## question about a quantum field.

 Quote by mpv_plate Thanks for this explanation. When the $x$ operates on a state, the state (vector in a Hilbert space) becomes sort of multiplied by $x$. In the QFT, is a state (vector in a Fock space) also "multiplied" by the field operator? I'm actually trying to understand what is mathematically changed in the field $\Phi$, to make it behave like an operator. Because multiplying a state by a field (function that assigns a number to each coordinate) does not seem to do the job of being an operator action.
Okay, here's what I hope is an enlightening discussion about the relationship between quantum field theory and quantum mechanics.

Suppose we have the following weird situation: You have a line of masses (labeled by consecutive integers) connected to each other by springs and also connected to a platform by springs. To simplify the motion (at the cost of a more complicated picture), I'm assuming that the masses are constrained to move vertically by rigid vertical rods running through them.

The energy associated with this system is given by:

$E = \sum_n \lbrace \frac{m}{2} \dot{y_n}^2 + \frac{\lambda}{2} (y_n)^2 + \frac{\lambda}{2} (y_{n+1} - y_n)^2\rbrace$

where $y_n$ is the height of the $n^{th}$ mass above its equilibrium height, $\dot{y_n}$ is $\dfrac{dy_n}{dt}$, $m$ is the mass of each of the masses, and $\lambda$ is the spring constant.

Now, if we assume that the spring constant is large enough (a very stiff spring) that the masses don't move far from their equilibrium positions, then we can assume that $y_n$ can be approximated by a continuous function $\Phi(x)$ where $x$ measures the horizontal location of the mass. If $L$ is the distance between masses, then the horizontal location of $n^{th}$ mass is $nL$. Then we can relate $y_n$ to $y_{n+1}$ by

$y_{n+1} = \Phi((n+1)L) = \Phi(nL) + \dfrac{\partial \Phi}{\partial x} \delta x + \ldots = \Phi(nL) + \Phi' (nL) L + \ldots$

where $\Phi' = \dfrac{\partial \Phi}{\partial x}$. In terms of $\Phi$, the energy can be written as

$E = \sum_n \lbrace \frac{m}{2}\dot{\Phi}(nL)^2 + \frac{\lambda}{2} \Phi(nL)^2 + \frac{\lambda L^2}{2} \Phi'(nL)^2 \rbrace$

This discrete sum can be approximated by an integral:
$E = \dfrac{1}{L} \int \lbrace \frac{m}{2}\dot{\Phi}(x)^2 + \frac{\lambda}{2} \Phi(x)^2 + \frac{\lambda L^2}{2} \Phi'(x)^2 \rbrace dx$

Now, the final bit of trickery is to define new constants:
$\mu = \sqrt{\frac{\lambda}{m}}$
$c = \sqrt{\frac{\lambda}{m}} L$

In terms of these constants, the energy for our system of oscillators looks like this:

$E = \dfrac{m}{L} \int \lbrace \frac{1}{2}\dot{\Phi}(x)^2 + \frac{\mu^2}{2} \Phi(x)^2 + \frac{c^2}{2} \Phi'(x)^2 \rbrace dx$

Mathematically, this is the relativistic hamiltonian for a free real scalar field in a spacetime with only one spatial dimension, although the constant $c$ isn't actually the speed of light, but is the speed that vibrational waves travel through the system of harmonic oscillators.

Now, if you quantize the original system of harmonic oscillators, by replacing each of the masses by a quantum particle obeying Schrodinger's equation, that should be the same (given the approximations) as if you treated $\Phi$ as a quantum field.

 Quote by stevendaryl Now, if you quantize the original system of harmonic oscillators, by replacing each of the masses by a quantum particle obeying Schrodinger's equation, that should be the same (given the approximations) as if you treated $\Phi$ as a quantum field.
Thank you very much. So the quantum field has uncertain value at every position - the field value in every point in space is in superposition of infinitely many values (some values are more probable than others). This is consistent with the perspective where the quantum field is in superposition of infinitely many configurations and each possible configuration is assigned a probability amplitude by the wave functional.

Now the field operator: is it correct to think of it as a collection of operators, assigning one operator to every point in space? So it is an operator depending on spatial (and time) coordinates? And at every point in space the field operator acts on the state of the field at that location (changing for example the expectation value of the field)?

