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I have a question regarding a homework problem for my quantum field theory assignment.

For the purposes of the question, we can just assume the Lagrangian is that of a real scalar field: $$\mathcal{L}=\frac{1}{2}\left( \partial _\mu \phi\right) ^2-\frac{1}{2}m^2\phi ^2$$.

The first part of the problem was to calculate the energy-momentum tensor corresponding to the translation symmetry $x\rightarrow x+a$. I obtained: $$ T_\mu ^\nu =(\partial _\mu \phi )(\partial ^\nu \phi )-\delta _\mu ^\nu \mathcal{L}. $$ So far, so good (at least I hope so).

The second part of the problem states: "Write the corresponding Ward Identity." What does he mean by this? I looked up "Ward Identity" in the index of our text (Peskin and Scroeder), but was unable to find something that seemed relevant to the problem. Where do I even begin?

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I'm not sure if I can wrap my head around this stuff well enough to explain it, but I would point you to the end of section 9.6 in P&S. It may be that the problem is asking for the Schwinger-Dyson equation for $T^{\nu}_{\mu}$. – David Zaslavsky♦ Nov 1 '11 at 6:55
I think for translations $a^\mu$, your conserved current is just $j^{\mu}(x)=a_{\nu}T^{\mu\nu}(x)$. The Ward identity is then just the vanishing of the derivative of the VEV of time ordered products of this current and the field (googling "Ward identity for continous symmetry" should turn something up). – twistor59 Nov 1 '11 at 8:10

2 Answers

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The Noether current is a conserved quantity in classical field theory. In quantum theory, it is mirrored by the Ward identity. You may possibly find this formula looking into the derivation for Schwinger-Dyson equations.

It is derived in chp 22 of Srednicki's book: http://web.physics.ucsb.edu/~mark/qft.html.

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In quantum field theory, a Ward (Takahashi) identity is an identity between correlation functions that follows from the global or gauged symmetries of the theory, and which remains valid after renormalization. For more setail you may check wikipwdia.

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While technically correct, this doesn't seem particularly helpful... – David Zaslavsky♦ Nov 1 '11 at 5:02
Also, it was copied directly from wikipedia.. – Bjorn Wesen Nov 1 '11 at 9:49
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I guess you love to see your name in print – Physiks lover Nov 1 '11 at 17:45

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