Dirac-Hamiltonian, Angular Momentum commutator

In summary: I just did it, amirite?Sorry, I think I didn't explain myself clearly. What I meant is that you should provide a summary of the content, not the entire conversation. Here is a summary of the conversation:In summary, the conversation is about showing that ##[\hat{\vec{H}}, \hat{\vec{L}}_T]=0##. The conversation discusses a guess that ##[\hat{\vec{H}}, \hat{\vec{L}}_T]=[\hat{\vec{H}}, \hat{\vec{L}}] + \frac{1}{2}[\hat{\vec{H}}, \vec{\sigma}]=0## must hold. The conversation then continues to evaluate ##
  • #1
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Homework Statement
Given the Dirac-Hamiltonian

$$\hat{ \vec H} = -i \gamma^0 \gamma^j \partial_{j} + m \gamma^0$$

And the Angular momentum operator

$$\hat{ \vec L}_T = -i \vec r \times \vec \nabla + \frac{1}{2} \vec \sigma$$

Show that

$$[\hat{ \vec H}, \hat{ \vec L}_T]=0$$
Relevant Equations
$$\vec \sigma := (\sigma^{23}, \sigma^{31}, \sigma^{12}) := (\sigma^1, \sigma^2, \sigma^3)$$

$$\sigma^{\mu \nu} = \frac{i}{2}[\gamma^{\mu}, \gamma^{\nu}]$$

$$\{\gamma^{\mu}, \gamma^{\nu} \} = 2 \eta^{\mu \nu}$$
We want to show that ##[\hat{ \vec H}, \hat{ \vec L}_T]=0##. I made a guess: we know that ##[\hat{ \vec H}, \hat{ \vec L}_T]=[\hat{ \vec H}, \hat{ \vec L}] + \frac 1 2 [\hat{ \vec H}, \vec \sigma]=0## must hold.

I have already shown that

$$[\hat{ \vec H}, -i \vec r \times \vec \nabla]= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}$$ Hence we expect to get

\begin{equation*}
[\hat{ \vec H}, \vec \sigma] = 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}
\end{equation*}

Let's look at ##[\hat{ \vec H}, \vec \sigma]## carefully. We first note that ##[\hat{ \vec H}, \vec \sigma] = [\hat{ \vec H}, \sigma^1] + [\hat{ \vec H}, \sigma^2] + [\hat{ \vec H}, \sigma^3]##. Let's evaluate ##[\hat{ \vec H}, \sigma^1]## explicitly

$$[\hat{ \vec H}, \sigma^1] = -i[ \gamma^0 \gamma^j \partial_{j} + m \gamma^0, i\gamma^2 \gamma^3] $$
$$= [ \gamma^0 \gamma^j \partial_{j}, \gamma^2\gamma^3] $$
$$= [ \gamma^0 \gamma^1 \partial_{1}, \gamma^2\gamma^3] + [ \gamma^0 \gamma^2 \partial_{2}, \gamma^2\gamma^3] + [ \gamma^0 \gamma^3 \partial_{3}, \gamma^2\gamma^3] $$
$$= \gamma^0 \gamma^1 \gamma^2\gamma^3\partial_{1} - \gamma^2\gamma^3\gamma^0 \gamma^1\partial_{1} + \gamma^0 \gamma^2 \gamma^2\gamma^3\partial_{2} - \gamma^2\gamma^3\gamma^0 \gamma^2\partial_{2} $$
$$+ \gamma^0 \gamma^3 \gamma^2\gamma^3\partial_{3} - \gamma^2\gamma^3\gamma^0 \gamma^3\partial_{3} $$
$$= \gamma^0 \gamma^1 \gamma^2\gamma^3\partial_{1} - \gamma^0 \gamma^1 \gamma^2\gamma^3\partial_{1} - \gamma^0 \gamma^3\partial_{2}- \gamma^0 \gamma^3\partial_{2} $$
$$+ \gamma^0 \gamma^2\partial_{3} + \gamma^0 \gamma^2\partial_{3} $$
$$= 2\gamma^0 (\gamma^2 \partial_3-\gamma^3 \partial_2) $$
$$= -2\gamma^0 (\gamma_2 \partial_3-\gamma_3 \partial_2)$$

Where I've used ##\sigma^1 := \sigma^{23}, \ \sigma^{\mu \nu} = \frac{i}{2}[\gamma^{\mu}, \gamma^{\nu}], \ \{\gamma^{\mu}, \gamma^{\nu} \}=2 \eta^{\mu \nu}## and ##\gamma_{\mu} = \eta_{\mu \nu} \gamma^{\nu}##, with the convention for the Minkowski metric ##(+,-,-,-)##

The funny thing is that I do not match the answer due to a sign (i.e. the right answer is ##[\hat{ \vec H}, \sigma^1]=2\gamma^0 (\gamma_2 \partial_3-\gamma_3 \partial_2)##).

