this particle would move if v(x)were of the forms shown in Fig. 1. 6, where the total energy e is denoted by the position of the horizontal line XR Figure 1.6. Three characteristic potentials showing left and right classical turning points at energies denoted by the horizontal lines A. The Classical Probability density I would like you to imagine what the probability density would be for this particle moving with total energy e and with V(x)varying as the above three plots illustrate. To conceptualize the probability density, imagine the particle to have a blinking lamp attached to it and think of this lamp blinking say 100 times for each time it takes for the particle to complete a full transit from its left turning point, to its right turning point and back to the former. The turning points x, and xg are the positions at which the particle, if 36
36 this particle would move if V(x) were of the forms shown in Fig. 1. 6, where the total energy E is denoted by the position of the horizontal line. Figure 1. 6. Three characteristic potentials showing left and right classical turning points at energies denoted by the horizontal lines. A. The Classical Probability Density I would like you to imagine what the probability density would be for this particle moving with total energy E and with V(x) varying as the above three plots illustrate. To conceptualize the probability density, imagine the particle to have a blinking lamp attached to it and think of this lamp blinking say 100 times for each time it takes for the particle to complete a full transit from its left turning point, to its right turning point and back to the former. The turning points xL and xR are the positions at which the particle, if xL xR
it were moving under Newton's laws, would reverse direction(as the momentum changes sign)and turn around. These positions can be found by asking where the momentum goes 0=p=(2m(EV(x) These are the positions where all of the energy appears as potential energy e=v(x)and correspond in the above figures to the points where the dark horizontal lines touch the V(x)plots as shown in the central plot The probability density at any value of x represents the fraction of time the particle spends at this value of x(i.e, within x and x+dx). Think of forming this density by allowing the blinking lamp attached to the particle to shed light on a photographic plate that is exposed to this light for many oscillations of the particle between x, and x, Alternatively, one can express this probability amplitude p(x) by dividing the spatial distance dx by the velocity of the particle at the point x P(x)=(2m(E-V(x))m dx Because E is constant throughout the particle's motion, P(x)will be small at x values where the particle is moving quickly (i.e, where V is low)and will be high where the particle moves slowly(where V is high). So, the photographic plate will show a bright region where V is high(because the particle moves slowly in such regions )and less brightness where V is low 37
37 it were moving under Newton’s laws, would reverse direction (as the momentum changes sign) and turn around. These positions can be found by asking where the momentum goes to zero: 0 = p = (2m(E-V(x))1/2 . These are the positions where all of the energy appears as potential energy E = V(x) and correspond in the above figures to the points where the dark horizontal lines touch the V(x) plots as shown in the central plot. The probability density at any value of x represents the fraction of time the particle spends at this value of x (i.e., within x and x+dx). Think of forming this density by allowing the blinking lamp attached to the particle to shed light on a photographic plate that is exposed to this light for many oscillations of the particle between xL and xR . Alternatively, one can express this probability amplitude P(x) by dividing the spatial distance dx by the velocity of the particle at the point x: P(x) = (2m(E-V(x))-1/2 m dx. Because E is constant throughout the particle’s motion, P(x) will be small at x values where the particle is moving quickly (i.e., where V is low) and will be high where the particle moves slowly (where V is high). So, the photographic plate will show a bright region where V is high (because the particle moves slowly in such regions) and less brightness where V is low
The bottom line is that the probability densities anticipated by analyzing the classical Newtonian dynamics of this one particle would appear as the histogram plots shown in Fig. 1.7 illustrate Figure 1. 7 Classical probability plots for the three potentials shown Where the particle has high kinetic energy(and thus lower V(x)), it spends less time and P(x)is small. Where the particle moves slowly, it spends more time and P(x)is larger For the plot on the right, V(x) is constant within the "box", so the speed is constant, hence P(x)is constant for all x values within this one-dimensional box. I ask that you keep these plots in mind because they are very different from what one finds when or solves the Schrodinger equation for this same problem. Also please keep in mind that these plots represent what one expects if the particle were moving according to class Newtonian dynamics (which we know it is not!)
