Solution of Schrodinger equation in axially symmetric case

In summary, the Schrodinger equation with rotational invariance around the ##z##-axis can be expanded into a series with continuum radial functions and angular momentum ##l##. The phase shift ##\delta_{l}## is determined by the requirement that ##R_{kl}(r)## is regular as ##r \rightarrow 0##. The factor of ##2M## in front of ##r^{2}## is most likely a typo, and the normalisation of ##R_{kl}(r)## at infinity is a standard practice in scattering theory to obtain stationary states.
  • #1
spaghetti3451
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The following extract is taken from Appendix A of the following paper: http://arxiv.org/abs/0810.0713.Any solution of the Schrodinger equation with rotational invariance around the ##z##-axis can be expanded as ##\psi_{k}=\Sigma_{l}A_{l}P_{l}(cos \theta)R_{kl}(r)##, where ##R_{kl}(r)## are the continuum radial functions associated with angular momentum ##l## satisfying

##-\frac{1}{2M}\frac{1}{r^{2}}\frac{d}{dr}\big(r^{2}\frac{d}{dr}R_{kl}\big)+\big(\frac{l(l+1)}{r^{2}}+V(r)\big)R_{kl} = \frac{k^{2}}{2M}R_{kl}##

The ##R_{kl}(r)## are real, and at infinity look like a spherical plane wave which we can choose to normalise as

##R_{kl}(r) \rightarrow \frac{1}{r}sin(kr-\frac{1}{2}l\pi+\delta_{l}(r))##

where ##\delta_{l}(r) << kr## as ##r \rightarrow \infty##.

The phase shift ##\delta_{l}## is determined by the requirement that ##R_{kl}(r)## is regular as ##r \rightarrow 0##. Indeed, if the potential ##V(r)## does not blow up faster than ##\frac{1}{r}## near ##r \rightarrow 0##, then we can ignore it relative to the kinetic terms, and we have that ##R_{kl} \sim r^{l}## as ##r \rightarrow 0##; all but the ##l=0## terms vanish at the origin.
I have a couple of questions regarding the extract.

Firstly, there is a factor of ##2M## in front of ##r^{2}## in the expression ##\frac{l(l+1)}{r^{2}}## in quantum mechanics textbooks. Why is this factor missing here? Is this a typo?

Secondly, why is ##R_{kl}(r)## at infinity normalised as ##R_{kl}(r) \rightarrow \frac{1}{r}sin(kr-\frac{1}{2}l\pi+\delta_{l}(r))## where ##\delta_{l}(r) << kr## as ##r \rightarrow \infty##?

Thirdly, what does it mean to normalise ##R_{kl}(r)## as above?
 
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  • #2
failexam said:
Firstly, there is a factor of ##2M## in front of ##r^{2}## in the expression ##\frac{l(l+1)}{r^{2}}## in quantum mechanics textbooks. Why is this factor missing here? Is this a typo?
I guess this is a typo, as otherwise there is a problem with dimensions.

failexam said:
Secondly, why is ##R_{kl}(r)## at infinity normalised as ##R_{kl}(r) \rightarrow \frac{1}{r}sin(kr-\frac{1}{2}l\pi+\delta_{l}(r))## where ##\delta_{l}(r) << kr## as ##r \rightarrow \infty##?

Thirdly, what does it mean to normalise ##R_{kl}(r)## as above?
As they say, this is a representation of a spherical wave, phase-shifted due to the interaction at r = 0. It is standard in scattering theory to normalize wave functions that way, in order to obtain stationary states, at the expense of wave function being no longer square integrable. You should find more details in a good book on quantum scattering. Sakurai's Modern Quantum Mechanics can be a good starting point.
 

What is the Schrodinger equation in axially symmetric case?

The Schrodinger equation in axially symmetric case is a mathematical equation used to describe the behavior of quantum systems in which there is cylindrical symmetry. It is a partial differential equation that takes into account both the wave-like and particle-like nature of matter.

How does the solution of the Schrodinger equation in axially symmetric case differ from the general solution?

The solution of the Schrodinger equation in axially symmetric case is a special case of the general solution, which takes into account the cylindrical symmetry of the system. This results in a simplified form of the equation and allows for easier solution techniques.

What are the physical implications of the solution of the Schrodinger equation in axially symmetric case?

The solution of the Schrodinger equation in axially symmetric case provides information about the energy levels and wavefunctions of a quantum system with cylindrical symmetry. It can also be used to determine the probability of finding a particle in a certain region of space.

What techniques are used to solve the Schrodinger equation in axially symmetric case?

There are several techniques that can be used to solve the Schrodinger equation in axially symmetric case, including separation of variables, perturbation theory, and numerical methods such as finite difference or finite element methods.

What are some applications of the solution of the Schrodinger equation in axially symmetric case?

The solution of the Schrodinger equation in axially symmetric case has many practical applications, such as in the study of atoms, molecules, and nuclei. It is also used in the development of quantum computing and in the design of devices such as lasers and transistors.

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