Connection Formulae. The WKB approximation falls apart near a turning point. Then E V 0 so 1

Transcription

1 P3 WKB I D.Rubin February 8, 8 onnection Formulae The WKB approximation falls apart near a turning point. Then E V so p. And because the momentum goes to zero the wavelength gets very long the approximation is only valid if the wavelength is short compared to the distance over which the potential changes. But if want to determine bound state energies, we need to be able to match wave functions at the turning points. The strategy to overcome this limitation of the WKB wave functions at the turning points is to. Linearize the potential at the turning point (x = ). V (x) = V ()x dv dx. Solve Schrodinger s equation exactly near the turning point for the linear potential There will be two cases to consider. One where dv > dx the particle has positive kinetic energy to the left of the turning point (x < ), negative kinetic energy to the right (x > ). The other case is when dv < then there is negative kinetic energy to dx the left positive to the right. We consider first dv >. dx h d ψ m dx (V () xv )ψ = Eψ d ψ dx = m h (E V () xv )ψ = p h ψ The turning point is at x = that is where V (x) = E. So V () = E we have d ψ dx = α3 xψ where α 3 = m V. Then if we define the dimensionless parameter h z = αx we have d ψ dz = zψ which is Airy s equation the general solution is ψ p = aai(x) bbi(x)

2 ψ p is referred to as the patching wave function since its sole purpose is to patch together the WKB wave functions on each side of the turning point. 3. Determine the WKB wave function in the region of the linear potential. It will be different on each side of the turning point. To the left of the turning point for case one, the WKB wave function is ψ l (x) = A p e iφ(x) B p e iφ(x) where φ(x) = h p(x) = p(x )dx = α 3 ( x) dx = 3 ( αx) 3 () m(e V ) m( xv ) = hα 3 ( x) (Since we are to the left of the turning point x < the argument of the square root is positive.) The WKB wave function to the left of the turning point where the slope of the potential is positive is A ψ l (x) = e i h ( α3 x) 3 ( αx) 3 B h ( α3 x) or in terms of the parameter z = αx A ψ l (x) = e i ( hα) ( z) 3 ( z) 3 B ( hα) ( z) e i 3 ( αx) 3 e i 3 ( z) 3 () (3) To the right of the turning point, (x > ), the WKB wave function is ψ r (x) = p eφ(x) p D e φ(x) φ(x) = h p(x ) dx = 3 (αx) 3 () the WKB wave function to the right of the crossing point is ψ r (x) = e h (α3 x) 3 (αx) 3 D h (α3 x) e 3 (αx) 3 (5)

3 or in terms of the parameter z ψ r (x) = e ( hα) z 3 z 3 D ( hα) z e 3 z 3 (6). Now we want to use the patching wave function to connect the WKB wave to the left of the turning point with the WKB wave function to the right. We use the large negatve z asymtotic form of the Airy functions Ai(z) sin [ ( z) ] 3 π π( z) 3 = π( z) i [ e i 3 ( z) 3 e i π e i 3 ( z) 3 e i π Bi(z) cos [ ( z) ] [ ] 3 π π( z) 3 = e i 3 ( z) 3 e i π e i 3 ( z) 3 e i π π( z) ] z Then setting we have ψ l (z) = aai(z) bbi(z) Evidently ψ l (x) = = π( z) A ( hα) ( z) e i 3 ( z) 3 B e i ( hα) ( z) 3 ( z) 3 [ ] (b ia)e i 3 ( z) 3 e i π (b ia)e i 3 ( z) 3 e i π B b ia = ( hα) / π π ei A b ia = ( hα) / π π e i Solving for a b in terms of A B we have that π π b = hα (Aeiπ/ Be iπ/ ) a = hα (Ae iπ/ Be iπ/ ) (7) that takes care of matching the WKB wave function to the left of the turing point to the patching wave function. Now let s do the same thing to the right of the turning point. Here we use the large positive z asymtotic form of the Airy functions. Namely Ai(z) e π(z) / 3 z 3 z Bi(z) π(z) / e 3 z 3 3

