$$ \lim_{N\to\infin} \left [ 1+\frac{x}{N} \right ]^N=e^x $$
We want an operator that preserve the overlap (scalar product) between any two wavefunctions
💼 Case: lets consider such an operator $\hat U$
Applying it to a quantum state gives
$$ \ket{\psi_U}=\hat U |\psi\rang \qquad \bra{\psi_U}=\bra{\psi}\hat U $$
We want the scalar product to be conserved thus we need
$$ \braket{\phi_U|\psi_U} =\bra{\phi}\hat U ^\dag \hat U |\psi\rang = \braket{\phi|\psi} $$
Since this must hold for any $|\psi\rang$ and $|\phi\rang$ we thus have
$$ \hat U ^\dag \hat U =\hat I $$
so $\hat U ^\dag = \hat U^{-1}$
<aside> 🌱
Unitary operator: an operator that satisfied $\hat U^\dag \hat U =\hat U \hat U^\dag =\hat I$
</aside>
We require out symmetry transformation operators derived in the previous lecture to be unitary, as under rotation and translation, length remains invariant
Thus for a general transformation operator $\hat T$ in position space we have
$$ \begin{aligned} \int \text d\vec r\,\psi^(\vec r,t )\psi(\vec r,t)=1&=\int \text d\vec r\,(\hat{T}\psi(\vec r,t))^ \hat{T}\psi(\vec r,t) \\ &=\int \text d\vec r\, \psi^* (\vec r,t)\hat{T}^\dag \hat{T} \psi(\vec r,t) \end{aligned} $$
where this is equivalent to saying $\hat T^\dag \hat T=\hat I$ as expected
💼 Case: lets now apply these properties
🧠Remember: the generator is $\hat p =-i\hbar \vec \nabla$
💼 Case: consider a finite translation in which $\psi(\vec r)\to\psi(\vec r-\vec a)$ where we don't assume $\vec a \ll 1$
To do this lets consider that we are taking infinite, infinitely small steps ie
$$ \vec a = N \frac {\vec a}N =\sum^N_{i=1} \frac{a}{N} $$
Plugging this in and taking the limit to infinity so that the steps are infinitesimal;
$$ \begin{aligned} \psi(\vec r-\vec a) & = \lim_{N\to\infin} \psi \left (\vec r - \sum^N_{i=1} \frac{\vec a}{N} \right ) \\ &=\lim_{N\to\infin} \psi \left ( \vec r - \sum^{N-1}_{i=1} \frac{\vec a}{N}-\frac{\vec a}{N}\right ) \end{aligned} $$
Now we can plug in our previous result $\hat T_{\vec a/N}=\hat I - (i/\hbar)(\vec a/N)\cdot \hat p$
$$ \begin{aligned} \psi(\vec r-\vec a)&=\lim_{N\to\infin} \psi \left ( \vec r - \sum^{N-1}{i=1} \frac{\vec a}{N}-\frac{\vec a}{N}\right ) \\ &=\lim{N\to\infin} \left [ \left ( \hat I -\frac i \hbar \frac {\vec a}N \cdot \hat p \right )\psi\left ( \vec r - \sum^{N-1}{i=1} \frac{\vec a}{N} \right ) \right ] \\ &=\lim{N\to\infin} \left [ \left ( \hat I -\frac i \hbar \frac {\vec a}N \cdot \hat p \right )\psi\left ( \vec r - \sum^{N-2}{i=1} \frac{\vec a}{N}-\frac{\vec a}{N} \right ) \right ] \\ &=\lim{N\to\infin} \left [ \left ( \hat I -\frac i \hbar \frac {\vec a}N \cdot \hat p \right )^2 \psi\left ( \vec r - \sum^{N-2}{i=1} \frac{\vec a}{N} \right ) \right ] \\ &=\lim{N\to\infin} \left [ \left ( \hat I -\frac i \hbar \frac {\vec a}N \cdot \hat p \right )^N \psi ( \vec r ) \right ] \\ &=e^{-i\vec a \cdot \hat p /\hbar}\psi(\vec r) = \hat U_{\vec a}\psi(\vec r) \end{aligned} $$
💎 Conclusion: the finite translation $\psi(\vec r - \vec a)=\hat U_a \psi(\vec r)$ is given by