Symmetries in quantum mechanics constrain theories, ensure consistency with formalism, and are linked to continuous transformations and reference frame independence.
Motion of a particle has 6 components 3 of position $\vec r(t)$ and momentum $\vec p(t)=m\dot {\vec r} (t)$ with $3$ dof
💼 Case: consider a translation in space by displacing the position vector $\vec r(t)\to \vec r(t)+\vec a$
We can write it in terms of a new position $\vec r'(t)$ as
$$ \vec r'(t)=T_a \vec r(t) = \vec r(t) + \vec a $$
Where he introduced the translation operator $T_a$ which displaced $\vec r(t)$ to $\vec r(t)+\vec a$
Now lets use our new found operator on a general function $f$ and a general point $\vec r_0$
$$ \begin{aligned} f'(\vec r)&= T_a f(\vec r) \\ \vec r_0 '&=T_a\vec r_0 = \vec r_0 + \vec a
\end{aligned} $$
By transforming both the function and the position we expect nothing to happen ie
$$ f'(\vec r_0')=f(\vec r_0) $$
thus we can find what the translation operator does of $f(\vec r)$
$$ f'(\vec r)=T_a f(\vec r)=f(T_a^{-1} \vec r)=f(\vec r - \vec a) $$
💎 Conclusion: if $f(\vec r)=f'(\vec r)=f(\vec r -\vec a)$ we say the function is translationally invariant
For momentum we have $\dot {\vec r'} =\dot {\vec r}$ with $\vec p'=T_a\vec p=\vec p$ so the fully transformed state is
$$ (\vec r,\vec p )\to (T_a\vec r ,T_a\vec p )= (\vec r+\vec a ,\vec p )=(\vec r',\vec p ) $$
💎 Conclusion: in inertial frames of reference $\vec p$ is conserved so $\vec F= \dot {\vec p}$ is invariant, ie particles respond to force the same way everywhere
💼 Case: lets now consider the total energy under the same transformation
For a single particle we express the total energy as the Hamiltonian
$$ H=\frac{p^2}{2m}+V(\vec r) $$
where $p^2 =\vec p \cdot \vec p$
💎 Conclusion: $V$ is the only $\vec r$ dependence so if $V(\vec r -\vec a )=V(\vec r)$ then $T_aH=H$
We can write a general result for $N$ particles, the Hamiltonian is
$$ H = \sum_{i=1}^N \frac{p_i^2}{2m_i} + \sum_{i < j} V(\mathbf{r}_i, \mathbf{r}_j) $$
The condition for translational invariance is thus $V(\mathbf{r}_i + \mathbf{a}, \mathbf{r}_j + \mathbf{a}) = V(\mathbf{r}_i - \mathbf{r}_j)$ $\forall(i,j)$
so the $n$-body system only depends on the relative distances of each particle not the absolute
Active: vector translated directly
Passive: coordinate system translated
💼 Case: we looked at active transformations ie $\vec r +\vec a$ lets now look at passive transformation
We thus define a new coordinate system $\hat i',\hat j',\hat k'$ such that $\vec r$ in this coordinate is
$$ \vec r= (r_1+a_1)\hat i'+(r_2+a_2)\hat j'+(r_3+a_3)\hat k' $$