WKB Approximation: Difference between revisions
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Revision as of 14:38, 14 August 2013
The WKB (Wentzel-Kramers-Brillouin method) method is a technique for finding approximations to certain differential equations, including the one dimensional Schrödinger equation. It was developed in 1926 by Wenzel, Kramers, and Brillouin, for whom it was named. The logic behind this method is that, as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\rightarrow 0,\!} the wavelength, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=2\pi\hbar/ p,\!} tends to zero where the potential is smooth and slowly varying. Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} can be thought of as a local quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda(x).\!} This is a semiclassical method of solving the Schrödinger equation.
However, for the classical turning points of the potential (i.e., where the energy of the particle becomes equal to the potential), Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} becomes infinite, and we can no longer say that the potential changes slowly. Here, the whole theory is bound to fail. A proper handling of the turning points is the most difficult aspect of the WKB approximation. In this case, the potential at the turning point is approximated as linear and slowly varying (almost constant).
By doing this, we can solve the Schrödinger equation and obtain the wave function in this "region". We can then use this wave function to connect the WKB wave functions at the two sides of the turning point.
For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x) = 0\!} (or constant) the solutions to the Schrödinger equation are simply plane waves of the form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{\pm ikx} } . If the potential varies smoothly, and the energy of the particle is fixed, the wave function can be described locally by writing its plane wave form Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{iu(x)}\!} . If we substitute this form into the Schrödinger equation, then we obtain
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left (\frac{du}{dx}\right )^2=\left [\frac{p(x)}{\hbar}\right ]^2+i\frac{d^2u}{dx^2},}
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(x)=\sqrt{2m[E-V(x)]}.} Let us now assume that the second derivative on the right is small compared to the rest of the terms and neglect it as a first approximation. Let us call this first approximation to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)\!} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x).\!} The equation satisfied by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x)\!} is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du_0}{dx}=\pm\frac{p(x)}{\hbar}.}
Solving this equation, we obtain
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x)=\pm\frac{1}{\hbar}\int_{x_0}^x dx'\,p(x')+C_0.}
We now use this first approximation to solve the full differential equation by an iterative process. We obtain the next approximation by substituting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x)\!} into the right-hand side and solving for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u(x)\!} on the left. If we call the result Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1(x),\!} then the equation that we solve is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du_1}{dx}=\pm\sqrt{\left [\frac{p(x)}{\hbar}\right ]^2\pm\frac{i}{\hbar}\frac{dp}{dx}}.}
We must assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1(x)\!} has the same sign as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x)\!} in order for this iterative process to converge, and we must assume that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dp}{dx}\ll \frac{[p(x)]^2}{\hbar}} so that the deviation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1(x)\!} from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_0(x)\!} is small. If this is the case, then we may assume that any further corrections introduced in further iterations are even smaller, and that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u_1(x)\!} is a good approximation to the exact solution. Under this assumption, we may expand the square root on the right-hand side of this equation, obtaining
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du_1}{dx}\approx\pm\frac{p(x)}{\hbar}+\frac{i}{p(x)}\frac{dp}{dx}.}
Solving this equation, we find that, for the classically accessible region, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E>V(x),\!} the WKB wave function is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\approx \frac{C_+}{\sqrt{p(x)}}\exp\left [\frac{i}{\hbar}\int_{x_0}^x p(x')\,dx'\right ]+\frac{C_-}{\sqrt{p(x)}}\exp\left [-\frac{i}{\hbar}\int_{x_0}^x p(x')\,dx'\right ],}
and for the classically forbidden region, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E<V(x),\!}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\approx \frac{C_+}{\sqrt{|p(x)|}}\exp\left[\frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right] + \frac{C_-}{\sqrt{|p(x)|}}\exp\left[- \frac{1}{\hbar}\int_{x_0}^x |p(x')|\,dx'\right].}
This is an exact solution if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)\!} is constant, otherwise it is a local solution for a locally-defined wavelength. In order for this approximation to be valid, we must assume that the wavelength is slowly varying as a function of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x.\!} In other words, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left |\frac{d\lambda}{dx}\right |=\left |\frac{d}{dx}\left(\frac{\hbar}{p(x)}\right)\right | \ll 1.}
