Time independence of a Noether charge in QFT?

In summary, the conversation discusses the issue of time independence for Noether charges in quantum field theory. It is noted that there are two different approaches, canonical quantization and the path integral, which yield different conclusions. The speaker is struggling to understand this contradiction and presents a contradiction in the 2nd approach using a time-ordered correlation function. The conversation concludes by mentioning the importance of the quantum Ward identities in understanding Noether charges in QFT.
  • #1
JunhoPhysics
11
0
In classical field theories, I believe I understood how to derive a Noether charge that corresponds to a symmetry of action. And there is no problem in understanding its time independence.

But in quantum field theory, it looks like the two different approaches,
1) Canonical quantization ##\to## Noether formalism (just like classical theory, but with quantum fields)
2) Path integral ##\to## Schwinger-Dyson equation
yield different conclusions on the time independence of a Noether charge in QFT.

Of course that means I am still not fully understanding the story. So let me show you what kind contradiction is driving me crazy and if you find any step that doesn't look correct, please let me know.

1st approach. Consider a classical action with some symmetry. Derive the corresponding Noether current ##j^\mu## charge ##Q##, which is time independent. Then canonically quantize the theory. Now the quantum version of Noether charge would be
$$\hat Q(t)=\int d^3\vec x\,\hat j^0(t,\vec x).$$
Of course the ordering issue remains in general since ##\hat j^0## is a composite operator. But at least for a simple example such as the spatial momentum operator ##\hat{\vec P}## of the Klein-Gordon theory, we have
$$\hat{\vec P}(t)=\int d^3\vec{x}\,\partial^0\hat\phi(t,\vec x)\vec{\nabla}\hat\phi(t,\vec x)=\int d^3\vec{p}\,\vec{p}\,\hat a_{\vec p}^\dagger\hat a_{\vec p}$$
so it looks truly time independent. Here, I would guess, in general, we can construct a time independent Noether charge in QFT with an appropriate ordering.

2nd approach. But if we consider ##\hat Q(t)## above in path integral formalism, something goes wrong. For example, consider a time ordered correlation function (t_1>t_2) as follows:
$$\begin{align}
\langle T(\hat Q(t_1)-\hat Q(t_2))\prod_k\hat\phi(y_k)\rangle&=\int_Vd^3\vec{x}\langle T\partial_\mu\hat j^\mu(x)\prod_k\hat\phi(y_k)\rangle\nonumber\\&=-i\int_Vd^3\vec{x}\sum_{k_0}\delta^4(x-y_{k_0})\langle T\delta\hat\phi(y_{k_0})\prod_k^{k\neq k_0}\hat\phi(y_k)\rangle\nonumber\\&=-i\sum_{k_0}^{y_{k_0}\in V}\langle T\delta\hat\phi(y_{k_0})\prod_k^{k\neq k_0}\hat\phi(y_k)\rangle\nonumber
\end{align}.$$
Here ##V## is the region surrounded by ##t=t_1##, ##t=t_2##, and the spatial infinity and in the 1st line I used the divergence theorem assuming ##\hat \phi(x)## vanishes fast enough at this spatial infinity. In the 2nd line I used the Schwinger-Dyson equation, which is closely related to the Ward identity.
Now the key observation is, this is not zero! You may think this is because the time ordering split ##\hat Q(t_1)## and ##\hat Q(t_2)## so even though they are the same operator the result might not vanish. But I think this does not explain the issue since when I guess ##\hat Q(t)## is time independent in the 1st approach, I mean ##\hat Q(t_1)\equiv\hat Q(t_2)## as an operator. So the LHS of the above equation is something like ##\langle T(\hat A-\hat A)\rangle=0## in my mind. It vanishes even before taking the time ordering.

Hence I have two possible conclusions:
1) ##\hat Q(t_1)\equiv\hat Q(t_2)##, as an operator, is true. I was missing something in the above 2nd approach.
2) ##\hat Q(t_1)\equiv\hat Q(t_2)##, as an operator, is NOT true. Then what about the simple example above? ##\hat{\vec P}## for KG theory looks truly satisfying ##\hat Q(t_1)\equiv\hat Q(t_2)##... Furthermore, if ##\hat Q(t_1)\equiv\hat Q(t_2)## is NOT true, all the charges like ##\hat P^\mu,~\hat M^{\mu\nu}, \hat D, \hat K^\mu,## in numerous QFT textbooks must have a subscript "##t##" implicitly to represent the hypersurface on which they are computed... right?I would really appreciate if you can help me out with this issue. Even though the explicit form of this question has varied over 3 years, this is still haunting me...
 
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  • #2
The reason is that with the path-integral formulation you always have the time-ordering implemented by construction, and that's wanted, because you use the path integral exactly to (formally) deal with vacuum-expectation values of time-ordered operator product. For a simple treatment of Noether's theorem in QFT in the path-integral formulation, see my QFT manuscript (Sect. 4.6.5)

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
 
  • #3
JunhoPhysics said:
But in quantum field theory, it looks like the two different approaches,
1) Canonical quantization ##\to## Noether formalism (just like classical theory, but with quantum fields)
2) Path integral

Maybe the only illusion more shaky than the path integral is that of "canonical quantization". What actually exists is formal deformation quantization of quantum field theories, and that is what yields the results that are motivated by informal arguments in the traditional literature, this is referenced in prop. 15.25 of the PhysicsForums QFT notes.

Notably the Schwinger-Dyson equation follows this way, without invoking a would-be path integral, (last section of chapter 14.).

But what you really want if you are looking into the quantum observables of Noether currents (prop. 6.7) is the generalization of Schwinger-Dyson to interacting fields, called the quantum Ward identities. These are discussed in the last section of chapter 15.
 

Related to Time independence of a Noether charge in QFT?

1. What is a Noether charge in quantum field theory (QFT)?

A Noether charge in QFT is a conserved quantity associated with a continuous symmetry in the theory. It is named after mathematician Emmy Noether, who showed that for every continuous symmetry in a physical system, there exists a corresponding conserved quantity.

2. What is the significance of time independence in a Noether charge?

Time independence in a Noether charge means that the value of the charge remains constant over time. This is important because it reflects the underlying symmetry of the system, and allows us to make predictions and calculations about the behavior of the system.

3. How is the time independence of a Noether charge related to the conservation of energy?

The time independence of a Noether charge is directly related to the conservation of energy. In physics, Noether's theorem states that for every continuous symmetry in a system, there exists a corresponding conserved quantity. In the case of time translation symmetry, the corresponding conserved quantity is energy.

4. Can the time independence of a Noether charge be violated?

In theory, the time independence of a Noether charge cannot be violated. This is because it is a fundamental property of a system with time translation symmetry and cannot be broken. However, in certain situations, such as in the presence of external forces or non-conservative systems, the time independence of a Noether charge may appear to be violated.

5. How is the time independence of a Noether charge experimentally verified?

The time independence of a Noether charge can be experimentally verified through various methods, such as measuring the energy of a system over time and ensuring that the values remain constant. This can also be done through direct measurements of other conserved quantities, such as momentum or angular momentum, which are related to the Noether charge through Noether's theorem.

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