# The use of the commutators in quantum mechanics: explanations [duplicate]

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.

• So the question is why $[\partial_x , x ] =1$? – FGSUZ Aug 5 '19 at 20:30
• @FGSUZ I don't understand how the commutators work. I have given three examples. – Sebastiano Aug 5 '19 at 20:33
• Okay, I've added an answer. I hope I understood your question well. I don't know if that was what you were asking. – FGSUZ Aug 5 '19 at 20:50
• Possible duplicate of Heisenberg Uncertainty Principle scientific proof – AccidentalFourierTransform Aug 5 '19 at 20:55

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}

• Hi, excuse me for asking, but did you remove the comments or didn't you add them? Thank you for your kind answer. – Sebastiano Aug 7 '19 at 12:28
• @Sebastiano No, I didn't remove the comments. I guess a moderator did it. – Thomas Fritsch Aug 7 '19 at 13:36

First of all, there's one idea that some people forget at the beginning: commutators are operators

That is, they act on a wavefunction, like all operators.

A commutator is something that needs an input of two operators: $$A$$ and $$B$$, and the output is another operator, which is $$(AB-BA)$$. That is an operator.

For me, it is dangerous to write $$[\frac{\partial}{\partial x} , x] =1$$, because a commutator is not a number. That "number 1" refers to the "operator one", or the identity operator. I prefer writing

$$\left[\frac{\partial}{\partial x} , x\right] =\mathbb{I}$$

What's more, it is better to write capital letters, or hats on the letters, because here $$x$$ is not a variable, it is the $$X$$ operator. We should write:

$$\left[\frac{\partial}{\partial X} , X\right] =\mathbb{I}$$

And this is important because this reminds you that operators make sense when they are applied on wavefunctions.

So, if you want to evaluate $$[X, P_x]$$, by definition, it is $$X\ P_x - P_x\ X$$

But this makes sense when you apply it to a wavefunction, that is, you should take an arbitrary wavefunction $$\psi$$ and compute

$$[X, P_x]\psi=X\ P_x \psi - P_x \ X\ \psi =X\ (P_x \psi) - P_x \ (X\ \psi)$$

And that's how you prove the different relations.

You have to know how each operator acts on the wavefunction. For example $$X\psi = x\psi$$, it is just multiplying it by the $$x$$ variable.

However, the way $$P$$ acts is $$P\psi = -i\hbar \frac{\partial \psi}{\partial x}$$

So, in sum, you demonstrate commuting relations by appliying them to an arbitrary $$\psi$$.

A nice and simple proof can be found in Introduction to Quantum Mechnics by David J. Griffiths, section 3.5. This book is a classical introductory book to quantum mechanics.

Basically, for any observable A, you have $$\sigma_A^2 = \langle (\hat{A} - \langle A \rangle) \Psi | ((\hat{A} - \langle A \rangle ) \Psi \rangle$$, from which it shown in the book that $$\sigma_A^2 \sigma_B^2 \geq \left(\frac{1}{2i} \langle [\hat{A}, \hat{B}] \rangle \right) .$$

Now you had the correct idea to calculate the commuator, but to see it correctly, you need to apply it to a test function, say $$f$$. Then, the derivative acts according to the product rule, that is $$[i\hbar \frac{\partial}{\partial x}, x]f = i\hbar \frac{\partial}{\partial x} xf - x \frac{\partial}{\partial x}f = i\hbar f + i\hbar x \frac{\partial f}{\partial x} - i\hbar x \frac{\partial f}{\partial x} = i\hbar f.$$ Thus, as an operator, $$[\hat{p}_x, \hat{x}] = i\hbar \mathbb{1}$$ where $$\mathbb{1}$$ is the identity which you can ignore.

• I am not sure what is the remaining of your question! – gingras.ol Aug 5 '19 at 21:42
• Your answer along with the others have responded to my request. :-) I have rewarded in addition all of them. Your answers are pieces that I have to put in order in my mind and all are fine. – Sebastiano Aug 5 '19 at 21:45
• Great! Good luck with quantum mechanics! :) If you have a background in maths, you should manage quite well. Also, I think the book I mentioned could really help you out if you want to learn well. – gingras.ol Aug 5 '19 at 21:47