- #1
MadMax
- 98
- 0
I have:
\int d^3 \mathbf{q} d^3 \mathbf{q'} K_{ij} K_{ji} f(-\mathbf{q}+\mathbf{q'})f(-\mathbf{q}+\mathbf{q'})
and f(\mathbf{q})=\frac{i}{q_z}\int d^2 \mathbf{x} e^{i \mathbf{q_\perp} \cdot \mathbf{x}} [a e^{i q_z[H+h_2(\mathbf{x})]} - b e^{i q_z h_1(\mathbf{x})}]
What does f(-\mathbf{q} + \mathbf{q'}) equal?
x is real space and q is Fourier space. I'm thinking I can simply substitute the q for -q + q' , but what about the integral over x and the x values? do i substitute x for -x + x' too? I can't see that turning out the way its supposed to...
Any help would be much appreciated.
\int d^3 \mathbf{q} d^3 \mathbf{q'} K_{ij} K_{ji} f(-\mathbf{q}+\mathbf{q'})f(-\mathbf{q}+\mathbf{q'})
and f(\mathbf{q})=\frac{i}{q_z}\int d^2 \mathbf{x} e^{i \mathbf{q_\perp} \cdot \mathbf{x}} [a e^{i q_z[H+h_2(\mathbf{x})]} - b e^{i q_z h_1(\mathbf{x})}]
What does f(-\mathbf{q} + \mathbf{q'}) equal?
x is real space and q is Fourier space. I'm thinking I can simply substitute the q for -q + q' , but what about the integral over x and the x values? do i substitute x for -x + x' too? I can't see that turning out the way its supposed to...
Any help would be much appreciated.
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