Local and global U(1) gauge symmetries of Hamiltonian

Question

This question is about understanding the basic ideas behind gauge transformations as I am fairly new to this!

I learned that the Hamiltonian is invariant under global U(1) gauge transformations $\Psi\rightarrow e^{i\lambda}\Psi$, which makes sense, as the phase factors just cancel out on both sites of the Schrödinger equation:

$\hat{H}e^{i\lambda}\Psi=Ee^{i\lambda}\Psi$

Now, am I save to say that the Hamiltonian is, quite generally, not invariant under local U(1) gauge transformations $\Psi\rightarrow e^{i\lambda(x)}\Psi$ (just looking at one spatial dimension for simplicity)? My thinking is that the derivatives from the Hamiltonian act on $\lambda(x)$ like

$\partial_x (\partial_x e^{i\lambda(x)})=i\lambda''e^{i\lambda(x)}-(\lambda')^2 e^{i\lambda(x)}$

which is responsible for the invariance.

Is that correct or did I get the whole concept of gauge transformations wrong and the Hamiltonian is, in fact, invariant under both local and global U(1) gauge transformations?

Answer 1

Yes, you missed the whole point of gauge invariance which clearly has to be a symmetry – the Hamiltonian has to be and is invariant under these transformations. In the case of a global symmetry, a Hamiltonian may afford "not to be symmetric" under it. But once a group of transformations is said to be a gauge symmetry, the failure of $H$ to be invariant would mean that the theory is mathematically inconsistent.

Your calculation of the second derivative of the exponential is also incorrect, the second term should have $(\lambda')^2$, not the factor with double primes. The Hamiltonians of gauge theories are invariant – but the extra terms from the transformation of the gauge fields $A_\mu = (\varphi,\vec A)$ are needed to cancel the terms with $\lambda'$ and $\lambda''$ that you noticed. But they do so. That's why the gauge fields have to exist in gauge invariant theories.