- #1
JunhoPhysics
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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...
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...
