3 Schroedinger Equation

Transcription

1 3. Schroedinger Equation 1 3 Schroedinger Equation We have already faced the fact that objects in nature posses a particle-wave duality. Our mission now is to describe the dynamics of such objects. When the De Broglie wave length is not important, the dynamics is described by the classical mechanics laws 1, but when it matters, classical mechanics does not work anymore. 3.1 Schroedinger equation Let us start from what we know. Objects as the electrons, can behave as a wave. We know a lot about waves, including the differential equation that represents the corresponding equations of motion. The general wave equation is given by 2 ψ 1 2 ψ = 0, (3.1) v 2 t2 where v is the velocity of the front wave and it is related to the wave length and frequency as v = λν = λω 2π. The wave function ψ = ψ( r, t), and an stationary solution is given in general by (3.2) ψ( r, t) = e iωt φ( r). (3.3) By substituting the above relations in the wave equation 2 and the De Broglie relation, we obtain 2 φ + p2 φ = 0. (3.4) 2 Now, since the energy of the particle is related to the momentum by E = p 2 /2m+V ( r), we finally deduce that 2 2m 2 ψ + (E V ( r))ψ = 0, (3.5) which is the Schroedinger equation. Observe that we have assumed that the relation between energy and momentum is still valid at the level of quantum mechanics, although 1 At this point we are not considering thermal or electromagnetic effects. 2 I will skip some intermediate steps. You must not. It is your duty to reproduce all of them.

2 3.1 Schroedinger equation 2 we still must understand what that means. Meanwhile, notice as well that some physical quantities are actually differential operators acting on the wave function. Here, this is reflected in momentum. Schroedinger equation reduces to Eφ = (p 2 /2m + V )φ by establishing that the momentum is actually a proper value of the complex differential operator i, this is i φ( r) = p φ( r), (3.6) or, if we want to write it as a proper value equation, it reads ˆ P φ( r) = p( r) φ( r), (3.7) where φ denotes a vector in a suitable vectorial space. The operator ˆ P acts on it. We must be very careful to distinguish between a wave function φ( r) and a vector representation of such wave function, denoted here as the ket 3 φ. There are many important points to stress out. First, notice that one is tempted to say that there are operators for each physical quantity we want to measure. For instance, we can say that the position of the particle is actually a proper value under the operator of position, denoted ˆX. Hence, we should have an equation of the form ˆX φ(x) = x φ(x). We shall see that indeed, this is the case. Second, by assuming the above proper value equation, it is straightforward to show that the commutator operator [ ˆX, ˆP ] does not vanish, since [ ˆX, ˆP ] φ(x) := ( ˆX ˆP ˆP ˆX) φ(x) = i φ(x), (3.8) a result which is always written as [ ˆX i, ˆP j ] = i δ ij, (3.9) where X = (X i, X 2, X 2 ) and similarly for P. So, in general, we write [ ˆ X, ˆ P ] = i Î, (3.10) 3 The notation for the vectors in the corresponding vectorial space, is called Dirac notation. It denotes vector as ψ. It is called a Ket vector. There are also Bra vectors, which are denoted as ψ. The reason is self-explained by noting that the internal product of two vectors is given by the bracket (bra-c-ket) ψ φ.

3 4. Infinite one-dimensional well 3 where Î is the identity operator represented by the unitary 3 3 matrix. This relation has a profound physical meaning. Naively we can see that measuring momentum and position depends on which quantity is first measured. Later on, we shall see that this is the cause that do not allows to measure with an infinity accuracy the position and momentum of a given particle. Schroedinger equation is then a wave equation. It gives us the dynamics of a quantum particle, and already has told us that measuring is not as trivial as we thought, but interacts with the physical properties we want to measure. Another feature we must make clear at this point, concerns tha physical meaning of the wave function ψ. According to our experience in previous experiments, all that we can say about ψ is that its squared is interpreted as the probability to find the particle in such place. By alone, ψ has no physical meaning, being a complex function. 4 Infinite one-dimensional well We are now ready to apply Schrödinger equation to a very simple, but important, system: the one-dimensional (1D) infinite well. Consider a particle immersed in a 1D box potential given by, x < 0; V (x) = 0, 0 < x < L,, x > L. (4.1) Hence, between the classically impenetrable wall at x = 0, L, the particle is free to move. But we know that under boundary conditions the wave-packet is going to be described by a Fourier series. This also can be deduced by looking for a solution of Schrödinger equation. Let us first of all consider a more general case, in which the particle is immersed in a finite well potential. This is V 0, x < 0; V (x) = 0, 0 < x < L, V 0, x > L. (4.2)

4 4. Infinite one-dimensional well 4 The one-dimensional Schrödinger equation reads ψ + 2m (E V )ψ = 0, (4.3) 2 and let us define the quantities κ 2 = 2m 2 E, q2 = 2m 2 (V 0 E)ψ, 0 < E < V 0. (4.4) There are 3 regions determined by the three values of the potential V (x). In the region I, for x > 0, the Schrödinger equation is given by ψ 1 q 2 ψ 1 = 0, (4.5) and its solution is ψ 1 = A 1 e qx + B 1 e qx. (4.6) For the region 2, i.e., for 0 < x < L, the corresponding equation is ψ 2 + κ 2 ψ 2 = 0, which solution is ψ 2 = A 2 sin κx + B 2 cos κx, (4.7) while in region 3, the solution reads ψ 3 = A 3 e q(x a) + B 3 e q(x a). (4.8) Now, not all solutions have physical meaning. For instance, in region 1 x can grow negatively till, for which ψ 1 is not bounded. That restricts the coefficient A 1 to vanish. SImialrly, A 3 = 0. Therefore, the set of solutions are ψ 1 = B 1 e qx, ψ 2 = A 2 sin κx + B 2 cos κx, ψ 3 = B 3 e q(x a). (4.9) However, there must be a unique and continuous solution all over the 3 regions. That means that we must match these solutions on the boundary points. The continuity conditions are that ψ 1 (0) = ψ 2 (0), ψ 2 (L) = ψ 3 (L), ψ 1(0) = ψ 2(0), ψ 2(L) = ψ 3(L). (4.10)

5 4. Infinite one-dimensional well 5 Let us now go back to the infinite well. In that case, ψ 1 and ψ 3 are not bounded, from which both must vanish. From the continuity conditions we get that, ψ 2 (0) = B 2 = 0, ψ 2 (L) = 0 = A 2 sin κx + B 2 cos κx, (4.11) and the solution reduces to ψ 2 = A 2 sin κx. From the boundary condition on L it follows that sin κl = 0 fro which κ = nπ L, n Z, (4.12) where n 0. Therefore, the energy is given by E n = π2 2 2mL 2 n2. (4.13) This tells us that the energy spectrum is discrete and that the particle will have energy in packets. The wave functions are given then by ψ n (x) = A n sin nπ L x, (4.14) where the coefficient A n needs to be fixed. This is accomplished by the normalization procedure ψ n 2 dx = 1. (4.15) In the present case this reduces to L 0 ψ 2 dx = L 2 A n 2 = 1, (4.16) which implies that A n = 2/L. Finally, the wave function is given by 2 ψ(x) = L sin nπ x. (4.17) L