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Quantum Mechanics & Uncertainty Principle: Physics HW 7 for Uni Students, Assignments of Quantum Physics

Four problems from a university-level physics homework assignment, focusing on topics in quantum mechanics and the uncertainty principle. Students are asked to derive various relationships and solve for probabilities and energies. The problems are based on textbooks by griffiths and shankar.

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Pre 2010

Uploaded on 03/16/2009

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Download Quantum Mechanics & Uncertainty Principle: Physics HW 7 for Uni Students and more Assignments Quantum Physics in PDF only on Docsity! Physics 486 Homework 7 Due March 11, 2005 1) This is Griffiths 3.27 (page 125) [10 points] ψ1 and ψ2 are eigenstates of operator  (observable A) with eigenvalues a1 and a2. φ1 and φ2 are eigenstates of operator B̂ (observable B) with eigenvalues b1 and b2. The eigenstates are related by: ψ1 = (3φ1 + 4φ2)/5 and ψ2 = (4φ1 - 3φ2)/5 a) A is measured, obtaining result a1. What is the state of the system immediately after this measurement? b) If B is now measured, what are the possible results, and what are their probabilities? c) Immediately after step b), A is measured again. What is the probability of obtaining a1? Note that the result of the B measurement was not specified. 2) Similar to Griffiths 3.15 (page 113) and Shankar 9.4.4 (page 244) [10 points] a) Derive the energy-position uncertainty relation, ! x ! E " ! 2m p , for a particle moving in an arbitrary potential, V(x). b) Why is this relation not useful for stationary bound states? 3) This is similar to Griffiths 3.31 (page 126) [10 points] Griffiths’ statement of the problem has more detail, which might be useful. a) Derive the virial theorem: For stationary states, KE = 1 2 x dV dx . This is a very useful relationship. For example, if V = xn, then x dV dx = n V , and KE = n 2 V . It holds classically as well, if one is careful with averages. b) Does the virial theorem hold when the system is not in a stationary state? Why or why not? 4) This is Shankar 4.2.3 (page 139) [5 points] Show that if ψ(x) has mean momentum <p>, then eik0 x! (x) has mean momentum <p>+p0. (p0 = k0)
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