homework and exercises - The use of the commutators in quantum mechanics: explanations

Considering that I've never studied quantum mechanics before I have need to understand the operator commutator. My start is: $[A,B]=AB-BA \tag{a}$

Now, why must be

$$\left[\frac{\partial }{\partial x},x\right]\stackrel{?}{=}1 \tag{1}$$ I have thought, from the rule (a),

This identity

$$\left[x,\frac{\partial }{\partial x}\right]=-1 \tag{2}$$ is easy because $[A,B]=-[B,A]$. I have not understood, also, (3) and (4) $$\left[i\hslash\frac {\partial}{\partial x},x\right]=i\hslash \tag{3}$$

$$[p_x,x]=i\hslash \tag{4}$$ where $p_x$ is the momentum on $x-$ axis.

Answer

Equations (a), (1), (2), (3) and (4) all are operator equations. Therefore you need to understand what an operator equation actually is.

Now, why must be

$$\left[\frac{\partial }{\partial x},x\right]\stackrel{?}{=}1 \tag{1}$$

That means, the operators on the left-hand-side and on the right-hand-side always yield the same result when applied to arbitrary functions.

Hence, here you must prove that

$$\left[\frac{\partial}{\partial x},x\right] \psi(x) = 1\cdot \psi(x)$$ for every function $\psi(x)$.

The proof is a long sequence of very elementary steps:

$$\begin{align} &\left[\frac{\partial }{\partial x},x\right] \psi(x) \\ = &\left(\frac{\partial}{\partial x} x - x \frac{\partial }{\partial x}\right) \psi(x) \\ = &\frac{\partial}{\partial x} x \psi(x) - x \frac{\partial }{\partial x} \psi(x) \\ = &\frac{\partial x}{\partial x}\psi(x) + x \frac{\partial \psi(x)}{\partial x} - x \frac{\partial \psi(x)}{\partial x} \\ = &\frac{\partial x}{\partial x}\psi(x) \\ = &1\cdot \psi(x) \\ \end{align}$$