Normalisation constant expansion of spinor field

[Josh1079]

Hi, I'm reading about the wave packet solution to the dirac equation but in the book I'm reading it states that $$\int \frac {d^3p} {(2\pi)^3 2E} [a u e^{-ipx} + b^\dagger \bar{v} e^{ipx}$$

The normalisation constant confuses me. I guess the 2pi^3 is reasonalbe. However, the 1/2E seems a bit weird to me. I thought it should always be something of the order of (-1/2) of E. Did I misunderstand something about the normalisation?

Thanks!

 

[stevendaryl]

Hi, I'm reading about the wave packet solution to the dirac equation but in the book I'm reading it states that $$\int \frac {d^3p} {(2\pi)^3 2E} [a u e^{-ipx} + b^\dagger \bar{v} e^{ipx}$$

The normalisation constant confuses me. I guess the 2pi^3 is reasonalbe. However, the 1/2E seems a bit weird to me. I thought it should always be something of the order of (-1/2) of E. Did I misunderstand something about the normalisation?

Thanks!

This might be the reason:

• Start with a 4-D Fourier transform: $\int d^4 p A(p_\mu) u e^{- i p_\mu x^\mu}$
• Realize that the only values of $p_\mu$ that can contribute satisfy: $(p_0)^2 - \vec{p}^2 - m^2 = 0$ (ignoring factors of $c$ and $\hbar$)
• This implies that $A(p_\mu)$ has the form $a(\vec{p}) \delta((p_0)^2 - \vec{p}^2 - m^2)$
• Just a fact about the delta-function: $\int dx Q(x) \delta(f(x)) = \frac{Q(x_0)}{f'(x_0)}$ where $x_0$ is a zero of $f(x)$ and $f'(x)$ means the derivative of $f$ with respect to $x$. (If $f(x)$ has multiple zeros, then there is a term like that for each zero)
• So if we let $E$ be the value of $p_0$ satisfying $(p_0)^2 - \vec{p}^2 - m^2 = 0$, then integrating over $dp_0$ gives: $\frac{1}{\frac{d}{dp_0}( (p_0)^2 - \vec{p}^2 - m^2)} = \frac{1}{2 p_0} = \frac{1}{2E}$

 

[Josh1079]

Hi stevendaryl, thanks for the reply! I think I roughly understand what your writing and it seems reasonable. There is just one more question that I'm a bit curious about. So are you trying to say that the reason why there is a factor 1/2E in the expression is because it's doing a 4D Fourier transform? Does this mean that the expressions with 1/√(2E) is doing a 3D Fourier transform? I'm also a bit curious about the difference.

Thanks!