⚽ Goal: represent classical Gauge transformation in quantum form
🧠 Remember: for classical electromagnetic we define gauge transformations as following
$$ \begin{aligned} \Phi(\vec r,t)&\to \Phi_\lambda(\vec r,t)=\Phi(\vec r,t)+ \varphi (\vec r,t) \\ \vec A(\vec r,t)&\to \vec A_\lambda (\vec r,t) = \vec A (\vec r,t)+\vec a(\vec r,t)
\end{aligned}\quad \text{where} \quad \begin{aligned} \vec \nabla \varphi (\vec r,t) + \frac{\partial \vec a (\vec r,t)}{\partial t} &=0 \\ \vec \nabla \times \vec a(\vec r,t)&=0 \end{aligned} $$
We can write the allowed solutions depending on $\lambda (\vec r,t)$ as follows
$$ \vec a(\vec r,t)=\vec \nabla \lambda (\vec r,t) \qquad \varphi (\vec r,t) =- \frac{\partial \lambda (\vec r,t)}{\partial t} $$
From which we can recover the typical form
$$ \boxed{ \begin{aligned} \Phi(\vec r,t)&\to \Phi_\lambda (\vec r,t)=\Phi(\vec r,t)-\frac{\partial \lambda (\vec r,t)}{\partial t} ) \\ \vec A(\vec r,t)&\to \vec A_\lambda (\vec r,t) = \vec A (\vec r,t)+ \vec \nabla \lambda (\vec r,t) \end{aligned} } $$
Our Hamiltonian we see it depends on $\Phi(\vec r,t)$ and $\vec A(\vec r,t)$ directly.
💎 Conclusion: The Hamiltonian is not a gauge invariant operator
We can verify this by applying our previous equations
$$ \begin{aligned} \hat H \to \hat H_\lambda &=\frac{1}{2m} [\hat{\vec p} -q \vec A_\lambda (\hat{\vec r},t)]^2 +q\Phi_\lambda (\hat{\vec r},t) \\ &=\frac{1}{2m} [\hat{\vec p} -q(\vec A(\hat{\vec r},t)+\vec \nabla \lambda (\hat{\vec r},t))]^2+q \left ( \Phi(\hat{\vec r},t)-\frac{\partial \lambda (\hat{\vec r},t)}{\partial t} \right )
\end{aligned} $$
Which we see is not invariant
💼 Case: lets try to define a unitary transformation operator that applies the gauge transformations
We define the unitary operator for the Gauge as
$$ \hat G _\lambda \equiv \exp [ iq\lambda (\hat{\vec r},t)/\hbar] $$
Which has properties
$$ \hat G_\lambda \hat{\vec r} \hat G_\lambda ^\dag =\hat{\vec r} \qquad \hat G \lambda \hat{\vec p} \hat G^\dag\lambda = \hat{\vec p} -q\vec \nabla \lambda (\hat{\vec r},t) $$
where $\hat G_\lambda \hat G_\lambda ^\dag=\hat I = \hat G_\lambda^\dag \hat G_\lambda$ for $\lambda (\hat{\vec r},t)\in \R$
We now try to transformation the time-dependent Schrodinger equation to see how it is affected
We can apply it to both sides of the TDSE and add another $\hat G_\lambda^\dag \hat G_\lambda$
$$ \hat G_\lambda i\hbar \frac{\partial}{\partial t} \ket{\psi} = \hat G_\lambda\hat H \hat G_\lambda^\dag \hat G_\lambda\ket{\psi} $$
Lets look at the left hand side, for that we can do the following
$$ \begin{aligned} \frac{\partial}{\partial t} \hat{G}\lambda |\psi\rangle &= \frac{i q}{\hbar} \frac{\partial \lambda (\hat{\vec{r}}, t)}{\partial t} \hat{G}\lambda |\psi\rangle + \hat{G}\lambda \frac{\partial}{\partial t} |\psi\rangle \\ \Rightarrow \qquad \hat{G}\lambda \frac{\partial}{\partial t} |\psi\rangle &= \frac{\partial}{\partial t} \hat{G}\lambda |\psi\rangle - \frac{i q}{\hbar} \frac{\partial \lambda (\hat{\vec{r}}, t)}{\partial t} \hat{G}\lambda |\psi\rangle \end{aligned} $$
Now looking at the right hand side we can expand to get
