4. Important theorems in quantum mechanics

Transcription

1 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 1 TILLEGG 4 4. Important theorems in quantum mechanics Before attacking three-imensional potentials in the next chapter, we shall in chapter 4 of this course iscuss a few important quantum-mechanical theorems. 4.1 Compatible observables, simultaneous eigenfunctions an commuting operators When two observables of a system can have sharp values simultaenously, we say that these two observables are compatible. As an example, if we consier a free particle with mass m, then the momentum p x an the kinetic energy K x = p 2 x/2m are compatible observables: In the state ψ p (x) = (2π h) 1/2 e ipx/ h, both p x an K x are sharp, with the values p an p 2 /2m: p x ψ p = p ψ p, K x ψ p = p2 2m ψ p. ( K x = p2 x 2m = h2 2m 2 x 2 ) We then say that ψ p (x) is a simultaneous eigenfunction; it is simultaneously an eigenfunction of both p x an K x. We shall see that this implies that the two operators must commute; [ p x, K x ] = 0. It is of course easy to check the last relation irectly, but it is more interesting to show that it oes in fact follow from the compatibility of the corresponing observables, as we shall now see. Suppose that the observables F an G are compatible. From the example above we then unerstan that there must exist a simultaneous set of eigenfunctions of the corresponing operators: F Ψ n = f n Ψ n, Ĝ Ψ n = g n Ψ n. (T4.1) It is then easy to prove the following rule: A. If the observables F an G are compatible, that is, if there exists a simultaneous set of eigenfunctions of the operators F an Ĝ, then these operators must commute; (T4.2) [ F, Ĝ] = 0. This rule can also be turne aroun : B. If the operators F an Ĝ commute, then it is possible to fin a simultaneous set of eigenfunctions. (T4.3)

2 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 2 From rule A it follows irectly that C. If the operators F an Ĝ o not commute, that is, if [ F, Ĝ] 0, then there oes not exist a common set of eigenfunctions of the two operators. This means that the corresponing observables can not have sharp values simultaneously; they are not compatible. (T4.4) Proof of rule A: When the two observables F an G are compatible, we must as mentione have a simultaneous set of eigenfunctions; cf (T4.1). From these eigenvalue equations it follows that ( F Ĝ Ĝ F )Ψ n = (f n g n g n f n )Ψ n = 0 (for all n). Because the operators are hermitian, the set Ψ n is complete. Then, for an arbitrary function Ψ = n c n Ψ n we have: ( F Ĝ Ĝ F )Ψ = n We have now prove the following operator ientity: F Ĝ Ĝ F [ F, Ĝ] = 0, q.e.. c n ( F Ĝ Ĝ F )Ψ n = 0. Proof of rule B: Let Ψ n be one of the eigenfunctions of the operator F, with the eigenvalue f: F Ψ n = f Ψ n. Since the two operators F an Ĝ are suppose to commute, we then have F (Ĝ Ψ n) = Ĝ F Ψ n = f(ĝ Ψ n). This equation states that the function ĜΨ n is an eigenfunction of the operator F with the eigenvalue f, together with the function Ψ n. For the remaning part of this proof we have to istinguish between two possibilities: (i) One of the possibilities is that the eigenvalue f of the operator F is non-egenerate, which means that there is only one eigenfunction with this eigenvalue. In that case the function ĜΨ n can iffer from the function Ψ n only by a constant factor g: Ĝ Ψ n = g Ψ n. This equation expresses that Ψ n is an eigenfunction also of Ĝ, an that is what we wante to prove. (ii) The other possibility is that the eigenvalue f is egenerate, so that there is more than one egenfunction of F with this eigenvalue. The proof of rule B then is a little bit more complicate, as you can see in section 4.1 in Hemmer, or in section 5.4 in B&J. Example 1 From the important commutator [x, p x ] = i h an rule C it follows that the observables x an p x can not have sharp values simultaneously; they are not compatible. This is is also expresse in Heisenberg s uncertainty relation, x p x 1 2 h. On the other han, y an p x are compatible, since [y, p x ] = 0. (T4.5)

