<aside> 🥪
Generator: For continuous system we can consider an infinitesimally small transition, we can then analyse this transition using Taylor expansion giving a generator of the transformation
</aside>
💼 Case: Consider $\hat T_{\vec \epsilon}$ the translation operator over a very small vector distance $\vec \epsilon$
We expand around $\vec r$ 🧽 Assume: no phase change
$$ \begin{aligned} \psi(\vec r) \to \psi'(\vec r) &=\hat T_{\vec \epsilon} \psi(\vec R) \\ &=\psi( T_{\vec \epsilon}^{-1} \vec r) \\ &=\psi(\vec r -\vec \psi)\\ &\approx\psi(\vec r)-\vec \epsilon \cdot \vec \nabla \psi(\vec r) \\ &=\psi(\vec r)-\frac{i}{\hbar} \vec \epsilon\cdot \hat p \psi(\vec r)
\end{aligned} $$
where we used the position space representation of the momentum $\hat p = -i\hbar \vec \nabla$
💎 Conclusion: thus for an infinitesimal spatial translation we recognise the operator as
$$ \boxed{\hat T_{\vec \epsilon} =\hat I -\frac{i}{\hbar} \vec \epsilon \cdot \hat p} $$
The momentum operator $\hat p$ is the generator of spatial translations. Since we are making a continuous translation expect than it makes sense that this equivalent to adding momentum
💼 Case: Consider when an observable is invariant under symmetry transformation, this time for small translation $\vec \epsilon$,
we can write this as
$$ \begin{aligned} \int \text d \vec r \,\psi_n^* (\vec r)\hat O \psi_m (\vec r ) & \to \int \text d \vec r \, (\hat T_{\vec \epsilon} \psi_n (\vec r)) ^* \hat O \hat T_{\vec \epsilon} \psi_m (\vec r) \\ &=\int \text d \vec r \,\psi^*_n (\vec r -\vec \epsilon) \hat O \psi_m (\vec r - \vec \epsilon)
\end{aligned} $$
We can expand the wave functions to first order giving
$$ \begin{aligned} \psi_m (\vec r - \vec \epsilon) &\approx \psi_m(\vec r) -\frac i \hbar \vec \epsilon\cdot\hat p \psi_m (\vec r) \\ \psi^_n (\vec r- \vec \epsilon)&\approx \psi_n^(\vec r) +\frac{i}{\hbar} \psi ^* (\vec r) \vec \epsilon \cdot \hat p \end{aligned} $$
where we used $(\vec p\psi(\vec r))^* =\psi^* (\vec r) \hat p ^\dag =\psi^*(\vec r) \hat p$ since $\hat p^\dag = \hat p$
Subbing in we get
$$ \begin{aligned} \int \text d \vec r \,\psi_n^* (\vec r)\hat O \psi_m (\vec r ) &=\int \text d \vec r \,\psi^_n (\vec r -\vec \epsilon) \hat O \psi_m (\vec r - \vec \epsilon) \\ &\approx \int\text d \vec r\,\psi^_n(\vec r) \left ( 1+\frac{i}{\hbar} \vec \epsilon \cdot \hat p \right )\hat O \left ( 1-\frac{i}{\hbar}\vec \epsilon \cdot \hat p \right )\psi_m(\vec r) \\ &\approx \int \text d \vec r \,\psi^_n (\vec r) \hat O \psi_m(\vec r) +\frac{i}{\hbar} \int \text d \vec r \,\psi_n^ (\vec r) (\vec \epsilon\cdot\hat p \hat O-\hat O \vec \epsilon \cdot \hat p )\psi_m(\vec r) \\ &=\int \text d \vec r \, \psi^_n (\vec r ) \hat O \psi_m(\vec r) + \frac{i}{\hbar} \sum_i \epsilon_i \int \text d \vec r \, \psi^_n (\vec r ) (\hat p_i\hat O -\hat O\hat p_i ) \psi_m (\vec r) \end{aligned} $$
where $i=x,y,z$ and non linear terms have been removed since $\vec\epsilon \ll \vec 1$ so $\vec \epsilon^2\sim0$
From there the condition for the Operator to be invariant to transitions in space we need
$$ \int \text d \vec r \,\psi_n^* (\vec r) (\hat p_i \hat O-\hat O \hat p_j)\psi_m (\vec r) = \int \text d \vec r\, \psi^*_n(\vec r)[\hat p_i ,\hat O] \psi_m (\vec r)=0 $$
where $[\hat p_i,\hat O]=\hat p_i \hat O -\hat O \hat p_i$
Since the states are arbitrary we can simplify the expression to
$$ [\hat p_i ,\hat O]=0 $$
💎 Conclusion: for $\hat O$ to be compatible with the symmetry it must commute with the generator
🗒️ Note: this is true for the $\rm KE$ part of $\hat H$ ie $[\hat p_i ,\hat p^2/2m]=0$ however not necessarily for the $\rm PE$ $V(x)$ part
💃 Example: Consider
💼 Case: Consider a translation in time of a quantum state over a small interval $\delta t$
Expanding about $t$ we get
$$ \begin{aligned} \psi(\vec r,t)\to \psi'(\vec r,t)&=\hat T_{\delta t} \psi(\vec r,t) \\ &=\psi(\vec r,T^{-1}_{\delta t} t) \\ &=\psi(\vec r,t-\delta t) \\ &\approx \psi(\vec r,t)-\delta t \frac{\partial}{\partial t}\psi(\vec r,t) \\ &=\psi(\vec r,t) - \delta t \frac{i}{\hbar} \hat H \psi(\vec r,t)
\end{aligned} $$
where we used the TDSE $\hat H \psi(\vec r,t)=i\hbar \frac{\partial}{\partial t} \psi(\vec r,t)$
💎 Conclusion: for infinitesimal time intervals we have