Ward-Takahashi identity

From Physics wiki

Consider the time-ordered vacuum expectation value of some operator $\mathcal{O}(\phi)\,$ ,

$\left\langle \mathcal{O} (\phi)\right\rangle = \int\!\mathcal{D}\phi\, \mathcal{O}(\phi) e^{i S[\phi]}\,$ ,

and consider some infinitesimal transformation of the fields $Q[\phi(x)] = \phi'(x) - \phi(x) = \delta\phi(x)\,$ , e.g., a gauge transformation. We may write $Q[\phi(x)] = i \epsilon_a(x) G_a \phi(x) \,$ where $G_a\,$ is the generator of the transformation, and $\epsilon_a(x)\,$ is some infinitesimal parameter. Then

$\left\langle \mathcal{O} (\phi')\right\rangle = \int\!\mathcal{D}\phi\, \mathcal{O}(\phi') e^{i S[\phi']}\,$ ,

but for most cases under consideration, the functional measure is invariant, i.e., $\mathcal{D}\phi' = \mathcal{D}\phi\,$ . Thus

$0\,$	$= \left\langle \mathcal{O} (\phi') - \mathcal{O} (\phi)\right\rangle\,$ ,
	$= \int\!\mathcal{D}\phi\, \left( \mathcal{O}(\phi') e^{i S[\phi']} - \mathcal{O}(\phi) e^{i S[\phi]}\right)\,$ ,
	$\approx \int\!\mathcal{D}\phi\, \left[ (\mathcal{O}(\phi) + Q[\mathcal{O}(\phi)]) e^{i S[\phi]} ( 1 + i Q[S[\phi]] + ...) - \mathcal{O}(\phi) e^{i S[\phi]}\right]\,$ .

If the transformation is a classical symmetry, i.e., $S[\phi] \to S[\phi] - \int_M \epsilon_a(x) \, \partial_\mu J_a^\mu(x)\,$ (see Noether current)

$\int\!\mathcal{D}\phi\, e^{i S[\phi]} ( Q[\mathcal{O}(\phi)] + i \mathcal{O} (\phi) Q[S[\phi]] ) = 0\,$ ,

$\left\langle Q[\mathcal{O}(\phi)]\right\rangle - i \int_{ M}\! \epsilon_a(x) \partial_\mu \left\langle \mathcal{O} (\phi) J_a^\mu(x) \right\rangle = 0\,$ ,

$\left\langle Q[\mathcal{O}(\phi)] \right\rangle = i \int_{ M} \epsilon_a(x)\, \partial_\mu \left\langle \mathcal{O} (\phi) J_a^\mu(x) \right\rangle \,$ .

Conserved charges

Suppose $\mathcal{O} (\phi) = \phi(x_1)\phi(x_2)...\phi(x_n)\,$ . Then

$i \int_{ M} \epsilon_a(x)\, \partial_\mu \left\langle \phi(x_1)\phi(x_2)...\phi(x_n) J_a^\mu (x)\right\rangle\,$	$= i \sum_i \left\langle \phi(x_1)... G_a \phi(x_i) \epsilon_a(x_i)...\phi(x_n)\right\rangle\,$ ,
	$= i \int_{ M} \epsilon_a(x) \sum_i \delta^{(d)}(x-x_i)\left\langle \phi(x_1)... G_a \phi(x_i) ...\phi(x_n)\right\rangle\,$ ,

and, since $\epsilon_a(x)\,$ is an arbitrary parameter,

$i \partial_\mu \left\langle \phi(x_1)\phi(x_2)...\phi(x_n) J_a^\mu (x)\right\rangle = i \sum_i \delta^{(d)}(x-x_i) \left\langle \phi(x_1)... G_a \phi(x_i) ...\phi(x_n)\right\rangle$ .

Now suppose $\mathcal{O}(\phi) = \phi(x_1) \mathcal{A}(\phi)\,$ where the time-component $x_1^0\,$ differs from all those in $\mathcal{A}(\phi)\,$ . Then, suppose we integrate the previous expression over an infinitesimally thin volume bounded by times $t_- < x_1^0 < t_+\,$ and by spatial infinity such that none of the points in $\mathcal{A}(\phi)\,$ lie in this volume. Then, by the divergence theorem, defining

$Q_a = \int\!d^{d-1}x\,J_a^0 (\mathbf{x})\,$ , we have

$i \left\langle \mathcal{A}(\phi) G_a \phi(x_1) \right\rangle$	$= \lim_{t_- \to t_+} i \left.\int\!d^{d-1}x\, \left\langle \mathcal{A}(\phi) \phi(x_1) J_a^0 (\mathbf{x}, t)\right\rangle\right\|_{t_-}^{t_+}\,$
	$= \lim_{t_- \to t_+}i \left\langle \mathcal{A}(\phi) \phi(x_1) Q_a(t_+)\right\rangle - i \int\!d^{d-1}x\, \left\langle \mathcal{A}(\phi) \phi(x_1) Q_a(t_-)\right\rangle\,$
	$= \lim_{t_- \to t_+} i \left\langle \mathcal{A}(\phi) Q_a(t_+) \phi(x_1)\right\rangle - i \int\!d^{d-1}x\, \left\langle \mathcal{A}(\phi) \phi(x_1) Q_a(t_-)\right\rangle\,$ (since the correlation function is time-ordered),
	$= i \left\langle \mathcal{A}(\phi) \left[ Q_a(x_1^0) , \phi(x_1) \right]\right\rangle\,$ .

Since $\mathcal{A}(\phi)\,$ is arbitrary, we may further argue, by generalizing the path integral to go between arbitrary initial and final states, that

$G_a \phi(x) = \left[ Q_a , \phi \right]\,$ .

Thus conserved charges are the generators of the symmetry transformations.

Ward-Takahashi identity

From Physics wiki

Conserved charges

References

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