Understanding Peskin Eq 3.50-3.53 and Dirac Spinor

[h-mina]

Hi, all

I'm reading peskin by myself.
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?

Thanks in advance!

 

[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...

 

[h-mina]

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

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.

 

[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...

 

[h-mina]

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...

I understand.
Thank you!

 

[ChrisVer]

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

 

[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}

 

[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).