How to Solve Commutators Using the Jacobian?

[Rorshach]

Homework Statement

Hell My problem is as follows:
Determine the following commutators: [p_x²,x],[p_xx²],[p_x²,x²],[]. The calculation can be done in two ways, either by inserting a test function, and using the explicit expressions for the operators, or by utilizing Jacobi identity and using the fact that [x,p_x]=ihbarred . You should master both methods.

Homework Equations

[A,B]=AB-BA
[x,p_x]=ihbarred
[A,BC]=B[A,C]+[A,B]C
p_x=-ihbarred*d/dx
p²=-hbarred²*d²/dx²

The Attempt at a Solution

Here's the problem- I just cannot find good paper with proper explanation how to do it and examples with solutions. I am also not familiar with term and meaning of jacobian- in the textbook I am using there is no explanation of jacobian, or how to solve commutators, not even mentioning solving commutators with jacobians.

 

[MisterX]

The Jacobi identity is not the same thing as the Jacobian. They are both named after the same person, but they are not the same thing. Here is the Jacobi identity.

[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0

It applies to the bracket operation for Lie algebras, so it applies to cross products and Poisson brackets as well as commutators.

 

[MisterX]

Also I have no idea why you should use the Jacobi identity with the commutators given to you. The unnamed identities [A,BC] = [A,B]C + B[A,C] and [AB,C] = A[B,C] + [A,C]B seem like a more appropriate things to use.

 

[Rorshach]

I don't know that either, but that is the content of the problem, so I have to do it. I have no idea how to solve commutator if there is a second derivative (p_x²).

 

[vela]

Use $\hat{p}_x^2 = \hat{p}_x\hat{p}_x$. You need to show some effort at working the problem out on your own.

 

[Rorshach]

ok, but what if there is a factor in front of the commutator? for example like [p_x²,x]ψ?
Is it supposed to be like this:

[p_x²,x]ψ=p[p_x,x]ψ+[p_x,x]ψp_x?

 

[vela]

I don't see a factor in front of the commutator. Could you elaborate?

 

[Rorshach]

One of the ways of solving commutators was to put a function after commutator, like this
[p,x]ψ - and then solve it. But unfortunately no matter if I put the function there or not- I can't get the correct value. I couldn't find anywhere how to solve commutators [p²,x], or [p²,x[/SUP]2[/SUP]] step by step. When I try do solve this on my own, I just keep getting confused by the order of operators in the equation.

 

[vela]

Show what you tried.

 

[Rorshach]

ok, so I did the calcuation as follows:
[p_xp_x,x]=p_x[p_x,x]+_x[p_x,x]p_x
p_x(p_xx-xp_x)+(p_xx-xp_x)*p_x
-ihbarred(∂/∂x)(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))+(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
-ihbarred(∂/∂x)(-ihbarred+ihbarredx(∂/∂x))+(-ihbarred+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
-ihbarred^2(∂/∂x)+xhbarred^2(∂/∂x)-hbarred^2(∂/∂x)+xhbarred^2(∂/∂x)(∂/∂x)

and that is pretty much the result that I obtain every single time I calculate this. Where am I making mistake?

 

[vela]

ok, so I did the calcuation as follows:
[p_xp_x,x]=p_x[p_x,x]+[p_x,x]p_x

Once you get to this point, use the fact that $[x,p_x]=i\hbar$.

p_x(p_xx-xp_x)+(p_xx-xp_x)*p_x
-ihbarred(∂/∂x)(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))+(-ihbarred(∂x/∂x)+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
-ihbarred(∂/∂x)(-ihbarred+ihbarredx(∂/∂x))+(-ihbarred+ihbarredx(∂/∂x))*(-ihbarred(∂/∂x))
-ihbarred^2(∂/∂x)+xhbarred^2(∂/∂x)-hbarred^2(∂/∂x)+xhbarred^2(∂/∂x)(∂/∂x)

and that is pretty much the result that I obtain every single time I calculate this. Where am I making mistake?

 

[Rorshach]

But that is what I want to do with the jacobi identity, so the demands in the problem are satisfied. I tried to do this the way I can see step by step what is coming from where, that is why I insisted on step by step explanation.

