Schrödinger's Equation and Quantum Mechanics: A Plausibility Argument and Consequences, Assignments of Physics

Download Schrödinger's Equation and Quantum Mechanics: A Plausibility Argument and Consequences and more Assignments Physics in PDF only on Docsity! • Need for a differential wave equation – limitations of the de Broglie postulate • Plausibility argument leading to Schrödinger’s equation • Born’s Interpretation of Wave Functions – complex character of wave functions – probability density – statistical predictions • Expectation Values – operators • The Time-Independent Schrödinger Equation (separation of variables) • Eigenfunctions • Energy Quantization in the Schrödinger Theory – geometrical properties – bound vs unbound Chapter 5: Outline Schrödinger’s Theory of Quantum Mechanics HW #5 due 10/4/07 questions: 20, 28 problems: 3, 4, 9, 10, 11, 12, 13, 14, 15, 27, 28, 29, 32 The general solution to this problem is the Schrödinger equation, which is a partial linear differential equation whose solution is the wavefunction Ψ(x,t) Does everyone know what a partial differential equation is ? 2 2 2 2 ),( ),( ),( ),( t tx x tx t tx x tx ! "! ! "! ! "! ! "!eqn. containing terms of the form Example 5-1: Evaluate the partial derivatives (w.r.t. x and t) up to second order for our free particle wavefunction: )sin(2sin),( tkxt x tx !" # $ %=& ' ( ) * + %=, Depending on the potential or boundary and initial conditions, solution may be 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. 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 2 2 dt xd m dt dp F == or of Maxwell’s equation )E( 00 ! " ! " =•#= $ $ + $ $ + $ $ z E y E x E zyx 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 can’t expect to derive the quantum mechanical wave equation from classical physics. We will get some direction from the de Broglie-Einstein relations λ = h/p and ν = E/h The text lists four assumptions which should be met by our equation: (1) It must be consistent with the de Broglie-Einstein postulates: λ = h/p and ν = E/h (2) It must be consistent with the equation E = p2/2m +V (3) It must be linear in Ψ(x,t), so that if Ψ1(x,t) and Ψ2(x,t) are solutions of the equation, then c1Ψ1(x,t) + c2Ψ2(x,t) is also a solution → so we can add wavefunctions to produce constructive and destructive interferences (basic characteristic of all waves and demonstrated by the various experiments (i.e. Davisson-Germer) to be valid for particle waves as well. (4) The solution of the equation for a free particle, where λ = h/p and ν = E/h, should be a linear combination of sinusoidal traveling wave functions. In other words, the equation we seek must be consistent with these postulates, at least for free particles. Note: we are not deriving Schrödinger’s equation, only providing a plausibility argument for it; that equation itself is a postulate x Let’s look at the consequences of these assumptions putting (1) into (2) ! " htxV m h txV m p E =+#+= ),( 2 ),( 2 2 22 Now recall Example 5-1 [and assumption (4)]: 2nd derivative of the traveling sine wave w.r.t. the spatial variable x: factor of –k2 1st derivative w.r.t. the time t: factor of –ω introduce k = 2π/λ and ω=2πν !h h =+" ),( 2 22 txV m k note factors k2 and ω t tx txtxV x tx ! "! ="+ ! "! ),( ),(),( ),( 2 2 #$ where the constants α and β are still to be determined. Seems ok ? Let’s test it with a constant potential V(x,t) = V0 (free particle) Suspect that Schrödinger’s equation must contain the 2nd spatial and 1st temporal derivatives of Ψ(x,t). However, V(x,t) must also appear, and since every term in the equation must contain Ψ(x,t) or its derivative (to be linear), the equation should have the form 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. We will 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 (albeit tedious) to verify that they do solve the equation for that 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 where A is real. Verify that this wavefunction is a solution