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How do derivatives of operators work? Do they act on the terms in the derivative or do they just get "added to the tail"? Is there a conceptual way to understand this?

For example: say you had the operator $\hat{X} = x$. Would $\frac{\mathrm{d}}{\mathrm{d}x}\hat{X}$ be $1$ or $\frac{\mathrm{d}}{\mathrm{d}x}x$? The difference being when taking the expectation value, would the integrand be $\psi^*\psi$ or $\psi^*(\psi+x\frac{\mathrm{d}\psi}{\mathrm{d}x})$?

My specific question is about the band effect in solids. To get a better understanding of the system, we've used Bloch's theorem to express the wavefunction in the form $\psi = e^{iKx}u_K(x)$ where $u_K(x)$ is some periodic function. With the fact that $\psi$ solves the Schrodinger equation, we've been able to derive an "effective Hamiltonian" that $u_K$ is an eigenfunction of, $H_K = -\frac{\hbar^2}{2m}(\frac{\mathrm{d}}{\mathrm{d}x}+iK)^2+V$. My next problem is to find $\left\langle\frac{\mathrm{d}H_z}{\mathrm{d}K}\right\rangle$, which led to this question.

Some of my reasoning: An operator is a function on functions, so like all other functions we can write it as $f(g(x))$. When you take the derivative of this function, you get $f'(g(x))*g'(x)$. So looking at the operator, $\hat{X}$, we can say that it is a function on $\psi(x)$, $\hat{X}(\psi)= x\psi$. So taking the derivative gives us: $$\frac{\mathrm{d}\hat{X}}{\mathrm{d}x} = \psi+ x\frac{\mathrm{d}\psi}{\mathrm{d}x}$$ but you could also say that $\hat{X}=x$ (not a function), so $$\frac{\mathrm{d}\hat{X}}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}x = 1$$ Now I'm inclined to say that $\hat{X}$ is a function, but it seems like for this question, it is better to just treat is as a constant and naively (in my opinion) take its derivative. So which way do I do it?

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If we leave out various subtleties related to operators, the core of OP's question (v4) seems to boil down to the following.

What is meant by $$\tag{0}\frac{d}{dx}f(x)?$$ Do we mean the derivative $$\tag{1} f^{\prime}(x),$$ or do we mean the first-order differential operator that can be re-written in normal-ordered$^1$ form as $$\tag{2} f^{\prime}(x)+f(x)\frac{d}{dx}?$$

The answer is: It depends on context. Different authors mean different things. One would have to trace carefully the author's definitions to know for sure. However, if it is written as $\frac{df(x)}{dx}$ instead, it always means $f^{\prime}(x)$, or equivalently, $[\frac{d}{dx},f(x)]$.

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$^1$ A differential operator is by definition normal-ordered, if all derivatives in each term are ordered to the right.

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The rule of thumb is that you can differentiate any function with respect to its argument.

In your second example, the Hamiltonian $$H_K=-\frac{\hbar^2}{2m}\left(\frac d{dx}+iK\right)^2+V$$ is a function of $K$: for each real $K$, you get a differential operator $H_K$. Thus (assuming everything is nice and regular and differentiable, of course) you can differentiate $H_K$ with respect to $K$. In cases like these differentiation is really noble: the chain rule and the product rule usually hold, and even most of vector calculus remains applicable. You do need to be careful, of course, when noncommuting observables are present: for example, while $\frac d {dt} e^{t\hat{A}}=\hat{A}e^{t\hat{A}}$ is true, $\frac d {dt} e^{\hat{B}+t\hat{A}}$ must be treated with some care when $\left[\hat{A},\hat{B}\right]\neq0$.

Your first example, on the other hand, doesn't quite fly. $X=\hat{x}$ is an operator that does not depend on any parameter; therefore you cannot differentiate it and $\frac{dX}{dx}$ is meaningless. Note, though, that objects like $\langle x|\psi\rangle$ are functions of $x$ and can therefore be differentiated with respect to $x$. So could, for instance, $\langle x|\hat{x}|\psi\rangle$, but only because of the dependence of the bra itself on the parameter $x$. You can also differentiate the bra: $$-i\frac{d}{dx}\langle x|=\langle x|\hat{p}.$$ But $\hat{x}$ itself, however, you can't.

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  • $\begingroup$ Doesn't $\hat{X}$ depend on $x$? So when you take the derivative with respect to $x$, it should mean something? $\endgroup$ – Mike Flynn Mar 4 '13 at 2:05
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    $\begingroup$ No. $\hat{x}:\mathcal{H}\rightarrow\mathcal{H}$ is just one operator and in a sense encompasses all $x\in\mathbb{R}$. The point is that there is one bra $\langle x|$ per real number $x$, which is not the case for $\hat{x}$. $\endgroup$ – Emilio Pisanty Mar 4 '13 at 10:25

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