Question 7.3: PCAC and the Quark Condensate Calculate ∫∫〈0|[Ab0^b(0, x),[......
PCAC and the Quark Condensate
Calculate
\int\int{\bigl\langle0\bigr|}\left[A_{0}^{b}(0,x), \left[A_{0}^{a}(0,0),m_{u}\bar{u}(y)u(y)+m_{u}\bar{d}(y)d(y)\right] \right]\left|\,0\right\rangle\,\mathrm{d}^{3}x\,\mathrm{d}^{3}y (1)
with
A_{0}^{b}(x)=\bar{q}(x)\gamma_{0}\gamma_{5}\frac{\tau^{b}}2q(x)\ ,\quad q(x)=\left(\begin{array}{l}{{u(x)}}\\ {{d(x)}}\end{array}\right)\ . (2)
First we write
m_{\mathrm{u}}\bar{u}(y)u(y)+m_{\mathrm{d}}\bar{d}(y)d(y)=\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}\bar{q}(y)q(y)+\frac{m_{\mathrm{u}}-m_{\mathrm{d}}}{2}\bar{q}(y)\tau^{3}q(y) (3)
and then we write, with simultaneous Lorentz and isospin indices α, β, γ, δ,
\left[{\bar{q}}(0){\frac{\tau^{a}}{2}}\gamma_{0}\gamma_{5}q(0),\,{\bar{q}}(y)q(y)\right]=\left[q_{\alpha}^{+}(0)q_{\beta}(0),q_{\gamma}^{+}(y)q_{\delta}(y)\right]\left(\frac{\tau^{a}}2\gamma_{5}\right)_{\alpha\beta}(\gamma_{0})_{\gamma\delta}=\left(-q_{\alpha}^{+}q_{\gamma}^{+}\left\{q_{\beta},q_{\delta}\right\}+q_{\alpha}^{+}\left\{q_{\beta}q_{\gamma}^{+}\right\}q_{\delta}-q_{\gamma}^{+}\left\{q_{\alpha}^{+},q_{\delta}\right\}q_{\beta} +\left\{q_{\alpha}^{+},q_{\gamma}^{+}\right\}q_{\delta}q_{\beta}\right)\left(\frac{\tau^{a}}2\gamma_{5}\right)_{\alpha\beta}\left(\gamma_{0}\right)_{\gamma\delta} (4)
as
(\cdot\cdot\cdot)=-q_{\alpha}^{+}q_{\gamma}^{+}q_{\beta}q_{\delta}-q_{\alpha}^{+}q_{\gamma}^{+}q_{\delta}q_{\beta}+q_{\alpha}^{+}q_{\beta}q_{\gamma}^{+}q_{\delta}+q_{\alpha}^{+}q_{\gamma}^{+}q_{\beta}q_{\delta} \\ -q_{\gamma}^{+}q_{\alpha}^{+}q_{\delta}q_{\beta}-q_{\gamma}^{+}q_{\delta}q_{\alpha}^{+}q_{\beta}+q_{\alpha}^{+}q_{\gamma}^{+}q_{\delta}q_{\beta}+q_{\gamma}^{+}q_{\alpha}^{+}q_{\delta}q_{\beta}\\ =q_{\alpha}^{+}q_{\gamma}^{+}q_{\delta}q_{\beta}(+1-1-1+1+1-1)+\left[q_{\alpha}^{+}q_{\beta},q_{\gamma}^{+}q_{\delta}\right]\ . (5)
Now we use
\left\{q_{\beta}(0,0),q_{\delta}(0,y)\right\}=0=\left\{q_{\alpha}^{+}(0,0),q_{\gamma}^{+}(0,y)\right\}\ , \\\left\{q_{\beta}(0,0),q_{\gamma}^{+}(0,y)\right\}=\delta_{\beta\gamma}\delta^{3}(y)\,\,\,,\\\left\{q_{\alpha}^{+}(0,0),q_{\delta}(0,y)\right\}=\delta_{\alpha\delta}\delta^{3}(y) (6)
to get
[\cdot\cdot\cdot,\cdot\cdot\cdot]=\delta^{3}(y)\biggl[q_{\alpha}^{+}(0)q_{\delta}(0)\left(\frac{\tau^{a}}2\gamma_{5}\gamma_{0}\right)_{\alpha\delta}-q_{\gamma}^{+}(0)q_{\beta}(0)\left(\gamma_{0}\frac{\tau^{a}}2\gamma_{5}\right)_{\gamma\beta}\biggr]\\ =\delta^{3}(y)q_{\alpha}^{+}(0)q_{\delta}(0)\left(\tau^{a}\gamma_{5}\gamma_{0}\right)_{\alpha\delta}\ . (7)
