Basic Theoretical Physics A Concise Overview

25.1 Bound Systems in a Box (Quantum Well); Parity 225 Next we assume E < 0 (i.e., for bound states). Then we have the followingsolutions of the Schro¨dinger equationu(x) = B−(1) exp(−κ|x|) + B+(1) exp(+κ|x|) , −x ≥ a , (25.3) (25.4)u(x) = C1 cos(k · x) + C2 sin(k · x) , |x| < a , (25.5)u(x) = B−(2) exp(−κx) + B+(2) exp(+κx) , +x ≥ a .a) Firstly, we recognise that the coefficents B+(1) and B+(2) must vanish, since otherwise it would not be possible to satisfy the condition ∞ dx|u(x)|2 =! 1 . −∞b) The remaining coefficients are determined (apart from a common factor, where only the magnitude is fixed by the normalization condition) from the continuity conditions for u and u at the potential steps, x = ±a.The calculation is thus much easier for symmetrical potentials, V (x) =V (−x), since then all solutions can be divided into two different classes:even parity, i.e., u(−x) = u(x) , (⇒ B−(1) = B−(2) ; C2 = 0) and odd parity, i.e., u(−x) = −u(x) , (⇒ B−(1) = −B−(2) ; C1 = 0) . One then only needs one continuity condition, i.e. the one for u /u atx = +a. From this condition (see below) one also obtains the discrete energyvalues E = En, for which continuity is possible (for k > 0 and κ > 0): κ(E) = tan(k(E) · a) , for even parity, (25.6) k(E) for odd parity . (25.7)− k(E) = tan(k(E) · a) , κ(E) These equations can be solved graphically (this is a typical exercise), byplotting all branches of tan(k·a) as a function of k·a (these branches interceptthe x-axis at k · a = n · π, where n is integer, and afterwards they diverge to+∞ from −∞, at k± · a = (2n ± 1) · π/2 ∓ 0+). Then one can determine theintersections of this multi-branched curve with the line obtained by plottingthe l.h.s. of (25.6) or (25.7) as a function of k(E) · a. In this way one obtains the following general statements which are truefor a whole class of similar problems: Existence I : There is always at least one bound state. (This statement is true forsimilar problems in one and two dimensions, but not in three dimensions1. 1 In d=3 dimensions one can show, see below, that for so-called s-states, i.e., if the state does not depend on the angular coordinates ϑ and ϕ, the wave-function

226 25 One-dimensional Problems in Quantum MechanicsFor example, for the analogous three-dimensional “potential box model” forthe mutual binding of a neutron and proton in the deuteron nucleus the depthV0 is just deep enough to generate a bound state, whereas for a “di-neutron”it is just not deep enough.) Nodal theorem: The ground state, ψ0, has no “nodes” (i.e., no zeroes) atall (between the interval limits, i.e., here for −∞ < x < ∞). In contrast, aneigenstate ψn, for n = 1, 2, . . ., if existent, has exactly n nodes. If parity is a “good” quantum number, i.e., for symmetric potentials,V (x) ≡ V (−x), the following principle is additionally true: Alternating parity: The ground state, ψ(n=0), has even parity, the firstexcited state odd parity, the 2nd excited state again even parity, etc.. Existence II : Quantitatively one finds that the nth bound state, n = 1, 2, . . ., exists iffthe quantum well is sufficiently deep and broad, i.e., for the present case iff2m|V0|a2 > n· π . 2 225.2 Reflection and Transmission at Stepsin the Potential Energy; UnitarityFor simplicity we assume firstly thatV (x) ≡ 0 for x < 0 and ≡ ΔV (x)(> 0) for all x ≥ 0 ,with a barrier in a finite range including x = 0. Consider the reflection andtransmission of a monochromatic wave traveling from the left. We assumebelow that E is sufficiently high (e.g., E > V (∞) in Fig. 25.2, see below.Otherwise we have total reflection; this case can be treated separately.) We thus have, with ω := E/ :ψ(x, t) = A · ei(k−x−ωt) + r · ei(−k−x−ωt) for x < 0 , and = A · t · ei(k+x−ωt) for x > a ; (25.8)is quasi one-dimensional in the following sense: The auxiliary quantity w(r) :=r · ψ(r) satisfies the same “quasi one-dimensional” Schr¨odinger equation as notedabove, see the three equations beginning with (25.3). As a consequence, one onlyneeds to put x → r and u(x) → w(r), and can thus transfer the above “one-dimensional” results to three dimensions. But now one has to take into accountthat negative r values are not allowed and that w(0) =! 0 (remember: w(r) =r · ψ(r)). The one-dimensional solutions with even parity, i.e. u(x) ∝ cos kx, arethus unphysical for a “three-dimensional quantum box”, i.e. for V (r) = −V0 forr ≤ a, V ≡ 0 otherwise. In contrast, the solutions of odd parity, i.e., w(r) =r · ψ(r) ∝ sin kr, transfer to d=3. – This is a useful tip for similar problems inwritten examinations.