 

[vela]

Your mistake is when dealing with terms like $p_x x$. It doesn't simplify to 1, like you have. The operator $p_x$ acts on everything to the right of it, so not just $x$ but the test function $\psi$ as well, i.e.,
$$p_x x \psi = -i\hbar \frac{\partial}{\partial x} (x\psi) = \cdots$$

 

[Rorshach]

so in case of [p_x²,x] (if I want to do it step by step) I am supposed to put the test function after the operator p?

 

[vela]

Yes. Why wouldn't it be at the end?

 

[Rorshach]

It just confuses me so I wasn't sure if it was supposed to be before or after the p operator. I tried with using ψ function, unfortunately it didn't help. first bracket zeroes:
-ihbarred(∂/∂x)(-ihbarred(∂x/∂x)ψ+ixhbarred(∂ψ/∂x))
but the second one in no way gives me the value -2ihbarredp_x, which is supposed to be an answer for this problem:
(-ihbarred(∂x/∂x)+ixhbarred(∂/∂x))(-ihbarred(∂ψ/∂x))
where am I making mistake?

 

[vela]

It just confuses me so I wasn't sure if it was supposed to be before or after the p operator.

Using $[AB,C] = A[B,C] + [A,C]B$, you have $[p_x^2,x] = p_x[p_x,x] + [p_x,x]p_x$. Now it's basic substitution:
$$[p_x^2,x]\psi = (p_x[p_x,x] + [p_x,x]p_x)\,\psi.$$ You can't just move $\psi$ past $p_x$.

I tried with using ψ function, unfortunately it didn't help. first bracket zeroes:
-ihbarred(∂/∂x)(-ihbarred(∂x/∂x)ψ+ixhbarred(∂ψ/∂x))

How'd you get 0? Also, you're missing a term. You should have
$$p_x [p_x, x] \psi = p_x (p_x x - x p_x)\psi = -i\hbar\frac{\partial}{\partial x} \left[-i\hbar\frac{\partial}{\partial x} (x\psi) - x \left(-i\hbar\frac{\partial}{\partial x}\right) \psi\right].$$ Now use the product rule to expand the first term in the brackets.

 

[Rorshach]

You were right, I confused the order of calculations, hence the zeroing of the first bracket. My result of the first bracket came out to be -2(hbarred)^2(∂ψ/∂x). The second one comes to -(hbarred)^2(∂ψ/∂x)+x(hbarred)^2(∂/∂x)(∂ψ/∂x). unless I miss something similar to what I did in the first bracket.

 

[Rorshach]

and also- if the Jacobi identity is [A,[B,C]]+[B,[C,A]]+[C,[A,B]]=0, then for [p²_x,x] it would be [p_x,[p_x,x]]+[p_x,[x,p_x]]+[x,[p_x,p_x]]=0, right? But how do I do this if there is commutator like [p_x²,x²]?

 

[Rorshach]

I tried calculation with Jacobi identity according to the scheme I wrote above, but the commutators just zero themselves, so it seems kind of pointless. What am I doing wrong?

 

[vela]

Like MisterX, I don't see how the Jacobi identity is very helpful here.

 

[Rorshach]

Do You mean it is not possible to solve those commutators with jacobi identity? And also, could You please tell me what am I doing wrong with second bracket? The result of the second bracket that I obtained is equal to -(hbarred)^2(∂ψ/∂x)+x(hbarred)^2(∂/∂x)(∂ψ/∂x)

 

[Rorshach]

could somebody please answer?

 

[vela]

It's kind of hard to offer help when all you do is post your incorrect answer and ask what went wrong. How are we supposed to tell if we can't see the steps you took?

 

[Rorshach]

I' sorry, here are my steps:
from the first bracket p_x(p_xx-xp_x)ψ I obtained the following result: -2\hbar²\frac{∂ψ}{∂x}. In the second bracket I made following steps:
<br /> (p_xx-xp_x)p&lt;sub&gt;x&lt;/sub&gt;ψ&lt;br /&gt; (-i/hbar\frac{∂x}{∂x}+xi/hbar\frac{∂}{∂x})(-i/hbar\frac{∂ψ}{∂x})&lt;br /&gt; -/hbar&lt;sup&gt;2&lt;/sup&gt;\frac{∂ψ}{∂x}+x/hbar&lt;sup&gt;2&lt;/sup&gt;r\frac{∂ψ}{∂x}r\frac{∂}{∂x}&lt;br /&gt;<br /> <br /> It seems that LaTeX does not want to play, no idea why. Tried to describe more clearly and visibly, if You know why it is like that please tell me

 

[Rorshach]

anybody?

