# Dimensions of the pion decay constant

anthony2005
The pion decay constant $f_{\pi}$ is defined by
$$<0|\overline{d}\gamma^{0}\gamma^{5}u|\pi^{+}>=f_{\pi}m_{\pi}$$
where I have set the momentum $\boldsymbol{p}=0$ (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?

Gold Member
2022 Award
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

https://arxiv.org/abs/nucl-th/9706075

anthony2005
Thanks, that perfectly solved my problem. I wasn't thinking about the delta function carrying a dimension, but indeed if we think of $\delta\left(\boldsymbol{p}\right)\propto\int d^{3}xe^{i\boldsymbol{p}\cdot\boldsymbol{x}}$ then the measure will lead to -3.

Staff Emeritus
Thanks, that perfectly solved my problem. I wasn't thinking about the delta function carrying a dimension, but indeed if we think of $\delta\left(\boldsymbol{p}\right)\propto\int d^{3}xe^{i\boldsymbol{p}\cdot\boldsymbol{x}}$ then the measure will lead to -3.