Perturbation theory is the most widely used approximate method. It requires we have a set of exact solutions to a Hamiltonian which is close to the realistic one.

Non-degenerate perturbation theory

🧽 Assume: the Hamiltonian $\hat H$ can be split into two parts, with the first part being exactly solvable and the second being small compared to the first one

🖊️ Notation:

• First part $\hat H^{(0)}$

  • Eigenstates $|n^{(0)}\rang$
  • Energy $E_n^{(0)}$
  • Wave function $\phi^{(0)}$

• Second part $\lambda\hat H^{(1)}$

  • Eigenstates $\lambda|n^{(1)}\rang$
  • Energy $\lambda E_n^{(1)}$
  • Wave function $\phi^{(1)}$

• Complete $\hat H$

  • Eigenstates $|n\rang$
  • Energy $E_n$
  • Wave function $\phi$

🧽 Assume: No degeneracies in eigenstates and energies

🗒️ Note: we can have over 2 terms ie $E_n=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\ldots$ $m$ denotes the order

We take $\lambda$ to be real so that $\lambda \hat H^{(1)}$ is Hermitian

• Starting with the Schrödinger equation for the full Hamiltonian

  $$ (\hat H^{(0)}+\lambda \hat H^{(1)})\ket{n} =E_n \ket{n} $$

• Subbing in the following Taylor expansions

  $$ \begin{aligned} E_n&=E_n^{(0)}+\lambda E_n^{(1)}+\lambda^2 E_n^{(2)}+\ldots \\ \ket{n} &=\ket{n^{(0)}}+\lambda \ket{n^{(1)}}+\lambda^2 \ket{n^{(2)}}+\ldots \end{aligned} $$

  we can match the terms on each side by the power of $\lambda$ to get

  $$ \begin{aligned} &\lambda^0: & \hat{H}^{(0)} \ket{n^{(0)}} &= E_n^{(0)} \ket{n^{(0)}} \\ &\lambda^1: & \hat{H}^{(0)} \ket{n^{(1)}} + \hat{H}^{(1)} \ket{n^{(0)}} &= E_n^{(0)} \ket{n^{(1)}} + E_n^{(1)} \ket{n^{(0)}} \\ &\lambda^2: \;\; & \hat{H}^{(0)} \ket{n^{(2)}} + \hat{H}^{(1)} \ket{n^{(1)}} + \hat{H}^{(2)} \ket{n^{(0)}} &= E_n^{(0)} \ket{n^{(2)}} + E_n^{(1)} \ket{n^{(1)}} + E_n^{(2)} \ket{n^{(0)}} \end{aligned} $$

  🗒️ Note: we divided by $\lambda^m$ afterwards which is why you don't see them in the equations

  These need to solved sequentially with our assumptions we assumed the first is known so the second will give us $E^{(1)}_n$ and $\ket{n^{(1)}}$, the third $E^{(2)}_n$ and $\ket{n^{(2)}}$ and so on

• Using the fact that the unperturbed states $\{ \ket{n^{(0)}}\}$ form a basis, we write the change in states

  $$ \ket{n^{(1)}}=\sum_m c_m \ket{m^{(0)}} =\sum_{m} \braket{m^{(0)}|n^{(1)}}\,\ket{m^{(0)}} $$

  To solve for the shift in energy we take the inner product with $\bra{n^{(0)}}$ whereas for the shift in state, we use $\bra{m^{(0)}}$

• Using $\bra{m^{(0)}} \hat H^{(0)} =E^{(0)}_m \bra{m^{(0)}}$ and $\braket{m^{(0)}|n^{(0)}}$=$\delta _{mn}$ we can get the first order term

  $$ \begin{aligned} E_n^{(1)}&=\braket{n^{(0)}|\hat H^{(1)}|n^{(0)} } \\ \braket{m^{(0)}|n^{(1)}}&=\frac{\bra{m^{(0)}}\hat H^{(1)}\ket{n^{(0)}}}{E_n^{(0)}-E^{(0)}_m} \quad \forall m\ne n

  \end{aligned} $$

  Where the second equation tells us the overlap of $\ket{n^{(1)}}$ with all the other $\ket{m^{(0)}}$ but not with $\ket{n^{(0)}}$

• Set $\braket{n^{(0)}|n^{(1)}}=0$ since the quantity is not constrained by the Schrödinger equation giving

  $$ \ket{n^{(1)}}=\sum_{m\ne n}\frac{\bra{m^{(0)}}\hat H^{(1)}\ket{n^{(0)}}}{E_n^{(0)}-E^{(0)}_m}\ket{m^{(0)}} $$

  If the spectrum of $\hat H^{(0)}$ is degenerate then the denominator could vanish however assumptions

• The second order is then

  $$ E_n^{(2)}=\bra{n^{(0)}}\hat H^{(1)}\ket{n^{(1)}}=\sum_{m\ne n}\frac{|\bra{m^{(0)}}\hat H^{(1)}\ket{n^{(0)}}|^2}{E_n^{(0)}-E^{(0)}_m} $$

🗒️ Note: finding higher terms ie $\ket{n^{(2)}}$ or $E^{(3)}_n$ becomes too complicated

Since the final equation is independent of $\lambda$ we will from now on set it to $1$ for convenience ie

$$ E_n=E_n^{(0)}+\lambda E_n^{(1)}+ E_n^{(2)}+\ldots $$

Connection to variational approach