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I've been thinking about the possibility of the proton evolving without external interactions and came up with the following problem. Let's say we have an isolated proton and therfore the Lagrangian for its dynamics is simply QCD's:

$$ {\cal L}_{int} = \sum_{a = 1}^8\bar{q}\gamma_\mu G^{a, \mu}T^a q $$

where $q$ is the quark field with internal sum on flavour and colour degrees of freedom, $G$ is the gluon field and $\{T^a, a = \{1, 2, ..., 8\}\}$ is the set of generators for $SU(3)$.

Thus, according to the QFT there exist the possibility for the proton to evolve with scattering matrix $\hat{S}$ such that

$$ \hat{S} = \exp\left( i{\cal L}_{int} \right) \approx 1 + i{\cal L}_{int} $$

so if initial proton state is a momentum state $|p\rangle$ and final is $|p^\prime\rangle$, then

$$ \langle p^\prime | \hat{S} |p\rangle \approx \langle p^\prime |p\rangle + \langle p^\prime | i{\cal L}_{int}|p\rangle \tag1 $$

Energy conservation provides $p = p^\prime$, hence $\langle p | i{\cal L}_{int}|p\rangle = 0$ which provides $i{\cal L}_{int}|p\rangle = |q\rangle \neq |p\rangle$, a momentum state different from the initial one.

But my quandary is: is there any way to determine in case $p \neq p^\prime$ that $\langle p^\prime | i{\cal L}_{int}|p\rangle \equiv 0$ or different from zero?

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  • $\begingroup$ @ChiralAnomaly None of the states are single-particle, understanding single-particle as elementary state of the theory. They are proton momentum states and proton is an hadron made out of quarks, antiquarks and gluons $\endgroup$
    – Vicky
    Commented May 2, 2021 at 15:46
  • $\begingroup$ @ChiralAnomaly $|p\rangle$ is an unknown state for the proton that satisfies to be a momentum eigenstate. But it doesn't imply that all quarks and gluons have the same momentum $\endgroup$
    – Vicky
    Commented May 2, 2021 at 20:04
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    $\begingroup$ Okay, I think I was getting distracted by details that don't affect the core of the question (so I deleted my comment). Suppose $L_{int}$ is invariant under translations, so that it commutes with the momentum operator $P$. If $P|p\rangle=p|p\rangle$, then $P L_{int}|p\rangle=L_{int}P|p\rangle=pL_{int}|p\rangle$, which implies that $L_{int}|p\rangle$ is orthogonal to $|p'\rangle$ for $p\neq p'$. Is that what you're looking for? $\endgroup$
    – Chiral Anomaly
    Commented May 2, 2021 at 21:38
  • $\begingroup$ The interaction term you wrote down is an interaction term of quarks with gluons, which ultimately would contribute to interaction terms for protons in obviously inaccessible ways, in a notional effective lagrangian. But, ultimately, they also shape the proton wave functions, propagating without interactions, which are also the major sticky point of QCD, even in lattice gauge simulations thereof. What is the point of spatchcocking that term inside a sandwich of proton states of undisclosed wavefunction? What do you imagine it represents? $\endgroup$
    – Cosmas Zachos
    Commented May 2, 2021 at 22:22
  • $\begingroup$ @CosmasZachos I'm trying to justify DGLAP equations for the evolution of the parton distribution function. This correction is of order $e^2 g_s^2$ so it comes from the 4th power in the Taylor expansion of scattering matrix, i.e. $\hat{S} \approx 1 + \hat{S}^{(1)} + \hat{S}^{(2)} + \hat{S}^{(3)} + \hat{S}^{(4)}$ with $\hat{S}^{(4)} \simeq {\cal L}^4 \simeq e^2 g_s^2$. If this 4th power is the 1st correction, that means that $\hat{S}^{(3)} \simeq e^2 g_s ({\cal L}_{int}^{QED})^2 \times {\cal L}_{int}^{QCD}$ doesn't contribute and since QCD Lagrangian... (goes on) $\endgroup$
    – Vicky
    Commented May 3, 2021 at 11:17

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