Commutators: solving with jacobian

In summary: You are trying to use the jacobi identity with the commutator but you are not getting the correct answer. You should use the unnamed identity [A,BC] = [A,B]C + B[A,C] instead.
  • #36
Then it is [itex]\hat{x}[/itex], but I don't know its formula, or if it even has one besides [itex]\hat{x}[/itex]
 
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  • #37
In the coordinate basis, which is the basis you're working in, the operator ##\hat{x}## when acting on a function ##f(x)## simply multiplies that function by ##x##. That is to say, ##\hat{x}f(x) = xf(x)##.
 
  • #38
So in this case it does nothing that would influence its right side, except for multiplication, right? Then I only see multiplication of two operators, [itex]\hat{p}x[/itex], which is just [itex]-i\hbar[/itex], and [itex]x\hat{p}\psi[/itex]. What can I do with this?
 
  • #39
No. Say A and B are matrices and x is a vector. What you keep doing is equivalent to saying ABx = (Ax)(Bx). That's not how it works.

Just work right to left and remember that ##\hat{p}## acts on everything to its right.
 
  • #40
the only thing I see is [itex]-i\hbar\frac{∂x}{∂x}\frac{∂ψ}{∂x}[/itex]. Where am I making mistake?
 
  • #41
##\hat{p}## acts on everything to its right.
 
  • #42
[itex]-i\hbar\frac{∂x\frac{∂\psi}{∂x}}{∂x}[/itex]?
 
  • #43
Gosh, I finally understood my mistake:P I just treated the whole thing as simple standard multiplication, just like I should do in normal mathematical case, however in case of operators I should just multiply every operator by the variable on its right side:D In normal situation it seems illogical, looking like this: ##(A+B)CD=(AD+BD)CD##, but in case of operators it should look like this ##(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi i\hbar x\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar\psi)(-i\hbar\frac{\partial \psi}{\partial x})=-\hbar^2\psi \frac{\partial \psi}{\partial x}=-\hbar^2\frac{\partial }{\partial x}##
 
  • #44
Unfortunately, I still do not know how to solve it using Jacobi identity... I am not even sure if I interpret it correctly. I tried to use the commutator ##[p^2_x,x]=[p_x p_x,x]##, use the given fact ##[p_x,x]=-i\hbar##, or ##[x,p_x]=i\hbar## and substitute it to Jacobi identity formula ##[p_x,[p_x,x]]+[p_x,[x,p_x]]+[x[p_x,p_x]]=0##, but those commutators zero themselves even before I sum them up. How to do this?
 
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  • #45
Rorshach said:
Gosh, I finally understood my mistake:P I just treated the whole thing as simple standard multiplication, just like I should do in normal mathematical case, however in case of operators I should just multiply every operator by the variable on its right side:D In normal situation it seems illogical, looking like this: ##(A+B)CD=(AD+BD)CD##, but in case of operators it should look like this ##(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})##
Unfortunately, you haven't understood your mistake. You're still doing the same thing you did before. It doesn't work at all like (A+B)CD = (AD + BD)CD.

After you apply the rightmost ##\hat{p}##, you have ##\hat{p}\hat{x}\,\left(-i\hbar \frac{\partial\psi}{\partial x}\right)##. Now ##-i\hbar\frac{\partial\psi}{\partial x}## is just another function. Let's call it ##f##. So now you have ##\hat{p}\hat{x}f##. What is ##\hat{x}f## equal to? Call that result ##g##. Then apply ##\hat{p}## to ##g##. What do you get?
 
  • #46
##-i\hbar\frac{\partial g}{\partial x}##?
 
  • #47
Ignoring the constants, yes, that's what you get. So what's the derivative of g equal to?
 
  • #48
##-i\hbar\frac{\partial \psi}{\partial x}##?
 
  • #49
No. What is ##g## equal to?
 
