<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/539a6e75-8765-40ad-97a7-9f572d852722/Operator.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/539a6e75-8765-40ad-97a7-9f572d852722/Operator.png" width="40px" /> Operator: An operator is anything that acts upon a wavefunction and produces another wavefunction, it is denotated using hat $\hat A$
$$ \widehat{A}\Psi(x,t)=\Phi(x,t) $$
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🧨 Kinetic energy operator:
$$ \widehat{T}=-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2} $$
🌍 Potential energy operator:
$$ \widehat{V}=V(x) $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/837c4627-9c36-4f86-ad12-6c7ee0dd305f/Hamiltonian_operator.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/837c4627-9c36-4f86-ad12-6c7ee0dd305f/Hamiltonian_operator.png" width="40px" /> Hamiltonian operator is the total energy operator and is given by
$$ \widehat{H}=-\frac{\hbar^2}{2m}\,\frac{\partial^2}{\partial x^2}+V(x) $$
the stationary states are the eigenfunctions of the Hamiltonian operator
$$ \widehat{H}\,\psi_i(x) = E_i\,\psi_i(x) $$
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❓ We assume the wavefunction $\Psi$ is only in one state $\psi_n$
In this case we get a definite value $E_n$ ( assuming you are just measuring one quantity)
The expectation is:
$$ \begin{aligned} \langle E\rangle &=E_n \\ &= E_n\overbrace{\int_{-\infty}^{\infty}\Psi_n(x,t)^\,\Psi_n(x,t)\,\mathrm{d}x}^{=\,0} \\ &= \int_{-\infty}^{\infty}\Psi_n(x,t)^\,E_n\,\Psi_n(x,t)\,\mathrm{d}x \\ \langle E\rangle&= \int_{-\infty}^{\infty}\Psi_n(x,t)^*\,\widehat{H}\,\Psi_n(x,t)\,\mathrm{d}x \end{aligned} $$
💡 In quantum mechanics a system needs to be in one definite definite energy whenever a measurement is taken
If we take a linear combination of two stationary states
$$ \Psi(x,t) = A_1\,\psi_1(x)\,\mathrm{e}^{-iE_1t/\hbar} + A_2\,\psi_2(x)\,\mathrm{e}^{-iE_2t/\hbar} $$
Then we can calculate the expectation (average) value of the energy:
$$ \begin{aligned}\langle E\rangle&= \int_{-\infty}^{\infty}\Psi(x,t)^\,\widehat{H}\,\Psi(x,t)\,\mathrm{d}x \\&= \int_{-\infty}^{\infty}\Bigl(A_1^\,\psi_1^*(x)\,\mathrm{e}^{+iE_1t/\hbar}+\ldots\Bigr)\Bigl(A_1\,E_1\,\psi_1(x)\,\mathrm{e}^{-iE_1t/\hbar}+\ldots\Bigr)\mathrm{d}x \\&= E_1|A_1|^2+E_2|A_2|^2\,.\end{aligned} $$
where $|A_1|^2$ and $|A_2|^2$ are the probabilities of getting $E_1$ and $E_2$
Thus it makes sence that $|A_1|^2+|A_2|^2 =1$ as the particle has to be in a state
$$ \langle E^2\rangle= \int_{-\infty}^{\infty}\Psi(x,t)^*\,\widehat{H}\,\Bigl(\widehat{H}\,\Psi(x,t)\Bigr)\,\mathrm{d}x= E_1^2|A_1|^2+E_2^2|A_2|^2 $$
🍎 In classical physics we would expect $\langle E^2\rangle=\langle E\rangle^2$
🔬 In quantum mechanics:
$$ \begin{aligned}\langle E\rangle^2 &= \left ( E_1|A_1|^2+E_2|A_2|^2 \right )^2\\ \langle E^2\rangle &= E_1^2|A_1|^2+E_2^2|A_2|^2\end{aligned} $$
💡 This behaviour means that when you measure the energy of a system its energy will remain the same for each subsequent measurement this can be linked back to Heisenberg's uncertainty principle $\Delta E \Delta t \ge \hbar /2$, if we know for certain the energy of the energy then $\Delta t\to \infin$ so the state transition to another energy level can never happen
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/916f9df6-f48f-4f26-86fc-5d0b1b364f2a/theorem.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/916f9df6-f48f-4f26-86fc-5d0b1b364f2a/theorem.png" width="40px" /> For any observable $A$ there is an operator $\hat A$ whose eigenvalues $\{{a_n}\}$ give the only possible values for measurements of $A$
$$ \widehat{A}\,\psi_n = a_n\,\psi_n $$
if the system is in state $\psi_n$ the measurement $A$ will always give the value $a_n$ and stay in the same state $\psi_n$
if the system is in a mixed state $\psi$ the measurement of $A$ gives one of the values $\{ a_n \}$ with expectation value
$$ \langle A\rangle = \int \psi^*\,\widehat{A}\,\psi\,\mathrm{d}x $$
and the system is left in energy state $\psi_n$
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🗒️ Note: the wavefunction contains more information than can be measured in any one measurement and any measurement will collapse the state which is why we can measure multiple properties using a wavefunction while we wouldn't be able to achieve that experimentally