Let us now suppose that we let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_0=a,\!} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a\!} is a classical turning point at a given value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E\!} and thereby separates a region where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E > V(x)\!} from a region where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E < V(x).\!} Let the classically inaccessible region be to the right of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a.\!} For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\!} sufficiently far away from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a,\!} the WKB wave function is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\approx \frac{A}{\sqrt{|p(x)|}}\exp\left[- \frac{1}{\hbar}\int_{a}^x |p(x')|\,dx'\right]+ \frac{B}{\sqrt{|p(x)|}}\exp\left[\frac{1}{\hbar}\int_{a}^x |p(x')|\,dx'\right] }
for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \gg a, \! } and
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)\approx \frac{C}{\sqrt{p(x)}}\exp\left[- \frac{i}{\hbar}\int_{a}^x p(x')\,dx'\right]+ \frac{D}{\sqrt{p(x)}}\exp\left[\frac{i}{\hbar}\int_{a}^x p(x')\,dx'\right] }
for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \ll a. \!}
Note that, at the classical turning point, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p(a) = 0\!} and the WKB solution diverges, meaning that it is no longer a valid approximation because the true wave function will not exhibit such divergent behavior at the turning points. Thus, around each turning point, we need to splice the two WKB solutions on either side of the turning point with a "patching" function that will straddle each turning point. Because we only need a solution for this function in the vicinity of the turning points, we can approximate the potential as being linear. If we center the turning point at the origin, i.e. we take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a=0,\!} the potential is approximately
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)\approx E+V'(0)x.\!}
Solving the Schrödinger equation with this linearized potential leads to the Airy equation whose solutions are Airy functions. Our patching wave function, which we denote by is then:
where and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C_2\!} are complex number coefficients and
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha=\left(\frac{2m}{\hbar^2}V'(0)\right)^{1/3}.}
The key to patching the wavefunction in the region of the turning point is to asymptotically match the patching wavefunction to the wavefunctions far away from the classical turning point. In the vicinity of the classical turning point,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p^2=2m(-V'(0)x)\Rightarrow 2p\frac{dp}{dx}=-2mV'(0)\Rightarrow \frac{dp}{dx}=-\frac{m}{p}V'(0) }
Since the region of applicability of the WKB approximation is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1\gg\frac{1}{2\pi}\left|\frac{d\lambda}{dx}\right|,}
we find that, near the turning point,
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |p|^3 \gg \hbar m|V'(0)|,}
or
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x|\gg \tfrac{1}{2}\hbar^{2/3}|mV'(0)|^{-1/3}.}
This implies that the width of the region around the classical turning point vanishes as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar^{2/3}.} Thus, we can come as close to the turning point as we wish with the WKB approximation by taking the limit as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\!} approaches zero, as long as the distance from the classical turning point is much greater than Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar^{2/3}.} Thus, by extending the patching function towards the singularity in the WKB wave function, while simultaneously extending the WKB wave function toward the classical turning point, it is possible to match the asymptotic forms of the wave functions from the two regions, which are then used to patch them together.
For this purpose, the following asymptotic forms of the Airy functions Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{Ai}(z)\!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{Bi}(z)\!} as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\rightarrow\pm\infty} will be useful:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\rightarrow \infty: \mathrm{Ai}(z)\rightarrow \frac{1}{2\sqrt{\pi}}z^{-\frac{1}{4}}e^{-\frac{2}{3}|z|^{\frac{3}{2}}}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\rightarrow \infty: \mathrm{Bi}(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}e^{\frac{2}{3}|z|^{\frac{3}{2}}}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\rightarrow -\infty: \mathrm{Ai}(z)\rightarrow \frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\cos\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z\rightarrow -\infty: \mathrm{Bi}(z)\rightarrow -\frac{1}{\sqrt{\pi}}z^{-\frac{1}{4}}\sin\left(\frac{2}{3}|z|^{\frac{3}{2}}-\frac{\pi}{4}\right)}
Noticing that in the vicinity of the turning point (for negative Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\!} ),
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\hbar}\int_{x}^0p(x')dx'=\sqrt{\frac{2mV'(0)}{\hbar^2}} \int_x^0\sqrt{-x'}dx'=\tfrac{2}{3}\sqrt{\frac{2mV'(0)}{\hbar^2}}|x|^{3/2}=\tfrac{2}{3}|\alpha x|^{3/2} }
and
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{\sqrt{p(x)}}=\left(2mV'(0)\right)^{-1/4}|x|^{-1/4},}
it becomes apparent that our WKB approximation of the wavefunction is the same as the patching function in the asymptotic limit. This must be the case, since as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\rightarrow 0} the region of invalidity of the semiclassical wavefunction in the vicinity of the turning point shrinks, while the solution of the linarized potential problem depends only on the accuracy of the linearity of the potential, and not on Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar.} The two regions must therefore overlap.