Could you please help me out spotting my mistake?

Thank you! :biggrin:
 
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  • #2
Well, first of all, you should really be more careful about your notation and put vectors only on the vector quantities, then don't sum over indices that are not summed.
And for your computation, I didn't check but I get the same answer as you for ##[H,\vec{\sigma}]## and a negative sign in ##[H, \vec{L}]##
 
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  • #3
Hi Gaussian97, nice to chat with you again!

Gaussian97 said:
And for your computation, I didn't check but I get the same answer as you for ##[H,\vec{\sigma}]## and a negative sign in ##[H, \vec{L}]##

Did you mean you got the following results?

$$[\hat{ \vec H}, -i \vec r \times \vec \nabla]= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}$$

\begin{equation*}

[\hat{ \vec H}, \vec \sigma] = 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}

\end{equation*}

If that is the case, could you please explain how did you get the latter? And particularly, how did you compute
##[\hat{ \vec H}, \sigma^1]## and if you did it similarly as I did?

Thank you.
 
  • #4
Well, first things first, note please that the Hamiltonian is not a vector, it's a scalar, so you shouldn't put a ##\vec{}##. Secondly, ##\vec{L}## is indeed a vector, i.e. has an index, therefore, something like ##[H, \vec{L}]## must also have an index, so something like
$$
[\hat{H}, -i \vec r \times \vec \nabla]= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}
$$
or
$$[\hat{H}, \vec \sigma] = 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}$$
cannot be right (there's no free index in it).
After doing the computation I obtain that the second one is
$$\left[\hat{H}, \frac{\vec{\sigma}}{2}\right] = \gamma^0 \vec{\gamma} \times \vec{\nabla}$$
Which for the first component is:
$$\left[\hat{H}, \frac{\sigma^1}{2}\right] = \gamma^0 (\gamma^2 \partial_3 - \gamma^3 \partial_2)$$
which coincides with your expression.
Maybe you should show us how you obtain your first expression and we can look for any error.

PD: Also nice to chat with you again.
 
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  • #5
Evaluating ##[\hat{ \vec H}, \hat{L}_1]## I get\begin{align}
[\hat{ \vec H}, \hat{L}_1] &= [-i \gamma^0 \gamma^j \partial_{j} + m \gamma^0, -i (r^2 \partial^3 - r^3 \partial^2)] \nonumber \\
&= - [\gamma^0 \gamma^j \partial_{j}, r^2 \partial^3]+[ \gamma^0 \gamma^j \partial_{j}, r^3 \partial^2] \nonumber \\
&= [ \gamma^0 \gamma^1 \partial_{1} + \gamma^0 \gamma^2 \partial_{2} + \gamma^0 \gamma^3 \partial_{3}, r^3 \partial^2] \nonumber \\
&- [\gamma^0 \gamma^1 \partial_{1} + \gamma^0 \gamma^2 \partial_{2} + \gamma^0 \gamma^3 \partial_{3}, r^2 \partial^3] \nonumber \\
&= [\gamma^{0}\gamma^{1}\partial_{1}, r^3\partial^2] + [\gamma^{0}\gamma^{2}\partial_{2}, r^3\partial^2] + [\gamma^{0}\gamma^{3}\partial_{3}, r^3\partial^2] \nonumber \\
&- [\gamma^{0}\gamma^{1}\partial_{1}, r^2\partial^3] - [\gamma^{0}\gamma^{2}\partial_{2}, r^2\partial^3] - [\gamma^{0}\gamma^{3}\partial_{3}, r^2\partial^3] \nonumber \\
&= [\gamma^{0}\gamma^{1}\partial_{1}, r^3]\partial^2 + r^3 [\gamma^{0}\gamma^{1}\partial_{1}, \partial^2] + [\gamma^{0}\gamma^{2}\partial_{2}, r^3]\partial^2 \nonumber \\
&+ r^3 [\gamma^{0}\gamma^{2}\partial_{2}, \partial^2] + [\gamma^{0}\gamma^{3}\partial_{3}, r^3]\partial^2 + r^3 [\gamma^{0}\gamma^{3}\partial_{3}, \partial^2] \nonumber \\
&- [\gamma^{0}\gamma^{1} \partial_{1}, r^2]\partial^3 - r^2[\gamma^{0}\gamma^{1} \partial_{1}, \partial^{3}]- [\gamma^{0}\gamma^{2} \partial_{2}, r^2]\partial^3 \nonumber \\
&- r^2[\gamma^{0}\gamma^{2} \partial_{2}, \partial^{3}]- [\gamma^{0}\gamma^{3} \partial_{3}, r^2]\partial^3 - r^2[\gamma^{0}\gamma^{3} \partial_{3}, \partial^{3}] \nonumber \\
&= \gamma^0(\gamma^2\partial_3 -\gamma^3 \partial_2) \nonumber \\
&= -\gamma^0(\gamma_2 \partial_3-\gamma_3 \partial_2)
\end{align}

