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


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)?

  • $\begingroup$ Your "on the mathematical side" doesn't make much sense to me. What exactly is wrong with the derivatives? $\endgroup$
    – Ryan Unger
    Apr 12, 2017 at 13:39
  • $\begingroup$ @0celouvsky Hi, doesn't the averaging of the derivatives in (1) demand $H$ to take definite values? Implying that for arbitrary (coherent) states $\psi$ the energy derivarives are not well-defined. If this is true, I'm trying to understand how we calculate the derivative under such requirements. $\endgroup$
    – user929304
    Apr 12, 2017 at 13:48
  • $\begingroup$ The sentence "Implying that for arbitrary (coherent) states $\psi$ the energy derivarives are not well-defined" does not make sense to me in the sense that I don't know what you mean. $\endgroup$
    – Ryan Unger
    Apr 12, 2017 at 14:19
  • $\begingroup$ Part of the point of the theorem is that mere existence of the eigenstates suffices, and no detailed knowledge of them: the theorem is a shortcut to bypass them in practice. Possibly related, 226174. $\endgroup$ Apr 12, 2017 at 14:25
  • $\begingroup$ Hi, how r u lately? The answer is perturbation theory. Or else, the Feynmann-Hellman formula is a particular way of taking advantage of lowest order perturbation theory. Taking the parameter derivative of the Hamiltonian and of its eigenvalues means writing the Hamiltonian as $H(\alpha) = H + \alpha\,V$ for $\alpha << 1$, doing perturbation theory in $\alpha$, and writing the perturbed eigenvalues as $E_n(\alpha) = E_n + \alpha \delta E_n $, where $\delta E_n = \langle \psi_n |V | \psi_n\rangle$. $\endgroup$
    – udrv
    Apr 12, 2017 at 22:33

2 Answers 2


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 derivative for 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).


As noted in answer by @fs137 a variation of the parameter $\alpha$ in the Hamiltonian will, in general, lead to a change in eigenvectors as well. However, the contribution to change in energy by these changing eigenvectors cancel out \begin{equation} \frac{\partial}{\partial \alpha} \langle\psi|\psi\rangle = 0 \Rightarrow \frac{\partial \langle \psi |}{\partial \alpha}|\psi\rangle = -\langle \psi | \frac{\partial |\psi \rangle}{\partial \alpha} \end{equation}


Feynman-Hellman theorem is a way to determine how the energy spectrum would change if an external parameter in a system is tuned slowly. Examples of such external parameters may include electromagnetic fields, width or overall size of potential wells, etc.


To answer your question as to "Why by arbitrarily introducing parameters into $H$ we can still estimate the correct expectation values using FHT?" - there is a catch here. To obtain the energy $E_n(\alpha)$ as a function of $\alpha$, you are putting in as much effort as much you would have to put in to compute the expectation value $\langle B \rangle$.

Your next question "On a mathematical side..." is a bit confusing. To clarify, $H$ is the hamiltonian which has a functional dependence on some parameter $\alpha$, just like the example you give $$H = A + \alpha B$$ or perhaps even more complicated form. It does not make sense when you say "$H$ is in the eigenstate ..." - a system can be in eigenstate, but $H$ is just an operator and cannot be in an eigenstate. And since $\alpha$ is a continuous parameter, the derivative is no way linked to the finite or discrete eigenvalues or eigenvectors and you just write $$\frac{\partial H}{\partial \alpha} = B$$ for the above hamiltonian.


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