# From observable to operators in QFT

#### lucas_

from the relativity forum https://www.physicsforums.com/threads/spacetime-in-qm-or-qft.802721/ Sonderval stated (transferred here so not off topic):

Second quantization is a somewhat misleading term to me (and many others) because it seems to imply that you do two steps of quantisation, which is not correct.

What happens is this:
In QM, you have a wave function that assigns a prob. amplitude to each point in space (and time). The Dirac equation or Klein-Gordon equation was initially conceived as an equation of a wave function like the Schroedinger equation.

Then people realised that this is not correct (because particle number is not conserved due to creation of particles/antiparticles etc.) and that you need in fact a field theory.
In a field theory, you use the same equation (for example the Klein-Gordon equation), but you now interpret this as the field equation of a classical field. Solutions of this equations look like they did before (when you thought that you were dealing with a prob. amplitude), but now the field has to be interpreted as a classical quantity (for example, the displacement of a membrane as a intuitive example).

In the second step, you then quantise this field, using the standard rules of quantum theory, converting observables to operators. Since the classical field itself is an observable, it becomes a field operator.

What makes things confusing is that you now usually do not deal with wave functions anymore - QFT is usually not phrased that way (laudable exception in the book of Hatfield "QFT of point particles and strings"). If you think in terms of wave functions, what QFT does is to assign a probability amplitude to each possible field configuration. (Since this is not easy to do because of the infinity of possible functions, people prefer other ways of describing QFT.)

The standard way of explaining this quantisation - turning wave functions into creation/annihilation operators - is also confusing for another reason, in my opinion: Before you have a set of possible solutions, with coefficients a and b that can have any possible value. Then you do the second quantisation and these coefficients turn into creation/annihilation operators, which are precisely defined objects. Where did all the possibilities for the solution go?
Answer: The 2nd quantised solution that contains operators has to be applied to a state vector, and the freedom you have is now in the state vector, not in the solution anymore. Unfortunately, this is not always explained.

Hope this helps - if you can read German, you can also look at my blog where I explain many aspects of QFT - see here and look for the QFT series:
http://scienceblogs.de/hier-wohnen-drachen/artikelserien/
[/PLAIN] [Broken]
So the standard Schroedinger Equation can be used for both particles and fields in that it is stated above that "but you now interpret this as the field equation of a classical field", really? can you use wave equation as field equation of a classical field? why, does a classical field have waves? When you second quantize it, why do you need to convert observables to operators?

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#### atyy

"Classical field" is jargon in this context.

Second quantization is badly named, and is essentially just a way to write down a quantum field theory.

But what has quantum field theory got to do with Schroedinger's equation? The answer is that the usual quantum mechanics with Schroedinger's equation for many identical particles can be exactly rewritten as a quantum field theory. The same theory can be expressed in two different ways.

#### bhobba

Mentor
OK - lets go back to basics.

Standard QM is based on the Galilean transformations with time assumed as an absolute - you will find the detail in Chapter 3 of Ballentine.

Now lets move beyond that and think in terms relativity which treats space and time on equal footing. This conflicts with standard QM - time is a parameter and position is an observable. To get around it QFT assumes both time and position is a parameter. Basically this means you have a field. But in QM you don't have values of anything - you have observables - its the basic foundational axiom of QM I harp on about in some of my other posts - but the gory detail can be found here (see post 137):

So what you have, instead of assigning a value to a point in space, you assign an operator or operators. Now how do we get equations like Schroedinger's equation that describes the dynamics of the field. Its done by first deciding on a Lagrangian for the classical version of the field, We find that symmetry considerations put constraints on these Lagrangisns eg if you want U(1) symmetry you get EM. Then from Lagrangian field theory you figure out the conjugate momentum of the field. Now we do the standard quantisation procedure that dates back to Dirac - you replace both the field and its conjugate momentum by operators and use the commutation relations from standard QM at each point in space and time. The value of a field in classical field theory is sort of like a particles position in classical mechanics and the conjugate momentum like its momentum so its a reasonable assumption on how to quantise the field especially when you think back to the basics of classical field theory where the field is considered a large number of lumps each with its own Lagrangian from particle physics. The gory detail can be found in standard books on QFT eg:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

The above book is accessible after a first exposure to proper QM eg Susskinds book:
https://www.amazon.com.au/Quantum-Mechanics-The-Theoretical-Minimum-ebook/dp/B00IFTT8GA

It will require attention and will be slow going - but is doable.

The quote you gave is correct.

Second quantisation is a misleading characterisation dating back to when we didn't understand this as well as we do now. To start with best to forget it.