4 Setting we have e ( hα) z 3 z 3 D ( hα) z from which we see that a = ψ r (z) = aai(z) bbi(z) e 3 z 3 = a π(z) e / 3 z 3 b e 3 z 3 π(z) / π hα D b = π hα (8) Finally we equate equations 7 8 to eliminate a b yielding the connection formulae π hα D = π hα (Ae iπ/ Be iπ/ ) (9) π π hα = hα (Aeiπ/ Be iπ/ ) () In summary, for a barrier to the right, the connection formulae are D = (Ae iπ/ Be iπ/ ) = Ae iπ/ Be iπ/ barrier to right A = De iπ/ e iπ/ B = De iπ/ eiπ/ Note that in the connection formulae, there is no mention of the linearized potential, the parameter α, V or Airy functions. The linearization procedure served only as a mechanism to relate the constants A B to the left of the turning point with the constants D to the right of the turning point. Having established that relationship, which by the way is good for any arbitrary potential, (as long as it is not too nonlinear), we no longer need the patching wave function. Now we know how to deal with a turning point to the right (namely with V > ). We need an equivalent set of relations to deal with turning points to the left (V < ) they can be derived by a similar procedure. The derivations are nearly identical. The only difference is that the phase integrals change sign. Equation will become φ(x) = h p(x )dx = α 3 ( x) dx = 3 ( αx) 3 ()

5 equation will become φ(x) = h p(x ) dx = 3 (αx) 3 () The result is that we interchange A for B for D. The connection formulae for a barrier to the left are given here for reference. D = Ae iπ/ Be iπ/ = (Aeiπ/ Be iπ/ ) barrier to left A = Deiπ/ e iπ/ B = De iπ/ e iπ/ 5. Now that we know how to match WKB wave functions at the turning points we can derive the quantization condition. Suppose that we are trying to determine the energy of a bound state. Imagine something like the harmonic oscillator potential with V < at the left turning point (x = x ) V > at the right turning point (x = x ). Then the region x < x is classically forbidden, (we refer to it as region I) ψ I (x) = p e x h x p(x ) dx D p e x h x p(x ) dx (3) In region II the WKB wave function for x > x is ψ II (x) = A e ī x h x p(x )dx B e i x h x p(x )dx () p p In order that ψ I (x) be finite for x, it must be that D =. Then connection formulae for a barrier to the left yield = Ae iπ/ Be iπ/, A = ib, A = e iπ/, B = e iπ/ (5) And using Equation 5 to substitute for A B, Equation becomes ψ II (x) = e ī x h x p(x )dx iπ/ p e i x h x p(x )dx iπ/ (6) p Next we have to connect WKB wave functions for region II III at x = x. In order to apply the connection formulae for a barrier to the right we need to have the wave function in the form ψ = A p e ī h x p(x )dx B e i h p x p(x )dx (7) 5

6 We can write Equation 6 in the same form as follows ψ II (x) = p e ī h ( x p(x )dx x p(x )dx iπ/) p e i h ( x p(x )dx x p(x )dx iπ/) omparing the previous equation with 7 we have that where e i(θ π/) = A, e i(θ π/) = B θ = h x p(x )dx Now we can use the connection formulae for a barrier to the right to connect ψ II with ψ III. ψ III (x) = p e x h x p(x ) dx D p e x h x p(x ) dx (8) We know that = since is the coefficient of the exponentially growing term for the WKB wave function in region III. The connection formulae for a barrier to the right give = A e iπ/ B e iπ/ = e i(θ π/) e iπ/ e i(θ π/) e iπ/ = cos θ But = so cos θ =, θ = (n )π (9) p(x)dx = hπ(n x ) () Equation determines the allowed energies of the system. 6