Schrodinger's Theory of Quantum mechanics - Lecture Notes | PHYS 321, Study notes of Physics

Download Schrodinger's Theory of Quantum mechanics - Lecture Notes | PHYS 321 and more Study notes Physics in PDF only on Docsity! 1 PHYSICS 321 CHAPTER 5 SCHRÖDINGER’S THEORY OF QUANTUM MECHANICS 5-1. INTRODUCTION The foregoing experiments and models demonstrated wave-particle duality conclusively, but they only dealt with specific situations. Schrödinger’s equation offers a general recipe to predict the behavior of the particles of any microscopic system: For each system, it specifies the equation that the wavefunction of the system must satisfy, and the connection between the behavior of the wavefunction and the behavior of the particles. Naturally, it reduces to Newtonian physics in the macroscopic limit, just as the theory of relativity reduces to Newtonian physics in the limit c → ∞. After introducing Schrödinger’s equation, we will identify some of its essential points and use them to treat several important situations such as transmission through and reflection from particle waves incident on various potential barriers, and tunneling, which is a purely quantum mechanical phenomenon. Then we shall treat the structure and properties of atoms in detail, to be followed by a relatively brief treatment of molecules. Over time, the student will find that (s)he is developing a “quantum mechanical intuition,” which will bear a striking resemblance to the intuition on EM phenomena, for the obvious reason that in both cases we will are dealing with waves. However, the student will have to be patient as this exposition will unfold gradually… We start from de Broglie’s postulate, which states that the motion of a microscopic particle is governed by the propagation of an associated wave, but it does not provide the law(s) governing that propagation. It does predict the wavelength λ successfully, but only when λ is essentially constant. Hence we first describe Schrödinger’s equation (1925), which determines the behavior of any wave in terms of the wavefunction (wf), if we provide the potential energy function for the particle associated with the wave, and the initial and boundary conditions of that wave-particle. Then we shall describe the relation between the wf and the associated particle, developed by Max Born (1926). While particular solutions to Schrödinger’s equation for some particular potential energy functions and boundary conditions are described in the next chapter(s), some general features of the solutions are described in this chapter, including quantization of energy and other properties. To gain some clues as to the required nature of the equation, we revisit the free particle. In Chapter 3, we used a simple sinusoidal traveling wave Ψ(x,t) = sin2π(x/λ – νt) (5-1) or a linear combination of such simple sinusoidal waves to describe the wf of a free particle. But we arrived at this wf by guessing, based on the constant linear momentum p which entailed a 2 constant λ. But what do we do when a force is acting on a particle? Generally, p will change with position and time. The problem is that λ is not even well-defined if it changes very rapidly. For example, in Fig. 5-1 the separation between adjacent maxima is not equal to that between adjacent minima. The general solution to this problem is the Schrödinger equation, which is a partial linear differential equation whose