  • #50
##-i\hbar x\frac{\partial x}{\partial x}##
 
  • #51
Oh right, I confused just derivative with applying the operator:P
 
  • #52
it came down to ##-i\hbar(\frac{\partial g}{\partial x})=-i\hbar (-i\hbar\frac{\partial \psi}{\partial x})##, which is basically a half of the result, so it is correct, right?
 
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  • #53
Rorshach said:
##-i\hbar x\frac{\partial x}{\partial x}##
That's not correct. Where's you get ##\frac{\partial x}{\partial x}## from?
 
  • #54
I just made a mistake while writing the result in code, it should be ##-i\hbar x\frac{\partial \psi}{\partial x}##
 
  • #55
OK, so what do you get when you differentiate that?
 
  • #56
##-i\hbar\frac{\partial \psi}{\partial x}##
 
  • #57
How'd you get that?
 
  • #58
I differentiated it by x, but from my experience I again made a mistake somewhere...
 
  • #59
What if you wrote ##g=xf##? What do you get when you differentiate g?
 
  • #60
oh shoot... it is ##x\frac{\partial f}{\partial x}+f##... but it only complicates the matter, now I have ##-i\hbar(x\frac{\partial f}{\partial x}-i\hbar\frac{\partial \psi}{\partial x})##
 
  • #61
That's only the first term (with some typos, I hope). You still have to subtract off the result from the xp term.
 
  • #62
Now I'm completely lost. I repeated that calculation and got the same result, I cannot see the mistake. I thought that xp term was included in that ##-i\hbar\frac{\partial g}{\partial x}## combination?
 
  • #63
my mistake again, forgot about the second term in the bracket. But I cannot use g substitution, since x is on the left side of the operator, I can only use f. What can I do with this?
 
  • #64
Try starting over from the beginning now. You began with
$$[p^2,x]\psi = (p[p,x]+[p,x]p)\psi = p[p,x]\psi + [p,x]p\psi.$$ Now redo your calculations for ##p[p,x]\psi## and ##[p,x]p\psi##.
 
  • #65
the result for the first term was correct, it came out as ##-\hbar^2\psi\frac{\partial \psi}{\partial x}##. I just have problem with the second term, because I can't think of the solution with changed order of the operator...how to bite this?
 
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  • #66
I think I finally did it. Here it goes: ##-i\hbar\frac{\partial }{\partial x}(-i\hbar\frac{\partial x\psi}{\partial x}+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar\frac{\partial xf}{\partial x}+i\hbar x\frac{\partial f}{\partial x})##, where we let ##f=-i\hbar\frac{\partial \psi}{\partial x}##. Then it goes on:
##-i\hbar\frac{\partial }{\partial x}(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar(x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x})+i\hbar x\frac{\partial f}{\partial x})##, proceeding
##-i\hbar\frac{\partial }{\partial x}(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi+i\hbar x\frac{\partial \psi}{\partial x})+(i\hbar x\frac{\partial f}{\partial x}-i\hbar f+i\hbar x\frac{\partial f}{\partial x})##, and then
##-\hbar^2\frac{\partial \psi}{\partial x}+(-i\hbar f)=-\hbar^2\frac{\partial \psi}{\partial x}-(-i\hbar\frac{\partial \psi}{\partial x})=-\hbar^2\frac{\partial \psi}{\partial x}-\hbar^2\frac{\partial \psi}{\partial x}=-2\hbar\frac{\partial \psi}{\partial x}##.
Please tell me that this time I am right.
 
