Physics 15 Quantum Mechanics 1 Assignment 5 Logan A. Morrison February 10, 016 Problem 1 A particle of mass m is confined to a one-dimensional region 0 x a. At t 0 its normalized wave function is 8 πx πx ψx, t 0 1 + cos sin 1 5a a a Part a What is the wave function at a later time t t 0 The potential function for this problem is 0 0 x a Vx 0 and the Hamiltonian in the region 0 x a is just Ĥ are with the energy eigen values being x n ˆp. Recal the energy eigenfunctions for this potential m nπx L sin a E n n π ma 3
Logan A. Morrison Assignment 5 Problem 1 Page / 11 Now, if we look at our intial wave function, we can see that it is just the sum of the first two energy eigenstates: 8 πx 8 πx πx ψx, t 0 5a sin + a 5a cos sin 5 a a πx 5 a sin 1 πx + a 5 a sin 6 a 1 5 x 1 + x 7 5 8 and hence ψ; t 0 5 1 + 1 9 5 10 Since the Hamiltonian is time-independent, the time evolution operator is given by Therefore, the wavefunction at time t t 0 is Ut; 0 exp iht 11 ψ; t t 0 Ut 0 ; 0 ψ; t 0 1 exp iht 0 1 5 1 + 5 13 5 exp ie 1t 0 1 1 + 5 exp ie t 0 1 5 exp iπ t 0 1 1 + ma 5 exp iπ t 0 15 ma 16 Part b What is the average energy of the system at t 0 and at t t 0?
Logan A. Morrison Assignment 5 Problem 1 Page 3 / 11 The average energy is given by ψ H ψ. At time t 0, the average energy is 1 1 ψ; 0 H ψ; 0 5 1 + 5 H 5 1 + 5 17 1 1 5 1 + 5 5 E 1 1 + 5 E 18 5 E 1 + 1 5 E 19 π 5 ma + 1 5 The average energy at time t t 0 is iπ ψ; t H ψ; t 5 exp t 0 ma H 5 exp iπ 5 exp t 0 ma 5 exp π 5 ma + 1 5 Thus, the average energy doesn t change with time! π ma 0 π 5ma 1 1 + iπ t 0 1 + iπ t 0 ma 1 iπ 5 exp t 0 ma 3 1 1 + ma 5 exp iπ t 0 ma 1 iπ 5 exp t 0 ma 5 1 E 1 1 + 5 exp iπ t 0 E ma 6 π ma 7 π 5ma 8 9 Part c What is the probability that the particle is found in the left half of the box i.e. in the region 0 x a/ at t t 0?
Logan A. Morrison Assignment 5 Problem 1 Page / 11 The probability that the particle is in the left side of the box is calculated via P0 x a/; t a/ x ψ dx 30 iπ 5 exp t 0 πx 1 iπ ma a sin + a 5 exp t 0 πx ma a sin a 31 5 exp iπ t 0 πx 1 ma a sin + a 5 exp iπ t 0 πx ma a sin a dx 3 0 a/ 0 8 5a a/ 0 + 8 5a 1 + 16 15π cos πx sin dx + a 5a 3iπ t 0 exp + exp ma 3π t 0 ma where we use Mathematica to solve the integral. a/ 0 dx 33 πx sin a a/ πx sin a 3iπ t 0 ma 0 sin πx dx 3 a 35
Logan A. Morrison Assignment 5 Problem Page 5 / 11 Problem A particle of mass M is in a one-dimensional harmonic oscillator potential V 1 1 kx.it is initially in its ground state, x ψ ψ 0 x AM, ω 0 exp Mω 0x 36 Part a What is AM, ω 0 if the wavefunction is normalized? If the solution is normalized, then ψ ψ 1: 1 ψ ψ 37 where we have used the well-known result that dx ψ x x ψ 38 A M, ω 0 exp Mω 0x dx 39 A π M, ω 0 0 Mω 0 and hence, exp ax dx π a 1 AM, ω 0 Mω0 1/ π Part b The spring constant is suddenly doubled k k so that the new potential is V kx. The particle s energy is then measured. What is the probability for finding that particle in the ground state of the new potential V? If the systems characteristic spring constant is suddenly doubled, then the angular frequency will increase by a factor of since, by definition, ω 0 k/m. Thus, the new ground state wave function will be φ 0 x x φ Mω0 π 1/ Mω0 x exp 3
