On Quantum Mechanics

in First part of Bohr Sommerfeld analysis we show that we can by scaling \(m \to cm\), \(p\to cp\), \(W \to W, n_{\phi} \to n_{\phi}, n_{r} \to n_{r}\) and \(e \to e\) (for the electron part) reproduce the same Bohr Sommerfeld model and the to total energy is the same for the ensemble as for one particle and the same holds for \(p_{\phi},p_r\). In principle this means that we can consider any ensemble of non interacting Bohr Sommerdels Model's which is a satisfying result. There is another scaling also. If we consider the total energy \(W_{tot} = W+mc^2\) we can keep that constant, and keep all other parts scaled as the same way. This means that \(K\to K/c\) and hence

$$ r \to c r_0. $$

So we take the weight equal to \(r/r_0\) and keep the total energy constant and also preserves total mass, total momentum's and total charge. Hence the density over a circle is constant. As the length of the circle is \(1/2\pi r\) we have that the weight for radius \(r\) through,

$$ u(r,x) = \frac{1}{a_0 2\pi}\delta_{x = r} $$

Now for this model if we take the \(c=\exp(i\beta)\) we get \(p \to \exp(i\beta)p\), \(m \to \exp(i\beta)p\), \(e \to \exp(i\beta)e\) and \(W \to \exp(i\beta)W\). Again this preserves the radius but not only this, we get back the old energy and momenta by noting that we can fix this by multiplying with essentially the conjugate of the density for u. This especially means that by taking \(\beta = tE/\hbar + 2 \pi n_\phi x / a_0 + n_{\rho}r\) we can actually see that we have a way to encode mathematically the model so that the energy levels and momenta for the underlying physical system with the link being the 2D momentum operators and the energy operator. Now rotating the plane to all possible directions, will in the end lead to all the usual 3D momentum Cartesian operators, still this preserve the momentum operators and energy operators and total energy of the underlying physical ensemble system that we take the limit of as as the Laplacian. Hence we have generated the Klein Gordon equation now forcing the L2 solutions so that \(\langle 1 \rangle = 1\) we get the usual solutions and proper values of the energy.

One complaint are that the gradient in the \(x\) direction is

$$ \frac{1}{r}\frac{\partial}{\partial r} $$

But the momenta in the limit is \(mvr\) and hence the remaining part is \(mv\) whic is constant.

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