I have this linear algebra problem in the context of quantum mechanics. Let \mathbf{f}_\lambda be a family of linear operators so to each \lambda \in \mathbf{R} we have a linear operator \mathbf{f}_\lambda : \mathcal{H} \to \mathcal{H} where \mathcal{H} is a complex vector space if one is unfamiliar with functional analysis (like I am) or is a Hilbert space if one is. Let’s suppose that this family is differentiable.

Suppose further that \mathbf{f}_\lambda is always a Hermitian operator. Suppose that \mathbf{f}_\lambda has a discrete spectrum of eigenvalues f_1(\lambda), f_2(\lambda), \cdots. I need to show the following:

Theorem

D_\lambda f_n(\lambda) = \left\langle f_n(\lambda)\right|D_\lambda \mathbf{f}_\lambda \left|f_n(\lambda) \right\rangle

Now here is a “proof,” it is not quite rigorous since there are probably a lot of technical details regarding functional analysis that I’m missing out on but:

Proof We begin by differentiating the eigenvalue equation \mathbf{f}_\lambda \left| f_n(\lambda) \right\rangle = f_n(\lambda) \left| f_n(\lambda) \right\rangle with respect to \lambda using the product rule:

(D_\lambda \mathbf{f}_\lambda) \left|f_n(\lambda)\right\rangle + \mathbf{f}_\lambda (D_\lambda \left| f_n(\lambda) \right \rangle) = (D_\lambda f_n(\lambda)) \left| f_n(\lambda) \right \rangle + f_n(\lambda) (D_\lambda \left| f_n(\lambda) \right\rangle)

After multiplying by \left\langle f_n(\lambda) \right| and rearranging terms we have the following:

\left\langle f_n(\lambda)\right| D_\lambda f_n(\lambda) \left|f_n(\lambda)\right\rangle = \left\langle f_n(\lambda)\right| D_\lambda \mathbf{f}_\lambda \left| f_n(\lambda) \right\rangle + \left\langle f_n(\lambda)\right| \mathbf{f}_\lambda (D_\lambda \left|f_n(\lambda)\right\rangle) - \left\langle f_n(\lambda)\right| f_n(\lambda) (D_\lambda \left|f_n(\lambda)\right\rangle)

Now we can take the adjoint of both sides of the eigenvalue equation to get that \langle f_n(\lambda)| \mathbf{f}_\lambda = \langle f_n(\lambda)| f_n(\lambda) since f_n(\lambda)^* = f_n(\lambda) because the eigenvalues of a normal operator are real. This implies that the difference on the right hand side of the above equation is zero. Supposing the eigenvectors \left|f_n(\lambda)\right\rangle normalised, we have \left\langle f_n(\lambda) \right| D_\lambda f_n(\lambda) \left| f_n(\lambda) \right\rangle = D_\lambda f_n(\lambda) \left\langle f_n(\lambda) | f_n(\lambda) \right\rangle = D_\lambda f_n(\lambda). Putting everything together gives the desired result. Now I’m curious, if there are any experts on functional analysis reading this what details have I missed out on?

How clear is this post?