We also find that wave-functions, while still existing in the form of a Fock space, takes on a secondary importance:
http://en.wikipedia.org/wiki/Fock_space

Thanks
Bill

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#### lucas_

OK - lets go back to basics.

Standard QM is based on the Galilean transformations with time assumed as an absolute - you will find the detail in Chapter 3 of Ballentine.

Now lets move beyond that and think in terms relativity which treats space and time on equal footing. This conflicts with standard QM - time is a parameter and position is an observable. To get around it QFT assumes both time and position is a parameter. Basically this means you have a field. But in QM you don't have values of anything - you have observables - its the basic foundational axiom of QM I harp on about in some of my other posts - but the gory detail can be found here (see post 137):
Why does a field, even a classical field makes time and position as parameter? When Faradays invented the concept of field, did he consider it making time and position as parameter?

So what you have, instead of assigning a value to a point in space, you assign an operator or operators. Now how do we get equations like Schroedinger's equation that describes the dynamics of the field. Its done by first deciding on a Lagrangian for the classical version of the field, We find that symmetry considerations put constraints on these Lagrangisns eg if you want U(1) symmetry you get EM. Then from Lagrangian field theory you figure out the conjugate momentum of the field. Now we do the standard quantisation procedure that dates back to Dirac - you replace both the field and its conjugate momentum by operators and use the commutation relations from standard QM at each point in space and time.
In a few words, why does replacing the field (and its conjugate momentum) by operators produce quantisation?

Should you visualize the field as waving or boiling like in hot water? And maybe replacing each point by operator creates more complex numbers? So are the operators for adjacent points of the field somewhat similar or differ hugely? And what is it operators of? momentum? energy? spin?

The value of a field in classical field theory is sort of like a particles position in classical mechanics and the conjugate momentum like its momentum so its a reasonable assumption on how to quantise the field especially when you think back to the basics of classical field theory where the field is considered a large number of lumps each with its own Lagrangian from particle physics. The gory detail can be found in standard books on QFT eg:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

The above book is accessible after a first exposure to proper QM eg Susskinds book:
https://www.amazon.com.au/Quantum-Mechanics-The-Theoretical-Minimum-ebook/dp/B00IFTT8GA

It will require attention and will be slow going - but is doable.

The quote you gave is correct.

Second quantisation is a misleading characterisation dating back to when we didn't understand this as well as we do now. To start with best to forget it.

We also find that wave-functions, while still existing in the form of a Fock space, takes on a secondary importance:
http://en.wikipedia.org/wiki/Fock_space

Thanks
Bill

#### bhobba

Mentor
Why does a field, even a classical field makes time and position as parameter? When Faradays invented the concept of field, did he consider it making time and position as parameter?
Its trivial. If something is described by parameters of position and time then it has a certain value at a position and time. That by definition is a field.

In a few words, why does replacing the field (and its conjugate momentum) by operators produce quantisation?
Its in the link I gave.

Should you visualize the field as waving or boiling like in hot water? And maybe replacing each point by operator creates more complex numbers? So are the operators for adjacent points of the field somewhat similar or differ hugely? And what is it operators of? momentum? energy? spin?
A field is simply something assigned to a point in space at a specific time.

Thanks
Bill

#### lucas_

Its trivial. If something is described by parameters of position and time then it has a certain value at a position and time. That by definition is a field.

Its in the link I gave.

A field is simply something assigned to a point in space at a specific time.

Thanks
Bill
Bill or Atyy, what is the equivalent of decoherence, improper mixed, proper mixed in QFT when every point is no longer an observable but operator acting on hidden quantum state... since position no longer exist.. how does the Hamiltonian pointer state with position work on QFT in decoherence?

#### bhobba

Mentor
when every point is no longer an observable but operator acting on hidden quantum state... since position no longer exist..
A few things.

First the operator at each point in space is an observable - its the observable of the field value.

While position is now a parameter you can in some cases still construct a position observable.

I haven't seen too much on decoherence and QFT - one usually uses standard QM which is the 'dilute' limit of QFT.

Thanks
Bill

#### atyy

Bill or Atyy, what is the equivalent of decoherence, improper mixed, proper mixed in QFT when every point is no longer an observable but operator acting on hidden quantum state... since position no longer exist.. how does the Hamiltonian pointer state with position work on QFT in decoherence?
I don't know a rigourous answer. But heuristically QFT is just QM of many identical particles. The important thing is that the interactions are still local in QFT. Also, we know cases in which QM can simulate a gauge theory, so I don't think there are any conceptual differences.

