Solution Set # 5 Wavefunctions and Expectation Values

Solution set 5: Wavefunctions and Expectation ValuesPH-102 Solution Set # 5 October, 21, 2009Wavefunctions and Expectation ValuesAnswer 1.(a)A particle of mass m, moving freely inside an infinite potential well of length a, has thefollowing wavefunction,ψ(x, 0) =A √ asin( πxa ) + √35a sin(3πx a ) + √15a sin(5πx a ) ·Suppose that ψ(x) is written as the superposition of three wavefunctions,ψ(x) = ψ 1 (x) + ψ 3 (x) + ψ 5 (x) .A measurement of position is performed. The measurement collapses ψ(x) into either ψ 1 (x),ψ 3 (x) or ψ 5 (x) withe certain probabilities. First we find out the normalization constant usingthe fact that the total probability of finding the particle in a well of length a is 1. Therefore,P =∫ a/2−a/2∫ a/2−a/2ψ ∗ (x) ψ ( x) dx = 1( A√asin( πx√ √ )31a ) + 5a sin(3πx a ) + 5a sin(5πx a )( A√asin( πx√ √ )31a ) + 5a sin(3πx a ) + 5a sin(5πx a ) dx = 1∫ a/2−a/2( A2a sin2 ( πxa ) + 35a sin2 ( 3πxa ) + 15a sin2 ( 5πxa ) +√ )A 3√ a 5a sin(πx a ) sin(3πx a ) + . . . dx = 1.The integrand has 9 terms, 3 perfect squares and 6 mixed products. It is possible to showthat the integral of each of the mixed products is zero. We explicitly show this only for onecase, the sin (πx/a) sin (3πx/a) term. We use the trigonometric relation,sin X sin Y = 1 2(cos(X − Y ) − cos(X + Y )).Chapter 5 of Eisberg and Resnick. 1

Solution set 5: Wavefunctions and Expectation ValuesTherefore,∫ a/2−a/2= A2 √ a= A2 √ a= A2 √ a= A2 √ a( √ )A√a 35a sin(πx a ) sin(3πx a ) dx√ ∫ 3 a/2(cos( πx5a −a/2 a − 3πxa ) − cos(πx a + 3πx )a ) √ ∫ 3 a/2(cos( −2πx ) − cos( 4πx )5a −a/2 aa ) √ ∫ 3 a/2(cos( 2πx)5a −a/2 a ) − cos(4πx a )√ ( 3 sin(2πx) 4πxsin(a2π− ) )∣a ∣∣∣a/24π5aaa −a/2= A√ (34π √ sin( 2πx5 a ) − 1 )∣ ∣∣∣a/22 sin(4πx a ) −a/2= A√ 3((sin(π)4π √ + sin(π)) − 1 )52 (sin 2π + sin 2π)= 0.In this way, all the last six terms will be equal to zero, we left with,∫ a/2−a/2∫ a/2−a/2( A2a sin2 ( πxa ) + 35a sin2 ( 3πxa ) + 15a sin2 ( 5πx )a ) dx = 1( A21 − cos( 2πx)a+ 3 1 − cos( 6πx)a+ 1 1 − cos( 12πx)aa 2 5a 2 5a 2)dx = 1A 2a( x2−sin(2πx2 2π aa ))∣ ∣∣∣a/2+ 3 ( x−a/25a 2−sin(6πx2 6π aa ))∣ ∣∣∣a/2+ 1 ( x−a/25a 2−sin(12πx) a2 12πa)∣ ∣∣∣a/2−a/2= 1,where sin( πxa ) ∣ ∣∣∣a/2−a/2= 0,A 2a2a4 + 3 2a5a 4 + 1 2a5a 4 = 1A 22 + 310 + 110 = 1A 2 = 1210A =√65 ,Chapter 5 of Eisberg and Resnick. 2

