Moern Physics: Home work 5 Due ate: 0 March. 014 Problem Set 6: Workbook on Operators, an Dirac Notation Solution 1. nswer 1: a The cat is being escribe by the state, ψ >= ea > If we try to observe it we will fin it to be ea because it is in a efinite state ea >. So there is no probability of fining it alive. P alive = 0. By the same argument as in i we are absolutely certain that the cat will be foun ea, So the probability of fining it ea is 1. P ea = 1. b Now the cat is in the state, ψ >= 1 ea > 3 3 alive > Now probability of fining the cat alive is just mo-square of the amplitue associate with state alive > in ψ >. So P alive = 3 = 3 Thus the probability of fining cat alive is /3. Similarly probability of fining the cat ea is just mo-square of the amplitue associate with state ea > in ψ >. P ea = 1 = 1 3 3 c By efinition, probability of an event represents the relative frequency of occurrence of an outcome in a certain experiment. So if we have 1000 cats in lab an they are all in the same state which is given in 1b, we expect to see /3r of them alive an 1/3r of them ea. More precisely, No. of cats we expect to see alive = 1000 667 3 No. of cats we expect to see ea = 1 1000 333. 3 Spring semester 014 1
Moern Physics: Home work 5 Due ate: 0 March. 014 To fin average I must multiply 1 with number of times we fin cat alive, 1 with number of times we fin cat ea; a the two figures an ivie by the total number of cats in the lab. 6671 333 1 verage = 1000 = 334 1000 = 1 3 e If I observe it an fin it alive, I can be certain that I will fin it if I observe it again immeiately. So the probability that I will fin it alive is 1. If I observe it after an hour then the cat may be alive or ea. It epens on how its wavefunction evolve with time over this hour, an that epens on all the forces acting on cat, incluing internal an external forces. The conition of repetition of results only hols for observations repeate immeiately after the first one. Otherwise, the wavefunction changes with time accoring to Schroeinger equation an the probabilities of ifferent outcomes also change unless the system is in a stationary state. Since, in this case it is almost impossible to solve the schroinger equation which involves gazillions of cat particles an their forces, we can t really be sure what will happen at least through application of quantum mechanics. If I observe it after two centuries, I can be certain that the cat will be ea. No cat in history has live that long an quantum mechanics shoul give the answers which are consistent with observations as oppose to intuitionfingers crosse, or, in other case, we shoul think about its valiity.. nswer : a Let s consier the two vertical alignments of the magnet as vertical arrows as shown in the figure below. N S = = S N Spring semester 014
Moern Physics: Home work 5 Due ate: 0 March. 014 Since there are only two possibilities an. Which have energy values E 0 an E 0 respectively. There are only two possible outcomes of energy: E 0 an E 0. b Magnet 1 is being escribe by the state, ψ >= 1 10 3 > 3 eiπ/4 >. This state is normalize. It is each to check: 1 3 10 1 3 eiπ/4 = 1 5 6 6 = 1 6 5 6 = 1 So there is probability 1/6 that magnet 1 is in state an thus it has energy E 0 an probability 5/6 that magnet 1 is in state an therefore, has energy E 0. So out of 10, 000 magnets, number of magnets with energy E 0 is expecte to be 1 10000/6 = 1667 an number of magnets with energy E 0 is expecte to be 5 10000/6 = 8333. c verage value by the same proceure as in problem 1 will be, 3. nswer 3: verage = 1667E 0 8333 E 0 10, 000 = 0.67 E 0. a State E > is the state corresponing to efinite energy value E. So upon measurement we will fin, b If the electron is in state, It has three possible values of energy, E = 13.6 = 3.4 ev. ψ >= a E 1 > b E 3 > c E 4 >. E 1 = 13.6 ev E 3 = 13.6 = 1.51 ev 9 an E 4 = 13.6 = 0.85 ev, 16 Spring semester 014 3
