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 following solutions of the Schr¨odinger equation u(x)=B (1) − exp(−κ|x|)+B (1) + exp(+κ|x|) , −x ≥ a, (25.3) u(x)=C 1 cos(k ·x)+C 2 sin(k · x) , |x| <a, (25.4) u(x)=B (2) − exp(−κx)+B (2) + exp(+κx) , +x ≥ a. (25.5) 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) − ; C 2 =0) and odd parity, i.e.,u(−x)=−u(x) , (⇒ B (1) − = −B (2) − ; C 1 =0). One then only needs one continuity condition, i.e. the one for u /u at x =+a. From this condition (see below) one also obtains the discrete energy values E = E n , for which continuity is possible (for k>0andκ>0): κ(E) k(E) =tan(k(E) ·a) , for even parity, (25.6) − k(E) κ(E) =tan(k(E) ·a) , for odd parity . (25.7) These equations can be solved graphically (this is a typical exercise), by plotting all branches of tan(k·a) as a function of k·a (these branches intercept the x-axis at k ·a = n ·π,wheren is integer, and afterwards they diverge to +∞ from −∞,atk ± ·a =(2n ± 1) ·π/2 ∓0 + ). Then one can determine the intersections of this multi-branched curve with the line obtained by plotting the 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 true for a whole class of similar problems: Existence I : There is always at least one bound state. (This statement is true for similar problems in one and two dimensions, but not in three dimensions 1 . 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 Mechanics For example, for the analogous three-dimensional “potential box model” for the mutual binding of a neutron and proton in the deuteron nucleus the depth V 0 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, ψ 0 , has no “nodes” (i.e., no zeroes) at all (between the interval limits, i.e., here for −∞ <x<∞). In contrast, an eigenstate ψ n ,forn =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) ,haseven parity, the first excited state odd parity, the 2nd excited state again even parity, etc Existence II : Quantitatively one finds that the nth bound state, n =1, 2, ,existsiff the quantum well is sufficiently deep and broad, i.e., for the present case iff 2m|V 0 |a 2 2 >n· π 2 . 25.2 Reflection and Transmission at Steps in the Potential Energy; Unitarity For simplicity we assume firstly that V (x) ≡ 0forx<0and ≡ ΔV (x)(> 0) for all x ≥ 0 , with a barrier in a finite range including x = 0. Consider the reflection and transmission of a monochromatic wave traveling from the left. We assume below that E is sufficiently high (e.g., E>V(∞) in Fig. 25.2, see below. Otherwisewehavetotal reflection; this case can be treated separately.) We thus have, with ω := E/: ψ(x, t)=A · e i(k − x−ωt) + r · e i(−k − x−ωt) for x<0 , and = A · t · e i(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 noted above, see the three equations beginning with (25.3). As a consequence, one only needs 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 account that 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,are thus unphysical for a “three-dimensional quantum box”, i.e. for V (r)=−V 0 for r ≤ 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 in written examinations. 25.2 Reflection and Transmission at a Barrier; Unitarity 227 k − and k + are the wave numbers on the l.h.s. and r.h.s. of the barrier (see below). The amplitude A is usually replaced by 1, which does not lead to any restriction. The complex quantities r and t are the coefficients of reflection and transmission (not yet the reflectivity R and transmittivity T ,seebelow). The coefficients r and t follow in fact from the two continuity conditions for ψ(x)and dψ(x) dx .Thereflectivity R(E)andthetransmittivity T (E)them- selves are functions of r(E)andt(E), i.e., R = |r| 2 ,T= k + k − |t| 2 . (25.9) The