# First order time-independent perturbation theory: How to practically calculate the perturbed wave-function

This is one of the problems that draws the line between academically learning something, and having to use it. While I learned the formulas relevant to this, I just want to make sure I'm using them right.

How can I calculate the perturbed wave-function up to first order?

### Calculating the expectation value from a Hamiltonian

Say we have an atomic system, with a Hamiltonian $H_0$, and we have some time-independent perturbation $H_1$, which consists of some spin interactions:

$$H_1= \hbar \gamma_S \vec{B}\cdot\hat{\vec{S}}+\hbar \gamma_I \vec{B}\cdot\hat{\vec{I}}+\hbar A \vec{I}\cdot\vec{S}$$

With $\gamma$ being the gyromagnetic ratio of that spin system and $A$ is the linear spin-spin coupling (which is not isotropic in general, but doesn't matter for my question). From what I understand, the energy of the first order perturbation is simply the expectation value of this system. So in the basis $\left| Sm_S Im_I \right>$, the energies perturbation up to first order will be:

$$E_1=\left< Sm_S Im_I \right|H_1\left| Sm_S Im_I \right>$$

"1" stands for first order perturbation. Now this can be represented by a $\left(2S+1\right)\times \left(2I+1\right)$ matrix. So the matrix elements will look like (where my states are written in the $\left| m_S m_I \right>$ for simplicity, and assuming $S=\frac{1}{2}$ and $I=\frac{1}{2}$):

$$\small E_{1,mn}=\left( \begin{matrix} \left< \frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right> && \left< \frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}-\frac{1}{2} \right> && \left< \frac{1}{2}\frac{1}{2} \right|H_1\left| -\frac{1}{2}\frac{1}{2} \right> && \left< \frac{1}{2}\frac{1}{2} \right|H_1\left| -\frac{1}{2}-\frac{1}{2} \right> \\ \left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right> && \left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}-\frac{1}{2} \right> && \left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| -\frac{1}{2}\frac{1}{2} \right> && \left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| -\frac{1}{2}-\frac{1}{2} \right> \\ \left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right> && \left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}-\frac{1}{2} \right> && \left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| -\frac{1}{2}\frac{1}{2} \right> && \left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| -\frac{1}{2}-\frac{1}{2} \right> \\ \left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right> && \left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}-\frac{1}{2} \right> && \left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| -\frac{1}{2}\frac{1}{2} \right> && \left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| -\frac{1}{2}-\frac{1}{2} \right> \end{matrix} \right)$$

These matrix elements are just numbers, or functions of $\gamma$, $\vec{B}$ and $A$.

### Calculating the first order perturbed wave-function

Looking in textbooks, the formula one sees in Schwabl's quantum mechanics (also the same in Wikipedia) is:

$$\left| n^1 \right>=\sum_{n\neq m} \frac{\left< m^0 \right|H_1\left| n^0 \right>}{E_n^0-E_m^0}\left| m^0 \right>$$

My question is: How can I use what I derived above to get the wave-function?

My answer: Say I want to get the first order perturbed wave-function of $\left| \frac{1}{2}\frac{1}{2} \right>$. Is this the right formula for it?

$$\left| \frac{1}{2}\frac{1}{2} \right>+\left| \frac{1}{2}\frac{1}{2}^1 \right>=\left(\begin{matrix} \left| \frac{1}{2}\frac{1}{2} \right> + \frac{\left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>}{\left< \frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>-\left< \frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}-\frac{1}{2} \right>} \left|\frac{1}{2}-\frac{1}{2} \right> + \frac{\left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>}{\left< \frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>-\left< -\frac{1}{2}\frac{1}{2} \right|H_1\left| -\frac{1}{2}\frac{1}{2} \right>} \left|-\frac{1}{2}\frac{1}{2} \right> + \frac{\left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>}{\left< \frac{1}{2}\frac{1}{2} \right|H_1\left| \frac{1}{2}\frac{1}{2} \right>-\left< -\frac{1}{2}-\frac{1}{2} \right|H_1\left| -\frac{1}{2}-\frac{1}{2} \right>} \left|-\frac{1}{2}-\frac{1}{2} \right> \end{matrix}\right)$$

where all these expectation values are taken from the matrix elements above.

### Is this correct?

Thanks.

• The eigenvectors of the perturbation matrix (your $E_{1,mn}$) in the degenerate subspace are the zeroth-order corrections to the wave function. In degenerate perturbation theory one needs to also determine the correct starting point for the series, since it is ambiguous to which superposition of degenerate states the system should return if the perturbation is switched off. The formula you show is only valid for nondegnerate perturbation theory. The first-order corrections to the wave function in degenerate perturbation theory are a bit harder to find. See sec.5.2 of Sakurai Modern QM. – Praan May 26 '16 at 17:27
• @Praan Thank you for the response. In my case there's no degeneracy actually; I have high magnetic fields. In that case, I gather from what you said that it's correct, right? – The Quantum Physicist May 26 '16 at 17:54
• The degeneracy relates to $H_0$. The energies in your formula for the wave function correction correspond to the unperturbed system. (the superscript 0 indicates this) It breaks down in case of degeneracy since the denominator becomes zero. Without $H_1$ all the spin states are presumably degenerate. – Praan May 26 '16 at 18:05
• @TheQuantumPhysicist What Praan said. And actually notice that the matrix elements in the denominators of your formula should't be $H_1$ matrix elements, but $H_0$ eigenvalues. This is why there is trouble if $H_0$ is degenerate, regardless of whatever $H_1$ might do to lift it. – udrv May 26 '16 at 18:27