Logan A. Morrison Assignment 5 Problem Page 6 / 11 Now, we want to calculate the probability that the state is in the new ground state immediately after the spring constant is doubled. Immediately after the spring constant is doubled, we know that the state of our system hasn t yet changed. In other words, it will still be in the old ground state. To calculate the probability of measuring the groundstate energy, we must then compute the overlap of the old ground state and the new: φ ψ φ ψ Mω0 π 1/8 Mω0 π dx φ x x ψ 1/ Mω0 π 1/ exp 1/ 1/8 Mω0 1/ π π Mω 0 + 1 1/8 + 1 Mω0 x exp exp Mω 0x dx 5 dx 6 + 1Mω 0 x Thus, the probability the energy of the particle will be measured to be the new ground state energy will be 7 8 9 φ ψ 5/ + 1 0.985 50 and hence, the probability that the energy will be measured to be the groundstate energy is 98.5%. Part c The spring constant is suddenly doubled as in part, so that V 1 suddenly becomes V, but the energy of the particle in the new potential is not measured. Instead, after a time T has elapsed since the doubling of the spring constant, the spring constant is suddenly restored back to the original value. For what values of T would the initial ground state in V1 be restored with 100% certainty? Let { n } be the set of new energy eigenstate of the system. Let 0 be the old groundstate. Now suppose we double the spring constant and then let the system evolve for T seconds. Then, the state of the system will be given by ψ; T U ψ; 0 51 exp ih T 0 5 where ψ; 0 is the state of the system at t 0 and and H ˆp m + k ˆx. What we wish to cacluate is the probability that the old groundstate energy will be measured once the system is reverted back after T seconds. This is simply 0 ψ; t. If we require that this is equal to 1, then this is equivalent to saying ψ; t 0 this is because the old energy eigenstates are non-degenerate, so if we are to measure the old groundstate energy with 100% efficienty,
Logan A. Morrison Assignment 5 Problem Page 7 / 11 that means the state has to be the old ground state. Let s re-write the state of our system at time T in the basis of new energy eigenstates: ψ; T exp ih T 0 53 exp n 0 n 0 ih T exp ie n T n n 0 5 n n 0 55 The old ground state, written in terms of the new energy eigenstates is simply, 0 n n 0 56 n 0 Now since we want ψ; T and 0 to be equal, the coefficients in the expansion in the new energy eigenstate basis for these two states have to be the same.the only expection to this is that the states are equal up to an overall phase. In that case, 0 e iα ψ; T and then each coefficient would differ by the phase factor e iα. That is, for every n, as long as n 0 0. However, note that e iα exp ie n T n 0 n 0 exp ie n T n 0 + iα 1 57 dx n x x 0 58 and n x is an even function for even n and an odd function for n odd this is because the Hermite polynomials follow this rule. Also, x 0 is an even function. Recall that the product of and even and odd function is an odd function and the integral of an odd function over and even interval is zero. Therefore, n 0 0 for odd n. It is also non-zero for even n. Thus for even n, we have E n T Let n l where l is an integer. Recall that E n ω 0 n + 1/. Now, + α mπ 59 mπ E n T + α 60 mπ ω 0 Tl + 1/ + α 61 mπ ω 0 Tl + ω 0T + α 6 Since we can choose α to be anything, choose it to be α ω 0T this isn t necessary, but is convenient. Then our requirement that we will measure 0 with 100% efficiency is mπ ω 0 Tl 63 To have a solution to this equation, Tω 0 has to be an integer multiple of π. If it is, then there is a solution for every l. Therefore, we will have 100% efficientcy when where k is a non-negative integer. T π ω0 k 6
