💼 Case: look at linear operators that act on a quantum state to transform them in space or time i.e.

$$ \ket{\psi(\vec r , t)} \to \ket{\psi' (\vec r,t)} \qquad \psi(\vec r ,t )\to \psi '(\vec r,t) $$

🧠 Remember: States and observables

• We write a wavefunction as $\psi(\vec r) = \braket{\vec r|\psi}$ where $|\vec r\rang$ is the position eigenstate

• We define an operator $\hat r$ which acts on a state, if it is an eigenstate we get $\hat r |\vec r \rang = \vec r|\vec r\rang$

• We define the Hermitian conjugate, ie the conjugate transpose as $\lang \vec r|=(|\vec r\rang )^\dag$

• Any observable is written as an operator ie $\hat O$, for it to be physical (ie measurable) we need $\hat O^\dag =\hat O$ ie we need it to be Hermitian 💃 Example: $\hat r^\dag=\hat r,\; \hat p^\dag = \hat p$

• The expectation value of an operator in the state $|\psi\rang$ is given by

  $$ \braket{\hat O}=\bra{\psi} \hat O\ket{\psi} \xrightarrow{\text{position space}}\int\text d \vec r\,\psi^*(\vec r)\hat O \psi(\vec r) $$

• In position representation we have: $\hat r=\vec r$ and $\hat p =-i\hbar \vec \nabla$

• We can represent any operator in any similar size basis $\{|\psi_n\rang \}$ with $n\in \N$ using the completeness relation $\hat I=\sum _n \ket{\psi_n}\bra{\psi_n}$ where $\hat I$ is the identity operator $I\ket{\psi_n}=\ket{\psi_n}$

  $$ \hat O=\hat I \hat O\hat I=\sum_{nm} \bra{\psi_n} \hat O \ket{\psi_m} \ket{\psi_n}\bra{\psi_m} $$

  where $c_{nm}=\bra{\psi_n} \hat O \ket{\psi_m}$ are the matrix elements

Transformation operators: active and passive views

🧗 Active:

• Lets define the transformation operator $\hat T_S$ through its action on the system state $\psi(\vec r,t)$

  $$ \hat T_s\psi(\vec r,t)=\psi'(\vec r,t)=\psi(T^{-1}_S(\vec r,t)) \qquad \hat T_s\psi(\vec r,t)=\ket{\psi'(\vec r,t)} $$

  We call this an active transformation as it has the direct effect of transforming the wavefunction

🛌 Passive:

• To do this we use our previous integral to change of basis thus upon transforming we have

  $$ \begin{aligned} \int \text d\vec r \, \psi_n^(\vec{r}, t) \hat{O} \psi_m(\vec{r}, t) &\to \int \text dr \, \psi_n^{\prime }(\vec{r}, t) \hat{O} \psi_m'(\vec{r}, t) \\ &= \int \text dr \, ( \hat{T}_S \psi_n(\vec{r}, t) )^ \hat{O} \hat{T}_S \psi_m(\vec{r}, t) \\ &= \int \text dr \, ( \psi_n^(\vec{r}, t) \hat{T}_S^\dagger ) \hat{O} \hat{T}_S \psi_m(\vec{r}, t) \\ &= \int \text dr \, \psi_n^*(\vec{r}, t) ( \hat{T}_S^\dagger \hat{O} \hat{T}_S ) \psi_m(\vec{r}, t) \end{aligned} $$

  for any state $\psi_n (\vec r)$ and $\psi_m (\vec r)$ and we used $(\hat T_S \psi_n (\vec r, t))^=\psi_n^ (\vec r, t)\hat T^\dag _S$

• We thus get that in quantum passive view we transform operators according to

  $$ \hat O \to \hat T_{S}^\dag \hat O \hat T_{S} $$

  while keeping the wavefunctions unchanged

💎 Conclusion: passive and active transformations are equivalent

<aside> 🥽

Invariance: a quantum system is invariant under a particular symmetry when the observables are unchanged after the associated symmetry transformation i.e.

$$ \int \text d \vec r \,\psi'^_n (\vec r ,t )\hat O \psi'_m (\vec r ,t )=\int \text d \vec r \,\psi^_n (\vec r ,t )\hat O \psi_m (\vec r ,t ) $$

</aside>

• Using our previous result we see that this condition implies that

  $$ \hat O = \hat T_S^\dag \hat O \hat T_S $$