 Quote by mpv_plate Now the field operator: is it correct to think of it as a collection of operators, assigning one operator to every point in space? So it is an operator depending on spatial (and time) coordinates? And at every point in space the field operator acts on the state of the field at that location (changing for example the expectation value of the field)?
I think that's right. In the 1-D quantum harmonic oscillator, there are "raising" and "lowering" operators $a^\dagger_n$ and $a_n$, where

$y_n^\dagger = \sqrt[4]{\frac{1}{4 \lambda m}} (a^\dagger_n +a_n)$
$p_n = +i \sqrt[4]{\frac{m \lambda}{4}}(a^\dagger_n - a_n)$

The raising operator $a^\dagger_n$ raises the quantum number of particle number $n$ by 1, and the lowering operator $a_n$ lowers the quantum number by 1. There is a corresponding operator $N_n$ defined by:

$N_n = a^\dagger_n a_n$

which returns the quantum number when acting on an eigenstate of the hamiltonian.

When you go over to the continuum limit, the $a^\dagger_n$ and $a_n$ are reinterpreted as "creation" and "annihilation" operators, $a^\dagger(x)$ and $a(x)$ which create or annihilate a particle at location $x$. The number operator $N_n$ goes over to $N(x)$, which counts the number of particles at position $x$.

 Quote by stevendaryl I think that's right. In the 1-D quantum harmonic oscillator, there are "raising" and "lowering" operators $a^\dagger_n$ and $a_n$, where $y_n^\dagger = \sqrt[4]{\frac{1}{4 \lambda m}} (a^\dagger_n +a_n)$ $p_n = +i \sqrt[4]{\frac{m \lambda}{4}}(a^\dagger_n - a_n)$ The raising operator $a^\dagger_n$ raises the quantum number of particle number $n$ by 1, and the lowering operator $a_n$ lowers the quantum number by 1. There is a corresponding operator $N_n$ defined by: $N_n = a^\dagger_n a_n$ which returns the quantum number when acting on an eigenstate of the hamiltonian. When you go over to the continuum limit, the $a^\dagger_n$ and $a_n$ are reinterpreted as "creation" and "annihilation" operators, $a^\dagger(x)$ and $a(x)$ which create or annihilate a particle at location $x$. The number operator $N_n$ goes over to $N(x)$, which counts the number of particles at position $x$.
I should warn you that this interpretation is off-the-cuff, and I should really check the literature to see if what I'm saying is completely correct. So it's only meant to be suggestive, rather than definitive.
 Thank you. Is it possible for a field to be not in superposition of various configurations? Such a quantum filed would have a definite shape, like a classical field. Such a state should be an eigenstate of some operator. Is that the a(x) operator?

 Quote by mpv_plate Thank you. Is it possible for a field to be not in superposition of various configurations? Such a quantum filed would have a definite shape, like a classical field. Such a state should be an eigenstate of some operator. Is that the a(x) operator?
The analogy with single-particle quantum mechanics is for a particle to have a definite location, instead of being in a superposition of different locations. What that means, in terms of the wave function is that the wave function is singular, like a delta-function; it is zero everywhere except at one point, $x_0$ and is infinite there.

There's actually a simple expression for the delta-function $\delta(x-x_0)$ in terms of harmonic oscillator eigenstates:

$\delta(x-x_0) = \sum_n \Psi^{*}_n(x_0) \Psi_n(x)$

So to have a particle with a definite position requires a superposition of all possible energy eigenstates $\Psi_n(x)$.

I think the same thing is true of the quantum field theory case. To have a definite value of the field requires a superposition of all possible numbers and positions of particles.

The operator $a^\dagger(x)$ that creates a particle at position $x$ does not make the field have a definite value. The uncertainty principle applies to quantum fields, too.
 Recognitions: Science Advisor One should, however, emphasize that a $\delta$ distribution never represents a quantum state, because it is not a square integrable function. Only these true Hilbert-space vectors represent states. The generalized eigenstates of essentially selfadjoint operators in the continuous part of the spectrum belong to a larger space, i.e., the dual of the operator's domain. Have a look at Ballentines book, where the rigged-Hilbert space formulation of quantum theory is used.
 A "classically" looking field (for example electromagnetic wave) is a superposition of all number eigenstates - search for the keyword "coherent state". My blog is in German, butperhaps the pictures may help nevertheless to get the idea: http://scienceblogs.de/hier-wohnen-d...rstehen-alles/