Here is an example of QM simulating QFT:

http://arxiv.org/abs/1205.0496
Optical Abelian Lattice Gauge Theories
L. Tagliacozzo, A. Celi, A. Zamora, M. Lewenstein

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#### lucas_

I have been reading qft book like M.Y. Han From Photons to Higgs but I can only understand the verbal part and I'd like to know something. In neighbor points of the fields which become operators. Do the values vary smoothly from neighboring points like spacetime manifold or vary greatly (sporatically)? I know the operators act on the quantum state. We know fields are defined to vary smoothly.. so if they vary sporatically. then instead of operator fields, it's more like boiling water.. maybe Quantum Operator Boil Theory would be more accurate than Quantum Field Theory, isn't it (so as not to confuse generations of students).

#### bhobba

Mentor
Do the values vary smoothly from neighboring points like spacetime manifold
Yes.

Thanks
Bill

#### lucas_

bhobba or atty.. need your input here. I read the thread https://www.physicsforums.com/threads/what-are-the-states-in-qft.388556/ "What are the states in QFT" one day long but the answer still not clear. That is because someone inside called Eugene who has another idea of QFT tried to mix up the issues confusing many and the original question is not directly answered. It is also a question I want to know. Someone called Asrael84 wrote:

Hello,

this is quite a basic question I know, but something I'm not sure I've fully got my head around. In classical particle mechanics the dynamical variable is the position vector x, and in classical field theory the dynamical variable becomes the field
ϕ(x)
\phi(x) , with x being relagated to just a label to each of the infinitity of these dynamical variables
ϕ
\phi at each point in space if you like.

Then in regualar QM, the things that were dynamical variables in classical particle mechanics (position, momentum etc...) get promoted to operators, with the actual states being abstract vectors in the hilbert space of these operators. In QFT the dynamical variable from classical field theory, the field itself, gets promoted to an operator. But what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?

Also I noticed that in QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states? are they somehow abstract vectors of the field operators?

1. But what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?
2. Also I noticed that in QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?

Eugene also wrote:
QFT is not fundamentally different from quantum mechanics. In QFT we also have a Hilbert space (it is called the Fock space) and states of the system are represented by vectors in this Hilbert space. The most significant difference is that the number of particles is not fixed in QFT. So the Fock space is a direct sum of subspaces with 0 particles (vacuum), 1 particle, 2 particles, etc. Thus in the Fock space we can describe processes in which the number of particles can change (radiation, decays, etc.). So, basically QFT is the same as "QM with variable number of particles".

Quantum fields have nothing to do with wave functions or states. Quantum fields are just certain operators in the Fock space which are convenient "building blocks" for construction of other more physical operators. For example, interaction terms in the Hamiltonian are usually constructed as products of several field operators.

Eugene.
Please confirm if it is true when Eugene said "Quantum fields have nothing to do with wave functions or states.". You guys said you agreed with initial message #1 in this thread when Sonderval wrote: "The 2nd quantised solution that contains operators has to be applied to a state vector, and the freedom you have is now in the state vector, not in the solution anymore. Unfortunately, this is not always explained."

Eugene said quantum fields have nothing to do with states while you agreed with Sonderval the 2nd quantized solution containing operators has to be applied to state vector. There is some conflict, could you kindly resolve it.

Lastly. What does it mean the operator fields are the solutions of the wave equations. Why the word solution? what is the context of "solution"? Is it like black hole is a solution of GR.. and operator fields are a solution of QFT? solution of what?

Many thanks!

#### atyy

Have you mastered quantum mechanics at the level of say Griffiths or Shankar?

#### lucas_

Have you mastered quantum mechanics at the level of say Griffiths or Shankar?
the rigorous parts maybe 10 years from now.. but for now want to understand on a bird eye view... and I understand QM like operators, hamiltonians, observable, Hilbert space, eigenvalues.. this is why I'm asking "in QFT, what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?"? because I understood the basic math of basic QM.

#### atyy

the rigorous parts maybe 10 years from now.. but for now want to understand on a bird eye view... and I understand QM like operators, hamiltonians, observable, Hilbert space, eigenvalues.. this is why I'm asking "in QFT, what are the analogues of the eigenvalues of position/momentum/angular momentum etc from regualar QM?"? because I understood the basic math of basic QM.
Regular QM of many identical particles can be rewritten as a QFT, so the observables are the same in QM. This sort of QFT is non-relativistic, but there are no conceptual differences for relativistic QFT, just many more technical difficulties because of the relativistic symmetry.

http://hitoshi.berkeley.edu/221b/QFT.pdf
These notes quantize the Schroedinger equation as a classical field theory to get a QFT, which is then shown to be QM of many identical particles.

These notes take QM of many identical particles and rewrite it as a QFT.