to Schrödinger’s equation for the harmonic potential V(x,t) = Cx2/2. ! "(x, t) = Ae # i / 2( ) C /mt e # Cm / 2h( )x 2 Born’s Interpretation of Wave Functions 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) It is complex. The reason is obvious: the wave equation is a linear relation between the first time and second spatial derivatives of the wavefunction. This itself results from the linear relation between the energy and the square of the linear momentum. The fact that the wavefunction is complex implies that its physical interpretation is different from that of functions describing classical waves. So what is the physical meaning of Ψ(x,t) ? The postulate: 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 sufficient) condition for it to be a probability function. 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) where Ψ*(x,t) is the complex conjugate of Ψ(x,t). Example 5-4: Prove that Ψ*(x,t)Ψ(x,t) ≥ 0 Note the analogy – and the distinction – between the probability density of the particle wavefunction and the intensity of an EM wave, which is proportional to ε2(x,t), where ε is the amplitude of the electric field. Example 5-5: Evaluate the probability density of the ground state wavefunction of the simple harmonic oscillator given in Example 5-3. What’s so special about this result ? P is independent of time, even though Ψ(x,t) is a function of t Note also: QM predicts that the particle will most likely be found in an element dx around the equilibrium point (x = 0). However, there are no well-defined limits beyond which P = 0. wavefunction is time-independent ⇔ particle is in a well defined energy state, (i.e., its energy is a conserved quantity) Later we’ll see that 2( / 2 ) ( / 2)( / )( , ) Cm x i C m tx t Ae e! !" = h 2 2* 2 ( / 2 ) ( / 2)( / ) ( / 2 ) ( / 2)( / )Cm x i C m t Cm x i C m t A e e e e ! + ! ! " " = h h 22 ( / )Cm x P A e ! = h Modern formulation of quantum mechanics: 1. Every well defined observable in physics (A) has an associated operator (Â). An experiment measuring this observable can yield only those values (a) that are the eigenvalues of the operator Â: Âφa(x)=a φa(x) φa(x) - is eigenfunction of the operator  a - is an eigenvalue of the operator  Momentum operator: ! ˆ p x = "ih # #x ! px"(x) = #ih $ $x "(x) Energy operator: ! ˆ H = " h 2 2m # 2 # 2x +V (x) Position operator: ! ˆ X = x ! ˆ X "(x) = x "(x) ! ˆ H "(x) = # h 2 2m $ 2 $ 2x "(x) +V (x)"(x) Examples of operators: 2. A measurement of observable A that yields value a leaves the system in state φa(x). Each immediate, consecutive measurement of A will yield therefore value a. 3. The state of the system at any moment is represented by wavefunction Ψ(x,t). This function and its first derivative are finite, single valued and continuous. This wavefunction containes ALL physical information about the system. If the system is in state Ψ(x,t), then an averaged value obtained in measurement of observable A is: ! ˆ A = "* (x, t) ˆ A # "(x,t)dx ! ˆ A Is expectation value of observable A 4. Time evolution of the wavefunction associated with the system is described by equation: ! ih " "t #(x,t) = ˆ H #(x, t) Expanding the energy operator: ! ih " "t #(x,t) = - h 2 2m " 2 " 2x #(x,t) +V (x)#(x,t) What is this called? Cu on Cu(111) When we deposit additional atoms on a clean surface, electrons will reflect from them, and the wavefunctions will interfere with each other creating standing waves of electron densities. www.specs.de 9K => 12K What do we expect for <x> in these cases ? symmetric, centered at x = 0 symmetric, centered at x = x1 > 0 asymmetric (skewed slightly to the left), centered at x = 0 asymmetric (skewed more to the left), centered at x = 0 <x> = 0 <x> = x1 <x> slightly to right of equilibrium <x> more to right of equilibrium It should be clear that if we substitute the other side of our Schrödinger equation into our expression for <E>, we find ! "+ "# $%% & ' (( ) * + + + #$>=< dxtxtxV xm txE ),(),( 2 ),(* 2 22 h