MATH3385/5385. Quantum Mechanics. Handout # 5: Eigenstates of the Hamiltonian and the One-dimensional Stationary Schr¨odinger Equation Time-independent potentials In the case of a structureless particle subject to a potential V (r,t) (possibly time-dependent!) we have seen in Handout # 3 that the Hamiltonian can be represented by the differential operator: ~2 H = 2 + V (r,t) −2m∇ (i.e. from the replacement of the momentum vector p by the operator i~∇ in the classical b − Hamiltonian function H(r, p; t)). If now the potential V is actually time-independent, i.e. V (r,t)= V (r) only depending on the position vector r, then we can look for solutions of the Schr¨odinger (SE) equation in the separated form: ψ(r,t)= χ(t)φ(r) which inserted in the SE leads to 1 dχ 1 i~ = Hφ = E χ dt φ where the l.h.s. of the equation depends only onb t, whereas the r.h.s. depends only on r. Thus, both sides must be equal to a constant E, and we can thus separate the equations into Hφ = Eφ the time independent SE (1a) − dχ i~ = Eχ (1b) bdt The second equation (1b) can be readily solved to give: iEt/~ χ(t)= e− χ(0) leading to a simple time-evolution1. The first equation (1a), i.e. the stationary or time- independent Schr¨odinger equation, which for a single structureless particle reads: ~2 2φ + V (r)φ = Eφ (2) −2m∇ has the form of an eigenvalue problem –with eigenvalue E, the energy– of the differen- tial operator H. Solving this eigenvalue (or “spectral”) problem under certain boundary conditions will lead to a possible set of values that E can take, i.e. the eigenvalues of the b 1Obviously, the separated solution ψ = χφ is not the general solution of the time-dependent Schr¨odinger equation, but only a particular solution: see Handout # 4. 1 operator under the boundary condtions, and this set of values is called the spectrum2. From now on we will concentrate, in given problems, in developing the techniques to calculate the spectrum of the Hamiltonian: once we know this and the corresponding eigenfunctions φ = φE(r), the quantum mechanical problem is practically solved. All physical quantities (such as expectation values of observables) can then in principle be calculated. Since the Hamiltonian operator is Hermitian H† = H, the eigenfunctions for different eigenvalues are orthogonal w.r.t. the inner product: b b (φ , φ ′ )= dr φ∗ (r)φ ′ (r) = 0 provided E = E′ . E E E E 6 Z Recalling further that if φE(r) is a solution of the eigenvalue problem (1a), e.g. (2), with eigenvalue E, then so is any constant multiple cφE(r). We can encounter then the following situations: the spectrum is discrete: the eigenvalues of the stationary SE take on a discontinuous • set of values En, n = 1, 2,... In this case the corresponding eigenfunctions φn(r)= φEn (r) can be normalised, i.e. φ 2= dr φ (r) 2 = 1 k E k | E | Z The corresponding physical states of the system are called bound states. the spectrum is continuous: the eigenvalues can take on a continuous range of values, • say in an interval E0 <E<E1. In that case the eigenvalues cannot be normalised, but instead we can choose c to have an orthogonality relation of the form dr φ∗ (r)φ ′ (r)= δ(E E′) E E − Z The corresponding physical states of the system are called scattering states. The completeness relation in this case can be written as: E1 φ (r)φ∗ (r′) dE = δ(r r′) . E E − ZE0 the spectrum is partly discrete and partly continuous. In this mixed situation there • are in the system bound states as well as scattering states. It can happen that we have different (i.e. independent) eigenfunctions for one and the same eigenvalue E: in that case we say that the corresponding “energy level” is degenerate. The corresponding eigenspace is then no longer one-dimensional as would be the case if the eigenvalue were nondegenerate (i.e. when the eigenfunction is unique up to a multiplicative constant). In the degnerate case we need additional “quantum numbers” to characterise the eigenstate uniquely. As we have noted in Handout # 4, this means that we need to look for other physical observables whose operators commute with the Hamiltonian operator H, and consequently for which we can then impose an additional eigenvalue problem consistent with the one for the Hamiltonian. 2 b The name, which has become customary in mathematics, is clearly inspired by physics: it is the set of values of E that corresponds to the measured spectra of atoms in experiments. 