Question

Playing around with Unitary matrices

Answer 1

This comes from trying to invert a matrix, this is basically my original thoughts an ideas so it might kind of deviate from regular quantum mechanics stuff. My idea of matrix multiplication is basically two argument functions integrated over, just like the matrix multiplication: \(H_{ln}=F_{lm}G_{mn}\)
\[h(x,z)=\int f(x,y)g(y,z) dy\]
And inverse matrices:
\[\delta(x-z)=\int_{-\infty}^\infty f(x,y)f^{-1}(y,z) dy\]

So it all basically comes down to this identity:

\[\delta(x-z)=\int_{-\infty}^\infty \frac{e^{iy(x-z)}}{2 \pi} dy\]
We can split that middle thing up:

\[\delta(x-z)=\int_{-\infty}^\infty \frac{e^{iyx}}{ \sqrt{2 \pi}} \frac{e^{-iyz}}{ \sqrt{2 \pi}} dy\]

Then I define functions this way, and this is what I'm wanting to really know if this is most general or if this should be done differently:

\[A(x,y)=\frac{e^{ixy}}{ \sqrt{2 \pi}} [a(x)b(y)]^i\] with \(a(x)\) and \(b(y)\) purely real functions. That way I can define this conjugate transpose operation:

\[A^\dagger(x,y)=A^*(y,x)\] which is just a fancy way of saying take the complex conjugate and switch the arguments. For now we can clearly see that letting \[a(x)=b(y)=1\] will allow us to recover our original delta statement:

\[\delta(x-z)=\int_{-\infty}^\infty A(x,y)A^\dagger(y,z) dy = \int_{-\infty}^\infty \frac{e^{iyx}}{ \sqrt{2 \pi}} \frac{e^{-iyz}}{ \sqrt{2 \pi}} dy\]

But for most functions a(x) and b(y) this looks like it'll work as well, which I'll work out now:

\[\int_{-\infty}^\infty A(x,y)A^\dagger(y,z) dy = \int_{-\infty}^\infty \frac{e^{iyx}}{ \sqrt{2 \pi}} \frac{e^{-iyz}}{ \sqrt{2 \pi}} [a(x)b(y)]^i [a(z)b(y)]^{-i}dy\]
\[\left(\frac{a(x)}{a(z)} \right)^i \int_{-\infty}^\infty \frac{e^{iy(x-z)}}{2 \pi} dy =\left(\frac{a(x)}{a(z)} \right)^i \delta(x-z) \]

Since the delta function is 0 when \(x \ne z\) they're trivially equal, and when \(x=z\) the functions divide each other out leaving just the delta function, so we have:

\[\delta(x-z)=\int_{-\infty}^\infty A(x,y)A^\dagger(y,z) dy\]

Answer 2

Actually I just realized I'd really like to get time dependence in there, specifically I want to connect these unitary matrices up to this identity:

\[H \equiv i \hbar \frac{\partial A}{\partial t} A^\dagger \]

Answer 3

By just sorta naively trading out \(a(x,t)\) and \(b(y,t)\) in there we have (where primes denote time derivatives):

\[H = \frac{- \hbar}{2 \pi} \left(\frac{a'(x,t)}{a(x,t)} + \frac{b'(y,t)}{b(y,t)} \right) \]

So maybe at this point I have either fallen off the deep end or something cause I don't really know, should I just like maybe guess at a(x,t) being the kinetic energy and b(y,t) being the potential energy? Haha I'll go to bed I think and sort this out.