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I Dimensions of the pion decay constant

  1. Nov 25, 2016 #1
    The pion decay constant [itex]f_{\pi}[/itex] is defined by
    where I have set the momentum [itex]\boldsymbol{p}=0[/itex] (and used the temporal component of the axial vector).
    Now, at the right-hand-side the dimension is two (the decay constant and the mass are measured in MeV). At the left-hand side, each quark carries dimension 3/2, so the whole axial vector has dimensions 3.
    The dimensions do not match!! Where is the problem?
  2. jcsd
  3. Nov 25, 2016 #2


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    How is the single-particle state ##|\pi^+ \rangle## normalized? The standard definition is that
    $$|\vec{p} \rangle = \hat{a}^{\dagger}(\vec{p})=|\Omega \rangle, \quad [\hat{a}(\vec{p}_1,\hat{a}^{\dagger}(\vec{p}_2)]=(2 \pi)^3 2 \sqrt{m^2+\vec{p}_1^2} \delta^{(3)}(\vec{p}-\vec{p}').$$
    This implies that
    $$\langle \vec{p}_1|\vec{p}_2 \rangle=(2 \pi)^3 2 \sqrt{m^2+\vec{p}_1^2} \delta^{(3)}(\vec{p}_1-\vec{p}_2),$$
    i.e., ##|\vec{p} \rangle## as the dimension ##1/\text{energy}##, and thus your formula is dimensionally correct. You only should take care of the value of ##f_{\pi}##, which varies in the literature. Assuming that your source uses the same normalization of states your pion-decay constant is ##f_{\pi}=\sqrt{2} F_{\pi} \simeq 130 \; \text{MeV}##.

    For a very nice review on chiral symmetry in QCD see

  4. Nov 25, 2016 #3
    Thanks, that perfectly solved my problem. I wasn't thinking about the delta function carrying a dimension, but indeed if we think of [itex]
    \delta\left(\boldsymbol{p}\right)\propto\int d^{3}xe^{i\boldsymbol{p}\cdot\boldsymbol{x}}[/itex] then the measure will lead to -3.
  5. Nov 25, 2016 #4


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    ##\delta^{(3)}(p)## has dimension -3 in post #2. The prefactor ##\sqrt{p^2 - m^2}## has dimension one, which leads to the states having dimension -1.
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