2 Example 1: particle in an infinite potential well As a first example of a system with a discrete spectrum, i.e. physically representing bound states, we consider the case of a one-dimensional particle in an infinite potential well given by: 0 , a x a V (x)= − ≤ ≤ otherwise ∞ In view of the fact that outside the interval a x a the potential is infinite (and − ≤ ≤ thus we have a potential “barrier” inhibiting the particle from being found anywhere else than within the mentioned interval) we solve the eigenvalue problem by imposing: ~2 d2φ = Eφ , a x a −2m dx2 − ≤ ≤ φ(x) = 0 , x a or x a ≤− ≥ implying that the particle has zero probability of being found outside the mentioned in- terval. So, concentrating on the problem within the interval [ a, a], we note that we have − a simple linear homogenous differential equation with constant coefficients to solve. We distinguish two cases: the case that E > 0 and the one when E 0. ≤ E > 0 In this case the stationary Schr¨odinger equation takes the form: 2 d φ 2 iκx iκx + κ φ = 0 φ(x)= Ae + Be− , (3) dx2 ⇒ with κ = √2mE/~ (which is real in this case). imposing continuity of the eigenfunction we have that at the boundaries of the interval, at x = a the wave function vanishes: ± Aeiκa + Be iκa = 0 φ( a) = 0 − ± ⇒ Ae iκa + Beiκa = 0 − This coupled set of equations has only a nontrivial solution for the coefficients A and B if the following condition holds: 2iκa 2iκa e e− = 2i sin(κa) = 0 2κa = nπ (n N) . − ⇒ ∈ Thus, we find the following energy levels: ~2 n2π2 E = E = (4) n 2m 4a2 associated with the following eigenfunctions: 2iA sin( nπx ) , n even φ (x)= n 2a (5) n 2A cos( nπx ) , n odd n 2a The coefficients An are determined by the normalisation of the eigenfunctions: a a 2 2 nπx 2 2 2 nπx 2 4 An sin ( ) dx =1 resp. 4 An cos ( ) dx = 1 | | a 2a | | a 2a Z− Z− 3 leading in both cases to A 2 = 1/(4a) . Thus, we can take for the normalised eigenfunc- | n| tions: 1 nπx √a sin( 2a ) , n even φn(x)= 1 nπx (6) ( √a cos( 2a ) , n odd Exercise: Check that these eigenfunctions are orthogonal, i.e. that we have: a φn∗ (x)φm(x) dx = 0 , n = m . a 6 Z− E 0 ≤ In this case κ is purely imaginary, and we have instead to introduce the real quantity: √ 2mE k − ≡ ~ in which case the stationary Schr¨odinger equation takes the form: 2 d φ 2 kx kx k φ = 0 φ(x)= αe + βe− . (7) dx2 − ⇒ Again imposing continuity of the eigenfunction at the boundaries of the interval, the wave function vanishes: ka ka αe + βe− = 0 φ( a) = 0 ka ka ± ⇒ αe− + βe = 0 which only leads to a nontrivial solution for α, β if e4ka = 1 k = 0 . ⇒ Thus, in this case only the eigenvalue E = 0 is allowed, from which we conclude that there are no negative eigenstates! Finally, we have also to reject the eigenvalue E = 0, because it is easy to see that if k = 0 the above boundary conditions only lead to the case α = β = 0 which would lead to the trivial function φ(x) = 0 which is to be rejected as an eigenfunction. Thus, we can draw the following conclusions about the spectrum for the infinite potential well: ~2 n2π2 Eigenvalues: En = 2m 4a2 , (n positive integer) 1 nπx √a sin( 2a ) , n even 1 nπx √a cos( 2a ) , n odd Eigenfunctions: φn(x)= if : a<x<a − 0 otherwise 4 Remark: Since we have only a discrete spectrum in this example, we say that the particle can only exist in bound states, meaning that the particle is confined to such a discrete energy state, unless it gets a “kick” of enough energy to jump to the next state. In Example 1 we have investigated a siuation where the potential is unbounded, namely V = at the boundaries of the interval [ a, a]. In that case the derivatives of the solution ∞ − φ at the boundaries of the infinite potential well do not play a role. In the examples that follow now, V (x) is discontinuous but bounded in a neighborhood of the discontinuity. In that case we have to consider also the boundary values for the derivative φ′(x). In fact, from the stationary SE we have: ǫ d2φ dφ dφ ǫ V (x) E dx = = φ(x) dx =0 as ǫ 0 , 2 ~2 − ǫ dx dx x=ǫ − dx x= ǫ ǫ /(2m) → Z− − Z− whenever V (x) is bounded around x = 0 and since φ(x) in the last integrand is continuous, implying continuity of φ′(x) at x = 0. Example 2: Finite one-dimensional well In example 1, we have seen a situation where the spectrum is discrete leading to bound states only, whereas in example 2 we have seen the other extreme situation, where the spectrum is fully continuous leading to scattering states characterised by the reflection and transmission coefficients.