solution is the wf Ψ(x,t). As a first step, we practice our partial derivative skills: Example 5-1. Evaluate the partial derivatives up to second order of the sinusoidal wf given in Eq. (5-1) w.r.t. x and t. Solution: We recall that a partial derivative w.r.t. x is evaluated by treating all the other variables as fixed. Hence, if Ψ(x,t) = sin2π(x/λ – νt) = sin(kx – ωt) (5-2) then ∂Ψ(x,t)/∂x = kcos(kx – ωt), ∂2Ψ(x,t)/∂x2 = -k2sin(kx – ωt), ∂Ψ(x,t)/∂t = -ωcos(kx – ωt), ∂2Ψ(x,t)/∂t2 = -ω2sin(kx – ωt). (5-3) In solving Schrödinger’s equation for different potential or boundary and initial conditions, we shall often see that the solution is obtained by separation of variables: If the solution can be written as the product of single-variable functions, then the partial differential equation separates into ordinary (i.e., single-variable) equations. 5-2. PLAUSIBILITY ARGUMENT LEADING TO SCHRÖDINGER’S EQUATION We are now looking for the equation that is the quantum mechanical analog of Newton’s equation of motion F = dp/dt = md2x/dt2 (5-4) or of Maxwell’s equation ∂Ex/∂x + ∂Ey/∂y + ∂Ez/∂z = ρ/ε0 (5-5) which can also be written as ∇•E = ρ/ε0 (5-6) and is equivalent to ∫E•dA = Qenc/ε0 (5-7) by Gauss’s Theorem. However, while the wave equation for a stretched string can be derived from Newton’s law, and the EM wave equation can be derived from Maxwell’s equations, we cannot expect to derive the quantum mechanical wave equation from classical physics. Yet we will get help from the de Broglie-Einstein relations λ = h/p and ν = E/h. (5-8) 5 Note the following: 1. We were led to this equation by treating the special case of the free particle. Hence, we postulate it to be true for any V(x,t). Hence, this is the quantum-mechanical wave equation whose solution Ψ(x,t) is the wavefunction associated with a particle of mass m subject to forces resulting from the potential energy V(x,t). The validity of the postulate is, of course, determined by the agreement between its predictions and observations. 2. Schrödinger’s equation is nonrelativistic. In 1928 Dirac developed the relativistic quantum mechanics. He utilized the same postulates as Schrödinger but replaced the classical expression for the energy with the relation ( ) VcmpcE ++= 22022 As required, in the low-velocity limit the Dirac equation reduces to Schrödinger’s equation. The treatment of Dirac’s equation is beyond the scope of this course, but we will describe some interesting features of Dirac’s theory qualitatively. Indeed, one feature, namely pair production, has already been described. Example 5-2. Verify that Schrödinger’s equation is linear in Ψ(x,t), i.e., that if Ψ1(x,t) and Ψ2(x,t), are solutions of the equation, so is any linear combination c1Ψ1(x,t) + c2Ψ2(x,t), where c1 and c2 are any complex numbers. Solution: Straightforward. In the sequel, we shall at times pull solutions of the equation for a specific V(x,t) out of the hat, without showing how they were derived, but once given, it will be straightforward (if tedious) to verify that they do solve the equation for that potential. The next example does just that for the harmonic potential. Example 5-3. The wavefunction describing a particle of mass m acted on by a linear restoring force of force constant C in its lowest energy state as a SHO is ( ) ( ) 22//2/),( xCmtmCi eAetx h−−=Ψ where A is real. Verify that this wavefunction is a solution to Schrödinger’s