Due to the coupled nature of Schrödinger equation and the sheer difficulty of solving the differential equations analytically we often rely on numerical solutions

🗒️ Note: Computers can solve differential equations very efficiently but

• It is good to check results using numerical methods
• They don’t interpret the results
• Some interactions are simply too complicated even for supercomputers
• Quantum field theories have infinitely many degrees of freedom and thus can only be approximated

Variational methods: ground state

💼 Case: we know the Hamiltonian of a bound system but not the wave function or the ground state energy

💫 The variation principle: If we guess the wave function, the expectation value of the Hamiltonian in that wave function will be greater than the true ground state energy

$$ \frac{\braket{\Psi |\hat H|\Psi}}{\braket{\Psi|\Psi}}\ge E_0 $$

🧙‍♂️ Proof: If we consider expanding the normalized $|\Psi\rang$ in the true energy eigenstate $\ket{n}$ we get $\braket{\hat H} =\sum_n P_ n E_n$ since $P_n>0$ and all $E_n \ge E_0$ ($\sum_nP_n=1$)

🗒️ Note: better guesses will be closer to $E_0$

💃 Example: Consider the infinite square well with $V=0$ for $0<x<a$ and $V=\infin$ elsewhere

• As a trial function we use $\Psi (x)=x(a-x)$ for $0<x<a$ and $\Psi(x)=0$ otherwise

  $$ \frac{\bra{\Psi} \hat H \ket{\Psi} }{\braket{\Psi|\Psi}}=\frac{10\hbar^2}{2ma^2}=1.013\frac{\pi^2 \hbar^2}{2ma^2} $$

  🗒️ Notes:

  • The result here is very close to the real solution
  • However in practice we don’t know the real solution so we wouldn't know it was this close

In practice we run a minimization algorithm say a Bayesian optimization to try different parameters until we get a good result. Here a better guess would be $\Psi(x)=x(a-x)+bx^2(a-x)^2$ were we let an algorithm try values of $b$ until it reaches a convergence.

This method gives the following precision

🗒️ Note: again however a minimization algorithm could minimize to a local minima and there would be hard to tell

Variational methods: excited states

⚽ Goal: find a way to get a condition to find $E_1$ and potentially higher

Going back to the expression $\braket{\hat H} =\sum_n P_ n E_n$ here $P_n=|\lang \psi_n |\Psi \rang |^2$ represents the probability of the trial wavefunction $\Psi$ overlapping with the eigenstates $\psi_n$.