Solution set 5: Wavefunctions and Expectation Valueswhich is the required normalization constant. Hence the wavefunction in its normalized formis,ψ(x, 0) =√65a sin(πx a ) + √35a sin(3πx a ) + √15a sin(5πx a ) ·(b)As the wavefunction collapses to one of the states, the corresponding eigen energies will bemeasured. In order to calculate the energy of each state, we use Schrodinger’s equation− 22md 2 ψ(x)dx 2 + V (x) ψ(x) = E ψ(x) .Since the particle is free, V (x) = 0, resulting in,− 22md 2 ψ(x)dx 2 = E ψ(x) .For collapse into ψ 1 (x), the measured energy will be E 1 .By comparison,− 22m− 22m(√d 2 6dx 2− 22m√6 π5a a√+ 2 62m 5addxd 2 ψ 1 (x)dx 2)a )5a sin(πx(cos( πx )a )(π 2sin( πx )a 2 a )(√ ) 2 π 2 62ma 2 5a sin(πx a )= E 1 ψ 1 (x)= E 1 ψ 1 (x)= E 1 ψ 1 (x)= E 1 ψ 1 (x)= E 1 ψ 1 (x) 2 π 22ma 2 ψ 1(x) = E 1 ψ 1 (x) .E 1 = 2 π 22ma 2 ·Chapter 5 of Eisberg and Resnick. 3

Solution set 5: Wavefunctions and Expectation ValuesSimilarly for ψ 3 (x),resulting in,− 22m− 22m√35a+ 22m− 22m(√d 2 3dx 23πa√35addxd 2 ψ 3 (x)dx 2)a )5a sin(3πx(cos( 3πx )a )9π 2a 2 (sin( 3πxa ) )(√ )9 2 π 2 32ma 2 5a sin(3πx a )= E 3 ψ 3 (x)= E 3 ψ 3 (x)= E 3 ψ 3 (x)= E 3 ψ 3 (x)= E 3 ψ 3 (x)9 2 π 22ma 2 ψ 3(x) = E 3 ψ 3 (x) ,E 3 = 92 π 22ma 2 ·Similarly, energy corresponding to the next state is,E 5 = 252 π 22ma 2 ·If a measurement is carried out on the system, we would obtain either of these energies withthe corresponding probability P (E n ) . Since the initially given wavefunction contains onlythree eigen states namely ψ 1 (x), ψ 3 (x) and ψ 5 (x), the possible energy measurements are,E 1 = π2 22ma 2 ,E 3 = 9π2 22ma 2 ,E 5 = 25π2 22ma 2 ,andwith the corresponding probabilities,P 1 (E 1 ) =P 3 (E 3 ) =P 5 (E 5 ) =∫ a/2−a/2ψ ∗ 1(x) ψ 1 (x)dx = 3 5∫ a/2−a/2∫ a/2−a/2ψ ∗ 2(x) ψ 2 (x)dx = 310ψ ∗ 3(x) ψ 3 (x)dx = 110 ·Chapter 5 of Eisberg and Resnick. 4

Solution set 5: Wavefunctions and Expectation ValuesThe average or expectation value of the energy will be equal to,< E > = E avg = ∑ nP (E n ) = P 1 E 1 + P 3 E 3 + P 5 E 5(c)= 3 5 E 1 + 310 E 3 + 110 E 5E avg = 29π2 210ma 2 ·The expectation or average value of the energy is not equal to any of the measured values.This is very similar to the throws of a dice. The average value is 3.5 but this is never anoutcome of a single throw!Answer 2.The wavefunction of a free electron is given by,ψ(x) = A sin(2.5 × 10 10 x) .Since x is in meters, the constant 2.5 × 10 10 must have the dimensions of m −1 to keep theargument of the sin function dimensionless. To calculate the required quantities, we startwith the time-independent Schrodinger equation,− 22md 2 ψ(x)dx 2 + V (x) ψ(x) = E ψ(x) .Since the particle is free, V (x) = 0, leading to the simpler form,− 22mMaking use of the given wavefunction, we obtain,d 2 ψ(x)dx 2 = ddx ( ddx (A sin(2.5 × 1010 x)))d 2 ψ(x)dx 2 = E ψ(x) . (1)= (2.5 × 10 10 ) ddx (A cos(2.5 × 1010 x))= − (2.5 × 10 10 )(2.5 × 10 10 ) A sin(2.5 × 10 10 x)d 2 ψ(x)= − (2.5 × 10 10 ) 2 ψ(x) ,dx 2and the R.H.S becomes,− 22md 2 ψ(x)dx 2= 22m (2.5 × 1010 ) 2 ψ(x) ,Chapter 5 of Eisberg and Resnick. 5

