My question about the Feynman-Hellmann theorem (FHT) is two-fold, one regarding the physical interpretation, and the other related to simply how one analytically or numerically does the calculations.

The theorem is often stated as followed:

Given a quantum system with the Hamiltonian $H$ and its eigen-equation: $H|\psi_n\rangle = E_n |\psi_n \rangle,$ then for any parameter $\alpha$ on which the Hamiltonian depends, the theorem states:

$$

\frac{\partial E_n(\alpha)}{\partial \alpha} = \left\langle \psi_n \left|\ \frac{\partial H}{\partial \alpha} \ \right|\psi_n \right\rangle. \tag{1}

$$

**Questions:**

From a physical point of view, one way to read this theorem is that it connects the variations of the Hamiltonian's eigenvalues with the variations of the Hamiltonian itself. Moreover, it says that to know by how much an eigenvalue changes, one needs only to know the derivative of the Hamiltonian operator and the corresponding eigenvector. Is there a more fundamental interpretation at play here that I am missing?

It gets a bit stranger, when one considers even introducing parameters $\alpha$ into the Hamiltonian, in order to compute average of a term in the Hamiltonian, e.g., writing $H = A + B $ instead as $H=A+\alpha B$ and calculating $E_n (\alpha)$ we can express the expectation value of the $B$ terms as:

$$

\left. \frac{\partial E(\alpha)}{\partial \alpha}\right|_{\alpha=1} =^{FHT} \left\langle \frac{\partial H}{\partial \alpha} \right\rangle = \langle B \rangle \tag{2}

$$

Why by arbitrarily introducing parameters into $H$ we can still estimate the correct expectation values using FHT? I feel I am missing some potentially basic point here.

On a mathematical side: in Eq. $(1),$ the averaging over the eigenstate $|\psi_n\rangle$ implies that the derivative of $H$ is only valid as long as $H$ is in eigenstates, right? But how can we mathematically compute such derivatives when only eigenstates are allowed? Isn't this similar to attempting to define derivatives over discrete functions?

How do we actually compute such derivatives (analytically or numerically)?

## Best Answer

Short answer:In general, the eigenvectors $\psi_n$ will change for finite $\alpha$. Hence, to apply the Feynman Hellman theorem correctly in your second question, you need to determine the spectrum and eigenvectors at $\alpha=1$ and then compute expectation values with respect to these eigenvectors.Longer answer:You're asking a few different questions here, so let me know if I've missed one. Firstly, on a basic mathematical level it seems that you're asking how to define the derivative of $H$ with respect to a parameter $\alpha$. To do this, you can imagine choosing a fixed basis for the Hilbert space on which $H$ acts, evaluating the matrix elements of $H$ with respect to this basis, and differentiating each matrix element individually. If instead you had chosen a new basis at each value of $\alpha$, you would need to specify a particular

covariant derivativefor the Hamiltonian, to avoid ambiguities associated with your choice of basis at each value of $\alpha$.Your first question is not entirely clear to me, but I'd guess that you're asking why changes in the eigenvectors are irrelevant for computing $\partial_\alpha E_n(\alpha)$. Essentially this follows from the fact that eigenvectors are constrained to lie on the unit sphere of $\mathbb C^n$ (where $n$ is the dimension of the Hilbert space). Explicitly, \begin{align*} \partial_\alpha E_n(\alpha)&=\langle\psi_n(\alpha)|\partial_\alpha H(\alpha)|\psi_n(\alpha)\rangle+(\partial_\alpha\langle\psi_n(\alpha)|)H(\alpha)|\psi_n(\alpha)\rangle+\langle\psi_n(\alpha)|H(\alpha)\partial_\alpha(|\psi_n(\alpha)\rangle)\\ &=\langle\partial_\alpha H\rangle_n+E_n\partial_\alpha(\langle\psi_n(\alpha)|\psi_n(\alpha)\rangle)\\ &=\langle\partial_\alpha H\rangle_n \end{align*} Physically, it's related to the observation that changing the Hamiltonian of a system slowly over time tends to cause the system to move continuously along an eigenstate (as long as the spectrum is well separated).