Dirac-Hamiltonian, Angular Momentum commutator

[JD_PM]

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!

 

[Gaussian97]

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 $[H, \vec{L}]$

 

[JD_PM]

Hi Gaussian97, nice to chat with you again!

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.

 

[Gaussian97]

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.

 

[JD_PM]

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.

 

[Gaussian97]

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.

 

[JD_PM]

http://lehman.edu/faculty/anchordoqui/517_problems15-sol.pdf

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.

 

[Gaussian97]

$[\hat{ H}, \hat{L}_1]$ 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).

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?

 

[JD_PM]

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

$$\vec{A} = A_1 \hat{x}+A_2 \hat{y}+A_3 \hat{z}$$

 

[Gaussian97]

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?

 

[JD_PM]

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.

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?

 

[Gaussian97]

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.

 

[JD_PM]

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.

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*}

 

[Gaussian97]

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 $[H, \sigma^1]$.
$$[H, \sigma^1] = 2\gamma^0 (\gamma^2 \partial_3 - \gamma^3 \partial_2)$$

 

[JD_PM]

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

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

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.

 

[Gaussian97]

Both results are the same and are correct, I think you are having problems converting the partial derivatives to the $\nabla$ operator? Remember that $\nabla_i = -\partial_i$

 

[JD_PM]

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*}

 

[Gaussian97]

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 $$\vec{L}=-i\vec{r}\times \vec{\nabla}$$

 

[JD_PM]

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 $\nabla_i = \partial_i$.

 

[Gaussian97]

https://en.wikipedia.org/wiki/Four-gradient#Notation

Please, don't confuse the gradient, with a covariant derivative, same symbol, totally different things.

 

[JD_PM]

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!