I am pretty sure this is OK (the same calculation is done here; it is equation (13) http://lehman.edu/faculty/anchordoqui/517_problems15-sol.pdf; note I get the same result). That's why I thought that my mistake was in computing
##[H,\vec{\sigma}^1]##.

Note that in the link they prove it using a particular representation of the gamma matrices (the Dirac-Pauli representation, page 460 in Mandl & Shaw), while I am trying to prove it without using any of them.
 
  • #6
Well, I have no permissions to look your link, but I totally agree with your computation.
The problem is I totally disagree that this is equivalent to
$$[\hat{H}, -i \vec r \times \vec \nabla]= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}$$
also, I think your errors are because you are making mistakes using the index, so please try to be very careful with all the index and, please, stop writing an arrow over scalar quantities.
 
  • #7
http://lehman.edu/faculty/anchordoqui/517_problems15-sol.pdf

Gaussian97 said:
The problem is I totally disagree that this is equivalent to
$$[\hat{H}, -i \vec r \times \vec \nabla]= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}$$

##[\hat{ H}, \hat{L}_1]## is indeed not equivalent to ##[\hat{H}, -i \vec r \times \vec \nabla]##. In the former commutator, we only deal with a component of the 3D angular momentum.

We have

\begin{align*}
[\hat{ H}, \hat{\vec {L}}] &= [\hat{ H}, \hat{L}_1] \hat i + [\hat{ H}, \hat{L}_2] \hat j + [\hat{ H}, \hat{L}_3] \hat k \\
&= -\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \\
&= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}
\end{align*}

So I think that my mistake must be in computing ##[\hat{H}, \vec \sigma] = [\hat{ H}, \sigma^1] + [\hat{ H}, \sigma^2] + [\hat{ H}, \sigma^3]## and, in particular, ##[\hat{ H}, \sigma^1]##. I compared it to equation (21) in the link and I get the same up to a sign.

If the ##[\hat{ H}, \sigma^1]## computation is OK then I must be missing something conceptually speaking.

PS: The link should work now. If not, please google 'dirac hamiltonian commutes with angular momentum'; then click in the first result you get.
 
  • #8
JD_PM said:
##[\hat{ H}, \hat{L}_1]## is indeed not equivalent to ##[\hat{H}, -i \vec r \times \vec \nabla]##. In the former commutator, we only deal with a component of the 3D angular momentum.
Well, of course, one is a vectorial equation while the other is only one component, they are not the same, but I'm saying that they are not compatible (if one is right then the other is wrong).

JD_PM said:
We have

\begin{align*}
[\hat{ H}, \hat{\vec {L}}] &= [\hat{ H}, \hat{L}_1] \hat i + [\hat{ H}, \hat{L}_2] \hat j + [\hat{ H}, \hat{L}_3] \hat k \\
&= -\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \\
&= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k}
\end{align*}
Well, ok, I see why we disagree, in the document they are using the convention
$$\vec{A} = A_1 \hat{x}+A_2 \hat{y}+A_3 \hat{z}$$
while I was using the other convention
$$\vec{A} = A^1 \hat{x}+A^2 \hat{y}+A^3 \hat{z}$$
of course both differ in a ##-## sign and that's the reason my results differ in a sign with the ones they get. Your problem is that you are using both conventions, the first one to compute ##[H,\vec{L}]## and the second one to compute ##[H,\vec{\sigma}]##. Also the last equality you put is wrong, you go from a vector to a number, how is this posible?
 
