Hi I am struggling trying to see understand the basic propagator integral trick.
\int \frac{d^{3}p}{(2\pi^{3})}\left\lbrace \frac{1}{2E_{p}}e^{-ip.(x-y)}|_{p_{0}=E_{p}}+\frac{1}{-2E_{p}}e^{ip.(x-y)}|_{p_{0}=-E_{p}}\right\rbrace = \int \frac{d^{3}p}{(2\pi^{3})}\int...
Could you clarify again the sign choices, am I correct:
1)is \psi_- and \psi_{+} inside the barrier, hence why only transitted wave functions?
2)- means going to -\infty + mean going to +\infty That I was hoping it would be a known example. One has to solve the above equation giving...
Hi when using the WKB approx, is there a general method to find these Refelction and Transmission coefficients, I have tried looking in books and on the net and I can't find a 'general' formula, they tend tjust to be quoted. I know that |T|^{2}+|R|^{2}=1 .
And generally that T=...
The initial intergal was f(x)=\int^{\infty}_{-\infty} \sqrt{x^{2}+y^{2}}dx so I taylor expanded it to get f(x) \approx \int^{\infty}_{-\infty} x + \frac{y^{2}}{2x} dx
I thought one could then justify that the cauchy principle value of \int^{\infty}_{-\infty} x dx =0 and then what I have...
In trying to solve \int^{\infty}_{-\infty} x + \frac{1}{x} dx could it be split up and solved using the Cauchy Principle Value theorem and a contour integral along a semi-circle. Thus;
PV\int^{\infty}_{-\infty}x dx =0 +\int \frac{1}{x} dx = \int^{\pi}_{0} i d\theta
Is this valid reasoning?
I have almost cracked it, I think it should be should there also be
\frac{1}{8}( \gamma^{\alpha}\gamma^{\beta} - \gamma^{\beta}\gamma^{\alpha}) ( e_{\alpha}^{\nu}(\frac{\partial}{\partial x^{\mu}})e_{\beta\nu}+e_{\alpha\nu}e_{\beta}^{\sigma}\Gamma^{\nu}_{\sigma\mu})
The trouble is I get...
I didn't wamt to post a long answer but here goes,
So \Gamma^{0}_{ij}=\dot{a}a and \Gamma^{i}_{0j}=\frac{\dot{a}}{a}
e^{\nu}_{\alpha}=(1,1/a,1/a,1/a)
e_{\beta\nu}=(1,a,a,a)
When \alpha or \beta equals zero then \gamma^{0}\gamma^{\rho}-\gamma^{\rho}\gamma^{0}=0 so this isn't allowed...
(1,a^2,a^2,a^2)) from the action; \mathcal{S}_{D}[\phi,\psi,e^{\alpha}_{\mu}] = \int d^4 x \det(e^{\alpha}_{\mu}) \left[ \mathcal{L}_{KG} + i\bar{\psi}\bar{\gamma}^{\mu}D_{\mu}\psi - (m_{\psi} + g\phi)\bar{\psi}\psi \right]
I can show that, i\bar{\gamma}^{\mu}D_{\mu}\psi -...
thx I am still struggling with this sum term.
Surely when \alpha = 1 then \nu =1 then \beta =1 .
and the same for \beta = to 2 and 3.
So when \beta= 1 wouldn't \sum ^{11} =0 and the same for beta =2,3 thus all these sum terms would become zero?
Sorry I think I may have made a mistake isn't: e_{\alpha}^{\mbox{ }\nu} equal to (1,1/a,1/a,1/a) and so on. I understand these e terms but I can't see how to deal with \sum ^{\alpha\beta} because surely it would give a gamma matrix but it shouldn't be there
In trying to derive the Dirac equation in space-time (1,-a^{2},-a^{2},-a^{2}), I have read that the Dirac equation is (i\bar{\gamma}^{\mu}(\partial_{\mu}+\Gamma_{\mu})-m)\psi=0 where,
\Gamma_{\mu}=\frac{1}{2}\sum ^{\alpha \beta}e_{\alpha}^{\mbox{ }\nu}(\frac{\partial}{\partial...