# Peskin page 46 to 47

1. Sep 12, 2014

### h-mina

Hi, all

I can't understand from eq(3.50) to eq(3.53).

i) What should I interpret $\sqrt{p\cdot\sigma}$?
I guess below, but I can't understand $\sqrt{\;\;}$ of matrices.

\begin{eqnarray}
p\cdot\sigma=E \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) - p^3\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) \\
= \left(\begin{array}{cc} E-p^3 & 0 \\ 0 & E+p^3 \end{array}\right)
\end{eqnarray}

And why is it the same as (3.49)?

ii)How can I confirm (3.50) is a solution of the Dirac equation?

iii)What's meaning of "large boost" in (3.52) and (3.53)?
If I understand the Dirac spinor more, is it easy transform?
When so, where can I study Dirac spinor easily?

Last edited: Sep 12, 2014
2. Sep 12, 2014

### ChrisVer

i. In Peskin they immediately explain what they mean by sqrt of a matrix right after its use...
then it's just algebra...

ii. you just put it in dirac's equation...

iii. Large boost means that you are doing a large boost... a boost is described by the parameter $\eta$ (in the same way rotations are described by $\theta$). At the limit $\eta \rightarrow \infinity$ you get that result...

3. Sep 12, 2014

### h-mina

Thanks ChrisVer.
I'm not native, so a little difficult to understand that sentense. Sorry,but I ask in another words.

If $A=\left(\begin{array}{cc} a & b \\ c & d \end{array}\right)$,
$\sqrt{A}=\left(\begin{array}{cc} \sqrt{eigenvalue1} & 0 \\ 0 & \sqrt{eigenvalue2} \end{array}\right)$?
I want to know the component expression.

4. Sep 12, 2014

### ChrisVer

then you have to find the eigenvalues of your matrix A.... :) except for the case the A is in diagonal form...
for the eigenvalues of a general matrix, you find them by solving the characteristic polynomial of $\alpha$:
$det(A-I \alpha)=0$
$\alpha$ are the eigenvalues. $I$ the unitary matrix... det=determinant...

5. Sep 12, 2014

### h-mina

I understand.
Thank you!

6. Sep 12, 2014

### ChrisVer

CORRECTION
sorry I just saw Peskin did a boost only along the 3 direction ok...

Last edited: Sep 12, 2014
7. Sep 20, 2014

### ChrisVer

Hello, I am sorry to come up here again, but I just read these parts in Peskin.
I don't understand your question now about the "Easy to transform spinor", But I can answer better the question about the large boosts.
You have the quantities $\sqrt{E \pm p_3}$ multiplying the 2-component spinors.
Now a large boost means that you are letting $p_3$ become large... In this case $E = \sqrt{m^2 + p_3 ^2} \approx p_3$
So the $\sqrt{E - p_3}= 0$ and $\sqrt{E + p_3}= \sqrt{2E}$

8. Sep 20, 2014

### vanhees71

Just let me make some remarks about the square root of a matrix. It's not so simple! It's not even necessary for the issue it's applied in (3.50). Everything can derived with the Dirac-$\gamma^{\mu}$ matrices without taking roots.

To define the square root of matrices, let's discuss only hermitean positive semidefinite matrices. As you know from linear algebra, a hermitean matrix can also be diagonalized by a unitary transformation, and all eigenvalues are real. The eigenvectors are orthogonal to each other and can be normalized, so that you have a unitary transformation from the original basis to the so defined eigenbasis. The matrix is called positive semidefinite, if all eigenvalues are $\geq 0$.

Now to define $\sqrt{\hat{A}}$ for such a matrix, of course you like to have $(\sqrt{\hat{A}})^2=\hat{A}$. Now you can diagonalize the original matrix with a unitary transformation,
$$\hat{A}'=\hat{U} \hat{A} \hat{U}^{\dagger}=\mathrm{diag}(\lambda_1,\ldots,\lambda_n),$$
where $n$ is the dimension of our unitary vector space (Hilbert space of finite dimension).

Now for this diagonal matrix, it's easy to find $\sqrt{\hat{A}'}$, but it's not unique. One solution possibility is the one Peskin and Schroeder choose: Just take the positive roots of all the eigenvalues:
$$\sqrt{\hat{A}'}=\mathrm{diag}(\sqrt{\lambda_1},\ldots,\sqrt{\lambda}_n).$$
Of course you can also choose the negative roots or the positive and negative roots for the different eigenvalues. All together you have $2^n$ square roots of such a postive semidefinite diagonal matrix.

Each of these square roots is uniquely mapped back to the original basis by
$$\sqrt{\hat{A}}=\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U}.$$
Indeed you directly verify
$$(\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U})^2 = \hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U}\hat{U}^{\dagger} \sqrt{\hat{A}'} \hat{U} = \hat{U}^{\dagger} (\sqrt{\hat{A}'})^2 \hat{U} = \hat{U}^{\dagger} \hat{A} \hat{U} = \hat{U}^{\dagger} \hat{U} \hat{A} \hat{U}^{\dagger} \hat{U} = \hat{A}.$$
In this way you define arbitrary functions of hermitean matrices (or even operators in infininte-dimensional Hilbert space).