fraction k + k − in the formula for T is the ratio of the velocities on the r.h.s. and l.h.s. of the barrier in the potential energy; i.e., E = 2 k 2 − 2m ≡ 2 k 2 + 2m + ΔV. (25.10) T can be directly calculated from R using the so-called unitarity relation 2 R + T ≡ 1. (25.11) Fig. 25.2. Scattering of a plane wave by a barrier (schematically). A wave ∝ e ik·x traveling from −∞ with a certain energy meets a rectangular barrier, where it is partially reflected and transmitted (indicated by the straight lines with arrows, i.e., the corresponding complex amplitudes r and t are also associated with cosine-like behavior). The conditions determining r and t are that the wave function and its derivative are continuous. The velocities on each side are different. The energy is also 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 Mechanics 25.3 Probability Current All these statements follow explicitly (with ˆ v = m −1 ( ˆ p − e ˆ A)) from a gauge invariant relation for the probability current density: j w (r,t):=Re {ψ ∗ (r,t) ˆ vψ(r,t)} = 2im (ψ ∗ ∇ψ − ψ∇ψ ∗ ) − e m A|ψ| 2 . (25.12) Together with the scalar probability density w (r,t)=|ψ(r,t)| 2 , the current density j w satisfies (as one can show) the continuity equation ∂ w (r,t) ∂t +divj w (r,t) ≡ 0 . (25.13) As shown in Part II in the context of electrodynamics, this continuity equa- tion is equivalent to the conservation theorem of the total probability: ∞ d 3 r w (r,t) ≡ 1, ∀t. Ultimately it is this fundamental conservation theorem, which stands behind the unitarity relation (25.11). For a series of different steps in potential energy the complex coefficients r n and t n may be calculated sequentially. This gives rise to a so-called “trans- fer matrix”. 25.4 Tunneling In this section the probability of tunneling through a symmetrical rectangu- lar barrier of width a and height V 0 (> 0) will be considered. Assume that V (x)=0forx<0andx>a, but V (x)=V 0 for 0 ≤ x ≤ a;furthermore assume that the energy E is smaller than the barrier height, i.e., 0 <E<V 0 (but see below!). Classically, in such a situation a particle will be elastically reflected at the barrier. In contrast, quantum mechanically, with the methods outlined above, it is straightforward to show that one obtains a finite tunnel- ing probability, given by T (E)= 1 1+ 1 4 ( κ k + k κ ) 2 sinh 2 (κa)) . (25.14) 25.4 Tunneling 229 Here k(E)= 2m 2 E 1/2 , and κ(E)= 2m 2 (V 0 − E) 1/2 . For barriers with κa 1 the tunneling transmittivity is therefore exponen- tially small, T 1, but finite, as can be seen from sinh x =(1/2)(e x +e −x )=e x /2 for x 1: T (E)= 16 k κ + κ k 2 · e −2κa . (25.15) The factor in front of the exponential in (25.15) is of the order of O(4), if k and κ are comparable. Therefore the following result for a non-rectangular tunneling barrier is plausible (here we assume that V (x) >Eonly in the interval a<x<b). Then one obtains (as long as the result is 1): T (E)=O(4) ·exp −2 b a dx 2m 2 (V (x) − E) . (25.16) (The exponent in this expression is essentially proportional to the product of the width and the square-root of the height of the barrier. This yields a rough but systematic approximation for tunnelling through a barrier.) Furthermore, the factor 2 in the exponent and the factor O(4) in (25.16) have an obvious meaning. They result from the correspondence T ∝|ψ| 2 ; (i), the factor 2 in the exponent in front of the integral immediately follows from the exponent 2 in |ψ| 2 , and, (ii), the prefactor O(4) is obtained from the relation 4 = 2 2 by the fact that the wave-function must decay exponentially on both sides of the barrier. Furthermore, the reciprocal length appearing in the exponent of (25.16) is, as expected, proportional