3 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 3 Example 2 We saw in Tillegg 2 that the Cartesian components of the angular momentum operator L = r p o not commute; they satisfy the so-calle angular-momentum algebra, which consists of the commutator relation [ L x, L y ] = i h L z, (T4.6) together with two aitional relations which follow from (T4.6) by cyclic interchange of the inices. This means that the components of the observable L are not compatible. This has the following consequence: Even if the size L of the angular momentum has a sharp value, the components can not be sharp simultaneously. This means that accoring to quantum mechanics the irection of L cannot be sharply efine, contrary to what is the case in classical mechanics. We have also seen that each of the Cartesian components L x etc commute with the square L 2 of the angular-momentum operator, [ L 2, L x ] = [ L 2, L y ] = [ L 2, L z ] = 0. (T4.7) Accoring to the rules above, this means that L 2 is compatible with for example L z. Thus it is possible to fin simultaneous eigenfunctions of these two operators, as we shall see in the next chapter. 4.2 Parity The concept of parity is a simple an important remey of quantum mechanics. The parity operator an its eigenvalues: When the parity operator P acts on a function ψ(r), its effect is efine as Pψ(r) = ψ( r). Thus the parity operator correspons to reflection with respect to the origin, or space inversion, as we also call it. We have previously seen examples of wave functions that are either symmetric or antisymmetric with respect to the origin, e.g. for the harmonic oscillator; ψ( x) = ±ψ(x). Such examples occur also in three imensions: ψ( r) = ±ψ(r). For such a function, Pψ(r) = ψ( r) = ±ψ(r). A function which is symmetric (or antisymmetric) with respect to space inversion thus is an eigenfunction of the parity operator with the eigenvalue +1 (or 1). This eigenvalue is calle the parity of the wave function. It turns out that the eigenvalues ±1 are the only possible ones, which means that the spectrum of the parity operator is ±1. Proof: Let ψ be an eigenfunction of P with the eigenvalue p, Pψ(r) = pψ(r).

4 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 4 Then P 2 ψ(r) = P( Pψ(r)) = p Pψ(r) = p 2 ψ(r). Here, the operator P 2 correspons to two space inversions; we have that P 2 ψ(r) = Pψ( r) = ψ(r). This means that ψ(r) = p 2 ψ(r), that is, p 2 = 1. Thus, the only possible eigenevalues of P are { +1 (even parity) p = 1 (o parity) q.e.. For a symmetric potential, [ P, Ĥ] = 0 For a syymetric potential, PV (r) = V ( r) = V (r), we note that the Hamiltonian is left unchange by the parity operator. This means that the parity operator commutes with the Hamiltonian: V ( r) = V (r) = PĤ(r) = Ĥ( r) = Ĥ(r) = [ P, Ĥ] = 0. Proof: We have for an arbitrary function ψ(r) [ P, Ĥ(r)]ψ(r) = PĤ(r)ψ(r) Ĥ(r) Pψ(r) = Ĥ( r)ψ( r) Ĥ(r)ψ( r) = 0, q.e.. Accoring to rule B above, it is then possible to fin simultaneous eigenstates of the two operators P an Ĥ, that is, energy eigenfunctions ψ E(r) with well-efine parity (i.e, symmetric or antisymmetric): Following the proceure in the proof of rule B we have that Ĥ( Pψ E ) = PĤψ E = E( Pψ E ). Thus Pψ E is an eigenfunction with energy E, together with ψ E : (i) If this energy is non-egenerate, the function Pψ E must be proportinal to ψ E, Pψ E = pψ E. Here p is either +1 or 1, since these are the only possible eigenvalues of P. Thus, for the non-egenerate energy E the eigenfunction ψ E necessarily has a well-efine parity; it is either symmetric or antisymmetric. (ii) The other possibility is that the energy E is egenerate, with several energy eigenfunctions ψ Ei (i = 1,, g). Then these eigenfunctions o not nee to have well-efine parities, but rule B tells us that they can be combine linearly into parity eigenstates. To remember the contents of the above iscussion, it is sufficient to remember a couple of well-known examples: (i) the harmonic oscillator has non-egenerate energy levels an eigenstates with wellefine parity,