equation for the harmonic potential V(x,t) = Cx2/2. Solution: Schrödinger’s equation in this case is ),(),( 2 ),( 2 2 2 22 tx t itxCxtx xm Ψ ∂ ∂ =Ψ+Ψ ∂ ∂ − h h The derivatives of the solution given above are 6 Ψ−=Ψ ∂ ∂ m Citx t 2 ),( ∂x hh2 Ψ−=Ψ−=Ψ ∂ xCmxCmtx 2),(and so Ψ+Ψ−=      Ψ−−Ψ−=Ψ ∂ ∂ 2 22 2 ),( xCmCmxCmxCmCmtx x hhhhh Substuting these expressions into Schrödinger’s equation we get Ψ     −=Ψ+Ψ−Ψ m CiixCx m Cm m Cm 2222 22 2 22 h h h h h or Ψ     −=Ψ+Ψ−Ψ m CiixCxC m C 2222 22 h h which is obviously satisfied. The general solution to the Schrödinger equation for the SHO is given in the next chapter. 5-3. BORN’S INTERPRETATION OF WAVEFUNCTIONS If we now go back to Schrödinger’s equation for a free particle, we see that the wavefunction is Ψ(x,t) = cos(kx – ωt) + γsin(kx – ωt) = cos(kx – ωt) + isin(kx – ωt) (5-23) In other words, it is complex. The reason is obvious: the wave equation is a linear relation between the first time and second spatial derivatives of the wf. This itself results from the linear relation between the energy and the square of the linear momentum. The fact that the wf is complex implies that its physical interpretation is different from that of functions describing classical waves. Hence we must explore the new physical meaning of this new beast. Following Born’s postulate (1926), the basic connection between Ψ(x,t) and its associated particle is expressed in terms of the probability density P(x,t), which is P(x,t) = Ψ*(x,t)Ψ(x,t) (5-24) where Ψ*(x,t) is the complex conjugate of Ψ(x,t). The postulate is the following: If the location of the particle associated with Ψ(x,t) is measured at time t, then the probability that it will be found in [x, x + dx] is Ψ*(x,t)Ψ(x,t)dx. Note that Ψ*(x,t)Ψ(x,t) is real and nonnegative, which is a necessary (and almost, but not quite sufficien) condition for it to be a probability function. The relation between the position of the particle and the wf is shown schematically in Fig. 5-2. Example 5-4. Prove that Ψ*(x,t)Ψ(x,t) ≥ 0. 7 Proof: Let Ψ(x,t) = R(x,t) + iI(x,t) (5-25a) Then Ψ*(x,t) = R(x,t) - iI(x,t) (5-25b) and Ψ*(x,t)Ψ(x,t) = R2(x,t) + I2(x,t) ≥ 0. Note the analogy – and the distinction – between the probability density of the particle wf and the intensity of an EM wave, which is proportional to E2(x,t), where E is the amplitude of the electric field. Example 5-5. Evaluate the probability density of the ground state wf of the SHO given above. Solution: Since ( ) ( ) 22//2/),( xCmtmCi eAetx h−−=Ψ we get ( ) ( ) ( ) ( ) 22 /2//2//2/2),(),(*),( xCmxCmtmCitmCi eAeeeAtxtxtxP hh −−+− ==ΨΨ= Note that this P(x,t) is independent of time, even though Ψ(x,t) is a function of t. We shall see that the wf is indeed time-independent in any case in which the associated particle is in a single energy state, i.e., its energy is well-defined, i.e., its energy is a conserved quantity. The probability density of the ground state of the SHO P(x) given above is plotted vs x in the upper part of Fig. 5-3. It is obviously a Gaussian which peaks at x = 0. Hence, quantum mechanics predicts that the particle will most likely be found in an element dx around the equilibrium point, which is x = 0. However, there are no well-defined limits beyond which the probability is zero. We shall now see that the probability function predicted by quantum mechanics is radically different from the classical prediction. Example 5-6. Evaluate the