Quantum mechanics follow 6 assumptions which cannot themselves be derived
$\bold I$: The state of a particle is given by a vector $|\psi(t) \rang$ in a Hilbert space. The state is normalised: $\lang \psi(t) | \psi(t) \rang =1$
🍌 In other words: The wave function $\psi(x,t)$ contains all possible information about the particle
🗒️ Note: this postulate leads to superposition
$\bold {II}$: There is a Hermitian operator corresponding to each observable property of the particle. Those corresponding to position $\rm \hat x$ and momentum $\rm \hat p$ satisfy $[\hat {\rm x}_i, \hat {\rm p}j]=i\hbar \delta{ij}$
🗒️ Note: operator choice can be guided by classical physics but is ultimately verified experimentally
🗒️ Note: $[\hat {\rm x}_i, \hat {\rm p}j]=i\hbar \delta{ij}$ is a formal expression of Heisenberg’s uncertainty principle
$\bold {III}$: Measurement of the observable associated with the operator $\hat \Omega$ will result in one of the eigenvalues $\omega _i$ of $\hat \Omega$. Immediately after the measurement the particle will be in the corresponding eigenstate $| \omega_i \rang$
🗒️ Note: this ensures reproducibility of measurements
$\bold {IV}$: The probability of obtaining the result $\omega_i$ in the above measurement (at time $t_0$) is $|\lang \omega _i | \psi (t_0)\rang |^2$
🍌 In other words: if a particle is repeatedly prepared in the same state $| \psi (t_0)\rang$ , the outcome will generally vary unless the state is an eigenstate of $\hat \Omega$. If it is, the result will always be $\omega_i$
🗒️ Note: This postulate means that only the distribution of results can be predicted. The expectation value of $\omega$ is $\lang \psi (t_0)| \hat \Omega |\psi(t_0 ) \rang$ .
🗒️ Note: If we expand the state in the orthonormal basis $\{ | \omega_i \rang \}$ , we have $|\psi (t_0) \rang = \sum_i c_i | \omega_i \rang$ , the probability of obtaining the result $\omega_i$ is $|c_i|^2$, and $\lang \hat \Omega \rang = \sum _ i | c_i |^2 \omega_i$
$\bold V$: The time evolution of the state $|\psi (t) \rang$ is given by $i \hbar \frac{\text d}{\text dt} |\psi (t) \rang ={\rm \hat H} | \psi(t) \rang$ , where $\rm \hat H$ is the operator corresponding to the classical Hamiltonian
🗒️ Note: in most cases is just the energy and is expressed as ${\rm \hat p \cdot \hat p} /2m + V( { \rm \hat x } )$
$\bold {VI}$: The Hilbert space for a system of two or more subsystems (for example, several particles) is a product space
🍌 In other words: If two or more subsystems