to , i.e., this decay length vanishes in the classical limit → 0. Quantum mechanical reflection at a depression in the potential energy, i.e., at a barrier with negative sign (e.g., a “quantum well” as in figure 25.1) is also of interest: We now assume an incoming plane wave with E>0(instead of the problem of bound states, E<0); in any case (as already mentioned) for V (x) we have the same situation as in Fig. 19.1, i.e. V ≡ 0forx<0 and x>a, but V (x) ≡−|V 0 | for 0 ≤ x ≤ a. For such a “quantum well” (as already stated) there is at least one bound state for E<0. On the other hand, classically an electron with E>0 does not “see” the quantum well. Quantum mechanically, however, even in this case there is a finite reflection probability. In fact the transmittivity T for this case is analogous to (25.14), again without proof: T (E)= 1 1+ 1 4 k 0 k + k k 0 2 sin 2 (k 0 · a) , (25.17) 230 25 One-dimensional Problems in Quantum Mechanics where k 0 (E)= 2m 2 (E + |V 0 |) . Complete transmission (T = 1) is only obtained for the special energy values k 0 · a = nπ (where n is an integer); otherwise T<1, i.e., there is a finite reflectivity. 26 The Harmonic Oscillator I This is one of the most important problems of quantum mechanics. The Hamilton operator is ˆ H = ˆp 2 2m + mω 2 0 ˆx 2 2 . (26.1) The harmonic oscillator is important inter alia because a potential energy V (x) in the vicinity of a local minimum can almost always be approximated by a parabola (which is the characteristic potential energy of the harmonic oscillator), V (x)=V 0 + 1 2 V (x 0 )(x −x 0 ) 2 + We thus have mω 2 0 ≡ V (x 0 ) , and it is assumed that anharmonicities, i.e., the correction terms of higher order, which are denoted by the dots, + , can be neglected. In contrast, the assumptions x 0 =0andV 0 = 0 impose no restrictions on generality. In the position representation (wave mechanics), we thus have to solve the Schr¨odinger equation − i ∂ψ(x, t) ∂t = ˆ Hψ(x, t) . Using the ansatz for stationary states ψ(x, t) ! = u(x) ·e −iEt/ , we obtain the following time-independent Schr¨odinger equation: u (x) ! = 2m 2 · mω 2 0 x 2 2 − 2m 2 E · u(x)= mω 0 2 − 2m 2 E ·u(x) (26.2) It is now advantageous to use reduced variables without physical dimen- sion, i.e., ε := E/(ω 0 /2),ξ := x/ /(mω 0 )and ˜u(ξ):=u(x) mω 0 . 232 26 The Harmonic Oscillator I Equation (26.2) is thus simplified to d 2 ˜u(ξ) dξ 2 =(ξ 2 − ε) · ˜u(ξ) , (26.3) and the normalization condition ∞ ∞ |u(x)| 2 dx ! = 1 becomes ∞ −∞ |˜u(ξ)| 2 dξ ! =1. Using the ansatz ˜u(ξ)=:v(ξ) · e −ξ 2 /2 the (asymptotically dominating) exponential behavior ∼ e −ξ 2 /2 can now be separated out; the differential equation for v(ξ) (the so-called Hermitian dif- ferential equation) is solved by means of a power series, v(ξ)= ∞ ν=0,1, a ν ξ ν . Finally after straightforward calculations, for the coefficients a ν the following recursion relation is obtained: a ν+2 a ν = 2ν +1− ε (ν +1)(ν +2) . (26.4) Although Schr¨odinger’s differential equation (26.3) is now satisfied, one must additionally demand that the recursion relation terminates at a finite ν(= ν 0 ) , i.e., ε =2ν 0 +1. If it were not terminated, i.e., if for all non-negative integers ν the identity ε =2ν + 1 were violated, then v(ξ) would diverge for |ξ|1(forevenparity, i.e., for [a 0 =0,a 1 = 0], the divergency would be asymptotically as ∼ e +ξ 2 , whereas for odd parity, i.e., for [a 1 =0,a 0 =0],v(ξ) ∼ ξ · e +ξ 2 ). This would violate the condition that u(ξ) = v(ξ) · e − ξ 2 2 must be square-integrable. In contrast, if there is termination at a finite ν, then the question of the convergence of the ratio a ν+2 /a ν becomes obsolete. In conclusion, iff the power series for v(ξ) terminates