Logan A. Morrison Assignment 5 Problem 3 Page 8 / 11 Problem 3 A particle of mass m bound in a one-dimensional harmonic oscillator potential of frequency ω and in the ground state is subjected to an impulsive force pδt. Find the probability it remains in its ground state. Immediately after the impulse force, the space is the same, but it has obtained a momentum p. That is, immediately after the impulse force, the state is mω 1/ ψx exp ipx/ mωx 65 π But let s prove this. We know that the force acting on the system is F pδt. The potential resulting in this force is Vx, t pδtx. Thus, our Hamilitonian is H ˆp m + 1 mω ˆx pδt ˆx 66 Since the delta function is zero for all times except t 0, the Hamiltonian commutes with itself at different times. Therefore, the time evolution operator is given by U exp i t ˆp t 0 m + 1 mω ˆx pδt ˆx dt 67 exp i ˆp m + 1 mω ˆx t t 0 + ip ˆx 68 Now, ψx, t x Ut, t 0 ψ; t 0 70 x exp i ˆp m + 1 mω ˆx t t 0 + ip ˆx ψ; t 0 71 Now, for arbitrary time, this formula is a mess. But, we can see that if we let t, t 0 0, then, immediately after the impluse, the wave function is 69 ip ψx, t 0 x exp ˆx ψ; t 0 0 7 ip exp x x ψ; t 0 73 ip mω 1/ exp x exp mωx 7 mω π π 1/ ip exp x mωx 75 76
Logan A. Morrison Assignment 5 Problem 3 Page 9 / 11 The probability amplitude that the state is still in the ground state immediately after the impulse force is thus: ψ; t 0 0 ψ; t 0 ψ; t 0 0 x x ψ; t 0 77 mω 1/ exp mωx ipx exp π mωx 78 mω 1/ ipx exp π mωx 79 mω 1/ exp mω x ip p 80 π mω mω exp p mω 1/ exp mω x ip 81 mω π mω exp p mω 1/ 1/ π 8 mω π mω exp p 83 mω and hence the probability the ground state remains in the ground state immediately after the impulse force is ψ; t 0 0 ψ; t 0 exp p mω 8
Logan A. Morrison Assignment 5 Problem Page 10 / 11 Problem For electronic states in a one-dimensional system, a simple model Hamiltonian is H E 0 n n + n1 W n + 1 n + n n + 1 85 n1 where { n } is an orthonormal basis, n n δ nn and E 0 and W are parameters. Assume periodic boundary conditions so that N + j j. Calculate the energy levels and wave functions. Let s find the eigenfunctions of this Hamiltonian. Let E be the energy eigenket corresponding to energy E. Expanding these energy eigenkets in terms of the basis { n }, we have E Now, let s solve the energy eigenvalue problem E E E E H E E 0 n n + W n + 1 n + n n + 1 n1 n1 m1 m1 n1 E 0 n n m c E m E 0 c E m m + W and this has to be equal to E E. That is m1 m1 E 0 c E m m + W + W n0 n1 m1 C E m m 86 n0 C E m m 87 c E m n + 1 n m + c E m n n + 1 m 88 c E m 1 m + ce m+1 m 89 m1 c E m 1 m + ce Since the function n are orthonormal, we can conclude that m+1 m N m1 90 c E m n 91 E 0 c E m m + W c E m 1 m + ce m+1 m Ec E m 9 But recall that we have periodic boundary conditions. I.e. N + j j. This implies that c E N+ j c E j. Now recursion relations like the one we have above are typically solved by assumed that c E m a m where a is some complex number. From the periodic boundary conditions, we see that a N 1. From complex analysis, we know the the N th root of 1 is a e πik/n 93 where k {1,,..., N}. Note the k could be any integer, but, the solutions repeat after k N. Now, plugging in e πimk/n, we find c E m E 0 e πimk/n + W e πim+1k/n + e πim 1k/N Ee πimk/n 9
Logan A. Morrison Assignment 5 Problem Page 11 / 11 Dividing by e πimk/n and using the fact that cosθ e iθ + e iθ, we find that the N engery levels are E k W cosπik/n + E 0 for k {1,,..., N} 95 We can now write down the energy eigen fucntions as well: E k n1 c E k n n 96 e πikn/n n 97 n1 We have used E for our states here instead of E since these aren t properly normalized. The norm of these states is E k E k m1 n1 m1 n1 m1 n1 Thens, the properly normalized energy eigenstates are 98 e πikm/n e πikn/n m n 99 e πikn/n e πikn/n δ mn 100 101 N 10 E k n1 e πikn/n N n and k {1,,..., N} 103