38 The bottom line is that the probability densities anticipated by analyzing the classical Newtonian dynamics of this one particle would appear as the histogram plots shown in Fig. 1.7 illustrate. Figure 1. 7 Classical probability plots for the three potentials shown Where the particle has high kinetic energy (and thus lower V(x)), it spends less time and P(x) is small. Where the particle moves slowly, it spends more time and P(x) is larger. For the plot on the right, V(x) is constant within the “box”, so the speed is constant, hence P(x) is constant for all x values within this one-dimensional box. I ask that you keep these plots in mind because they are very different from what one finds when one solves the Schrödinger equation for this same problem. Also please keep in mind that these plots represent what one expects if the particle were moving according to classical Newtonian dynamics (which we know it is not!). xL xR
B The Quantum Treatment To solve for the quantum mechanical wave functions and energies of this sam problem, we first write the Hamiltonian operator as discussed above by replacing p by i t d/dx H=-A/2m d/dx (x) We then try to find solutions y(x)to Hy= Ey that obey certain conditions. These conditions are related to the fact that ly(x) is supposed to be the probability density for finding the particle between x and x+dx. To keep things as simple as possible, lets focus on the box potential V shown in the right side of Fig. B. 7. This potential, expressed a function of x is: V(x)=oo for x<0 and for x> L; v(x)=0 for x between 0 and L The fact that V is infinite for x<O and for x>L, and that the total energy e must be finite, says that y must vanish in these two regions (y=0 for x<0 and for x> L). This condition means that the particle can not access these regions where the potential is infinite. The second condition that we make use of is that y(x)must be continuous; this means that the probability of the particle being at x can not be discontinuously related to the probability of it being at a nearby point C. The energies and wave functions The second-order differential equation
39 B. The Quantum Treatment To solve for the quantum mechanical wave functions and energies of this same problem, we first write the Hamiltonian operator as discussed above by replacing p by -i h d/dx : H = - h2 /2m d2 /dx2 + V(x). We then try to find solutions y(x) to Hy = Ey that obey certain conditions. These conditions are related to the fact that |y (x)|2 is supposed to be the probability density for finding the particle between x and x+dx. To keep things as simple as possible, let’s focus on the “box” potential V shown in the right side of Fig. B. 7. This potential, expressed as a function of x is: V(x) = ¥ for x< 0 and for x> L; V(x) = 0 for x between 0 and L. The fact that V is infinite for x< 0 and for x> L, and that the total energy E must be finite, says that y must vanish in these two regions (y = 0 for x< 0 and for x> L). This condition means that the particle can not access these regions where the potential is infinite. The second condition that we make use of is that y (x) must be continuous; this means that the probability of the particle being at x can not be discontinuously related to the probability of it being at a nearby point. C. The Energies and Wave functions The second-order differential equation
h/2m dv/dx+ V(x)Y=Ey has two solutions(because it is a second order equation) in the region between x=0 and X= L. =sin(kx)and y =cos(kx), where k is defined as k=(2mE/h) Hence, the most general solution is some combination of these two Y=A sin(kx)+ B cos(kx) The fact that y must vanish at x=0(n b, y vanishes for x< 0 and is continuous, so it must vanish at the point x=0)means that the weighting amplitude of the cos(kx)term must vanish because cos(kx)=I atx=0. That is B=0 The amplitude of the sin(kx)term is not affected by the condition that y vanish at x=0, since sin(kx) itself vanishes at x=0. So, now we know that y is really of the form y(x)=A sin(kx)
40 - h 2/2m d2y/dx2 + V(x) y = E y has two solutions (because it is a second order equation) in the region between x= 0 and x= L: y = sin(kx) and y = cos(kx), where k is defined as k=(2mE/h 2) 1/2 . Hence, the most general solution is some combination of these two: y = A sin(kx) + B cos(kx). The fact that y must vanish at x= 0 (n.b., y vanishes for x< 0 and is continuous, so it must vanish at the point x= 0) means that the weighting amplitude of the cos(kx) term must vanish because cos(kx) = 1 at x = 0. That is, B = 0. The amplitude of the sin(kx) term is not affected by the condition that y vanish at x= 0, since sin(kx) itself vanishes at x= 0. So, now we know that y is really of the form: y (x) = A sin(kx)