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PHYS 3313 – Section 001 Lecture #13 Wednesday, Oct. 23, 2013 Dr. Jaehoon Yu • • • • • Wave Function Normalization Time-Independent Schrödinger Wave Equation Expectation Values Operators – Position, Momentum and Energy Infinite Square Well Potential Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 1 Announcements • Mid-term grade discussion Monday, Oct. 28 – – – – In Dr. Yu’s office, CPB 342 Last name begins with A – C: 12:50 – 1:20pm Last name begins with D – L: 1:20 – 1:50pm Last name begins with M – Z: 1:50 – 2:20pm • Colloquium today – 4pm today, SH101, Dr. X. Chu, U. of Colorado – Double extra credit for this colloquium Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 2 Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 3 Ex 6.4: Normalization Consider a wave packet formed by using the wave function that Ae- , where A is a constant to be determined by normalization. Normalize this wave function and find the probabilities of the particle being between 0 and 1/ , and between 1/ and 2/ . Y = Ae Probabilit y density ò +¥ -¥ -a x Y Ydx = ò * -¥ +¥ +¥ -¥ 0 = ò A2 e-2a x dx = 2 ò A= a Wednesday, Oct. 23, 2013 +¥ ( Ae ) ( Ae )dx = ò ( A e )( Ae )dx = -a x * +¥ -a x * -a x -a x -¥ 2 2A -2a x 2 -2a x e Ae dx = -2a Normalized Wave Function PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu +¥ = 0+ 0 A2 a = 1 Y = ae -a x 4 Ex 6.4: Normalization, cont’d Using the wave function, we can compute the probability for a particle to be with 0 to 1/ and 1/ to 2/ . Y = ae -a x For 0 to 1/ : P=ò 1a 0 Y Ydx = ò * 1a 0 a -2a x -2a x e a e dx = -2a 1a 0 1 -2 = - ( e -1) » 0.432 2 For 1/ to 2/ : P=ò 2a 1a Y Ydx = * 2a ò a ae 1 2a -2a x a -2a x 1 -4 -2 e = e - e ) » 0.059 dx = ( -2a 2 1a How about 2/ :to ∞? Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 5 Properties of Valid Wave Functions Boundary conditions To avoid infinite probabilities, the wave function must be finite everywhere. 2) To avoid multiple values of the probability, the wave function must be single valued. 3) For finite potentials, the wave function and its derivatives must be continuous. This is required because the second-order derivative term in the wave equation must be single valued. (There are exceptions to this rule when V is infinite.) 4) In order to normalize the wave functions, they must approach zero as x approaches infinity. Solutions that do not satisfy these properties do not generally correspond to physically realizable circumstances. 1) Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 6 Time-Independent Schrödinger Wave Equation • The potential in many cases will not depend explicitly on time. • The dependence on time and position can then be separated in the Schrödinger wave equation. Let, Y ( x,t ) = y ( x ) f ( t ) 2 ¶ f (t ) f ( t ) ¶2y ( x ) which yields: i y ( x ) =+ V ( x )y ( x ) f ( t ) 2 ¶t 2m ¶x 2 1 ¶ f (t ) 1 ¶2y ( x ) i =+ V ( x) 2 f ( t ) ¶t 2m y ( x ) ¶x Now divide by the wave function: • The left side of this last equation depends only on time, and the right side depends only on spatial coordinates. Hence each side must be equal to a constant. The time dependent side is 1 df i =B f dt Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 7 Time-Independent Schrödinger Wave Equation(con’t) We integrate both sides and find: i df ò f = ò Bdt Þ i ln f = Bt + C where C is an integration constant that we may choose to be 0. Bt Therefore ln f = i This determines f to be by comparing it to the wave function of a free particle f ( t ) = eBt i = e-iBt = e-iw t Þ B = w Þ B = w = E 1 ¶ f (t ) i =E f ( t ) ¶t This is known as the time-independent Schrödinger wave equation, and it is a fundamental equation in quantum mechanics. d 2y ( x ) + V ( x )y ( x ) = Ey ( x ) 2 2m dx 2 Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 8 Stationary State • Recalling the separation of variables: Y ( x,t ) = y ( x ) f ( t ) -iw t and with f(t) = e the wave function can be -iw t written as: Y ( x,t ) = y ( x ) e • The probability density becomes: Y Y=y * 2 ( x )( e iw t -iw t e ) =y ( x ) 2 • The probability distributions are constant in time. This is a standing wave phenomena that is called the stationary state. Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 9 Comparison of Classical and Quantum Mechanics • • • Newton’s second law and Schrödinger’s wave equation are both differential equations. Newton’s second law can be derived from the Schrödinger wave equation, so the latter is the more fundamental. Classical mechanics only appears to be more precise because it deals with macroscopic phenomena. The underlying uncertainties in macroscopic measurements are just too small to be significant due to the small size of the Planck’s constant Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 10 Expectation Values • In quantum mechanics, measurements can only be expressed in terms of average behaviors since precision measurement of each event is impossible (what principle is this?) • The expectation value is the expected result of the average of many measurements of a given quantity. The expectation value of x is denoted by <x>. • Any measurable quantity for which we can calculate the expectation value is called a physical observable. The expectation values of physical observables (for example, position, linear momentum, angular momentum, and energy) must be real, because the experimental results of measurements are real. N i xi å N x + N x + N x + N x + • The average value of x is x = 1 1 2 2 3 3 4 4 = i N1 + N 2 + N 3 + N 4 + åN i Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 11 i • Continuous Expectation Values +¥ We can change from discrete to xP ( x ) dx ò -¥ continuous variables by using the probability P(x,t) of observing the particle at a particular x. • Using the wave function, the x = expectation value is: • The expectation value of any function g(x) for a normalized wave function: +¥ x= ò +¥ -¥ +¥ ò ò xY ( x,t ) Y ( x,t ) dx -¥ +¥ -¥ P ( x ) dx * Y ( x,t ) Y ( x,t ) dx * g ( x ) = ò Y ( x,t ) g ( x ) Y ( x,t ) dx * -¥ Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu 12 Momentum Operator • To find the expectation value of p, we first need to represent p in terms of x and t. Consider the derivative of the wave function of a free particle with respect to x: ¶Y ¶ i( kx-w t ) éë e ùû = ikei( kx-w t ) = ikY = ¶x ¶x With k = p / ħ we have ¶Y = i p Y ¶x ¶Y ( x,t ) This yields p éë Y ( x,t ) ùû = -i ¶x • This suggests we define the momentum operator as • The expectation value of the momentum is p = pˆ = -i. ¶ ¶x ¶Y ( x,t ) ò-¥ Y ( x,t )pˆ Y ( x,t ) dx = -i ò-¥ Y ( x,t ) ¶x dx +¥ +¥ * Wednesday, Oct. 23, 2013 PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu * 13 Position and Energy Operators The position x is its own operator as seen above. The time derivative of the free-particle wave function is ¶ i( kx-w t ) ¶Y i( kx-w t ) ¶t éë e ¶t = Substituting ùû = -iw e = -iwY ¶Y ( x,t ) = E / ħ yields E éë Y ( x,t ) ùû = i ¶t ¶ ˆ E=i ¶t The energy operator is The expectation value of the energy is E = ò +¥ -¥ ˆ ( x,t ) dx = i Y ( x,t )EY Wednesday, Oct. 23, 2013 * PHYS 3313-001, Fall 2013 Dr. Jaehoon Yu ¶Y ( x,t ) ò-¥ Y ( x,t ) ¶t dx +¥ * 14