5 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 5 (ii) the free particle. In this case we can choose to work with symmetric or antisymmetric solutions (cos kx an sin kx), but we can also use asymmetric solutions (e ±ikx ). A thir example is the square well V (x) = { 0 for x < l, V 0 for x > l. For 0 < E < V 0, the energy levels are iscrete an non-egenerate, an the energy eigenfunctions have well-efine parity, as we have seen. For E > V 0, we have two inepenent eigenfunctions for each energy. Here, accoring to rule B, these can be chosen so that one is symmetric an the other one is antisymmetric. However, it is more interesting to work with an asymmetric linear combination of these, on the form Ae ikx + Be ikx for x < l, ψ E (x) = De iqx + F e iqx for l < x < l, Ce ikx for x > l. (T4.8) This is calle a scattering solution, because it contains an incoming an a reflecte wave to the left of the well, an a transmitte wave on the right. (It can be shown that the probability of transmission is T = C/A 2.) 4.3 Time evelopment of expectation values The time evelopment of a state Ψ(r, t) is etermine by the Schröinger equation, which of course also etermines how the expectation values of the various observables, F Ψ = Ψ F Ψ τ, behave as functions of t. It turns out that certain aspects of this time epenence can be clarifie without solving the Schröinger equation explicitly. This can be shown by taking the erivative (/t) of the left an right sies of the equation above. On the right, we can move the ifferentiation insie the integral, provie that we take the partial erivative with respect to t, that is, keep the other variables fixe. The partial erivative of the integran is t [Ψ ( F Ψ)] = Ψ t F Ψ + Ψ ( F Ψ). t If F oes not epen explicitly on t, it acts as a constant uner the last ifferentiation. If it oes epen on t, the last ifferentiation gives two terms, so that the result becomes Ψ t F Ψ = t Using the Schröinger equation we have F Ψτ + Ψ Ψ F t τ + Ψ F t Ψ τ. Ψ t = ī ĤΨ. h

6 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 6 The first an the secon terms in the above equation then become respectively ( ī ) h ĤΨ F Ψ τ = ī h Ψ Ĥ Ψ τ an ī h The resulting formula for the time evelopment of expectation values then is t F Ψ = ī [Ĥ, F ] F h + Ψ t Ψ F Ĥ Ψ τ.. (T4.9) Ψ Here, F will usually not epen explicitly on t, so that we can remove the last term. If F, in aition to being time inepenent, also commutes with Ĥ, that is, if the observable F is compatible with the energy E, we fin the interesting result that t F Ψ = 0. In such a case F Ψ thus is time inepenent, an we note that this hols for any state Ψ of the system. We then say that the observable F is a quantum-mechanical constant of the motion. All observables which are compatible with the energy thus are constants of motion. (Note the efinition of a constant of motion: It is the expectation value of the observable that stays constant.) Example 1 An example is the energy itself. With F = E we fin that t E Ψ = ī [Ĥ, h Ĥ] = 0. Ψ This hols provie that Ĥ/ t = 0, that is, if we have a conservative system. For such a conservative system we can thus state that the energy is a constant of the motion for an arbitrary state, whether it is stationary or nonstationary. Example 2: Parity conservation For a symmetric potential we have seen that [Ĥ, P] = 0. Then it follows that the parity is a constant of the motion, in the sense that P Ψ = Ψ P Ψ τ is time inepenent. As a special case we can assume that the state is symmetric (or antisymmetric) at some initial time, so that this integral is equal to 1 (or 1). The parity conservation then means that the symmetry is conserve for all times (also in cases where the state is non-stationary).

7 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 7 Example 3 For a free particle, with Ĥ = p 2 /2m, we have [Ĥ, p] = 0, so that (/t) p Ψ = 0. This means that the momentum is a constant of the motion (that is, p Ψ =constant) for an arbitrary time-epenent free-paticle state Ψ(r, t). This is the quantummechanical parallel to Newton s first law. How will the expectation value r Ψ of the position behave for a free-particle wave Ψ(r, t)? The answer is that t r Ψ = p Ψ m, which is the quantum-mechanical parallel to the classical relation r/t = v = p/m. It turns out that this relation hols also for a particle in a potential V (r). This is a part of Ehrenfest s theorem, which has more to say about the connection between classical an quantum mechanics: 4.4 Ehrenfest s theorem (The Ehrenfests theorem is escribe in section 5.11 in B&J, an in section 4.4 in Hemmer.) We shall see that (T4.9) can also be use to erive the Ehrenfests theorem. This therem is concerne with the connection between classical an quantum mechanics, or with the so-calle classical limit of the latter, if you like. We know that classical mechanics works perfectly for macroscopic objects (with certain exceptions; cf superfluiity), but we also know that this theory fails when applie to the really small things in nature, like molecules, atoms an sub-atomic particles. On the other han, the ynamics of these sub-microscopic objects is is escribe perfectly by quantum mechanics. The question then is: What happens if we apply quantum mechanics to larger objects? Can it be that it fails when the objects become large? The answer is that quantum mechanics has no such limitation. When the scale is increase, it turns out that the quantum-mechanical esription converges towars the classical one (except for a few cases where the classical esription in fact fails; cf superfluiity an superconuctivity, where quantum effects are observe even on macroscopic scales). We can therefore state that quantum mechanics contains classical mechanics as a limiting case. This graual convergence of quantum an classical mechanics can to some extent be illustrate by the Ehrenfest theorem. Let us consier a particle moving in a potential V (r). Classically the motion is escribe by the equations p t = V, r t = p. (Newton) (T4.10) m In quantum mechanics we escribe this motion in terms of a wave function Ψ(r, t) a wave packet which follows the particle on its journey in the potential V (r). What must then be compare with the classical position r an the momentum p are the quantum-mechanical expectation values r Ψ = Ψ r Ψ 3 r an p Ψ = Ψ p Ψ 3 r. (T4.11)

