I was given the following question:

Let $A(\lambda)$ be a Hermitian operator, which is dependent on some real parameter $\lambda$. Let us denote the eigenvalues and corresponding eigenstates of $A$ with $a_k(\lambda), |u_k(\lambda)\rangle$. Show that: $$\frac{da_k}{d\lambda}=\langle\frac{dA}{d\lambda}\rangle_{|u_k(\lambda)\rangle}$$

This is the solution I suggested:

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I wanted to know if my solution is correct. I'm not entirely sure that my final step is valid (since the normalized magnitude of the eigenvector is only a convention, it seems strange to use this in such a general context).

Moreover, I wanted to know if there is a name for this result, or if you know of a source where I can read more about it and its implications.

  • 4
    $\begingroup$ See the Hellmann–Feynman theorem. $\endgroup$ – Alexander May 7 '19 at 18:03
  • 5
    $\begingroup$ Add an aside, save yourself from using LyX & just use plain LaTeX. $\endgroup$ – Kyle Kanos May 7 '19 at 18:05

The last step is wrong, but the problem is not that you require that the $\left| u_k \right>$ be normalized which can always be done without loss of generality. The problem is that $\left< \partial_\lambda u_k \middle| u_k \right>$ can be non-zero (however, it will always be imaginary, and since $a_k$ is real the second term will vanish nevertheless).

Assuming normalized states, we get (dropping the index $k$ for brevity): $$ 0 = \partial_\lambda \left< u \middle| u \right> = \left< \partial_\lambda u \middle| u \right> + \left< u \middle| \partial_\lambda u \right> = \left< \partial_\lambda u \middle| u \right> + \left< \partial_\lambda u \middle| u \right>^*. $$ And for a complex number $a + a^* = 0$ does not imply that $a = 0$, but only that $\text{Re}\,a = 0$.

If the states were not normalized, you would have to adapt your expression for the expectation value from the beginning, and use $$ \left< A \right>_\psi = \frac{\left<\psi \middle| A \middle| \psi \right>}{\left< \psi \middle | \psi \right>}, $$ and then you can check that everything still works out.

Also note, that the normalization is used before the last step, for example $ \left< u_k \middle| A(\lambda) \middle| u_k \right> = a_k(\lambda)$ does only hold if the $\left|u_k\right>$ are normalized, in general: $$ \left< u_k \middle| A(\lambda) \middle| u_k \right> = \left< u_k \middle| a_k(\lambda) \middle| u_k \right> = a_k(\lambda) \left< u_k \middle | u_k \right>$$

Story time: The non-vanishing of the imaginary part of $\left< u_k \middle| \partial_\lambda u_k \right>$ plays a prominent role in the expression for the density of the Berry phase, when considering the evolution of a quantum system subject to adiabatically changing external parameters $\lambda$.

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