Field redefinition changing energy of one particle states

Question

Consider Free scalar field theory $$ S[\psi^*,\psi] = -\int dx^4 (\partial_\mu \psi^* \partial^\mu \psi + m^2 \psi^* \psi) $$ Upon usual quantisation $$ \hat{\psi}(x) = \int \frac{d^3p}{(2\pi)^3 2 E_p} \left( a_p e^{i x_\mu p^\mu} + b^\dagger e^{-i x_\mu p^\mu}\right), \quad \quad E_p=\sqrt{p^2+m^2},$$ one finds the usual Hamiltonian $$\hat{\mathcal{H}} = \Pi \dot{\psi} + \Pi^* \dot{\psi^*} - \mathcal{L} = \int d^3p E_p \left( a^\dagger_p a_p+ b^\dagger_p b_p \right).$$

Suppose, however, I redefined my fields as $\psi = e^{-i\alpha t}\psi'$, where $\alpha$ is a constant, so that $$ S[\psi^*,\psi] = -\int dx^4 \left[\eta^{\mu\nu}(\partial_\mu + i \alpha \delta_{\mu t})\psi^*(\partial_\nu -i\alpha \delta_{\nu t}) \psi + m^2 \psi^* \psi \right] .$$ Notice this theory now looks CPT violating. Quantising this theory I find that $$\hat{\psi}'(x) = \hat{\psi}(x) e^{i\alpha t}, \quad \quad \Pi'(x) = \Pi(x) e^{-i\alpha t}, $$ and have found (apologies for skipping many steps here; it's the usual canonical quantisation procedure and would clutter the page. I will give more details if requested)

$$\hat{\mathcal{H}}' = \Pi' \dot{\psi'} + \Pi'^* \dot{\psi'^*} - \mathcal{L'}$$ $$ =\Pi \dot{\psi} + i \alpha \Pi \psi + \Pi^* \dot{\psi^*} - i\alpha \Pi^* \psi^* - \mathcal{L}$$ $$ = \hat{\mathcal{H}} + i \alpha( \Pi \psi -\Pi^* \psi^*)$$ $$ =\int d^3p \left[ (E_p-\alpha)a^\dagger_p a_p + (E_p+\alpha)b^\dagger_p b_p \right].$$

So the field redefinition has seemingly changed the energies of the 1-particle states.

1) Are the energies of the canonically quantised 1 particle states supposed to be field redefinition invariant? 2) If so, is there something wrong with the redefinition $\psi = e^{-i\alpha t}\psi'$ or have I missed an extra piece that arises from this redefinition?

Answer 1

This is a very interesting question. Think about this: what happens if you add an electric potential to the original field (by minimal coupling)? You will see the same result! $\alpha$ in your expression can be actually regarded as $q\phi$.

This is like, say, we have a charge of $-3q$ somewhere in space and $2q$ at some other place. If we choose the uniform (trivial) electric potential to be $0$ then the energy of this configuration is just $0$. But is we set the potential all over the space to be $\phi_0$ then the energy would be $-3q\phi_0+2q\phi_0=-q\phi_0$.

The point is, there is nothing wrong with your calculation. What you did is actually a local gauge transformation. The original action obviously possesses a global $U(1)$ symmetry (invariant under $\psi\mapsto e^{i\theta}\psi$, where $\theta$ is independent both in space and time). But if we further require this symmetry to be local, that is, let $\theta$ depend on space time, we need some other terms to cancel terms like $\partial_\mu\theta$. In your case, $\theta(x,t)=-\alpha t$ and we are left with some additional term $\partial_t\theta=-\alpha$ to be canceled. This step is so-called gauging the global U(1) symmetry, and that's how we get electromagnetism interaction in QED. Redefinition of fields is fine but at the same time you should allow your Hamiltonian differing for a gauge transformation.

To make the theory invariant under $\psi\mapsto e^{i\theta(x,t)}\psi$, we will need other terms in the Lagrangian. The terms can be written is a compact form: $$ \begin{eqnarray} \mathcal{L}&=&-(\partial_\mu+iA_\mu)\psi^*(\partial^\mu+iA^\mu)\psi-m^2\psi^*\psi\\ &=&-D_\mu\psi^*D^\mu\psi-m^2\psi^*\psi, \end{eqnarray} $$ and $\psi, A_\mu$ change simultaneously as $$ \psi\mapsto e^{i\theta(x,t)}\psi\\ A_\mu\mapsto A_\mu + \partial_\mu\theta. $$

And in your case, $\theta=-\alpha t$, you see $A_0$ exactly cancels with the additional term you got.

For more details, take a look at

• Examples of “gauging a global symmetry”
• When can a global symmetry be gauged?
• When is a global symmetry a gauge symmetry? [duplicate]