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#### lucas_

Regular QM of many identical particles can be rewritten as a QFT, so the observables are the same in QM. http://hitoshi.berkeley.edu/221b/QFT.pdf
I've read that part a few days ago when you first shared it. But there seems to be differences. And I have the same questions as the OP in the other thread which is unanswered (instead Eugene tried to share his new views which just confused everything). Here is the specific question:

"I was already aware of the Fock space being the analogue of the Hilbert space in QFT when I originally posted. I guess what I was really wondering is what are the states in the "non-abstract sense"? e.g. in QM you could choose to work in the position rep, express the operators of your Hilbert space in their position rep, and have wavefunctions that are dependent on space. So what I'm wondering is can you do something equivalent in QFT....after so called "second quantization", the operators P, X from QM have been relagated to labels, and the things satisfying the Schroedinger equation (or the KG or Dirac equation), that were our states in QM (Wavefunctions or more abstract Hilbert vectors) have been promoted to operators. But the new states of these field operators, are at the moment just abstract Fock vectors in my mind at the moment, I can't quite see how to visualize them, and am still left with the question what actually are they?

They are not configurations of the field, since the field is the operator, but then what are they...." ?

#### atyy

I've read that part a few days ago when you first shared it. But there seems to be differences. And I have the same questions as the OP in the other thread which is unanswered (instead Eugene tried to share his new views which just confused everything). Here is the specific question:

"I was already aware of the Fock space being the analogue of the Hilbert space in QFT when I originally posted. I guess what I was really wondering is what are the states in the "non-abstract sense"? e.g. in QM you could choose to work in the position rep, express the operators of your Hilbert space in their position rep, and have wavefunctions that are dependent on space. So what I'm wondering is can you do something equivalent in QFT....after so called "second quantization", the operators P, X from QM have been relagated to labels, and the things satisfying the Schroedinger equation (or the KG or Dirac equation), that were our states in QM (Wavefunctions or more abstract Hilbert vectors) have been promoted to operators. But the new states of these field operators, are at the moment just abstract Fock vectors in my mind at the moment, I can't quite see how to visualize them, and am still left with the question what actually are they?

They are not configurations of the field, since the field is the operator, but then what are they...." ?
Are you asking what Fock space is? Fock space is essentially the Hilbert space for one particle, and the Hilbert space for two identical particles, and the Hilbert space for three identical particles, and the Hilbert space for four identical particles, and the Hilbert space for five identical particles, and .....

#### lucas_

Are you asking what Fock space is? Fock space is essentially the Hilbert space for one particle, and the Hilbert space for two identical particles, and the Hilbert space for three identical particles, and the Hilbert space for four identical particles, and the Hilbert space for five identical particles, and .....
I have rough idea what are fock space. But again my question is. In QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?

#### bhobba

Mentor
In QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?
They are states. The KG and Dirac equation have issues (eg negative probabilities in current densities) which is why we had to go to QFT. These problems all pointed to the one difficulty - if you include relativity the number of particles turns out not to be fixed (negative probabilities were particles of opposite charge which is why it was negative - the probabilities were still positive) - they can be created and destroyed. That's exactly what a Fock space handles.

The following explains it all carefully:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

Thanks
Bill

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#### atyy

I have rough idea what are fock space. But again my question is. In QM the solutions of the Schrodinger equation are the actual states themselves, but in QFT the solutions of the KG eqn or the Dirac equation say, are the field operators not the states. So what are the actual states?
The states are elements of the Fock space.

#### lucas_

They are states. The KG and Dirac equation have issues (eg negative probabilities) which is why we had to go to QFT.
You mean the operator fields are the states themselves? Or you have to apply the operators to the state vector? But this is not directly like in QM. It's like after getting the operator fields, you have to apply another operation to get the states. Can you give any everyday example of this.. is this natural thing to do at all?

The following explains it all carefully:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

Thanks
Bill

#### lucas_

The states are elements of the Fock space.
you mean the field operators themselves are elements of Fock space? but fock space (like Hilbert space) is composed of vector, not operator.

#### bhobba

Mentor
You mean the operator fields are the states themselves?
No. The states are from the Fock space because the number of particles are not fixed.

Its just a title.

You will only understand QFT by putting in the hard yards and studying it.

Thanks
Bill

#### atyy

you mean the field operators themselves are elements of Fock space? but fock space (like Hilbert space) is composed of vector, not operator.
States are vectors.

Field operators are not states. Roughly, a field operator acting on a state in Fock space will create a state with one additional particle.

#### bhobba

Mentor
you mean the field operators themselves are elements of Fock space? but fock space (like Hilbert space) is composed of vector, not operator.
They are different things - the fields and the states its operates on.

The mathematical connection is it can be shown the field is the same as creation and annihilation operators.

Thanks
Bill

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