at a finite n, i.e. iff the reduced energy ε = E/ ω 0 /2 26 The Harmonic Oscillator I 233 is identical to one of the eigenvalues 2n +1with n =0, 1, 2, , ˜u(ξ) = v(ξ) · e −ξ 2 /2 is square-integrable. In this way the following energy levels for the harmonic oscillator are obtained: E n = ω 0 ·(n +1/2) , (26.5) for n =0, 1, 2, , with the eigenfunctions ˜u n (ξ) ∼ H n (ξ)e −ξ 2 /2 . (26.6) The H n (ξ) are the so-called Hermite polynomials; e.g., H 0 (ξ)=1,H 1 (ξ)=ξ, H 2 (ξ)=1− 2ξ 2 and H 3 (ξ)=ξ · 1 − 2 3 ξ 2 (only the first and second should be kept in mind). The missing normalization factors in (26.6) are also unimportant. Eigenfunctions of a Hermitian operator, corresponding to different eigen- values of that operator, are necessarily orthogonal, as can easily be shown. We thus have ˜u i |˜u j = u i |u j = δ i,j , as expected. The completeness 1 , i.e., the basis property of the function system {u n }, is also given, if all polynomial degrees n =0, 1, 2, are taken into account 2 . In fact, the probability that an oscillating particle of a given energy E n is found outside the “inner” region which is classically allowed, is very small, i.e., ∝ e −x 2 /x 2 0 , but finite even for large n. As we shall see in a later section (→ 28.1), the harmonic oscillator can also be treated in a purely abstract (i.e., algebraic) way. 1 We remember that the completeness property is not satisfied for the bound functions of an arbitary quantum well. 2 Here the following examination questions suggest themselves: (i) What do the eigenfunctions u n (x) of a harmonic oscillator look like (qualitatively!) for the following three cases: n =0;n =1;andn = 256? (ii) What would be obtained, classically and quantum mechanically, for the probability of a harmonically os- cillating particle of energy E to be found in a small interval Δx? 27 The Hydrogen Atom according to Schr¨odinger’s Wave Mechanics 27.1 Product Ansatz; the Radial Function The electron in a hydrogen atom is treated analogously to the previous cases. The Hamilton operator is written ˆ H = ˆ p 2 2m + V (r) . (27.1) The explicit expression for the potential energy, V (r)=− Z e 2 4πε 0 r , is not immediately important: one only requires that V (r) should be rota- tionally invariant. Next, the usual ansatz for stationary states is made: ψ(r,t)=u(r) · e −iEt/ . Then, in spherical coordinates a product ansatz (separation of variables)fol- lows: u(r)=R(r) ·Y lm (θ, ϕ) , with the spherical harmonics Y lm (θ, ϕ) (which have already been introduced in Part II 1 ). This product ansatz can be made essentially because the operators ˆ L 2 and ˆ L z commute with each other and (as conserved quantities) also with ˆ H, such that all three operators can be diagonalised simultaneously. As for the Y lm , here we only need the property that they are eigenfunctions of the (orbital) angular momentum operators ˆ L 2 and ˆ L z , i.e., ˆ L 2 Y lm (θ, ϕ)= 2 l · (l +1)Y lm (θ, ϕ) , (27.2) ˆ L z Y lm (θ, ϕ)=mY lm (θ, ϕ) . (27.3) 1 It is important to remind ourselves of other instances where the same or similar ideas or equations are used. . (26.1) The harmonic oscillator is important inter alia because a potential energy V (x) in the vicinity of a local minimum can almost always be approximated by a parabola (which is the characteristic. that V (r) should be rota- tionally invariant. Next, the usual ansatz for stationary states is made: ψ(r,t)=u(r) · e −iEt/ . Then, in spherical coordinates a product ansatz (separation of variables)fol- lows: u(r)=R(r). (qualitatively!) for the following three cases: n =0;n =1;andn = 256? (ii) What would be obtained, classically and quantum mechanically, for the probability of a harmonically os- cillating particle