A Hamiltonian Invariant Under Wavefunction Phase (or Gauge) Transformations

We want to investigate what it takes for the Hamiltonian to be invariant under a local phase transformation of the wave function.

$$\bgroup\color{black}\psi(\vec{r},t)\to e^{i\lambda(\vec{r},t)}\psi(\vec{r},t)\egroup$$

That is, we can change the phase by a different amount at each point in spacetime and the physics will remain unchanged. We know that the absolute square of the wavefunction is the same. The Schrödinger must also be unchanged.

$$\bgroup\color{black}\left(\vec{p}+{e\over c}\vec{A}\right)^2\psi=(E+e\phi)\psi\egroup$$

So let's postulate the following transformation then see what we need to keep the equation invariant.

$$\begin{eqnarray*} \psi(\vec{r},t) & \to & e^{i\lambda(\vec{r},t)}\psi(\vec{r},t)... ...to & \vec{A}+\vec{\Delta A} \\ \phi & \to & \phi+\Delta\phi \\ \end{eqnarray*}$$

We now need to apply this transformation to the Schrödinger equation.

$$\begin{eqnarray*} \left({\hbar\over i}\vec{\nabla}+{e\over c}\vec{A}+{e\over c}\... ...r\partial t}+e\phi+e\Delta\phi\right)e^{i\lambda(\vec{r},t)}\psi \end{eqnarray*}$$

Now we will apply the differential operator to the exponential to identify the new terms. Note that $\vec{\nabla}e^{i\lambda(\vec{r},t)}=e^{i\lambda(\vec{r},t)}i\vec{\nabla}\lambda(\vec{r},t)$.

$$\begin{eqnarray*} e^{i\lambda(\vec{r},t)}\left({\hbar\over i}\vec{\nabla}+{e\ove... ...hi- \hbar{\partial\lambda(\vec{r},t)\over\partial t}\right)\psi \end{eqnarray*}$$

Its easy to see that we can leave this equation invariant with the following choices.

$$\begin{eqnarray*} \vec{\Delta A} & = & -{\hbar c\over e}\vec{\nabla}\lambda(\vec... ... & = & {\hbar\over e}{\partial\lambda(\vec{r},t)\over\partial t} \end{eqnarray*}$$

We can argue that we need Electromagnetism to give us the local phase transformation symmetry for electrons. We now rewrite the gauge transformation in the more conventional way, the convention being set before quantum mechanics.

$$\begin{eqnarray*} \psi(\vec{r},t) & \to & e^{i\lambda(\vec{r},t)}\psi(\vec{r},t)... ...al t} \\ f(\vec{r},t) & = & {\hbar c\over e}\lambda(\vec{r},t). \end{eqnarray*}$$