Generals 2000 II 2

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\psi\left(t\right)=t^{2}\int_{0}^{1}dw\left(\frac{w}{1-w}\right)^{i\lambda}e^{-4i\lambda wt^{2}}


Writing this as:

\psi\left(t\right)=t^{2}\int_{0}^{1}dw\, e^{\phi\left(w,t\right)}


With:

\phi\left(w,t\right)=i\lambda\left[\ln\left(w\right)-\ln\left(1-w\right)-4wt^{2}\right]


Setting the derivative to zero to find saddles:

\phi^{\prime}=0=\frac{1}{w_{s}}+\frac{1}{1-w_{s}}-4t^{2}


Rearranging:

1-w_{s}+w_{s}-4t^{2}w_{s}\left(1-w_{s}\right)=0


w_{s}^{2}-w_{s}+\left(4t^{2}\right)^{-1}=0


Solving for ws:

w_{s}=\frac{1\pm\sqrt{1-4\cdot\left(4t^{2}\right)^{-1}}}{2}=\frac{1}{2}\pm\frac{1}{2}\sqrt{1-t^{-2}}


Which we will now call w_{\pm}.


[edit] t<<1 limit

The saddle points are off the axis. Looking at the second derivative:

\phi^{\prime\prime}\left(w\right)=i\lambda\left[-\frac{1}{w^{2}}+\frac{1}{\left(1-w\right)^{2}}\right]


So that:

\phi^{\prime\prime}\left(w_{+}\right)/i\lambda=-\frac{4}{\left(1+\sqrt{1-t^{-2}}\right)^{2}}+\frac{4}{\left(1-\sqrt{1-t^{-2}}\right)^{2}}=-\frac{4t^{2}}{t^{2}+2t\sqrt{t^{2}-1}+t^{2}-1}+\frac{4t^{2}}{t^{2}-2t\sqrt{t^{2}-1}+t^{2}-1}


Using t\ll1 we taylor expand:

\phi^{\prime\prime}\left(w_{+}\right)=i\lambda\left[2t^{2}\left(1+2it-2t^{2}\right)-2t^{2}\left(1-2it-2t^{2}\right)\right]\approx-8\lambda t^{3}


So that the direction \sqrt{-\phi^{\prime\prime}\left(w_{+}\right)} is along the real axis. For the other direction:

\phi^{\prime\prime}\left(w_{-}\right)/i\lambda=-\frac{2}{\left(1-\sqrt{1-t^{-2}}\right)^{2}}+\frac{2}{\left(1+\sqrt{1-t^{-2}}\right)^{2}}=-\frac{2t^{2}}{t^{2}-2t\sqrt{t^{2}-1}+t^{2}-1}+\frac{2t^{2}}{t^{2}-2t\sqrt{t^{2}-1}+t^{2}-1}


Using t\ll1 we taylor expand:

\phi^{\prime\prime}\left(w_{-}\right)=i\lambda\left[2t^{2}\left(1-2it-2t^{2}\right)-2t^{2}\left(1+2it-2t^{2}\right)\right]\approx8\lambda t^{3}


So that the direction \sqrt{-\phi^{\prime\prime}\left(w_{-}\right)} is purely imaginary and cannot be used. We then get from the saddle at w + :

\psi\left(t\right)=\frac{t^{2}\sqrt{2\pi}}{\sqrt{-\phi^{\prime\prime}\left(w_{+}\right)}}e^{\phi\left(w_{+}\right)}


Expressing \phi\left(w_{+}\right) approximately by taking t\ll1:

\phi\left(w_{+},t\right)=i\lambda\left[\ln\left(\frac{1+\sqrt{1-t^{-2}}}{1-\sqrt{1-t^{-2}}}\right)-4\left(1+\sqrt{1-t^{-2}}\right)t^{2}\right]\approx i\lambda\left[\ln\left(-1\right)-4it\right]=4\lambda t-\lambda\pi


Putting this in:

\psi\left(t\right)\approx\frac{t^{2}\sqrt{2\pi}}{\sqrt{8\lambda t^{3}}}e^{4\lambda t-\lambda\pi}\sim\sqrt{t}e^{4\lambda t}


[edit] t>>1 limit

For this limit, the saddle points are on the real axis. Finding the second derivative and using t\gg1:

\phi^{\prime\prime}\left(w_{+}\right)=-\frac{4i\lambda}{\left(1+\sqrt{1-t^{-2}}\right)^{2}}+\frac{4i\lambda}{\left(1-\sqrt{1-t^{-2}}\right)^{2}}\approx i\lambda\left(-1+\frac{4}{\left(t^{-2}/2\right)^{2}}\right)\approx16i\lambda t^{4}


So the direction of integration is \sqrt{-16i\lambda t^{4}}\sim e^{3\pi i/4}. For the other saddle point:

\phi^{\prime\prime}\left(w_{-}\right)=-\frac{4i\lambda}{\left(1-\sqrt{1-t^{-2}}\right)^{2}}+\frac{4i\lambda}{\left(1+\sqrt{1-t^{-2}}\right)^{2}}\approx i\lambda\left(-\frac{4}{\left(t^{-2}/2\right)^{2}}+1\right)\approx-16i\lambda t^{4}


So the direction of integration is \sqrt{16i\lambda t^{4}}\sim e^{\pi i/4}. We integrate through both saddles. Evaluating φ at the first saddle, with t^{2}\gg1:

\phi\left(w_{+},t\right)=i\lambda\left[\ln\left(\frac{1+\sqrt{1-t^{-2}}}{1-\sqrt{1-t^{-2}}}\right)-4\left(1+\sqrt{1-t^{-2}}\right)t^{2}\right]\approx i\lambda\left[\ln\left(\frac{2}{t^{-2}/2}\right)-8t^{2}\right]=2i\lambda\ln\left(2t\right)+8i\lambda t^{2}


So this is just a phase factor. At the other saddle:

\phi\left(w_{-},t\right)=i\lambda\left[\ln\left(\frac{1-\sqrt{1-t^{-2}}}{1+\sqrt{1-t^{-2}}}\right)-4\left(1+\sqrt{1-t^{-2}}\right)t^{2}\right]\approx i\lambda\left[\ln\left(\frac{t^{-2}/2}{2}\right)-8t^{2}\right]=-2i\lambda\ln\left(t/2\right)+8i\lambda t^{2}


So this is another phase factor. Then the saddles contributions are:

\psi\left(t\right)=\frac{t^{2}\sqrt{2\pi}}{\sqrt{-\phi^{\prime\prime}\left(w_{+}\right)}}e^{\phi\left(w_{+}\right)}+\frac{t^{2}\sqrt{2\pi}}{\sqrt{-\phi^{\prime\prime}\left(w_{-}\right)}}e^{\phi\left(w_{-}\right)}


Plugging in:

\psi\left(t\right)=\frac{t^{2}\sqrt{2\pi}}{\sqrt{-16i\lambda t^{4}}}e^{2i\lambda\ln\left(2t\right)+8i\lambda t^{2}}+\frac{t^{2}\sqrt{2\pi}}{\sqrt{16i\lambda t^{4}}}e^{-2i\lambda\ln\left(t/2\right)+8i\lambda t^{2}}


Which simplifies to:

\psi\left(t\right)=-\frac{t^{2}\sqrt{2\pi}}{\sqrt{16\lambda t^{4}}}e^{8i\lambda t^{2}}2i\sin\left(2\lambda\ln\left(2t\right)-\frac{\pi}{4}\right)\sim const
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