  • #67
Just a few minor errors, probably typos, and you're not quite done.
Rorshach said:
I think I finally did it. Here it goes:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar\frac{\partial x\psi}{\partial x}+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar\frac{\partial xf}{\partial x}+i\hbar x\frac{\partial f}{\partial x}),$$ where we let ##f=-i\hbar\frac{\partial \psi}{\partial x}##. Then it goes on:
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar x\frac{\partial \psi}{\partial x})+(-i\hbar(x\frac{\partial f}{\partial x}+f\frac{\partial x}{\partial x})+i\hbar x\frac{\partial f}{\partial x})$$
$$-i\hbar\frac{\partial }{\partial x}(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi+i\hbar x\frac{\partial \psi}{\partial x})+({\color{red}-}i\hbar x\frac{\partial f}{\partial x}-i\hbar f+i\hbar x\frac{\partial f}{\partial x})$$
$$-\hbar^2\frac{\partial \psi}{\partial x}+(-i\hbar f)
= \hbar^2\frac{\partial \psi}{\partial x}-{\color{red} {(i\hbar)}}(-i\hbar\frac{\partial \psi}{\partial x})
= -\hbar^2\frac{\partial \psi}{\partial x}-\hbar^2\frac{\partial \psi}{\partial x}
= -2\hbar^{\color{red}2} \frac{\partial \psi}{\partial x}$$.
Please tell me that this time I am right.
Now use the fact that ##-\hbar^2 = (-ih)^2## and rewrite the final result in terms of ##\hat{p}_x## to show that ##[\hat{p}_x^2,\hat{x}] = (-i\hbar)2\hat{p}_x##.
 
  • #68
I don't quite understand what do You mean by presenting it in terms of ##p_x##, I thought that my calculations (without those stupid typos) are enough to show the correct result?
 
  • #69
Well, ask yourself what you're trying to show. The problem asks you to calculate ##[\hat{p}_x^2,\hat{x}]## which is equal to ##-2i\hbar \hat{p}_x##. You haven't shown that. You've shown that when you work in the position basis, the action of ##[\hat{p}_x^2,\hat{x}]## on some arbitrary function ##\psi(x)## is to differentiate it and multiply it by ##-2\hbar^2##. You need to show that that's equivalent to applying ##\hat{p}## and multiplying by ##-2i\hbar##.
 
  • #70
So basically I should take function ##\psi(x)##, act on it with ##\hat{p}## and multiply it by ##-2i\hbar##?
 
<h2>1. What is a commutator in the context of solving with Jacobian?</h2><p>A commutator is a mathematical operation that involves taking the difference between two quantities and multiplying it by a third quantity. In the context of solving with Jacobian, commutators are used to calculate the rate of change of a function with respect to its variables.</p><h2>2. How do commutators help in solving with Jacobian?</h2><p>Commutators are useful in solving with Jacobian because they allow us to express the rate of change of a function in terms of its partial derivatives. This makes it easier to solve systems of equations involving multiple variables.</p><h2>3. Can commutators be used to solve any type of problem with Jacobian?</h2><p>Yes, commutators can be used to solve a wide range of problems involving Jacobian, including optimization, differential equations, and physics problems.</p><h2>4. Are there any limitations to using commutators in solving with Jacobian?</h2><p>One limitation of using commutators is that they can only be used for differentiable functions. Additionally, they may not always provide the most efficient or accurate solution to a problem.</p><h2>5. How can I learn more about using commutators in solving with Jacobian?</h2><p>There are many resources available online that provide detailed explanations and examples of using commutators in solving with Jacobian. You can also consult textbooks or seek guidance from a mathematics tutor or professor.</p>

1. What is a commutator in the context of solving with Jacobian?

A commutator is a mathematical operation that involves taking the difference between two quantities and multiplying it by a third quantity. In the context of solving with Jacobian, commutators are used to calculate the rate of change of a function with respect to its variables.

2. How do commutators help in solving with Jacobian?

Commutators are useful in solving with Jacobian because they allow us to express the rate of change of a function in terms of its partial derivatives. This makes it easier to solve systems of equations involving multiple variables.

3. Can commutators be used to solve any type of problem with Jacobian?

Yes, commutators can be used to solve a wide range of problems involving Jacobian, including optimization, differential equations, and physics problems.

4. Are there any limitations to using commutators in solving with Jacobian?

One limitation of using commutators is that they can only be used for differentiable functions. Additionally, they may not always provide the most efficient or accurate solution to a problem.

5. How can I learn more about using commutators in solving with Jacobian?

There are many resources available online that provide detailed explanations and examples of using commutators in solving with Jacobian. You can also consult textbooks or seek guidance from a mathematics tutor or professor.

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