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  • #9
Gaussian97 said:
Your problem is that you are using both conventions, the first one to compute ##[H,\vec{L}]## and the second one to compute ##[H,\vec{\sigma}]##.

That's a helpful comment! However, please note that

$$\sigma^{23} = i \gamma^{2} \gamma^{3} = i \gamma_{2} \gamma_{3} = \sigma_{23}$$

So I do not see where I pick the change in sign, as I end up with all indices lowered and using again

$$\vec{A} = A_1 \hat{x}+A_2 \hat{y}+A_3 \hat{z}$$
 
  • #10
Ok, I think you're right, sorry I'm getting a little bit confused with the notations. Let's start again, if I'm not wrong the two computations must give the following results (notation independent):
$$[H, \vec{L}] = -\gamma^0 \vec{\gamma}\times \vec{\nabla}$$
$$[H, \vec{\sigma}] = 2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$
Do we agree on that?
 
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  • #11
Gaussian97 said:
Ok, I think you're right, sorry I'm getting a little bit confused with the notations.

No worries. At first it is a bit annoying but everything falls into place once one gets used to it.

Gaussian97 said:
Let's start again, if I'm not wrong the two computations must give the following results (notation independent):
$$[H, \vec{L}] = -\gamma^0 \vec{\gamma}\times \vec{\nabla}$$
$$[H, \vec{\sigma}] = 2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$
Do we agree on that?

I do agree with your results. I see how to get the first one. I am indeed struggling with the latter.

I think it is best to go step by step, instead of jumping into the computation. I think we have

\begin{align*}
[\hat{H}, \vec \sigma] &= [\hat{ H}, \sigma^1] + [\hat{ H}, \sigma^2] + [\hat{ H}, \sigma^3] \\
&= [\hat{ H}, \sigma^{23}] + [\hat{ H}, \sigma^{31}] + [\hat{ H}, \sigma^{12}] \\
&= i[\hat{ H}, \gamma^{2}\gamma^{3}] + i[\hat{ H}, \gamma^{3}\gamma^{1}] + [\hat{ H}, \gamma^{1}\gamma^{2}]
\end{align*}

Are we on the same page so far?
 
  • #12
Nop, ##[H, \vec{\sigma}]## is a vector while ##[H, \sigma^1]+[H, \sigma^2]+[H, \sigma^3]## is not, but i would agree on
\begin{align*}
[\hat{H}, \vec \sigma] &= [\hat{ H}, \sigma^1] \vec{x}+ [\hat{ H}, \sigma^2]\vec{y} + [\hat{ H}, \sigma^3]\vec{z} \\

&= [\hat{ H}, \sigma^{23}]\vec{x} + [\hat{ H}, \sigma^{31}]\vec{y} + [\hat{ H}, \sigma^{12}]\vec{z} \\

&= i[\hat{ H}, \gamma^{2}\gamma^{3}]\vec{x} + i[\hat{ H}, \gamma^{3}\gamma^{1}]\vec{y} + i[\hat{ H}, \gamma^{1}\gamma^{2}]\vec{z}
\end{align*}
I suppose that the ##i## factor in the last term was intended to be there.
 
  • #13
Gaussian97 said:
Nop, ##[H, \vec{\sigma}]## is a vector while ##[H, \sigma^1]+[H, \sigma^2]+[H, \sigma^3]## is not, but i would agree on
\begin{align*}
[\hat{H}, \vec \sigma] &= [\hat{ H}, \sigma^1] \vec{x}+ [\hat{ H}, \sigma^2]\vec{y} + [\hat{ H}, \sigma^3]\vec{z} \\

&= [\hat{ H}, \sigma^{23}]\vec{x} + [\hat{ H}, \sigma^{31}]\vec{y} + [\hat{ H}, \sigma^{12}]\vec{z} \\

&= i[\hat{ H}, \gamma^{2}\gamma^{3}]\vec{x} + i[\hat{ H}, \gamma^{3}\gamma^{1}]\vec{y} + i[\hat{ H}, \gamma^{1}\gamma^{2}]\vec{z}
\end{align*}

Indeed, I forgot to include the basis vectors.

