Answer Happy

Example 3.8 Imagine a system in which there are just two linearly independent states:32 (1) = (6) and 12) = () The most general state is a normalized linear combination: |S) = a1) +b|2) = (8). with a2 + b² = 1. The Hamiltonian can be expressed as a (hermitian) matrix (Equation 3.83); suppose it has the specific form g H= (1² h 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 equation³3 says d iħ |S(t)) = Ĥ|S(t)). (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 Ĥ. The characteristic equation determines the eigenvalues: h - E 8 det = (h-E)²-g² = 0⇒h-E = Fg ⇒ E=h±g. g h-E Evidently the allowed energies are (h+g) and (hg). To determine the eigenvectors, we write 8 (a) = (h± 8) (a). g) ⇒ha + gß = (hg)a ⇒ B= ±α, B so the normalized eigenvectors are |$+) = = √/2 (+1). Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian: 1 |S(0)) = -(1) = √ √ 2 (1+) - √√2 -
Example 3.8 Imagine a system in which there are just two linearly independent states:32 (1) = (6) and 12) = () The most general state is a normalized linear combination: |S) = a1) +b|2) = (8). with a2 + b² = 1. The Hamiltonian can be expressed as a (hermitian) matrix (Equation 3.83); suppose it has the specific form g H= (1² h where g and h are real constants. If the system starts out (at t = 0) in state [1), what is its state at time t? Solution: The (time-dependent) Schrödinger equation³3 says d iħ |S(t)) = Ĥ|S(t)). (3.87) dt As always, we begin by solving the time-independent Schrödinger equation: Ĥ\s) = E\s); (3.88) that is, we look for the eigenvectors and eigenvalues of Ĥ. The characteristic equation determines the eigenvalues: h - E 8 det = (h-E)²-g² = 0⇒h-E = Fg ⇒ E=h±g. g h-E Evidently the allowed energies are (h+g) and (hg). To determine the eigenvectors, we write 8 (a) = (h± 8) (a). g) ⇒ha + gß = (hg)a ⇒ B= ±α, B so the normalized eigenvectors are |$+) = = √/2 (+1). Next we expand the initial state as a linear combination of eigenvectors of the Hamiltonian: 1 |S(0)) = -(1) = √ √ 2 (1+) - √√2 -
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.