Solutions to Problems 1-6 in Quantum Mechanics: Uncertainty, Eigenvectors, and Operators, Assignments of Health sciences

Download Solutions to Problems 1-6 in Quantum Mechanics: Uncertainty, Eigenvectors, and Operators and more Assignments Health sciences in PDF only on Docsity! PHY 389K QM1, Homework Set 2 Solutions Matthias Ihl 09/21/2006 Note: I will post updated versions of the homework solutions on my home- page: http://zippy.ph.utexas.edu/~msihl/PHY389K/ 1 Problem 1 Let us determine the state W in terms of the heights hk of the peaks. W = 7 ∑ k=0 wk|Ek >< Ek|, (1) with |Ek >< Ek| =: Λk and wk = hk ∑7 k=0 hk . (2) From fig. 4.2(b), I determined the wk to be: w0 = 0.627, w1 = 0.134, w2 = 0.068, w3 = 0.051, w4 = 0.031, w5 = 0.024, w6 = 0.031, w7 = 0.034. Moreover, ∆E = ~ω ≈ 0.26 eV for CO. (a) Now we can calculate the uncertainty for H : ∆W H = √ < H2 >W − < H >2W = √ Tr(H2W ) − (Tr(HW ))2 = ~ω √ √ √ √ 7 ∑ k=0 (k + 1 2 )2wk − ( 7 ∑ k=0 (k + 1 2 )wk )2 ≈ 1.885~ω ≈ 0.49eV . 1 (b) Let us first look at the uncertainty of H in an energy eigenstate W = Λn: ∆ΛnH = √ < H2 >Λn −(< H >Λn)2 = √ Tr(H2Λn) − (Tr(HΛn))2 = √ E2n − E2n = 0. This result is not very surprising, because this is what it means to be in an energy eigenstate (cf. page 76 of the textbook). The uncertainty of P in an energy eigenstate, however, should not vanish, because the |En > are not simultaneously eigenstates of P or Q: ∆ΛnP = √ < P 2 >Λn −(< P >Λn)2 = √ Tr(P 2Λn) − (Tr(PΛn))2 = √ ~µω(n + 1 2 ) 6= 0, where < P >Λn= 0 as usual. 2 Problem 2 First, let us investigate what eigenvectors we get by acting with N = a†a on a|n > and a†|n >: N(a|n >) = a†aa|n >= (aa†−1)a|n >= a(a†a−1)|n >= (n−1)a|n > . (3) Therefore, for n = 0, we have a|n >= 0 and for n 6= 0, we get a|n >∝ |n − 1 >,i.e., an eigenvector to eigenvalue (n − 1). Similarly, N(a†|n >) = a†aa†|n >= a†(a†a + 1)|n >= (n + 1)a†|n > . (4) Thus, we get an eigenvector to eigenvalue (n+1), a†|n >∝ |n+1 >. Next, we want to determine the corresponding proportionality constant. Let a|n >= cn|n − 1 > and a†|n >= dn|n + 1 >. Then, n =< n|a†a|n >=< n − 1|c∗ncn|n − 1 >⇒ |cn| = √ n, (5) and n + 1 =< n|aa†|n >=< n + 1|d∗ndn|n + 1 >⇒ |dn| = √ n + 1. (6) Again, these coefficients are only determined up to a complex phase which can be conveniently set to 1. 5 Problem 5 We start from the canonical commutation relation [Q, P ] = i~, which can be used to define creation and annihilation operators, a = 1 2~ (√ cQ + i√ c P ) , a† = 1 2~ (√ cQ − i√ c P ) . (17) The constant c will be determined below. These operators satisfy the usual relation [a, a†] = 1. Conversely, we may express Q and P in terms of these operators: Q = √ ~ 2c (a + a†), P = −i √ ~c 2 (a − a†). (18) Plugging this into H yields H = −~c 4µ (a − a†)2 + ~k 4c (a + a†)2 + k1I. (19) In order to ensure that both terms have the same dimensionality (units of energy), we choose c = √ kµ. (20) Thus, H reads H = −~ 4 √ k µ (a− a†)2 + ~ 4 √ k µ (a + a†)2 + k1I = ~ √ k µ (a†a + 1 2 I) + k1I. (21) Since H is again diagonal, we can introduce discrete energy eigenstates |En > (as before), H|En >= En|En >, where En is given by En = ~ √ k µ (m + 1 2 ) + k1. (22) Note, for later convenience, that the oscillator frequency ω = √ k µ . 6 Problem 6 (a) We want to express k3 and k4 in terms of k1, k2, k4 and k: 1 2 kQ′2 + k3I = 1 2 k ( Q2 + 2k4Q + k 2 4I ) + k3I = 1 2 kQ2 + kk4Q + ( 1 2 kk24 + k3)I, from which we read off k4 = k1 k , and k3 = k2 − 1 2 kk24 = k2 − 1 2 k21 k . (23) (b) Now let us calculate the new commutation relation. [P, Q′] = [P, Q + k1 k I] = [P, Q] + k1 k [P, I] = [P, Q] = −i~. (24) (c) Since [P, Q′] = [P, Q] = −i~, one can again define creation and annihila- tion operators a = 1 2~ (√ µωQ′ + i√ µω P ) , a† = 1 2~ (√ µωQ′ − i√ µω P ) , (25) satisfying [a, a†] = 1. Then one can use the result of Problem 5 to find the eigenvalues of H on the eigenstates |Em >. The only difference is the constant (energy) shift, which is now k3 = k2 − 12 k2 1 k : H|Em >= Em|Em >, with Em = ~ √ k µ (m + 1 2 ) + k2 − 1 2 k21 k . (26) (d) The eigenstates |n >:= |En > can be constructed from the ground state as follows |n >= 1√ n! a†n|0 > . (27) Here, a† is given by a† = 1 2~ (√ µωQ′ − i√ µω P ) = 1 2~ (√ µω(Q + k1 k I) − i√ µω P ) . (28) Therefore, |n >= 1√ n! ( 1 2~ )n [√ µω(Q + k1 k I) − i√ µω P ]n |0 > . (29)