Acting the derivative operator on a bra to its left

Question

Let $f(x)$ be a ket, and $\langle f(x)|$ be the corresponding bra.

Start with

$$\frac{d}{dx} |f(x)\rangle = |f'(x)\rangle$$

Take transpose conjugate of both sides. Since derivative is anti Hermitian, we get:

$$\langle f(x)|(-\frac{d}{dx})=\langle f'(x) |$$

I don't really understand the above equation. It says that the negative of the derivative of the row vector $\langle f(x)|$ is the row vector $\langle f'(x)|$. I think that the row $\langle f'(x)|$ should simply be the derivative of the row vector $\langle f(x)|$, because its corresponding column vector $|f'(x)\rangle$ is the derivative of the column vector $|f(x)\rangle$. Why should things change when we write the vector as a row vs as a column?

Answer 1

I believe what's happening in terms of the delta function is that:

The derivative of a function is given by the integral:

$$f'(x)=\int _{-\infty} ^{\infty} f(x')\delta '(x-x')dx'$$

However, the notation $\langle f(x)| \frac{d}{dx}$ is not a shorthand for the above integral. Because of the new order of the matrix multiplication, this notation instead refers to the integral:

$$\int _{-\infty} ^{\infty} f(x')\delta '(x'-x)dx'$$

Now using $\delta ' (x-x')=-\delta ' (x'-x)$, we get

$$=-\int _{-\infty} ^{\infty} f(x')\delta '(x-x')dx'$$

$$=-f'(x)$$

So, in short, the notation $\langle f(x)|\frac{d}{dx}$ means the negative of the derivative of $f(x)$, because of the order of matrix multiplication.

Answer 2

Most often you'll be working in some $L^2$ space, just because of how Quantum Mechanics is naturally formulated. In the case of $L^2(\mathbb{R})$, for example, integration by parts gives you the dual or transpose map. In the case where evaluation terms vanish, you have \begin{align} \langle \frac{dg}{dx}|f\rangle&=\int f \frac{dg}{dx}^*dx \\ &=-\int \left(\frac{df}{dx}\right)g^*dx+\mbox{(possible evaluation terms)} \\ &=\langle g|-\frac{df}{dx}\rangle. \end{align} The Physicist's notation is a little awkward for me; so correct if necessary. The relevant tool is integration by parts, where you assume that evaluation terms vanish.