# I Klein-Gordon in QFT

1. Jun 10, 2018

### 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!

2. Jun 10, 2018

### Mentz114

Where did you read this ? I ask because I had a similar difficulty until I realised 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.

3. Jun 10, 2018

### Silviu

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?

4. Jun 10, 2018

### Mentz114

I can't help much because I'm no further than you in this. But I can quote these snippets,
I don't understand what that means. It seems to contradict itself.

5. Jun 10, 2018

### stevendaryl

Staff Emeritus
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.

6. Jun 10, 2018

### Mentz114

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

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

Last edited: Jun 10, 2018
7. Jun 11, 2018

### Demystifier

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

8. Jun 11, 2018

### DrDu

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

9. Jun 12, 2018

### A. Neumaier

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.

10. Jun 12, 2018

### Mentz114

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

11. Jun 12, 2018

### Silviu

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)?

12. Jun 13, 2018

### Demystifier

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.

13. Jun 13, 2018

### Silviu

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).

14. Jun 13, 2018

### A. Neumaier

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.

15. Jun 13, 2018

### stevendaryl

Staff Emeritus
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.

16. Jun 13, 2018

### Demystifier

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

17. Jun 13, 2018

### A. Neumaier

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

18. Jun 13, 2018

### Demystifier

How is that different from my statement?

19. Jun 13, 2018

### A. Neumaier

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.

20. Jun 13, 2018

### Silviu

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.