ph40109final

# ph40109final - Physics 401 Final Exam 1 A certain quantum...

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Unformatted text preview: Physics 401 Final Exam May 16, 2009 1. A certain quantum system has only three energy levels, and three corre- sponding eigenstates. Even if it has nothing to do with rotation and angular momentum, we can identify the states with the states (m = 1,0, —1) of an E = 1 “effective” angular momentum system, because the angular momen— tum operators E provide us with a convenient set of operators. In terms of these the Hamiltonian for the system is (a) Verify that the states |m) are eigenstates of H. (b) Find the energy eigenvalues of the three states. (c) The system is initially in the superposition state —|— | — 1)), and evolves according to the time—dependent Schrodinger equation. Evaluate the expectation value (11\$) as a function of time. 2. In this problem we ﬁrst consider a system with an arbitrary angular momentum, which is called L but may be orbital or spin (S From part (b) on we specialize to the angular momentum of a system of two spin % particles. (a) The state vector (wave function) 1b has angular momentum E (L2 = €(E + New state vectors are generated by acting on 1b with I: that is Liw and Lil/jib, where i,j : 51:, y, or ,2. Show that these new state vectors also have angular momentum E. (Hint: Use commutators.) (b) For the system of two spin % particles, show that there is no function —) —) of the S-operators (5(1), 5(2) or if that will change a singlet state into a triplet state. (You may assume that the general function of the S operators is a polynomial.) (c) For the same system of two particles, let Q+ be the operator that changes the singlet state into the m : 0 triplet state, and vanishes on all other states: one 0) = l1 0) 6241 m) = 0 form = —1,0,+1 Is this operator hermitian? If not, ﬁnd its hermitian adjoint Q_, deﬁning it by equations like the one above, or as a matrix in the is m) basis. (d) From what you proved in (c) it follows that Q: cannot be expressed as a function of the S operators. However, Q_Q+ and Q+Q_ can be so 1 expressed. Do so. (One way is ﬁrst to ﬁnd the eigenstate for each of these operators that has non—zero eigenvalue; then construct an operator out of the 3’s that vanishes on all states except that eigenstate.) 3. The (time—independent) wave function for the nth energy state of a Hydro— gen atom with maximal angular momentum has the (not normalized) form wnn_1n_1(7“, 97 : Tn—le—r/na Sinn—le €i(n—1)gb where a is the Bohr radius, and the sequence of subscripts on 1b is 71,6, m, (a) What is the most probable value of 7“ in this state? Compare with the value 7“ = @712 given by the simple Bohr model. (Hint: ﬁrst you must ﬁgure out the probability that the electron would be found between 7‘ and 7“ —1— dr.) (b) For the corresponding time—dependent state @039, gb,t), show that the probability density, and hence the expectation value of all operators, is constant in time. The energy of these states is E, 2 E1 / n2. (c) Time-dependent probability density occurs for superposition states of time—dependent stationary states. Let Q! : \lfnn_1n_1 —|— \llann and show that the probability density consists of time—independent terms as in (b) plus time—dependent “interference terms.” Using the above expression for 1b and its time dependence, and in the limit of large n, write this inter— ference term explicitly as a function of 7“, 0, gb and 75. (Do not worry about normalization.) Show that it really depends only on a linear combination gb—wt, and ﬁnd to. This means that the probability density rotates rigidly with angular frequency to. Compare with the Bohr theory value of this frequency, h/ma2n3, with which the electron rotates about the nucleus. Formulas Time—dependent Schrodinger equation thawéf’t) = H (Mac) Time—independent Schrodinger equation H (Maj) = Ew(\$) Hamiltonian operator H = % —|— V(\$) Momentum operator p : %% Expansion I Z l€n><€n|¢> Commutator [ ,93] : E. Expectation value <Q> = WlQlW Matrix elements an = <€m|Q|€n> 2 2 Allowed energies of H atom En = — (4:60) 1 73—2 Angular momentum 1—: = 77 X 15’ [Lb Ly] = z'hLZ; [L3], L2] = ihL\$; [L2,Lx] zany; [L2,Li] =0 Singlet state |0 0) : ﬁﬁl — lT) Triplet state \1 1) :M |1 0) : ﬁm + H) \1 —1) :tt ...
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## This note was uploaded on 12/29/2011 for the course PHYSICS 402 taught by Professor Anlage during the Spring '09 term at Maryland.

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ph40109final - Physics 401 Final Exam 1 A certain quantum...

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