Solution set 5: Wavefunctions and Expectation Valuescomparing with the R.H.S of expression (1), we obtain, 22m (2.5 × 1010 ) 2 ψ(x) = E ψ(x)E = (2.5 × 1010 ) 2 22mwhich is the required energy of the free electron. In case of the free particle, the total energyis just equal to the kinetic energy and contribution of the potential energy is zero.(a)To compute the momentum of free electron, we use,p2Total Energy = E = K.E =2mp 2 = 2mEThis is the momentum of the free electron.(b)The wavelength of the free electron is,(c)λ = h p =h(2.5 × 10 10 ) h2π=p = ± √ 2mE√= ± 2m (2.5 × 1010 ) 2 22m= ±(2.5 × 10 10 ) (Kg m s −1 ).2π(2.5 × 10 10 ) = 2.5 × 10−10 = 2.5 Å .As we have find out the exact value of momentum, therefore the uncertainty in momentumis zero. On the other hand, uncertainty in position will be infinite (∆x = ∞).Answer 3.(a)For the problem of the finite square well potential, when the total energy is greater than thepotential energy, (E > V o ), the wavefunction will display an oscillating pattern. This canbe seen from the Schrodinger equation,,d 2 ψdx 2= 2m 2 (V − E o) ψ .This equation show that ψ(x) will be concave downwards when ψ is positive and concaveupwards when ψ is negative. The bending is sharper when |V − E o | is larger; this is theChapter 5 of Eisberg and Resnick. 6

Solution set 5: Wavefunctions and Expectation Valuesregion labeled II. In regions I and III, the bending is gentler, the maximum amplitude ishigher and the separation between the nodes is also larger._ I0.0003_ II_ III0.00020.0001-4 -2 2 4-0.0001-0.0002-0.0003It is also clear from the figure that the behavior of wavefunction is symmetric about x = 0and identical in regions I and III.(b)The energy will not be quantized because the particle is not confined and is free to movein any direction. The energy is only quantized when the particle is confined within someregion and its energy is less than the barrier potential (E < V o ).(c)Solving equation (1), the wavefunction isψ(x) = A exp(ikx) + B exp(−ikx) .The value of the potential is non zero in region I (x < −a/2) and III (x > a/2), so the wavenumbers in these two regions will be,k =√2m(E − Vo )·Chapter 5 of Eisberg and Resnick. 7

Solution set 5: Wavefunctions and Expectation ValuesIn region II, the wavenumber will be,k =√2mE·Answer 4.Correct option: EWhen an electron is in the superposition of two or more wavefunctions and you make ameasurement of the energy of this electron, then the wavefunction collapses into an energyeigenstate. The probability of finding energy E 1 will be simply equal to modulus squaredof the amplitude accompanying ψ 1 (x). So, the probability of obtaining the value E 1 is 9/25and the probability of obtaining the value E 2 is 16/25.Answer 5.Correct option: AAfter the first measurement, you are again interested in measuring the energy of the sameelectron. Since the wavefunction is already collapsed and the state becomes an eigen stateof energy, the consecutive measurement will yield same value of energy as in the first case,(with probability one).Answer 6.Correct option: AAfter precisely measuring the position, if you measure the energy again, you will get thesame value of energy.Chapter 5 of Eisberg and Resnick. 8