# Partial Derivative and Dirac Notation [duplicate]

Does the partial derivative of $\langle x'|\alpha\rangle$ with respect to $x'$ equal $|\alpha\rangle$? Why?

Note: $|\alpha\rangle$ is an arbitrary ket, $x'$ is an eigenvalue, and $\langle x'|$ is an eigenket corresponding to $x'$.

In particular, I do not understand how these two expressions are equivalent:

$$\langle x' - \Delta x' | \alpha \rangle = \langle x' |\alpha\rangle - \Delta x' \frac {\partial} {\partial x'} \langle x' | \alpha \rangle$$

• is $\frac{\mathrm d}{\mathrm dx}\sin x=\sin$? Commented May 28, 2017 at 20:31
• The two expressions aren't quite equivalent: the RHS is the first two terms of a Taylor expansion of the LHS. Commented May 28, 2017 at 20:43

## 1 Answer

No it doesn't. First, we must clarify what is $|\alpha\rangle$. By the postulates of Quantum Mechanics, the states of a quantum system are described by unit vectors in a space of states $\mathcal{E}$ assumed to be a Hilbert space.

The $|\alpha\rangle$ is such a vector carrying the information of the state of the system.

On the other hand, the physical quantities are described by hermitian operators called observables. It is a postulate that for each physical quantity there will be a corresponding observable.

The possible values to be measured of the quantity are the ones from the spectrum of the operator. Furthermore, there's for each such value the states on which the value is certain to be measured.

In your example we must deal with position. Position $X$ is one observable with continuous spectrum, meaning that any real number could be measured. For each $x\in \sigma(X)$ you may consider the ket of definite position $|x\rangle$.

The postulates of Quantum Mechanics tells you that the "amplitude of probability" for measuring $x\in \sigma(X)$ in the state $|\alpha\rangle$ is $\varphi(x)=\langle x |\alpha\rangle$.

So this object which you differentiate has a precise meaning according to the postulates. It is a function $\varphi : \sigma(X)\to \mathbb{C}$ representing the probability amplitude for the values of $\sigma(X)$.

This means that $\rho(x)=|\varphi(x)|^2$ is the probaiblity density for the coordinate of position.

Now since $\varphi = \langle \cdot | \alpha\rangle$ is a function taking values in $\mathbb{C}$, its derivative clearly can't take values in $\mathcal{E}$, and hence can't be $|\alpha\rangle$ as you suggest.

Edit: Your issue seems to be with the infinitesimal translation transformation. Either way, what the author is computing is $\varphi(x-\Delta x)$. Assume $\Delta x$ to be small, such that $\Delta x^n \approx 0$ if $n \geq 2$. Taylor expand $\varphi$ around $x$, displacing $\Delta x$

$$\varphi(x-\Delta x)-\varphi(x)=\sum_{n=1}^\infty \dfrac{1}{n!}\dfrac{d^n\varphi}{dx^n}(x) (-\Delta x)^n$$

but by the assumption on $\Delta x$ we have

$$\varphi(x-\Delta x)\approx\varphi(x) - \dfrac{d}{dx}\varphi(x) \Delta x$$

Substitute the definition $\varphi$ to get

$$\langle x-\Delta x | \alpha\rangle \approx \langle x | \alpha\rangle - \Delta x\dfrac{d}{dx}\langle x | \alpha\rangle.$$

Again, you are just manipulating a function $\varphi : \sigma(X)\subset \mathbb{R}\to \mathbb{C}$ as in complex analysis.