Gaussian97 said:
I suppose that the ##i## factor in the last term was intended to be there.

You are right, that was a typo.

OK so the idea now is that ##[\hat{H}, \vec \sigma]## should cancel out with ##[\hat{H}, \hat{\vec L}]##. But notice that when I go to compute all three terms (I explicitly showed how to do so for ##[\hat{H}, \sigma^1]## at #1; we simply do the analogous computation for the other two) I get the expected answer up to a -ive sign.

To summarize:

We would expect to get

\begin{equation}
[\hat{ H}, \sigma^1] = 2\gamma^0 (\gamma_2 \partial_3-\gamma_3 \partial_2)
\end{equation}

\begin{equation}
[\hat{ H}, \sigma^2] = -2\gamma^0 (\gamma_1 \partial_3-\gamma_3 \partial_1)
\end{equation}

\begin{equation}
[\hat{ H}, \sigma^3] = 2\gamma^0 (\gamma_1 \partial_2-\gamma_2 \partial_1)
\end{equation}

Ending up with

\begin{align}
[\hat{ H}, \vec \sigma] &= [\hat{ H}, \sigma^1] \hat i + [\hat{ H}, \sigma^2] \hat j + [\hat{ H}, \sigma^3] \hat k \nonumber \\
&= 2\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \nonumber \\
&= 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \nonumber
\end{align}

So that

\begin{align*}
[\hat H, \hat{ \vec {L}}_T] &= [\hat H, \hat{ \vec {L}}] + \frac 1 2 [\hat H, \vec \sigma] \\
&= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} + \frac 1 2 \left( 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \right) \\
&= 0
\end{align*}

What I get

\begin{equation}
[\hat{ H}, \sigma^1] = - 2\gamma^0 (\gamma_2 \partial_3-\gamma_3 \partial_2)
\end{equation}

\begin{equation}
[\hat{ H}, \sigma^2] = 2\gamma^0 (\gamma_1 \partial_3-\gamma_3 \partial_1)
\end{equation}

\begin{equation}
[\hat{ H}, \sigma^3] = - 2\gamma^0 (\gamma_1 \partial_2-\gamma_2 \partial_1)
\end{equation}

Ending up with

\begin{align}
[\hat{ H}, \vec \sigma] &= [\hat{ H}, \sigma^1] \hat i + [\hat{ H}, \sigma^2] \hat j + [\hat{ H}, \sigma^3] \hat k \nonumber \\
&= - 2\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \nonumber \\
&= - 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \nonumber
\end{align}

So that

\begin{align*}
[\hat H, \hat{ \vec {L}}_T] &= [\hat H, \hat{ \vec {L}}] + \frac 1 2 [\hat H, \vec \sigma] \\
&= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} - \frac 1 2 \left( 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \right) \\
&\neq 0
\end{align*}
 
  • #14
JD_PM said:
Ending up with

\begin{align}
[\hat{ H}, \vec \sigma] &= [\hat{ H}, \sigma^1] \hat i + [\hat{ H}, \sigma^2] \hat j + [\hat{ H}, \sigma^3] \hat k \nonumber \\
&= 2\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \nonumber \\
&= 2 \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \nonumber
\end{align}
Again, you forget the vectors, and I totally disagree. This is not the result which both we have agreed
$$[H,\vec{\sigma}]=2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$
Actually, if we take the first component of this equation we get exactly the result you are claming you have for ##[H, \sigma^1]##.
$$[H, \sigma^1] = 2\gamma^0 (\gamma^2 \partial_3 - \gamma^3 \partial_2)$$
 
  • #15
Gaussian97 said:
Again, you forget the vectors, and I totally disagree. This is not the result which both we have agreed
$$[H,\vec{\sigma}]=2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$

That's indeed the result I want to get. The issue is that I have

$$[H,\vec{\sigma}]=-2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$

Gaussian97 said:
Actually, if we take the first component of this equation we get exactly the result you are claming you have for ##[H, \sigma^1]##.
$$[H, \sigma^1] = 2\gamma^0 (\gamma^2 \partial_3 - \gamma^3 \partial_2)$$

But notice that we need to lower all indices, so we end up with

$$[H, \sigma^1] = -2\gamma^0 (\gamma_2 \partial_3 - \gamma_3 \partial_2)$$

So we indeed picked up the -ive sign we wanted to avoid.

