Dimensions of the pion decay constant

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SUMMARY

The pion decay constant, denoted as f_{\pi}, is defined by the equation <0|\overline{d}\gamma^{0}\gamma^{5}u|\pi^{+}>=f_{\pi}m_{\pi}, where the dimensions on both sides must match. The left side has a dimension of 3 due to quark contributions, while the right side is dimension 2, indicating a mismatch. The normalization of the single-particle state |π+⟩ is crucial, with the standard definition involving the creation operator and the delta function, which carries a dimension of -3. The correct value of f_{\pi} is approximately 130 MeV, assuming consistent state normalization.

PREREQUISITES
  • Understanding of quantum field theory concepts, particularly pion decay.
  • Familiarity with the normalization of quantum states and creation operators.
  • Knowledge of dimensional analysis in particle physics.
  • Basic grasp of chiral symmetry in Quantum Chromodynamics (QCD).
NEXT STEPS
  • Study the normalization of states in quantum field theory.
  • Research the implications of dimensional analysis in particle physics.
  • Explore chiral symmetry and its role in QCD, referencing the review at https://arxiv.org/abs/nucl-th/9706075.
  • Investigate the variations of the pion decay constant f_{\pi} across different literature sources.
USEFUL FOR

Physicists, particularly those specializing in particle physics and quantum field theory, as well as students seeking to understand the intricacies of pion decay and dimensional analysis in theoretical frameworks.

anthony2005
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The pion decay constant f_{\pi} is defined by
&lt;0|\overline{d}\gamma^{0}\gamma^{5}u|\pi^{+}&gt;=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?
 
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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

https://arxiv.org/abs/nucl-th/9706075
 
Thanks, that perfectly solved my problem. I wasn't thinking about the delta function carrying a dimension, but indeed if we think of <br /> \delta\left(\boldsymbol{p}\right)\propto\int d^{3}xe^{i\boldsymbol{p}\cdot\boldsymbol{x}} then the measure will lead to -3.
 
anthony2005 said:
Thanks, that perfectly solved my problem. I wasn't thinking about the delta function carrying a dimension, but indeed if we think of <br /> \delta\left(\boldsymbol{p}\right)\propto\int d^{3}xe^{i\boldsymbol{p}\cdot\boldsymbol{x}} then the measure will lead to -3.

##\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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