25.2 Reflection and Transmission at a Barrier; Unitarity 227k− and k+ are the wave numbers on the l.h.s. and r.h.s. of the barrier (seebelow).The amplitude A is usually replaced by 1, which does not lead to anyrestriction. The complex quantities r and t are the coefficients of reflectionand transmission (not yet the reflectivity R and transmittivity T , see below).The coefficients r and t follow in fact from the two continuity conditionsfor ψ(x) and dψ(x) . The reflectivity R(E) and the transmittivity T (E) them- dxselves are functions of r(E) and t(E), i.e., R = |r|2, T = k+ |t|2 . (25.9) k−The fraction k+ in the formula for T is the ratio of the velocities on the k−r.h.s. and l.h.s. of the barrier in the potential energy; i.e., E = 2k−2 ≡ 2k+2 + ΔV. (25.10) 2m 2mT can be directly calculated from R using the so-called unitarity relation2 R + T ≡ 1. (25.11)Fig. 25.2. Scattering of a plane wave by a barrier (schematically). A wave ∝ eik·xtraveling from −∞ with a certain energy meets a rectangular barrier, where it ispartially reflected and transmitted (indicated by the straight lines with arrows, i.e.,the corresponding complex amplitudes r and t are also associated with cosine-likebehavior). The conditions determining r and t are that the wave function and itsderivative are continuous. The velocities on each side are different. The energy isalso allowed to be higher than that of the barrier 2 The name unitarity relation follows by addition of an incoming wave from the right, i.e., the incoming wave is now a two-component vector with indices l and r (representing left and right, respectively). This is also the case with the outgoing wave. The incoming and outgoing waves are related, as can be shown, by a unitary matrix (the so-called S-matrix ), which generalizes (25.11).

228 25 One-dimensional Problems in Quantum Mechanics25.3 Probability CurrentAll these statements follow explicitly (with vˆ = m−1(pˆ − eAˆ )) from a gaugeinvariant relation for the probability current density: jw(r, t) := Re {ψ∗(r, t)vˆψ(r, t)} = 2im (ψ∗∇ψ − ψ∇ψ∗) (25.12) − e A|ψ|2 . mTogether with the scalar probability density w(r, t) = |ψ(r, t)|2 ,the current density jw satisfies (as one can show) the continuity equation ∂ w(r, t) + divj w (r, t) ≡ 0 . (25.13) ∂tAs shown in Part II in the context of electrodynamics, this continuity equa-tion is equivalent to the conservation theorem of the total probability: d3r w(r, t) ≡ 1, ∀t . ∞Ultimately it is this fundamental conservation theorem, which stands behindthe unitarity relation (25.11). For a series of different steps in potential energy the complex coefficientsrn and tn may be calculated sequentially. This gives rise to a so-called “trans-fer matrix”.25.4 TunnelingIn this section the probability of tunneling through a symmetrical rectangu-lar barrier of width a and height V0(> 0) will be considered. Assume thatV (x) = 0 for x < 0 and x > a, but V (x) = V0 for 0 ≤ x ≤ a; furthermoreassume that the energy E is smaller than the barrier height, i.e., 0 < E < V0(but see below!). Classically, in such a situation a particle will be elasticallyreflected at the barrier. In contrast, quantum mechanically, with the methodsoutlined above, it is straightforward to show that one obtains a finite tunnel-ing probability, given by T (E) = 1 . (25.14) 1 + 1 ( κ + k )2 sinh2(κa)) 4 k κ