WKB approximation

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I was scared by this term once before. ( the approach an explanation from J.J. Sakurai’s book is not so good)  in fact, don’t panic, it is easy. Let me explain.

i just copy what written in Introduction to Quantum Mechanics by David Griffiths (1995) Chapter 8.

The approx. can be applied when the potential is varies slowly compare the wavelength of the wave function. when it expressed in Exp( i k x) , wavelength = 2 π / k, when it expressed in Exp( - \kappa x ) , wavelength = 1/κ.

in general, the wavefunction can be expressed as amplitude and phase:

\Psi(x) = A(x)Exp(i \phi(x))

where A(x) and \phi(x) are real function

sub this into the time-independent Schrödinger equation (TISE)

\Psi '' (x) = - \frac {2 m} {\hbar^2 } ( E - V(x) ) \Psi (x)

\Psi ''(x) = ( A''(x)- A(x) \phi'(x)^2 + 2 i A'(x) \phi'(x)+ i A(x)\phi''(x) ) Exp(i \phi (x) )

and separate the imaginary part and real part.

The imaginary part is can be simplified as:

2 A'(x) \phi '(x) + A(x) \phi ''(x) = 0 = \frac {d}{dx} ( A^2(x) \phi '(x)

A(x) = \frac {const.} {\sqrt {\phi '(x)}}

The real part is

A''(x) = \left ( \phi ''(x) - \frac {2m}{\hbar^2 } ( E - V(x) ) \right) A(x)

we use the approx. that A''(x) = 0  ,  since it varies slowly.

Thus,

\phi '(x) = \sqrt { \frac {2m}{\hbar^2} (E - V(x) ) }

\Rightarrow \phi(x) = \int \sqrt { \frac {2m}{\hbar ^2} ( E - V(x ) )} dx

if we set,

p(x) = \sqrt { \frac {2m}{ \hbar^2 } ( E - V(x) )}

for clear display and p(x) is the energy different between energy and the potential. the solution is :

\Psi(x) = \frac {const.}{\sqrt {p(x)}} Exp \left( i \int p(x) dx \right)

Simple! but one thing should keep in mind that, the WKB approx is not OK when Energy = potential.

This tell you, the phase part of the wave function is equal the square of the area of the different of Energy and the Potential.

when the energy is smaller then the potential, than, the wavefunction is under decay.

one direct application of WKB approxi is on the Tunneling effect.

if the potential is large enough, so, the transmittance is dominated by the decay, Thus, the probability of the tunneling is equal to

Exp \left( - 2 \sqrt { \frac {2m}{\hbar ^2 } A_{area} ( V(x) - E )} \right)

Therefore, when we have an ugly potential, we can approx it by a rectangular potential with same area to give the similar estimation.

Hydrogen Atom (Bohr Model)

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OK, here is a little off track. But that is what i were learning and learned. like to share in here. and understand the concept of hydrogen is very helpful to understand the nuclear, because many ideas in nuclear physics are borrow from it, like “shell”.

The interesting thing is about the energy level of Hydrogen atom. the most simple atomic system. it only contains a proton at the center, um.. almost center, and an electron moving around. well, this is the “picture”. the fact is, there is no “trajectory” or locus for the electron, so technically, it is hard to say it is moving!

why i suddenly do that is because, many text books said it is easy to calculate the energy level and spectrum for it. Moreover, many famous physicists said it is easy. like Feynman, Dirac, Landau, Pauli, etc… OK, lets check how easy it is.

anyway, we follow the usual say in every text book. we put the Coulomb potential in the Schrödinger equation, change the coordinate to spherical. that is better and easy for calculation because the coulomb potential is spherical symmetric. by that mean, the momentum operator (any one don’t know what is OPERATOR, the simplest explanation is : it is a function of function.) automatically separated into 2 parts : radial and angular part. The angular part can be so simple that it is the Spherical harmonic.

Thus the solution of the “wave function” of the electron, which is also the probability distribution of  the electron location, contains 2 parts as well. the radial part is not so trivial, but the angular part is so easy. and it is just Y(l,m) .

if we denote the angular momentum as L, and the z component of it is Lz, thus we have,

L^2 Y(l,m) = l(l+1) \hbar^2 Y(l,m)

L_z Y(l,m) = m \hbar Y(l,m)

as every quadratic operator, there are “ladder” operator for “up” and “down”.

L_\pm Y(l,m) =\hbar \sqrt{l(l+1) - m(m\pm 1)} Y(l,m \pm 1)

which means, the UP operator is increase the z-component by 1, the constant there does not brother us.

it is truly easy to find out the exact form of the Y(l,m) by using the ladder operator. as we know, The z component of the a VECTOR must have some maximum. so, there exist an Y(l,m) such that

L_+ Y(l,m) =0

since there is no more higher z-component.

by solve this equation, we can find out the exact form of Y(l,m) and sub this in to L2, we can knowMax(m) = l . and apply the DOWN operator, we can fins out all Y(l,m) , and the normalization constant is easy to find by the normalization condition in spherical coordinate, the normalization factor is sin(\theta) , instead of 1 in rectangular coordinate.

\int_0^\pi \int_0^{2 \pi} Y^*(l',m') Y(l,m) sin(\theta) d\theta d \psi = \delta_{l' l} \delta_{m' m}

more on here