# Transforming the Hamiltonian of a free quantum field

I have been trying to explicitly Lorentz transform the Hamiltonian of a free quantum field between two inertial observers, instead of reading it off the manifestly Lorentz invariant action. My reasoning was that from the Schrödinger equation

$$i\frac{d}{d t}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle,$$

applying the unitary Lorentz transformation matrix $U(\Lambda)$ on both sides,

$$i\frac{\partial}{\partial t}\Big(U(\Lambda)|\psi(t)\rangle\Big)=\Big(U(\Lambda)\hat{H}U^{-1}(\Lambda)\Big)\Big(U(\Lambda)|\psi(t)\rangle\Big)$$

and so that by definition

$$\hat H'=U(\Lambda)\hat{H}U(\Lambda)^{-1}$$

should be the Hamiltonian in a moving frame. Starting with $\hat{H} = \int \frac{d^3\vec{p}}{(2\pi)^3}E_{\vec{p}}\hat{a}^\dagger_\vec{p}\hat{a}_ \vec{p}$, I used the unitary transformations given in Peskin and Schroeder:

$$U(\Lambda)\hat{a}^\dagger_\vec{p}U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda\vec{p}}}{E_{\vec{p}}}}\hat{a}^\dagger_{\Lambda\vec{ p}}.$$

However, if you use this after using the Lorentz invariant measure of integration $\int d^3\vec{p}/E_\vec{p}$, you find

$$H' = \gamma \int \frac{d^3\vec{p'}}{(2\pi)^3}(E_\vec{p'}+\vec{v}\cdot\vec{p'})\hat{a}^\dagger_\vec{p'}\hat{a}_ \vec{p'},$$

where $v$ and $\gamma$ denote the velocity between the two frames and $\vec{p'}$ denotes the set of momenta in the moving frame. Now of course, the answer should be

$$H' = \int \frac{d^3\vec{p'}}{(2\pi)^3}E_{\vec{p}'}\hat{a}^\dagger_{\vec{p}'}\hat{a}_ {\vec{p}'},$$

i.e. identical in form in every frame.

I'm able to get this result only by also acting with the unitary on the field momentum operator and then using the transformation rules of tensor components: $$H' = \gamma (H-\vec{v}\cdot\vec{P}).$$

On the one hand this makes sense... the Hamiltonian is a component of a tensor and so it should mix with momentum when transforming. But from the Schrödinger equation, I would have expected that hitting it with the unitary might be enough (up to perhaps a factor of $\gamma$ since really I should have accounting for time dilation in the Schrödinger equation $\frac{d}{dt} = \frac{d\tau}{dt}\frac{d}{d\tau}$ )

What do you think? Should I have been able to get the right Hamiltonian using only $\hat H'=U(\Lambda)\hat{H}U(\Lambda)^{-1}$?

• What you get is the right way the fourth component of the 4-momentum, the energy, changes under Lorentz transformations. – Jon Sep 17 '18 at 9:12