through its effect on something else. The EM wave can be there even without charges, but you cannot measure it without charges. To observe the field, you need an interaction between field and something else, or sometimes the interaction of the field with itself.

The last statement is true for all fields, including $\phi$. The Klein-Gordon equation is incomplete because it does not describe the interaction. But if you read some QFT textbook, in later chapters you will see how the interactions are described.

 

[Silviu]

You are right, the electromagnetic field is measured through its effect on something else. The EM wave can be there even without charges, but you cannot measure it without charges. To observe the field, you need an interaction between field and something else, or sometimes the interaction of the field with itself.

The last statement is true for all fields, including $\phi$. The Klein-Gordon equation is incomplete because it does not describe the interaction. But if you read some QFT textbook, in later chapters you will see how the interactions are described.

So to make sure I understand what has been discussed up to now... In QM if you act on a (single particle) state with $\hat{x}$ and then you look at $\psi(x)$, you get a delta function at that x. In QFT, if you act with $\hat{\phi(x)}$ on the vacuum, you create a particle at the position x and when you look at $\psi[\phi ;t)$ you obtain still a (field valued) delta function which corresponds to the state of the field having a single excitation at position x? Is this right?

 

[Demystifier]

So to make sure I understand what has been discussed up to now... In QM if you act on a (single particle) state with $\hat{x}$ and then you look at $\psi(x)$, you get a delta function at that x. In QFT, if you act with $\hat{\phi(x)}$ on the vacuum, you create a particle at the position x and when you look at $\psi[\phi ;t)$ you obtain still a (field valued) delta function which corresponds to the state of the field having a single excitation at position x? Is this right?

No.

 

[Silviu]

No.

ok...then I am still confused. I thought this was the analogy with the position operator mentioned before... In the paper you suggested, the functional Schrodinger equation was identical to the QM one, except that the x was replaced with $\phi(x)$. So if what I said before is wrong, what is the relation between $\phi(x)$ and $\hat{\phi(x)}$?

 

[Demystifier]

ok...then I am still confused. I thought this was the analogy with the position operator mentioned before... In the paper you suggested, the functional Schrodinger equation was identical to the QM one, except that the x was replaced with $\phi(x)$. So if what I said before is wrong, what is the relation between $\phi(x)$ and $\hat{\phi(x)}$?

You were wrong already at the level of standard QM.

First, the operator $\hat{x}$ does not act on a particle. A particle is a physical object in a laboratory, while the operator is a mathematical object in the theory. You will never find $\hat{x}$ in the laboratory.

Second, the operator $\hat{x}$ may act on the wave function $\psi(x)$. The result of this action is
$$\hat{x}\psi(x)=x\psi(x)\neq\delta(x)$$
so you don't get a $\delta$-function.

 

[Silviu]

You were wrong already at the level of standard QM.

First, the operator $\hat{x}$ does not act on a particle. A particle is a physical object in a laboratory, while the operator is a mathematical object in the theory. You will never find $\hat{x}$ in the laboratory.

Second, the operator $\hat{x}$ may act on the wave function $\psi(x)$. The result is of this action is
$$\hat{x}\psi(x)=x\psi(x)\neq\delta(x)$$
so you don't get a $\delta$-function.

Wait, sorry, I said you act with $\hat{x}$ on a single particle state. By this i meant $|\psi>$. So you have $\hat{x}|\psi>=x|\psi>$. So now the state has a defined position. So if you are to write $|\psi>=\int \psi(x)|x>dx$, as the state has a defined position you have $|\psi>=|x>$ for some x. This means that the coefficient of that $|x>$, which is $\psi(x)$ is a delta function i.e. $\psi(x)=\delta(x)$ so that it can pick it up from inside the integral and ignore all the others. Isn't this correct?

 

[Demystifier]

Wait, sorry, I said you act with $\hat{x}$ on a single particle state. By this i meant $|\psi>$.

OK.

So you have $\hat{x}|\psi>=x|\psi>$.

That's wrong. A correct equation that you might have in mind is
$$<x|\hat{x}|\psi>=x<x|\psi>$$

So now the state has a defined position.

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
$$\hat{x}=\int dx\, x \hat{\pi}(x)$$
Note that
$$<x'|\hat{\pi}(x)|\psi>=<x'|x><x|\psi>=\delta(x-x')\psi(x)\propto \delta(x-x')$$

 

[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 $\langle 0|\hat{\phi}(x) = \langle 0|0$?