classical probability density of the SHO given above. Solution: Classically, the particle has a definite momentum p and velocity v for any given displacement from equilibrium x. The probability of finding it in element dx around x is proportional to the amount of time it spends in that element, i.e., inversely proportional to its speed at x. Hence P = B2/v where B2 is a constant. Hence, to find P(x) we need to express v in terms of x. This relation is contained in the expression for the energy E = K + V = mv2/2 + Cx2/2. Rearranging we get 10 For example, the expectation value of the potential energy of that SHO in the ground state is ( )∫ ∫ ∞+ ∞− ∞+ ∞− −=ΨΨ>=< dxexCdxtxtxVtxtV xCm 2/2 2 ),(),(),(*)( h ∫ +∞ ∞− ΨΨ>=< dxtxptxp ),(),(* Even though it can readily be done, we shall not evaluate this <V(t)> = <V> at present, since we wish to focus on other fundamental dynamical quantities, namely the momentum p and energy E. While the quantity is classically valid with p(x) = [2m(E – Cx2/2)]1/2, it is ruled out as a quantum mechanical quantity due to the uncertainty principle. So we turn to a new approach. Consider the free particle with the wavefunction Ψ(x,t) = cos(kx – ωt) + isin(kx – ωt) = ei(kx – ωt) Then Ψ=Ψ=Ψ=Ψ= ∂ Ψ∂ h pi h piiik x π λ π 22 In other words, p[Ψ(x,t)] = -ih∂Ψ(x,t)/∂x Therefore, there is an association between the linear momentum p and the differential operator - ih∂/∂x. We can derive a similar association for the energy: Ψ−=Ψ−=Ψ−= ∂ Ψ∂ h Eiii t πνω 2 I.e., E[Ψ(x,t)] = ih∂Ψ(x,t)/∂t Are these relations restricted to free particles? NO: Consider the general case where p2/(2m) + V(x,t) = E Replacing the dynamical quantities by the operators we get t itxV x i m ∂ ∂ =+      ∂ ∂ − hh ),( 2 1 2 t itxV xm ∂ ∂ =+ ∂ ∂ − h h ),( 2 2 22 t itxV xm ∂ Ψ∂ =Ψ+ ∂ Ψ∂ − h h ),( 2 2 22 x ipop ∂ ∂ −= h or This is an operator equation, so for any wavefunction Ψ(x,t) we have … which is Schrödinger’s equation… t iEop ∂ ∂ = h In summary, 11 Returning to <p> we get ∫∫ ∞+ ∞− ∞+ ∞− Ψ ∂ ∂ Ψ−=ΨΨ>=< dxtx x txidxtxptxp op ),(),(*),(),(* h Similarly, ∫∫ ∞+ ∞− ∞+ ∞− Ψ ∂ ∂ Ψ=ΨΨ>=< dxtx t txidxtxEtxE op ),(),(*),(),(* h Indeed, substituting Schrödinger’s equation for ih∂Ψ/∂t in the last expression we get ∫ ∞+ ∞− Ψ      + ∂ ∂ −Ψ>=< dxtxtxV xm txE ),(),( 2 ),(* 2 22h ∫ ∞+ ∞− Ψ ∂ ∂ −Ψ>=< dxtxt x ixftxf op ),(),,(),(* h Hence, if f(x,p,t) is any dynamical quantity, then its expectation value when the particle’s wf is Ψ(x,t) is Example 5-9. Consider a 1-d free particle confined to |x| < a/2, i.e., confined to the region [-a/2, +a/2]. We will see that the ground-state wf of the particle is 0 )cos(),( / hiEte a xAtx − =Ψ π for |x| < a/2 otherwise. where A is a normalization constant and E is its energy. Show that this Ψ(x,t) satisfies Schrödinger’s equation and find the lowest energy E. t i xm ∂ Ψ∂ = ∂ Ψ∂ − h h 2 22 2 Solution: Since the particle is free within the box |x| < a/2, V(x,t) = const which we can choose = 0. Hence Schrödinger’s equation is Substituting the given expression for Ψ(x,t) we get Ψ=Ψ⇒Ψ     −=Ψ−= ∂ Ψ∂ − E ma Eii maxm 2 22 2 22 2 22 222 h h h hh ππ which is indeed satisfied if E = π2h2/(2ma2) = h2/(8ma2) Example 5-10. Evaluate the expectation values (a) <x>, (b) <p>, (c) <x2>, and (d) <p2> of the particle-in-a-box described in the previous example, when it’s in its ground state. Solution: (a) As for the ground state of the SHO, we note that Ψ(x,t) = Acos(πx/a)exp(-iEt/h) is an