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# Quantum Physics PHYS 130A

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This 5 page Class Notes was uploaded by Mrs. Taryn Treutel on Thursday October 22, 2015. The Class Notes belongs to PHYS 130A at University of California - San Diego taught by Lu Sham in Fall. Since its upload, it has received 46 views. For similar materials see /class/226819/phys-130a-university-of-california-san-diego in Physics 2 at University of California - San Diego.

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Date Created: 10/22/15

LJ Sham April 24 2009 Physics 130A Problem Set 4 1 Positive energy eigenvalues a Show that the expectation value of kinetic energy is always nonnegative by proving A 7 52 war WKW dz 695 a 1 7 Hint take one differential operator and integrate by parts Specify conditions needed b If the potential energy of a particle V 2 0 show that its energy eigenvalues are always nonnegative Hint examine the expectation value of the Hamiltonian in an energy eigenstate as the sum of kinetic and potential energies 2 First excited state of SHO a Find the functional dependence on position in dimensionless form of the rst excited state from 1 cle starting with the normalized Gaussian form of the ground state b Without evaluating the integral prove that the wave function for the excited state you have obtained is normalized 0 Again without differentiation nor integration verify the eigenenergy of the excited state 3 Grif ths p 50 Problem 212 4 Deduction of Properties of the State A quantum onedimensional harmonic oscillator has mass m classical angular frequency w and the bottom of the potential at the origin of the coordinate axis At time t 0 it is in a state with 1 equal probability in the eigenstates with energies 71 and n 7 II zero mean position and III the mean momentum in the positive axis direction a Find a wave function of the oscillator at time t 0 satisfying all three conditons b Find the mean values of its position and momentum at a subsequent time t 0 Consider a classical oscillator with its total energy equal to the mean energy of the above quantum oscillator Does the classical amplitude of oscillation agree with the largest mean value of the position of the quantum oscillator in b LJ Sham April 24 2009 Physics 130A Problem Set 4 1 Positive energy eigenvalues a Show that the expectation value of kinetic energy is always nonnegative by proving A 52 3 6 ltle gt dz 5 1 Hint take one differential operator and integrate by parts Specify conditions needed b If the potential energy of a particle V 2 0 show that its energy eigenvalues are always nonnegative Hint examine the expectation value of the Hamiltonian in an energy eigenstate as the sum of kinetic and potential energies Solution a h away l Elmo nd 595 7 2m 700 h 3x7 provided that 7 and its derivative tend to zero as x a ioo For later use I would add an alternate periodic condition wz 7L wz L and zZm 7L WW L as L a 00 b 52 97 61 ltwlf1lwgt dz 6 a dz wmlzwx 2 0 3 If 7 is an energy eigenstate with eigenvalue E and normalized ltwl lwgt WEW E 2 o 4 2 First excited state of SHO a Find the functional dependence on position in dimensionless form of the rst excited state from 1 cle starting with the normalized Gaussian form of the ground state b Without evaluating the integral prove that the wave function for the excited state you have obtained is normalized c Again without differentiation nor integration verify the eigenenergy of the excited state Solution a Starting with 1 cle 1 a 1 14 162 1 14 1 1515 E 5 5 555 4 14 d 5562 5 7r b The proof of normalization is WillJD ltCl ol01 ogt d5 cworcwo d zbSCCTwO using integration by parts to switch cl to c7 d 3 clc l we 17 since cwo 0 6 c In terms of the operators sz1 710le Clo clwo clccl cl we clclc 1 cl we gwl 7 using from step 3 to step 4 Gel clc l 3 Gri iths p 50 Problem 212 Solution a Position expectation value of state 71 ltxgt wcmw 111 mm WWW 0 8 Or from symmetry consideration Similarly p 0 b 2 h 1 2 lt95gt 7lt nl00 WM 9 277m 2 Work on just the dimensionless part c cl2 02 cl2 ccl clc 02 cl2 25 17 10 using 07 cl on the third term ccl of the middle expression The expectation value of 02 is zero because the operator lowers 71 to 71 2 Similarly ltcl2gt 0 The state 71 is the eigenstate of the number operator clc clad ml 11 Hence ltwnlccl2lwngt 2nl 12 Finally 5 l ltz2gt 61 i 13 c We could work out ltp2gt in the same way as the previous part using c 7 cl2 Just for fun try this The total energy for state 71 is n is made up of the kinetic energy part and the potential energy part 7 L 2 l 2 2 ltHgt 2mltpgt27m2lt9cgt 14 1 1 1 n5w ltp2gt n m 15 Hence ltp2gt mm n 16 d Uncertainty relation AzAp ltZgtltp2gt 571 2 l7 4 Deduction of Properties of the State A quantum onedimensional harmonic oscillator has mass m classical angular frequency w and the bottom of the potential at the origin of the coordinate axis At time t 0 it is in a state with 1 equal probability in the eigenstates with energies 71 and n 7 II zero mean position and III the mean momentum in the positive axis direction a Find a wave function of the oscillator at time t 0 satisfying all three conditons b Find the mean values of its position and momentum at a subsequent time t 3 c Consider a classical oscillator with its total energy equal to the mean energy of the above quantum oscillator Does the classical amplitude of oscillation agree with the largest mean value of the position of the quantum oscillator in b Solution a The state at t 0 involves only 71 and 71 1 states Condition 1 tells us that both coef cients have equal moduli in the formula w 0 zz mile 18 The phase of one of them is arbitrary and dispensable but the relative phase of the other one to the former is essential Since we need it later W wne iEnt walewe iEwh 19 7inwt 1 Wm 20 7 5 E l 107L716 Then the position expectation value at time t is lt I tl1l1 tgt 0M e W W Wnill 95 WW aneiwww 7 Wnlmwnil 6iltwwt 5 i4 wtlt nillxl ngt 7 since ltwnlxl ngt 07 llt gt wnlmlwnAWKWWD using as a shorthand ltwnlzlwn1gt cosltp wt 56 O Hle 21 For the last line we have used ltwn4lxlwngt as a real integral ltwnllzbn1gt The amplitude 7i VmltwnKCCTgtl nilgt h tMwnl lwni n5 7 22 27mm Thus by condition 11 att 0 lt gt i m 0 23 m 7 cos 277m p 4

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