Similarly we obtain
\left[\bar{q}(0)\frac{\tau^{a}}{2}\gamma_{0}\gamma_{5}q(0),\,\bar{q}(0,y)\tau^{3}q(0,y)\right] \\ =\delta^{3}(y)\biggl[q_{\alpha}^{+}(0)q_{\delta}(0)\left(\frac{\tau^{a}}2\gamma_{5}\gamma_{0}\tau^{3}\right)_{\alpha\delta}-q_{\gamma}^{+}(0)q_{\beta}(0)\left(\gamma_{0}\tau^{3}\frac{\tau^{a}}2\gamma_{5}\right)_{\gamma\beta}\biggr]\\ =\delta^{3}(y)q_{\alpha}^{+}(0)q_{\delta}(0)\biggl(\gamma_{5}\gamma_{0}\,\underbrace{\frac{1}{2}\,\biggl\{\tau^{a},\tau^{3}\biggr\}}_{\delta a_3}\biggr)_{\alpha\delta} (8)
and thus
\int\,\mathrm{d}^{3}y\,\left[A_{0}^{a}(0,0),\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}\bar{q}(0,y) \mathbb{1} q(0,y)+\frac{m_{\mathrm{u}} m_{\mathrm{d}}}{2}\bar{q}(0,y)\,\tau^{3}\,q(0,y)\right] \\=\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}q_{\alpha}^{+}(0)q_{\delta}(0)\,\left(\gamma_{5}\gamma_{0}\tau^{a}\right)_{\alpha\delta}\\+{\frac{m_{\mathrm{u}}-m_d}{2}}q_{\alpha}^{+}(0)q_{\delta}(0)\,(\gamma_{5}\gamma_{0})_{\alpha\delta}\,\delta_{a3}=J(0)\ . (9)
The next step just repeats the preceding one:
\int\,\mathrm{d}^{3}x\left[A_{0}^{b}(0,x),\,J(0)\right] \\=\int\,\mathrm{d}^{3}r\,\delta^{3}(x)\biggl[\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}q_{\alpha}^{\dagger}(0)q_{\delta}(0)\biggl(\frac{\tau^{b}}{2}\gamma_{5}\gamma_{5}\gamma_{0}\tau^{a}-\gamma_{5}\gamma_{0}\tau^{a}\frac{\tau^{b}}{2}\gamma_{5}\biggr)_{\alpha\delta} +{\frac{m_{\mathrm{u}}-m_{\mathrm{d}}}{2}}q_{\alpha}^{\dagger}(0)q_{\delta}(0)\biggl({\frac{\tau^{b}}{2}}\gamma_{5}\gamma_{5}\gamma_{0}\delta_{a3}-\gamma_{5}\gamma_{0}\delta_{a3}{\frac{\tau^{b}}{2}}\gamma_{5}\biggr)_{\alpha\delta}\biggr] \\=q_{\alpha}^{\dagger}(0)q\delta(0)\bigg[\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}2\left(\gamma_{0}\frac12\{\tau^{b},\tau^{a}\}\right)_{\alpha\delta}+\frac{m_{\mathrm{u}}-m_{\mathrm{d}}}2\left(\gamma_{0}\tau^{b}\delta_{a3}\right)_{\alpha\delta}\bigg]\\=\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}\bar{q}(0)\,q(0)\,\delta_{a b}+\frac{m_{\mathrm{u}}-m_{\mathrm{d}}}{2}\bar{q}(0)\,\tau^{b}\,q(0)\,\delta_{a3}~. (10)
Finally we take the vacuum expectation value of this expression. As the vacuum cannot carry any quantum number, such as isospin, the second term does not contribute:
\int\mathrm{d}^{3}x~\mathrm{d}^{3}y \\\times\left\langle0\left|\left[A_{0}^{b}(0,x),\left[A_{0}^{a}(0,{ O}),m_{\mathrm{u}}\bar{u}(0,y)u(0,y)+m_{\mathrm{d}}\bar{d}(0,y)d(0,y)\right]\right]\right|0\right\rangle=\frac{m_{\mathrm{u}}+m_{\mathrm{d}}}{2}\langle0|\bar{u}(0)u(0)+\bar{d}(0)d(0)|0\rangle\,\delta_{a b}\,\,\,. (11)