even function of x, and so is Ψ*(x,t)Ψ(x,t) = A2cos2(πx/a). Therefore, Ψ*(x,t)xΨ(x,t) = A2xcos2(πx/a) is an odd function of x, so as expected ( ) 0/cos 2/ 2/ 22 =>=< ∫ + − dxaxxAx a a π 12 dxtx x txip a a ),(),(* 2/ 2/ Ψ ∂ ∂ Ψ−>=< ∫ + − h(b) . Since the integrand is now proportional to cos(πx/a)sin(πx/a), it is also an odd function of x, so as expected again, <p> = 0. dxaxxAdxaxxAdxtxxtxx aa a ∫∫∫ ++ − +∞ ∞− ==ΨΨ>=< 2/ 0 222 2/ 2/ 22222 )/(cos2)/(cos),(),(* ππ(c)       −>=⇒<                 = ∫ + 1 64 )/(cos2 2 2 3 22 2/ 0 2 23 2 π π πππ π aAx a xdax a xaA a As mentioned above, A is the normalization constant which we now proceed to evaluate: ( ) ( ) ===ΨΨ= ∫∫∫ ++ − +∞ ∞− axdaxaAdxaxAdxtxtx a a /)/(cos/2)/(cos),(),(*1 2/ 0 22 2/ 2/ 22 ππππ π ( ) ( ) a AaAaAdaAdaA 2 2 0 4 2 2 2cos1/2cos/2 2 2 2/ 0 2 2/ 0 22 =⇒=    +     = + == ∫∫ π π θθπθθπ ππ axxaaa a aAx rms 18.0033.0162 1 64 21 64 22 2 2 22 2 32 2 3 22 =><=⇒≈      −=      −=      −>=< π π π π π π Returning to <x2> we get Hence xrms is a measure of the fluctuation of x about <x> = 0. Generally, (<x2> - <x>2)1/2 is a measure of the fluctuation when <x> ≠ 0. Similarly for <p>, <p2>, and (<p2> - <p>2)1/2: (d) Now evaluate <p2> for the 1-d particle in the 1-d box: ( ) 2 2 2 2 2 2 2 2 2 22 ***      =ΨΨ+= ∂ Ψ∂ Ψ−= ∂ Ψ∂ −Ψ>=< ∫∫∫ +∞ ∞− +∞ ∞− +∞ ∞− a dx a dx x dx x ip ππ hhhh )2/(2 ahp =><⇒ …which is precisely the de Broglie value of p = h/λ = h/(2a)! So now we can see why <x2>1/2 = ∆x and <p2>1/2 = ∆p! In addition, ∆x∆p = <x2>1/2<p2>1/2 = 0.18a×(πh/a) = 0.57h ≥ h/2! 15 if ψ < 0, d2ψ/dx2 < 0 and ψ is concave downwards. In both cases, ψ is convex toward the x axis or concave away from the x axis. In region II the behavior of ψ is opposite, and it is concave toward the x axis. In region III it’s behavior is the same as in region I. Fig. 5-14 shows the consequences of these properties of ψ and its derivatives: Assume that ψ > 0 and dψ/dx < 0 at some x’ < x0 < x”. Then d2ψ/dx2 < 0 and ψ is concave downwards. If ψ is still > 0 at x = x”, it will start curving upwards at x > x” and grow without a bound, diverging as x → ∞. This behavior is obviously unacceptable. If ψ becomes negative at some x > x”, it will start curving downwards and diverge to -∞ as x → ∞, which is equally unacceptable. Hence only if ψ and dψ/dx have special values at x0 so that ψ curves upwards at x > x” but at a slower and slower rate, will ψ → 0 as x → ∞, to yield an acceptable solution. In general, we will also have similar problems at x < x’. Hence we will see that there are only special values of E for which ψ → 0 as x → ±∞, 0. This results in energy quantization “automatically,” so we find a set of eigenfunctions ψ1, ψ2, ψ3, …, with energy eigenvalues E1, E2, E3, … Fig. 5-15 shows ψ1, ψ2, and ψ3 for the lowest energies E1, E2, and E3. ψ1 behaves as function 3 of Fig. 5-14 for large x, and also → 0 for small x. ψ2(x0) = ψ1(x0) but the magnitude of its second derivative |d2ψ/dx2| is larger, so it crosses the x axis at some point x’ < x < x0. Then the sign of d2ψ/dx2 changes, and it turns from concave downwards to concave upwards. At x = x’ it reverses again, turning concave downwards and it approaches the x axis. The basic observation that should be noted here is that 122 1 2 2 2 2 00 EE dx d dx d xxxx >⇔      >      == ψψ Similarly, E3 > E1. Note that E3 – E2 and E2 - E1 are finite, so the energy levels are discrete. This is true as long as there are x’ and x” such that V(x) > E for x < x’ or x > x” (see Fig. 5-16). When the energy of the atom is E > Vl the situation changes: Classically, the atom is unbound; for all x > x’, V(x) - E < 0 so ψ will be concave towards the x axis for all x. In other words, it will oscillate! And since we can always cause ψ → 0 as x → ∞ by an appropriate choice of an initial value of dψ/dx, we will find an eigenfunction for any value of E. Thus the set of allowed eigenvalues forms a continuum. In summary, when the relation between E and V(x) is such that classically the particle is bound, then Schrödinger’s equation leads to quantized energies. If classically the particle is not bound, the set of eigenvalues forms a continuum. 16 Example 5-12. Use the foregoing arguments to describe the eigenfunctions of the SHO. Then compare the resulting P(x) with the classical SHO of the same E. Solution: Consider Figs. 5-17, 5-18, and 5-19. V(x) is such that for any E the particle is classically bound between x’ < 0 < x”, with x’ = -x”. Since [ ] )()(2 22 2 xExVm dx d ψψ −= h ψ will oscillate in the region where V(x) - E < 0. However, as we approach x’ and x” in the bounded region, |V(x) - E| decreases so oscillations weaken. In other words, the separation between the nodes of ψ increases. Hence, the larger E is, the more negative V(x) - E, the more rapid are the oscillations, the greater the number of nodes, the shorter the “effective λ,” the greater the momentum p = h/λ, the greater the energy E = p2/(2m) + V… We have come full circle in understanding the relation between the nature of the wavefunction and the energy of the particle. For the nth allowed energy En, there are n - 1 nodes. As the SHO particle approaches x’ or x”, the amplitude of the oscillations of ψ increases, since ψ itself must be larger when it “bends over”: Since |d2ψ/dx2| ∝ |V(x) – E||ψ| and |V(x) – E| is smaller, |ψ| must be larger in order to “bend over.” Hence P(x) = ψ *(x)ψ(x) = ψ2(x) is higher. This situation resembles the behavior of the classical SHO. Also note that as E increases, the oscillations become more rapid, so the decay of ψ outside the bounded region becomes more rapid, and the overall behavior of P(x) approaches the classical limit. 5-8. SUPERPOSITION OF WAVEFUNCTIONS For each allowed energy eigenvalue En we have an eigenfunction ψn(x) and a wavefunction Ψn(x,t) = ψn(x)exp[-iEnt/h]. n is called the quantum number. If the system is described by Ψn(x,t), it is said to be in state n. Since the Schrödinger equation is linear, Ψ(x,t) = ΣcnΨn(x,t) is also a solution of the equation. In fact, it is the general solution of the equation for a specific V(x) and specific boundary conditions. 17 Example 5-13. Consider a particle in a state such that measurement of its total energy E can result in either of two values E1 or E2, e.g., an e- making a transition from an excited state to the ground state. Then Ψ(x,t) = c1Ψ1(x,t) + c2Ψ2(x,t) = c1ψ1(x)exp[-iE1t/h] + c2ψ2(x)exp[-iE2t/h]. Show that P(x) oscillates with t and calculate the oscillation frequency. Solution: P(x,t) = Ψ*(x,t)Ψ(x,t) = = c1*c1ψ1*(x)ψ1(x) + c2*c2ψ2*(x)ψ2(x) + c2*c1ψ2*(x)ψ1(x)exp[-i(E2 – E1)t/h] + c.c. Since exp[iθ] = cosθ + isinθ, P(x) oscillates at ν = (E2 – E1)/(2πh) = (E2 – E1)/h. Note that if an e- in an atom is in an eigenstate, P(x) does not oscillate, which is a situation analogous to the classical stationary state. Hence, it should not radiate.