Undergrad 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!