# A Angular momentum of Dirac particle

1. Apr 6, 2016

### ShayanJ

I'm reading Sakurai's "Advanced Quantum Mechanics" (which is different from his "Modern Quantum Mechanics"). In chapter 3, which is about the Relativistic Quantum Mechanics of spin 1/2 particles, after discussing the covariance of the Dirac equation, he goes on to give some examples to clarify the significance of the operator S (as in $\psi'(x')=S \psi(x)$). In the first example, he considers an infinitesimal rotation around the z-axis and so he considers the transformation:

$\psi'(x')= \left[ 1+i\Sigma_3 \frac{\delta \omega}{2} \right] \psi(x) \ \ \ \ (1)$

where $\vec{x}'=\vec x+\delta \vec{x}$ and $\delta \vec x=(x_2 \delta \omega,-x_1 \delta \omega,0)$.

He then notes that:
$\psi'(x')=\psi'(x)+\delta x_1 \frac{\partial \psi'}{\partial x_1}+\delta x_2 \frac{\partial \psi'}{\partial x_2} \ \ \ \ (2)$

Then from all the above, he concludes that:

$\psi'(x)=\left[ 1+ i \Sigma_3 \frac{\delta \omega}{2}-\left(x_2 \delta \omega \frac{\partial}{\partial x_1}-x_1 \delta \omega \frac{\partial}{\partial x_2} \right) \right] \psi(x) \ \ \ \ (3)$

It seems he rearranged equation 2 and then used equation 1 to substitute for $\psi'(x')$. But there still remain the derivative terms which are derivatives of $\psi'(x)$ which should somehow be related to the derivatives of $\psi(x)$. But he seems to just assumes that $\frac{\partial \psi'(x)}{\partial x_i}=\frac{\partial \psi(x)}{\partial x_i }$ which doesn't seem justified at all! In fact using equation 1, you can easily show how far this is from truth! Am I missing something here?

Thanks

2. Apr 6, 2016

### vanhees71

You only take contributions up to order $\mathcal{O}(\delta \omega)$. So there's no mixing of derivatives with the spin matrix $\Sigma_3$.

3. Apr 6, 2016

### dextercioby

Sakurai uses the unfortunate x_4 = ict, the only flaw of this book. If you are into well-written but old books, you've got from the 60s either the monograph by S. Schweber, or the two books by Bjorken and Drell.

4. Apr 6, 2016

### ShayanJ

Yeah...I like Sakurai's too, but x=ict really hurts sometimes!
I just wish my first QFT course were with someone else so I didn't have to meet this book in this situation! I'm starting to feel I have a very bad luck regarding the professors I have to take my courses with!
And I appreciate any well-written book because I like to search for insights that are present only in a few books.

5. Apr 6, 2016

### dextercioby

Yes, unfortunately Japanese books are marred by this annoying issue, but most of them are still good. I particularly like Umezawa's QFT book, North Holland, 1956. It has nice sections on general relativistic equations.

6. Apr 6, 2016

### samalkhaiat

It is extremely important to know about the calculus of infinitesimals. When you do infinitesimal coordinate transformation $$\bar{x} = x + \epsilon \delta x ,$$
where $\epsilon \ll 1$ is the infinitesimal parameter, and $\delta x$ is some arbitrary function of x, the word infinitesimal means you can always put $\epsilon^{2} = 0$. This leads to many useful relations. The first and most important one is the chain rule operator:
$$\epsilon \frac{\partial}{\partial x} = \epsilon \frac{\partial \bar{x}}{\partial x} \frac{\partial}{\partial \bar{x}} = \epsilon \frac{\partial}{\partial x}(x + \epsilon \delta x) \frac{\partial}{\partial \bar{x}}.$$
Since $\epsilon^{2} = 0$, we get
$$\epsilon \frac{\partial}{\partial x} = \epsilon \frac{\partial}{\partial \bar{x}} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
Now, consider the following infinitesimal transformation of some object $\varphi (x)$
$$\bar{\varphi}(\bar{x}) = \varphi (x) + \epsilon \Sigma \varphi (x) ,$$ which is Eq(1) in your post. Using (1), and $\epsilon^{2} = 0$, you find the following important relation (no coordinate expansion yet)
$$\epsilon \frac{\partial}{\partial \bar{x}} \bar{\varphi}(\bar{x}) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (2)$$
Now, if we do the coordinate expansion
$$\bar{\varphi}(\bar{x}) = \bar{\varphi}( x + \epsilon \delta x ) = \bar{\varphi}(x) + \epsilon \delta x \ \frac{\partial}{\partial x} \bar{\varphi}(x) ,$$ substitute in (2) and use $\epsilon^{2} = 0$ again, we get
$$\epsilon \frac{\partial}{\partial \bar{x}} \bar{\varphi}(x) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (3)$$
On the left hand side, if we use (1) again, we obtain the following very important relation (which justify Sakurai’s action in replacing $\bar{\varphi}(x)$ with $\varphi (x)$)
$$\epsilon \frac{\partial}{\partial x} \bar{\varphi}(x) = \epsilon \frac{\partial}{\partial x} \varphi(x) . \ \ \ \ \ \ \ \ \ \ (4)$$
So, always remember to use the equations (1)-(4), when you deal infinitesimal transformations.