 

[MisterX]

I' sorry, here are my steps:
from the first bracket p_x(p_x x-xp_x)\psi I obtained the following result: -2\hbar^2 \frac{∂ψ}{∂x}. In the second bracket I made following steps:
(p_x x-xp_x)p_x ψ
(-i\hbar\frac{∂x}{∂x}+xi\hbar\frac{∂}{∂x})(-i\hbar\frac{∂ψ}{∂x})-\hbar^2\frac{∂ψ}{∂x}+x\hbar^2 r\frac{∂ψ}{∂x}r\frac{∂}{∂x}<br />

It seems that LaTeX does not want to play, no idea why. Tried to describe more clearly and visibly, if You know why it is like that please tell me

I modified the quote so the \LaTeX works. Don't use the BBCode tags like

[SUP]

inside the \LaTeX code.

 

[vela]

Reevaluate the first term, $p_x x p_x \psi$. Remember that an operator acts on everything to the right of it, so evaluate the expression from right to left.

 

[Rorshach]

In that case, is it supposed to be (-i/hbar\frac{∂xψ}{∂x})(-i/hbar\frac{∂ψ}{∂x})?

 

[Rorshach]

I still don't know why the LaTeX doesn't work:/

 

[vela]

You need to use \hbar, not /hbar.

I'm going to use hats over the operators so you can keep track of what's going on. The first thing on the right is $\hat{p}_x \psi = -i\hbar \frac{\partial \psi}{\partial x}$, so
$$\hat{p}_x \hat{x} \hat{p}_x \psi = \hat{p}_x \hat{x} (-i\hbar)\frac{\partial \psi}{\partial x} = -i\hbar\,\hat{p}_x \hat{x} \frac{\partial \psi}{\partial x}.$$
Now, again, evaluate the what the rightmost operator does.

 

[Rorshach]

if it is supposed to be \frac{∂ψ}{∂x}, then I don't know what to do with this. If it is -i\hbar \frac{∂}{∂x} x, then it is just -i\hbar

 

[vela]

What is the operator farthest on the right in
$$-i\hbar\,\hat{p}_x \hat{x} \frac{\partial \psi}{\partial x}?$$

 

[Rorshach]

It is \frac{∂ψ}{∂x}

 

[vela]

I meant a hatted operator. $\psi'(x)$ is just a function.

 

[Rorshach]

Then it is \hat{x}, but I don't know its formula, or if it even has one besides \hat{x}

 

[vela]

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, $\hat{x}f(x) = xf(x)$.

 

[Rorshach]

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, \hat{p}x, which is just -i\hbar, and x\hat{p}\psi. What can I do with this?

 

[vela]

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.

 

[Rorshach]

the only thing I see is -i\hbar\frac{∂x}{∂x}\frac{∂ψ}{∂x}. Where am I making mistake?

 

[vela]

$\hat{p}$ acts on everything to its right.

 

[Rorshach]

-i\hbar\frac{∂x\frac{∂\psi}{∂x}}{∂x}?

 

[Rorshach]

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 $(-i\hbar\frac{\partial }{\partial x}x+i\hbar\frac{\partial}{\partial x})(-i\hbar\frac{\partial }{\partial x})\psi=(-i\hbar(x\frac{\partial \psi}{\partial x}+\psi\frac{\partial x}{\partial x})+i\hbar\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar x\frac{\partial \psi}{\partial x}-i\hbar\psi i\hbar x\frac{\partial \psi}{\partial x})(-i\hbar\frac{\partial \psi}{\partial x})=(-i\hbar\psi)(-i\hbar\frac{\partial \psi}{\partial x})=-\hbar^2\psi \frac{\partial \psi}{\partial x}=-\hbar^2\frac{\partial }{\partial x}$

 

[Rorshach]

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?

 

[vela]

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?

 

[Rorshach]

$-i\hbar\frac{\partial g}{\partial x}$?

 

[vela]

Ignoring the constants, yes, that's what you get. So what's the derivative of g equal to?

 

[Rorshach]

$-i\hbar\frac{\partial \psi}{\partial x}$?

 

[vela]

No. What is $g$ equal to?

 

[Rorshach]

$-i\hbar x\frac{\partial x}{\partial x}$