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Lorentz transformation of Hamiltonian

Physics Asked on August 6, 2021

I’m really rusty on LT and hope to get some clarification. How would the Lorentz transformation of the following quantity turn out?

$$Lambda_mu{}^nu int d^3k a^dagger(k)a(k)$$

One Answer

The Hamiltonian, by construction singles out the time component, $$H = p dot{q} - L, quad text{with}quad p = frac{partial L}{partial dot{q}}$$ meaning it breaks Lorentz symmetry explicitly or in other words the Hamiltonian is not a Lorentz covariant quantity. As it is written it makes little sense since you cannot operate with $Lambda_mu^nu$ on a number which is not a Lorentz scalar, vector or tensor.

Alternatively, the Hamiltonian corresponds to the energy of the system and is therefore not a Lorentz invariant quantity either, you could perhaps understand it as the time-like component of a 4-vector, however referring to the Hamiltonian as such is not common.

You should ask yourself what are the elements of your theory and then understand how they transform, namely which representations of the Lorentz group they belong to. You can then address the question of how would the associated Hamiltonian transform.

Answered by ohneVal on August 6, 2021

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