Help with this Commutator question please

  • #1
Clifford Williams
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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:

Answers and Replies

  • #2
ergospherical
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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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  • #3
PeroK
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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.
 
Last edited:
  • #4
Clifford Williams
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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.
 
  • #5
PeroK
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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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  • #6
Clifford Williams
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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...
 
  • #7
ergospherical
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Can you post your professor's solution? We can then look for the mistake.
 
  • #8
Clifford Williams
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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.
 
  • #9
PeroK
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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
PhDeezNutz
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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
vela
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What did your professor mean by "that's not a single valued function so it won't work, remember the postulates!"?
 
  • #12
vanhees71
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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
Steve4Physics
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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
martinbn
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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
LCSphysicist
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This identity will give me nightmares.
 
  • #16
vanhees71
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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.
 
  • #17
kuruman
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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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