Let's have arbitrary half-integer spin $n + \frac{1}{2}$ representation: $$ \Psi_{\mu_{1}...\mu_{n}} = \begin{pmatrix} \psi_{a, \mu_{1}...\mu_{n}} \\ \kappa^{\dot {a}}_{\quad \mu_{1}...\mu_{n}}\end{pmatrix}, $$ $$ (i\gamma^{\mu}\partial_{\mu} - m)\Psi = 0, \quad \gamma^{\mu}\Psi_{\mu \mu_{1}...\mu_{n - 1}} = 0, $$ $$ \Psi_{\mu_{1}\mu_{i}...\mu_{j}\mu_{n}} = \Psi_{\mu_{1}\mu_{j}...\mu_{i}\mu_{n}}, \quad \partial^{\mu}\Psi_{\mu ...\mu_{n - 1}} = 0. $$ I need to find the general expression for the full energy of corresponding particles, but I don't know how to do that. I tried to do the following.

  1. The solution of equations above may be given as $$ \hat {\Psi}_{\mu_{1}...\mu_{n}} = \sum_{s}\int \left(\hat {a}^{\dagger}_{s}(\mathbf p)u^{s}_{\mu_{1}...\mu_{n}}(\mathbf p)e^{-ipx} + \hat {b}_{s}(\mathbf p)v^{s}_{\mu_{1}...\mu_{n}}(\mathbf p)e^{ipx}\right)\frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2E_{\mathbf p}}}, $$ where $s$ refers to the possible polarizations, $u, v$ are "spinor-tensor" waves which have some normalization law (as example) $u^{+}u = 2E_{\mathbf p}$.

  2. The Dirac equation for each $\mu_{1}...\mu_{n}$ may be rewritten in a form $$ i\partial_{0}\Psi_{\mu_{1}...\mu_{n}} = \hat {H}\Psi_{\mu_{1}...\mu_{n}} = E\Psi_{\mu_{1}...\mu_{n}}. $$ So for finding the full energy I must construct integral like $$ E = \langle \hat {H} \rangle = \int \Psi^{\dagger} \hat {H}\Psi d^{3}\mathbf r = i \int \Psi^{\dagger} \partial_{0}\Psi d^{3}\mathbf r. $$ The expression $$ \int \Psi^{\dagger}\Psi d^{3}\mathbf r $$ must be lorentz invariant. But how to make it invariant? My first idea is convolute $\Psi^{\dagger}\Psi$ as $\Psi^{\dagger}_{\mu_{1}...\mu_{n}}\Psi^{\mu_{1}...\mu_{n}}$, and I think that it is the only possible way to construct the "correct" lorentz invariant object (in vector indices space), but it is not strictly. After that I only use conservation law $\partial_{\alpha}(\bar {\Psi}\gamma^{\alpha}\Psi ) = 0$. But I'm not sure at this.

Can you help?

I know that in general the expression looks like $$ E = \sum_{s}\int (\hat {a}^{\dagger}_{s}(\mathbf p)\hat {a}_{s}(\mathbf p) - \hat {b}_{s}(\mathbf p) \hat {b}^{\dagger}_{s}(\mathbf p ))E_{\mathbf p} d^{3}\mathbf p . $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.