For example, one can take Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\sim \hbar^{1/3}} and then take the limit Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\rightarrow 0} . The semiclassical solution must hold as we are always in the region of its validity and so must the solution of the linearized potential problem. Note that the argument of the Airy functions at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\sim \hbar^{1/3}} goes to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm\infty} , which is why we need their asymptotic expansion.
For a classical turning point Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a\!} that separates a classically inaccessible region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>a\!} from a classically accessible region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<a,\!} we compare the WKB wave functions with the asymptotic expressions for the Airy functions and find that the WKB wave functions on the two sides of this turning point are connected as follows:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2A}{\sqrt{k(x)}}\cos\left (\int_{x}^{a}k(x\,)dx-\frac{\pi}{4}\right )-\frac{B}{\sqrt{k(x)}}\sin\left (\int_{x}^{a}k(x)\,dx-\frac{\pi}{4}\right ) \longleftrightarrow \frac{A}{\sqrt{|k(x)|}}\exp\left (-\int_{a}^{x}|k(x)|\,dx\right )+\frac{B}{\sqrt{|k(x)|}}\exp\left (\int_{a}^{x}|k(x)|\,dx\right ) }
Similarly, for a classical turning point Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=b,\!}
that separates a classically accessible region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>b\!}
from a classically inaccessible region Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x<b,\!}
we find that the connection between the WKB wave functions on either side of the turning point is
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2A}{\sqrt{k(x)}}\cos\left (\int_{b}^{x}k(x)\,dx-\frac{\pi}{4}\right )-\frac{B}{\sqrt{k(x)}}\sin\left (\int_{b}^{x}k(x)\,dx-\frac{\pi}{4}\right ) \longleftrightarrow \frac{A}{\sqrt{|k(x)|}}\exp\left (-\int_{x}^{b}|k(x)|\,dx\right )+\frac{B}{\sqrt{|k(x)|}}\exp\left (\int_{x}^{b}|k(x)|\,dx\right ). }
Bohr-Sommerfeld Quantization Rule
The quantized energy levels of a bound state can be approximated by the WKB method with an expression known as the Bohr-Sommerfeld quantization rule. A particle in a potential well is subject to bound states. This common example of the WKB method can be found in most undergraduate level quantum texts.
For a potential with no rigid walls the Bohn-Sommerfeld Quantization rule is:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \oint p(x)dx = 2\int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = \left(n+\frac{1}{2}\right)2\pi\hbar}
For a potential with one rigid wall:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = \left(n+\frac{3}{4}\right)\pi\hbar }
For a potential with two rigid walls:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_{x_1}^{x_2}\sqrt{2m(E_n - V(x))}dx = n\pi\hbar }
For a central potential:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p_r^2 = E - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}}
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_{r_1}^{r_2}p_r(r)dr &= \int_{0}^{\infty}\sqrt{2m\left(E_n - V(r) - \frac{\hbar^2}{2m}\frac{\ell(\ell+1)}{r^2}\right)}dr \\ &= \left(n + \frac{1}{2}\right)\pi\hbar \end{align} }
WKB method for the Coulomb Potential
For the coulomb potential, the potential is given by:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(r) = -\frac{-Ze^2}{r} }
Since the electron is bound to the nucleus, it can be veiwed as moving between two rigid walls at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = 0 \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r = a \!} with energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E = V(a), a = -\frac{-Ze^2}{E}\!} . Thus, the energy of the electron is negative.
The energies of the s-state (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ell = 0 \!} ) can be obtained from:
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr = n\pi\hbar }
Using the change of variable: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x = \frac{a}{r} }
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \int_0^a \sqrt{2m\left(E+\frac{Ze^2}{r}\right)}dr &= \sqrt{-2mE} \int_0^a dr \sqrt{\frac{a}{r} - 1} \\ &= a\sqrt{-2mE} \int_0^1 dx\sqrt{\frac{1}{x} - 1} \\ &= \frac{\pi}{2}a\sqrt{-2mE} \\ &= -Ze^2\pi\sqrt{-\frac{2m}{E}} \end{align} }
Where I have used the integral
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^1\sqrt{\frac{1}{x} -1} = \frac{\pi}{2} }
Thus we have the expression:
Where is the Bohr radius. Notice that this is the correct expression for the energy levels of a Coulomb potential.