Moern Physics: Home work 5 Due ate: 0 March. 014 with probabilities, a a b c, b a b c an c a b c respectively. It is the principle of superposition that is at han here. If a system is in a state that is linear combination of eigenstates of a measurement operator, then upon measurement it will collapse into either one of the eigenstates an the probability of oing so in a particular eigenstate is equal to the mo-square of amplitue corresponing to that eigenstate. c Let the total number of electrons be N. Then the number of electrons with energy E 1 is a N a b c, with energy E 3 is b N a b c an with energy E 4 is So the average energy is, E vr. = 1 [ ] a NE 1 N a b c b NE 3 a b c c NE 4 a b c = a E 1 b E 3 c E 4 a b c c N a b c. 4. nswer 4: For a free particle, energy operator or Hamiltonian is just is no potential present. So m x as there m x eikx = m x ikeikx = m k e ikx = k m eikx. So the plane wave e ikx is an eigenstate of the energy operator for a free particle. The eigenvalue corresponing to the energy of the free particle which is k /m. 5. nswer 5: If the particle is escribe by the wavefunction ψx = xe, it will have efinite energy without any uncertainty only if this wavefunction is an eigenstate of the energy operator. For a harmonic oscillator V x = 1 x. Spring semester 014 4
Moern Physics: Home work 5 Due ate: 0 March. 014 So the energy operator is m m x 1 x x e x 1 x. = m x e 1 x x 3 e e x e 1 x 3 e = m x = m x e ω = m xe ω 1 x 3 e x x e 1 x 3 e xe x3 e = ω xe ωxe 1 x 3 e So xe 1 x 3 e = 3 ω xe. is an eigenfunction of energy operator of Harmonic oscillator. The eigenvalue corresponing to this eigenfunction is 3 ω, which is the efinite energy value of the particle with this wavefunction. 6. nswer 6: m x x ψx = m x 1 x = m = m = m = m x e 1 x 4 e x e x x x e ω x e x e 1 x 4 e xe x3 e 1 x 4 e x x3 e 1 x 4 e ω 3x e x4 e = m e 5 ωx e 1 x 4 e 1 x 4 e = m e 5 ωx e x e for any real number. Hence ψx = x e is not an eigenfunction of energy. Now we cannot preict with certainty what value will be obtaine if we measure energy. Spring semester 014 5
Moern Physics: Home work 5 Due ate: 0 March. 014 7. nswer 7: a Energy operator for a free particle is just B sin kx is an eigenfunction of m m x m or not. cos kx B sin kx = x m Thus ψx is an eigenfunction of = k m x, so checking if ψx = cos kx k sin kx Bk cos kx x sin kx B cos kx x = k k cos kx Bk sin kx m = k cos kx B sin kx. m m particle with this wavefunction is k m. x. The value of the energy of the free b We can check if ψx is an eigenfunction of momentum operator ˆp = i x or not. i cos kx B sin kx = i k sin kx Bk cos kx x = i k sin kx B cos kx where D is some real number. D cos kx B sin kx, To see what are the possible outcomes we can write ψx as a linear combination of eigenfunctions of momentum operator ˆp. these eigenfunctions are of the form e ikx, where k is some real number relate to momentum eigenvalue p as: k = p. Now using Euler s formula: e ikx = cos kx i sin kx 1 an e ikx = cos kx i sin kx e ikx = cos kx i sin kx ition an subtraction of equation 1 an gives, an cos kx = eikx e ikx sin kx = eikx e ikx i Spring semester 014 6
Moern Physics: Home work 5 Due ate: 0 March. 014 respectively. ψx = eikx e ikx B i eikx e ikx = B e ikx i B e ikx. i ψx is a linear combination of e ikx an e ikx, which are eigenfunctions of ˆp = i x with eigenvalues k an k respectively. check! This implies that possible outcomes for momentum are k an k. Probabilities associate with these outcomes are mo-square of amplitues of the corresponing wavefunctions in ψx. So the probability of measuring k will be, B i P 1 = B i B i B i = = 1 = 1. B B i i B i B i B i B i B i B i Similarly the probability of measuring k will be, B i P = B i B i = 1. B i c Possible momentum values are k an k; both of which correspon to the same energy value: p = k m m. Physically, the ifference of minus sign in the two momentum values translate to the particle moving in right irection or in left irection with the magnitue of the momentum an thus the energy value Spring semester 014 7
Moern Physics: Home work 5 Due ate: 0 March. 014 staying the same. So we are certain about the energy of the particle but not its momentum. I check if ψx = xe is an eigenfunction of ˆp = i i x xe x = i e = i e F xe, x. x e ix e for any real number F. So for the wavefunction ψx = xe, momentum is not efinite. Spring semester 014 8