This is an example of the general property that if f(x,p,t) is any dynamical quantity, then its expectation value is given by ! "+ "# $ % % #$>=< dxtxt x ixftxf op ),(),,(),(* h We’ve just seen that the wave function contains more info than just the probability density Ψ*Ψ (i.e. <x>, <V>, <p>, <E>, and any dynamical quantity <f(x,p,t)>) Key point: The wavefunction contains all the information that the uncertainty principle will allow us to learn about about the associated particle 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 later that the ground-state wf of the particle is 0 )cos( ),( /hiEt e a x A tx ! =" # 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. 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 t i xm ! "! = ! "! # h h 2 22 2 putting in our expression for Ψ(x,t) we find 2 2 2 2 2 2 2 2 2 2 2 2 E i i E m x ma ma ! !" # $ % & = + # = & #' # = #( )" * + h h h h h which is indeed satisfied if 2 22 2ma E h! = 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. (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 # dxtx x txip a a ),(),(* 2/ 2/ ! " " !#>=< $ + # h(b) Since the integrand is 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 !!! ++ " +# #" ==$$>=< 2/ 0 222 2/ 2/ 22222 )/(cos2)/(cos),(),(* %%(c) !! " # $$ % & '>=<(! " # $ % & ! " # $ % & ! " # $ % & = ) + 1 64 )/(cos2 2 2 3 22 2/ 0 2 23 2 * * * * * * a Ax a x dax a xa A a using tables ( ) ( )axdaxaAdxaxAdxtxtx a a /)/(cos/2)/(cos),(),(*1 2/ 0 22 2/ 2/ 22 !!!! ! """ ++ # +$ $# ==%%= ( ) ( ) a A aAa AdaAdaA 2 2 0 4 2 2 2cos1 /2cos/2 2 2 2/ 0 2 2/ 0 22 =!="# $ %& ' +( ) * + , - = + == .. / / 0 0 /00/ // Before continuing, we need to evaluate our normalization constant A axxa aa a a Ax rms 18.0 033.01 62 1 64 2 1 64 22 2 2 22 2 32 2 3 22 =><=!"## $ % && ' ( )=## $ % && ' ( )=## $ % && ' ( )>=< * * * * * * Therefore, 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 ( ) 2 2 2 2 2 2 2 2 2 22 *** ! " # $ % & =''+= ( '( ')= ( '( )'>=< *** ++ +) ++ +) ++ +) a dx a dx x dx x ip ,, h hhh(d) If we define <x2>1/2 = Δx and <p2>1/2 = Δp, we see that ΔxΔp = <x2>1/2<p2>1/2 = 0.18a×(πħ/a) = 0.57ħ ≥ ħ/2! )2/(2 ahp =><! …which is precisely the de Broglie value of p = h/λ = h/(2a)! Energy Quantization in the Schrödinger Theory We shall now try to gain some intuitive understanding of the solutions of the Schrödinger equation by examining plots of V(x), Ψ(x), etc.. First, recall that ! d 2" dx 2 = 2m h 2 V (x) # E[ ]"(x) The properties of ψ(x) depend on those of V(x) since classically F = −dV/dx. This potential energy V(x) is typical for an atom bound to a similar atom in a diatomic molecule (e.g., H2, N2, O2, etc.). At the minimum in V(x), F = 0. For closer x, the atom will experience the “hard core” repulsion from the other atom. For farther x, the atom will experience attraction to the other atom. When the molecule dissociates, V(x) = V0, F = 0. Now consider an atom with a total energy E. Classically, x’ and x” are the turning points so x’ ≤ x ≤ x”. Recall that d2ψ/dx2 tells us how the slope of ψ is changing. Hence, if d2ψ/dx2 > 0 and ψ> 0, the slope is increasing and ψ is concave upwards, and vice versa. Now divide the x axis into three regions: In region I (x < x’) V(x) – E > 0 so: If ψ > 0, d2ψ/dx2 > 0 and ψ is concave upwards; 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. dψ/dx is increasing (becoming more positive), therefore, d2ψ/dx2 > 0 dψ/dx is decreasing (becoming more negative), therefore, d2ψ/dx2 < 0 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 → ∞ (#1) unacceptable If ψ becomes negative at some x > x”, it will start curving downwards and diverge to -∞ as x → ∞ (#2) also unacceptable 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, so ψ→ 0 as x → ∞, to yield an acceptable solution (#3) Let’s see if we can put some of these together into a solution: In general, we will also have similar problems at x < x’. 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, … V(x) ” > <—— ABiau3 Wox) P(x) —x] —x, se 3