# Demonstration of identities appearing in Dirac spinors in the chiral representation

Using the chiral representation of the gamma matrices, Peskin and Schroeder arrive in some expressions for the 4-component spinors $$u(p)$$ and $$v(p)$$ in terms of a square root of the Pauli matrices marices, pages 45-48. Doing some research online, I found out that we can express these square roots as

$$\sqrt{p.\sigma} \equiv \frac{E_p+m-{\bf \sigma}.\bf{p}}{\sqrt{2(E_p+m)}}$$ and $$\sqrt{p.\bar{\sigma}} \equiv \frac{E_p+m+{\bf \sigma}.\bf{p}}{\sqrt{2(E_p+m)}}.$$

But I could not prove this fact.

• Is the $p$ on the left different from the $\mathbf{p}$ on the right? If so, how is the $0$th sigma matrix defined? – G. Smith Jun 8 at 21:14
• On the left the $p$ is a four-vector, on the right it is the three momentum. The 0th sigma is the 2by2 identity matrix. – Slayer147 Jun 8 at 21:41

Here's the proof of the first identity; the second is similar.

To show that

$$\sqrt{p \cdot \sigma} = \frac{E + m - {\vec{p} \cdot \vec{\sigma}}}{\sqrt{2(E + m)}},$$

square both sides and multiply through by the denominator on the right side. This gives

$$2(E + m)(p \cdot \sigma) = (E + m - {\vec{p} \cdot \vec{\sigma}})^2.$$

Now use

$$p = (E, \vec{p})$$

and

$$\sigma = (I_2, \vec{\sigma})$$

to get

$$p \cdot \sigma = E - \vec{p} \cdot \vec{\sigma},$$

where, as in the equation we're trying to prove, we don't bother to write the $$I_2$$.

Substituting this gives

$$2(E + m)(E - \vec{p} \cdot \vec{\sigma}) = (E + m - {\vec{p} \cdot \vec{\sigma}})^2.$$

Expand this to get

$$2(E + m)E - 2(E + m)\vec{p} \cdot \vec{\sigma} = (E + m)^2 - 2(E + m)\vec{p} \cdot \vec{\sigma} + (\vec{p} \cdot \vec{\sigma})^2.$$

The middle terms on each side cancel, giving

$$2(E + m)E = (E + m)^2 + (\vec{p} \cdot \vec{\sigma})^2.$$

This simplifies to

$$E^2 = m^2 + (\vec{p} \cdot \vec{\sigma})^2.$$

Finally,

$$(\vec{p} \cdot \vec{\sigma})^2 = p_i p_j \sigma_i \sigma_j = \frac{1}{2} p_i p_j \{\sigma_i, \sigma_j\} =\frac{1}{2} p_i p_j (2\delta_{ij}) = \vec{p}^2$$

so the original equation is equivalent to

$$E^2 = m^2 + \vec{p}^2,$$

which is simply the relationship between energy and momentum.