Btw did you get

$$[H,\vec{\sigma}]=2\gamma^0 \vec{\gamma}\times \vec{\nabla}$$

?

I am sure I am missing something really trivial here.
 
  • #16
Both results are the same and are correct, I think you are having problems converting the partial derivatives to the ##\nabla## operator? Remember that ##\nabla_i = -\partial_i##
 
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  • #17
Gaussian97 said:
Remember that ##\nabla_i = -\partial_i##

OK I did not know about that rule. Let's assume it is true.

Then I do not see how you got at #10

$$[H, \vec{L}] = -\gamma^0 \vec{\gamma}\times \vec{\nabla}$$

Following such a rule I get

\begin{align*}

[\hat{ H}, \hat{\vec {L}}] &= [\hat{ H}, \hat{L}_1] \hat i + [\hat{ H}, \hat{L}_2] \hat j + [\hat{ H}, \hat{L}_3] \hat k \\

&= -\gamma^0\left[(\gamma_2 \partial_3-\gamma_3 \partial_2)\hat i-(\gamma_1 \partial_3-\gamma_3 \partial_1)\hat j + (\gamma_1 \partial_2-\gamma_2 \partial_1)\hat k \right] \\

&= - \gamma^{0} \sum_{i=1}^3 \varepsilon_{ijk}\gamma^{j}\partial^{k} \\

&= \gamma^0 \vec{\gamma}\times \vec{\nabla}

\end{align*}
 
  • #18
Ok, let's look at the computation of ##[H, L_1]## that you did in #5, there's an error in the first line, you wrote
$$L_1 = -i(r^2\partial^3 - r^3 \partial^2)$$
that differs in sign with the definition $$\vec{L}=-i\vec{r}\times \vec{\nabla}$$
 
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  • #19
Gaussian97 said:
Remember that ##\nabla_i = -\partial_i##

Could you please prove such statement?

I am not used at all to include a -ive sign when switching from partial to the del operator. For instance: in GR, when we introduce curvature, one of the steps is to switch from partial to covariant derivatives (i.e. del operator) and we do not use ##\nabla_i = -\partial_i## but simply ##\nabla_i = \partial_i##.
 
  • #21
OK I got it! 😍💪

All boils down to

$$\partial_i = \eta_{ij} \partial^j$$

As you pointed out, there was a mistake in #5: I missed the fact that ##\partial_2 = \eta_{22} \partial^2=- \partial^2## and ##\partial_3 = \eta_{33} \partial^3=- \partial^3##.

We indeed have

$$\left(\hat{ \vec L}\right)_i = \left(-i \vec r \times \vec \nabla\right)_i = + i \varepsilon_{i j k} r^{j} \partial^{k} \hat i$$

@Gaussian97 thank you, I really appreciate you patience! 🤝

I hope to come across you again very soon!
 
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1. What is the Dirac-Hamiltonian?

The Dirac-Hamiltonian is a mathematical operator used in quantum mechanics to describe the behavior of particles with spin. It is a combination of the Dirac equation, which describes the behavior of relativistic particles, and the Hamiltonian, which describes the total energy of a system.

2. What is the Angular Momentum commutator?

The Angular Momentum commutator is a mathematical expression that describes the relationship between the angular momentum and the position and momentum of a particle. It is used to calculate the uncertainty in the measurement of these quantities.

3. How is the Dirac-Hamiltonian related to the Angular Momentum commutator?

The Dirac-Hamiltonian and the Angular Momentum commutator are related through the Dirac equation. This equation includes the Angular Momentum commutator as one of its terms, which allows for the description of spin in addition to position and momentum.

4. What is the significance of the Dirac-Hamiltonian and Angular Momentum commutator in quantum mechanics?

The Dirac-Hamiltonian and Angular Momentum commutator are important concepts in quantum mechanics because they allow for the description of relativistic particles with spin. This is crucial for understanding the behavior of subatomic particles and their interactions.

5. How are the Dirac-Hamiltonian and Angular Momentum commutator used in practical applications?

The Dirac-Hamiltonian and Angular Momentum commutator are used in various applications, such as in the development of quantum computers and in the study of particle physics. They also play a role in the understanding of magnetic materials and the behavior of atoms in strong magnetic fields.

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