$$ \begin{aligned} \hat{G}\lambda \hat{H} \hat{G}\lambda^\dagger &= \frac{1}{2m} \hat{G}\lambda \left[ \hat{\vec{p}} - q \mathbf{A} (\hat{\vec{r}}, t) \right]^2 \hat{G}\lambda^\dagger + q \hat{G}\lambda \Phi (\hat{\vec{r}}, t) \hat{G}\lambda^\dagger \\ &= \frac{1}{2m} \hat{G}\lambda \left[ \hat{\vec{p}} - q \vec{A} (\hat{\vec{r}}, t) \right] \hat{G}\lambda^\dagger \hat{G}\lambda \left[ \hat{\vec{p}} - q \vec{A} (\hat{\vec{r}}, t) \right] \hat{G}\lambda^\dagger + q \Phi (\hat{\vec{r}}, t) \\ &= \frac{1}{2m} \left[ \hat{\vec{p}} - q \left( \mathbf{A} (\hat{\vec{r}}, t) + \boldsymbol{\nabla} \lambda (\hat{\vec{r}}, t) \right) \right]^2 + q \Phi (\hat{\vec{r}}, t) \end{aligned} $$
Now plugging in the left hand side into the equation we can write
$$ \begin{aligned} \left( i\hbar \frac{\partial}{\partial t}+ q\frac{\partial \lambda (\hat{\vec r},t)}{\partial t} \right )\hat G_\lambda \ket{\psi} &=\hat G_\lambda \hat H \hat G_\lambda ^\dag \hat G_\lambda \ket{\psi} \\
i\hbar \frac{\partial}{\partial t}\underbrace{\hat G_\lambda \ket{\psi}}{\ket{\psi\lambda}} &=\underbrace{\left ( \hat G_\lambda \hat H \hat G_\lambda ^\dag - q\frac{\partial \lambda (\hat{\vec r},t)}{\partial t} \right )}{\hat H\lambda }\underbrace{\hat G_\lambda \ket{\psi}}{\ket{\psi\lambda}} \\ i\hbar \frac{\partial}{\partial t} \ket{\psi_\lambda}&=\hat H_{\lambda} \ket{\psi_\lambda} \end{aligned} $$
Where we can use our right hand side calculation to get the expression for $H_\lambda$
$$ \begin{aligned} \hat H_\lambda &= \hat G_\lambda \hat H \hat G_\lambda ^\dag - q\frac{\partial \lambda (\hat{\vec r},t)}{\partial t} \\ &=\frac{1}{2m} \Big [ \hat{\vec{p}} - q \underbrace{\left( \mathbf{A} (\hat{\vec{r}}, t) + \boldsymbol{\nabla} \lambda (\hat{\vec{r}}, t) \right)}{\vec A\lambda (\hat{\vec r},t)} \Big ]^2 + q \underbrace{\left ( \Phi (\hat{\vec{r}}, t) - \frac{\partial \lambda (\hat{\vec r},t)}{\partial t}\right )}{\Phi{\lambda}(\hat{\vec r},t)} \\ &=\frac{1}{2m} \Big [ \hat{\vec{p}} - q \vec A_\lambda (\hat{\vec r},t) \Big ]^2 + q \Phi_{\lambda}(\hat{\vec r},t) \end{aligned} $$
💎 Conclusion: The Schrodinger equation has the same form in any gauge chosen
The matrix elements $\bra{\phi}\hat O \ket{\psi}$ and the expectation value are gauge invariant
We write the expectation as
$$ \bra{\phi}\hat O \ket{\psi}=\bra{\phi_\lambda}\hat O_\lambda \ket{\psi_\lambda }=\bra{\phi} \hat G_\lambda ^\dag \hat O_\lambda \hat G_\lambda \ket{\psi} $$
for arbitrary states $\ket{\phi},\ket{\psi}$
Now remember that $\hat O_\lambda$ not simply $\hat G_\lambda \hat O \hat G_{\lambda}^\dag$ as we saw for the Hamiltonian thus we define
$$ \hat O'=\hat G_\lambda \hat O \hat G_{\lambda}^\dag $$
We can write the expectation value as
$$ \bra{\phi}\hat O\ket{\psi}=\bra{\phi} \hat G^\dag {\lambda} \hat G{\lambda} \hat O \hat G^\dag \lambda \hat G\lambda \ket{\psi} =\bra{\phi_\lambda} \hat O'\ket{\psi_\lambda} $$
💎 Conclusion: if $\hat O_\lambda=\hat O'$ then its expectation value and observable are gauge invariant, these are considered true physical quantities (they don't dependent on gauge)