QUANTUM MECHANICS. Third edition Please refer to the result of Example 3.8 of the book to solve it.
Posted: Wed May 25, 2022 6:14 am
QUANTUM MECHANICS. Third edition
Please refer to the result of Example 3.8 of the book to solve it.
Y Using the result of Example 3.8 in the textbook, answer the following questions. (a) Calculate the energy expectation value, (E)(t) = (S(t)|Ĥ|S(t)) and describe its change in time. (b) Calculate the uncertainty of energy, AE(t) = √√(E²)(1) – ((E)(1))². -
3.6 Vectors and Operators Finally, we tack on the standard time-dependence (the wiggle factor) exp(-i Ent/h): |S(t)) = -i(h+g)t/h|s+) + e¯i(h-g)t/h|s_) √2 =/=/e-thi/h [e-181/h (1) e-igt/h 2 +eist/h (-:-)) 1 -iht/h (e-igt/h = +eigt/h e-igt/h-eigt/h -iht/h = 2 cos (gt/h) -i sin(gt/h) {)). If you doubt this result, by all means check it: Does it satisfy the time-dependent Schrödinger equation (Equation 3.87)? Does it match the initial state when t = 0?34
Example 3.8 Imagine a system in which there are just two linearly independent states:32 | 1) = (6) and (2): (). The most general state is a normalized linear combination: |S) = a1) +b|2) = with a² + b² = 1. The Hamiltonian can be expressed as a (hermitian) matrix (Equation 3.83); suppose it has the specific form g H = (½ where g and h are real constants. If the system starts out (at t = 0) in state 1), what is its state at time t? Solution: The (time-dependent) Schrödinger equation ³3 says iħ hd (S(1)) = Â|S(1)). (3.87) dt As always, we begin by solving the time-independent Schrödinger equation: Ĥ|s) = E|s); (3.88) that is, we look for the eigenvectors and eigenvalues of Â. The characteristic equation determines the eigenvalues: h- E 8 det (hE)²-g² = 0⇒h-E= Fg ⇒ Et = h±g. g h - E Evidently the allowed energies are (h+g) and (hg). To determine the eigenvectors, we write 8 a α (5)) = = (h± g) ⇒ha + gß = (hg)a⇒ B= ±ta, so the normalized eigenvectors are 1941-(4) = Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian: (S(0)) = ( ! ) = -(6)=√(18+) =([s+) + \s-)).
Please refer to the result of Example 3.8 of the book to solve it.
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3.6 Vectors and Operators Finally, we tack on the standard time-dependence (the wiggle factor) exp(-i Ent/h): |S(t)) = -i(h+g)t/h|s+) + e¯i(h-g)t/h|s_) √2 =/=/e-thi/h [e-181/h (1) e-igt/h 2 +eist/h (-:-)) 1 -iht/h (e-igt/h = +eigt/h e-igt/h-eigt/h -iht/h = 2 cos (gt/h) -i sin(gt/h) {)). If you doubt this result, by all means check it: Does it satisfy the time-dependent Schrödinger equation (Equation 3.87)? Does it match the initial state when t = 0?34
Example 3.8 Imagine a system in which there are just two linearly independent states:32 | 1) = (6) and (2): (). The most general state is a normalized linear combination: |S) = a1) +b|2) = with a² + b² = 1. The Hamiltonian can be expressed as a (hermitian) matrix (Equation 3.83); suppose it has the specific form g H = (½ where g and h are real constants. If the system starts out (at t = 0) in state 1), what is its state at time t? Solution: The (time-dependent) Schrödinger equation ³3 says iħ hd (S(1)) = Â|S(1)). (3.87) dt As always, we begin by solving the time-independent Schrödinger equation: Ĥ|s) = E|s); (3.88) that is, we look for the eigenvectors and eigenvalues of Â. The characteristic equation determines the eigenvalues: h- E 8 det (hE)²-g² = 0⇒h-E= Fg ⇒ Et = h±g. g h - E Evidently the allowed energies are (h+g) and (hg). To determine the eigenvectors, we write 8 a α (5)) = = (h± g) ⇒ha + gß = (hg)a⇒ B= ±ta, so the normalized eigenvectors are 1941-(4) = Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian: (S(0)) = ( ! ) = -(6)=√(18+) =([s+) + \s-)).