Help with this Commutator question please

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Homework Statement
Evaluate the commutator [π‘₯(πœ•/πœ•π‘¦), 𝑦(πœ•/πœ•π‘₯)] by applying the operators to an arbitrary
function 𝑓(π‘₯,𝑦).
Relevant Equations
[A , B] = AB - BA
Hello,
In QM class this morning my Prof claimed that the commutator [π‘₯(πœ•/πœ•π‘¦), 𝑦(πœ•/πœ•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:
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Your answers looks correct, yes. (Unless I have made the same mistake as you.)
\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*}
 
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You can also check for a specific function, e.g, ##f(x, y) = x## to show that the commutator cannot be zero.
 
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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 βˆ‚f(x,y)/βˆ‚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.

I suspect there is something about this problem you are not telling us.
 
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The problem was delivered exactly as phrased in the "Homework Statement":
"Evaluate the commutator [π‘₯(πœ•/πœ•π‘¦), 𝑦(πœ•/πœ•π‘₯)] by applying the operators to an arbitrary
function 𝑓(π‘₯,𝑦)."

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:
Capture.PNG

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.
 
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  • #10
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##.
 
  • #11
What did your professor mean by "that's not a single valued function so it won't work, remember the postulates!"?
 
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  • #12
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.
 
  • #13
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.

You might ask your professor if this is a mistake.
 
  • #14
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?
 
  • #15
This identity will give me nightmares.
 
  • #16
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.
 
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  • #17
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, $$[\partial f(x,y)/\partial x] = g(y)*[\partial h(x)/\partial x]+h(x)*[\partial g(y)/\partial x]=g(y)*[\partial h(x)/\partial x]+h(x)*0.$$
 
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