# Help with this Commutator question please

• Clifford Williams
In summary, the professor is claiming that the commutator of two functions, x(d/dx) - y(d/dy), is zero, but his own solution shows that this is not necessarily the case.
Clifford Williams
Homework Statement
Evaluate the commutator [π₯(π/ππ¦), π¦(π/ππ₯)] by applying the operators to an arbitrary
function π(π₯,π¦).
Relevant Equations
[A , B] = AB - BA
Hello,
In QM class this morning my Prof claimed that the commutator [π₯(π/ππ¦), π¦(π/ππ₯)] = 0.
However, my classmate and I arrived at x(d/dx) - y(d/dy).
Can someone explain how (or if) our professor is correct?

Last edited:
\begin{align*}
[x \partial_y, y\partial_x] &= (x \partial_y)(y\partial_x) - (y\partial_x)(x\partial_y) \\
&= x(\partial_x + y \partial_{xy} ) - y(\partial_y + x \partial_{xy}) \\
&= x\partial_x - y\partial_y
\end{align*}

wrobel and PeroK
You can also check for a specific function, e.g, ##f(x, y) = x## to show that the commutator cannot be zero.

Last edited:
I checked a specific function, but I picked the function f(x) = x^2 +y^2, and my prof replied that "that's not a single valued function so it won't work, remember the postulates!"

But even when I plug in single valued functions s.a. y=x-x^3, the commutator is not 0...

My professor's solution uses the identity:
\begin{align*}

[\partial f(x,y)/\partial x] &= [\partial g(y)/\partial x]*[\partial h(x)/\partial x]

\end{align*}

and the fact that
\begin{align*}
[\partial h(y)/\partial x]=0
\end{align*}
to show that βf(x,y)/βx goes to 0, and that the commutator is 0. Is this maybe for a special case?

One of those classes where I left more confused than I arrived.

None of that makes any sense to me. At least one of you and the professor is confused.

The problem was delivered exactly as phrased in the "Homework Statement":
"Evaluate the commutator [π₯(π/ππ¦), π¦(π/ππ₯)] by applying the operators to an arbitrary
function π(π₯,π¦)."

That feeling that you're missing something? That's exactly where I've been all day...

Can you post your professor's solution? We can then look for the mistake.

My professor's solution is as follows:

It relies on the identity
\begin{align*}
[\partial f(x,y)/\partial x] &= [\partial f(y)/\partial x]*[\partial f(x)/\partial x]
\end{align*}
to get from the penultimate line to the last line, however this is an identity I am neither familiar nor comfortable with.

Clifford Williams said:
to get from the penultimate line to the last line, however this is an identity I am neither familiar nor comfortable with.
That's obviously wrong. In fact, what that "proves" is that ##x\frac{\partial}{\partial y} \equiv 0## and ##y\frac{\partial}{\partial x} \equiv 0##. That's because he's actually shown that each term in the commutator is separately zero. Which can't be right.

And, in fact, a simple corollary is that ##\frac{\partial}{\partial y} \equiv 0## and ##\frac{\partial}{\partial x} \equiv 0##. Which is a bizarre conclusion, to say the least.

Sometimes people, even professors, get things wrong.

PhDeezNutz and ergospherical
Clifford Williams said:
My professor's solution is as follows:
View attachment 297188
It relies on the identity
\begin{align*}
[\partial f(x,y)/\partial x] &= [\partial f(y)/\partial x]*[\partial f(x)/\partial x]
\end{align*}
to get from the penultimate line to the last line, however this is an identity I am neither familiar nor comfortable with.

To me that would imply every function ##f(x,y)## has a 0 derivative with respect to both ##x## and ##y##.

What did your professor mean by "that's not a single valued function so it won't work, remember the postulates!"?

PeroK
It's proportional to
$$[\hat{x} \hat{p}_y,\hat{y} \hat{p}_x]=\hat{x}[\hat{p}_y,\hat{y} \hat{p}_x]+[\hat{x},\hat{y} \hat{p}_x]\hat{p}_y = \hat{x} [\hat{p}_y,\hat{y}] \hat{p}_x + \hat{y} [\hat{x},\hat{p}_x] \hat{p} y =-\mathrm{i} \hbar \hat{x} \hat{p}_x +\mathrm{i} \hbar \hat{y} \hat{p}_y \neq 0.$$
I've no clue, what's the meaning of the calculation presented as solution by the professor.

Clifford Williams said:
My professor's solution is as follows:
View attachment 297188
It relies on the identity \begin{align*} [\partial f(x,y)/\partial x] &= [\partial f(y)/\partial x]*[\partial f(x)/\partial x]
\end{align*} to get from the penultimate line to the last line, however this is an identity I am neither familiar nor comfortable with.
f(x,y) magically changes from being a function of two variables to being a function of a single variable.

Clifford Williams said:
My professor's solution is as follows:
View attachment 297188
It relies on the identity
\begin{align*}
[\partial f(x,y)/\partial x] &= [\partial f(y)/\partial x]*[\partial f(x)/\partial x]
\end{align*}
to get from the penultimate line to the last line, however this is an identity I am neither familiar nor comfortable with.
Are you sure you copied it correctly?

This identity will give me nightmares.

Which identity? The obviously wrong formula quoted in #14? Just forget it. It's wrong. I've no clue, how one could come to the idea to write it down in the first place.

PhDeezNutz
vanhees71 said:
Which identity? The obviously wrong formula quoted in #14? Just forget it. It's wrong. I've no clue, how one could come to the idea to write it down in the first place.
The original "identity" in post #4 is $$[\partial f(x,y)/\partial x] = [\partial g(y)/\partial x]*[\partial h(x)/\partial x]$$ By looking at it, one can assume that ##f(x,y)## is separable into ##h(x)## and ##g(y)##, i.e. ##f(x,y)=h(x)g(y)##.

If that assumption is the case, then we all know* that ##[\partial f(x,y)/\partial x] = g(y)*[\partial h(x)/\partial x].##
If that assumption is not the case, then how the &\$%@ are ##h(x)## and ##g(y)## related to ##f(x,y)##?

*Edit: That comes from application of the product rule, $$[\partial f(x,y)/\partial x] = g(y)*[\partial h(x)/\partial x]+h(x)*[\partial g(y)/\partial x]=g(y)*[\partial h(x)/\partial x]+h(x)*0.$$

PhDeezNutz and SammyS

## 1. What is a commutator?

A commutator is a mechanical or electrical device used to reverse the direction of current flow in an electric motor or generator.

## 2. Why is a commutator important in electric motors?

A commutator is important in electric motors because it allows for the continuous rotation of the motor by reversing the direction of current flow in the coils, which creates a magnetic field that interacts with the permanent magnets to produce motion.

## 3. How does a commutator work?

A commutator works by using a set of copper segments, called commutator bars, that are connected to the ends of the motor's armature coils. As the armature rotates, the brushes make contact with the commutator bars, switching the direction of current flow in the coils and allowing for continuous rotation.

## 4. What is the purpose of a commutator in a generator?

In a generator, the commutator serves to convert the alternating current (AC) produced by the rotating coils into direct current (DC) by reversing the direction of current flow at specific intervals. This allows for a steady output of electricity.

## 5. How do you troubleshoot problems with a commutator?

To troubleshoot problems with a commutator, you can check for signs of wear or damage, clean the commutator and brushes, and ensure that the brushes are making good contact with the commutator bars. If the problem persists, it may require professional repair or replacement.

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