First-Order Energy Corrections

Deriving the first-order shift

Start from the eigenvalue problem for the full Hamiltonian and insert the power series from the previous lesson:

\big(\hat{H}^{(0)} + \lambda \hat{H}'\big)\big(|n^{(0)}\rangle + \lambda|n^{(1)}\rangle + \cdots\big) = \big(E_n^{(0)} + \lambda E_n^{(1)} + \cdots\big)\big(|n^{(0)}\rangle + \lambda|n^{(1)}\rangle + \cdots\big).

Collect the terms order by order in $\lambda$ . The $\lambda^0$ equation is just the unperturbed problem, $\hat{H}^{(0)}|n^{(0)}\rangle = E_n^{(0)}|n^{(0)}\rangle$ . The $\lambda^1$ equation is

\hat{H}^{(0)}|n^{(1)}\rangle + \hat{H}'|n^{(0)}\rangle = E_n^{(0)}|n^{(1)}\rangle + E_n^{(1)}|n^{(0)}\rangle.

Now take the inner product with $\langle n^{(0)}|$ . Because $\hat{H}^{(0)}$ is Hermitian, $\langle n^{(0)}|\hat{H}^{(0)}|n^{(1)}\rangle = E_n^{(0)}\langle n^{(0)}|n^{(1)}\rangle$ , which cancels the matching term on the right. With the states normalized, $\langle n^{(0)}|n^{(0)}\rangle = 1$ , and we are left with the central result:

\boxed{\,E_n^{(1)} = \langle n^{(0)} | \hat{H}' | n^{(0)} \rangle\,.}

The shift is just an expectation value

The first-order energy correction is the expectation value of the perturbation in the unperturbed state. You do not need to solve the perturbed problem at all — you only evaluate one matrix element using the states you already know. Setting $\lambda = 1$ , the energy to first order is

E_n \approx E_n^{(0)} + \langle n^{(0)} | \hat{H}' | n^{(0)} \rangle.

A two-level example

Let the unperturbed Hamiltonian be diagonal in the computational basis, $\hat{H}^{(0)} = \mathrm{diag}(\varepsilon_0, \varepsilon_1)$ , with eigenstates $|0\rangle$ and $|1\rangle$ . Take the perturbation

\hat{H}' = \begin{pmatrix} a & b \\ b^* & d \end{pmatrix}.

The first-order shift of the ground level is the diagonal entry seen by $|0\rangle$ :

E_0^{(1)} = \langle 0 | \hat{H}' | 0 \rangle = a.

The off-diagonal element $b$ does not enter at first order — it only shifts the energy at second order. Concretely, if $\hat{H}' = 0.3\,\sigma_z$ , then in the basis where $\sigma_z = \mathrm{diag}(1, -1)$ we get $E_0^{(1)} = \langle 0|0.3\,\sigma_z|0\rangle = +0.3$ and $E_1^{(1)} = -0.3$ .

Try it

Take $\hat{H}' = 0.3\,\sigma_z + 0.5\,\sigma_x$ acting on a two-level system with unperturbed eigenstates $|0\rangle$ and $|1\rangle$ . Compute the first-order energy shift of the ground state $E_0^{(1)} = \langle 0 | \hat{H}' | 0 \rangle$ and return it as a number. (Recall $\sigma_z|0\rangle = |0\rangle$ and $\sigma_x|0\rangle = |1\rangle$ , and $\langle 0|1\rangle = 0$ .)

Run your code to see the quantum state.