8 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 8 If the particle is heavy (macroskopic), we must expect that these expectation values behave exactly like the classical quantities in (T4.10). For microscopic particles, on the other han, we cannot expect (T4.10) to be more than a very poor approximation to (T4.11). We shall now investigate if this actually is the case. From (T4.9) we have: t r = ī h [H, r] an t p = ī h [H, p], (T4.12) where 1 Ĥ = p 2 + V (r, t). 2m We note that r commutes with the potential, an that p commutes with the kinetic term p 2 /2m. Thus we nee the commutators [ p 2, r] an [V, p]. In the calculation of these an other commutators we can use the general rules (T2.24): [Â, BĈ] = [Â, B]Ĉ + B[Â, Ĉ] [Â B, Ĉ] = Â[ B, Ĉ] + [Â, Ĉ] B. We then fin that so that an Furthermore, so that [ p 2 x, x] = p x [ p x, x] + [ p x, x] p x = 2i h p x, 2 p [Ĥ, x] = [ 2m, x] = 1 2m [ p2 x, x] = i h p x m, p [Ĥ, r] = i h m. [V, ]Ψ = V Ψ (V Ψ) = ( V )Ψ, [Ĥ, p] = [V, p] = i h V. Inserting into (T4.12) we get the two equations (T4.13) (T4.14) (T4.15) (T4.16) (T4.17) which together are known as Ehrenfest s teorem: t r Ψ = p Ψ m, (T4.18) t p Ψ = V Ψ, (T4.19) The quantum-mechanical equations of motion for r an p are foun by replacing the quantities in the classical equations (T4.10) by their expectation values. Here, it may look as if the expectation values follow the same trajectories as the classical orbits r an p, but it is not quite as simple as that: The trajectories are the same if the right-han sie of (T4.19), which is the expectation value of the force F = V, is equal to the force precisely at the mean position r. In general, however, this conition is not 1 We can allow the potential to be time epenent.

9 TFY4215 Kjemisk fysikk og kvantemekanikk - Tillegg 4 9 satisfie. This is seen by expaning the force in a Taylor series aroun the mean position. In one imension, we have F (x) = F ( x ) + (x x )F ( x ) (x x )2 F ( x ) +, so that the right sie of (T4.19) correspons to F (x) = F ( x ) (x x ) 2 F ( x ) + = F ( x ) ( x)2 F ( x ) +. (T4.20) We can then summarize as follows: (i) For macroscopic particles, x an p x will be negligible compare to x an p x. At the same time, the variation of F over the extension of the wave packet will be insignificant, so that F (x) = F ( x ), to a very goo approximation. In this case, the expectation values x an p x satisfy Newton s equation (T4.10); thus classical mechanics is a limiting case of quantum mechanics. (ii) Even for microscopic particles we are in some cases allowe to set F (x) = F ( x ) an to use the classical equations. This requires that the variation of the potential over the wave packet can be neglecte. This is in fact very well satisfie e.g. in particle accelerators. (iii) The conition F (x) = F ( x ) can also be satisfie for particles moving in potentials of sub-microscopic extension. This requires that F (x) = 0, that is, the force must either be constant (as in the gravitational fiel) or vary linearly, as in the harmonic oscillator. For these kins of potentials we can thus conclue that x an p x will follow classical trajectories. In an oscillator, e.g., x will oscillate harmonically (as cos ωt).