Phys460.nb 81 ψ n (t) is still the (same) eigenstate of H But for tdependent H. The answer is NO. 5.5.5. Solution for the tdependent Schrodinger s equation If we assume that at time t 0, the electron starts out with spin along B(t 0), (i.e., we start from one of the two eigenstates of H at t 0), we get initial condition ψ(0) χ + (0) (5.170) In the same time, we know that the state need to follow the tdependent Schrodinger s equation i ħ d ψ(t) H(t) ψ(t) (5.171) This differential equation and the initial condition mentioned above, uniquely determine ψ(t). If we find the solution, we will see that ψ(t) χ + (t) for t > 0, although we start from ψ(0) χ + (0). The Schrodinger s equation: i ħ d β(t) H(t) β(t) γ(t) γ(t) (5.17) i ħ d β(t) i ħ d γ(t) ħ ω 1 cos α α e +i ω t α e i ω t cos α β(t) γ(t) ħ ω 1 cos α β(t) + α e i ω t γ(t) α e +i ω t β(t) cos α γ(t) (5.173) Two differential equations: i ħ dβ(t) i ħ dγt) ħ ħ cos α β(t) + α e i ω t γ(t) α e +i ω t β(t) cos α γ(t) (5.174) The initial condition β(t 0) γ(t 0) χ+ (t 0) which means β(0) cos α γ(0) α cos α α (5.175) (5.176) With this initial condition and the Shrodinger s equation, we get one and only one solution where ψ(t) cos t i ω t cos α ei ω t/ cos t i +ω t α (5.177) ei ω t/ ω + ω 1 ω ω 1 cos α (5.178) 5.5.6. How did we find the solution? (not required) i ħ dβ(t) i ħ dγt) dβ(t) dγt) ħ ħ cos α β(t) + α e i ω t γ(t) α e +i ω t β(t) cos α γ(t) i cos α β(t) + α ei ω t γ(t) i α e+i ω t β(t) cos α γ(t) (5.179) (5.180) Define
8 Phys460.nb β(t) f (t) ei ω t/ (5.181) and γ(t) g(t) e i ω t/ ei ω t/ e i ω t/ i cos α f (t) ei ω t/ + α e i ω t g(t) e i ω t/ i α e+i ω t f (t) e i ω t/ cos α g(t) e i ω t/ (5.18) (5.183) So we have e i ω t/ i ω f (t) ei ω t/ i (cos α f (t) + α g(t) ) ei ω t/ e i ω t/ + i ω g(t) ei ω t/ i ( α f (t) cos α g(t) ) ei ω t/ (5.184) i ω + i ω f (t) i g(t) i (cos α f (t) + α g(t) ) i ( α f (t) cos α g(t) ) i cos α f (t) i α g(t) α f (t) + i cos α g(t) (5.185) i i cos αω f (t) i α g(t) α f (t) + i cos αω g(t) (5.186) For these differential equations, all terms are proportional to f or g (or their derivative) and all the coefficients are constants, independent of time t. For this structures, the solution must be exponential functions of t f (t) A ei ω' t g(t) B e i ω' t i ω' A ei ω' t i i ω' B e i ω' t i α A ei ω' t + i ω' A cos αω cos αω A ω' B α B cos αω α A + B A e i ω' t i α B ei ω' t cos αω (5.187) B e i ω' t (5.188) Now we convert the differential equations to algebra equations. This equation is actually an eigenvalue problem for a matrix cos αω α α cos αω A B ω' A B where ω' is the eigenvalue for the matrix M cos αω α α cos αω The eigenvalues for of a matrix is the solution of equation det(ω' I M) 0 det ω' + cos αω There are two eigenvalues α cos αω α ω' ω' + ω 1 cos α ω, while A is the eigenvector B ω' ω 1 cos α ω (5.189) (5.190) ω 1 α 0 (5.191) ω' ± ± ω 1 α + ω 1 cos α ω ± ω + ω 1 ω 1 ω cos α (5.19) We can define ω + ω 1 ω 1 ω cos α (5.193) so the eigenvalues are ω' ± ± (5.194)
Phys460.nb 83 For ω + ' /, the eigenvector satisfies cos αω α α cos αω A B + A B (5.195) The solution is A+ B + α cos αω α ω cos α (5.196) For ω ' /, the eigenvector is A B α ω cos α+ (5.197) So any solution of the Schrodinger equation is ψ(t) X e i t A + ei ω t/ + Y e i B + e i ω t/ t A ei ω t/ B e i ω t/ (5.198) Ug the initial condition ψ(0) X A + B + + Y A B (X + Y) (X + Y) α ω cos α (X Y) cos α α (5.199) We know that X + Y cos α ω 1 α cos α ω 1 α cos α 1 ω 1 α (5.00) X Y (X + Y) ω ω 1 cos α So we find that X + ω ω 1 ω 1 α α ω ω 1 cos α ω 1 α ω 1 α ω 1 α ω ω 1 cos α + α ω 1 α ω ω 1 ω 1 α (5.01) (5.0) Y ω + ω 1 ω 1 α (5.03) So we have ψ(t) + ω ω 1 ω 1 α e i t α ei ω t/ ω cos α e i ω t/ + ω + ω 1 ω 1 α e i t α ei ω t/ ω cos α+ e i ω t/ +ω + cosα+ (1cos α)ω (1+cos α) 4 α cos t + i (ω ω 1) t α ei ω t/ α cos t + i ω t cos α ei ω t/ (1cos α) cos t + i +ω + α e i t + ω cosα (1cos α)+ω (1+cos α) 4 α cos αω (1+cos α) α cos t + i ω t cos α ei ω t/ (1cos α) cos t + i (1cosα)ω (1cos α) α α t ei ω t/ t ei ω t/ e i t e i ω t/ (5.04)
84 Phys460.nb cos t + i ω t cos α ei ω t/ cos t i +ω t 1cos α e i ω t/ α cos t i ω t cos α ei ω t/ cos t i +ω t α ei ω t/ Exactly the same as the solution that we shown above ψ(t) cos t i ω t cos α ei ω t/ cos t i +ω t α (5.05) ei ω t/ 5.5.7. Probability for the system to change to another state For H(t) at time t, the two eigenstates are χ + cos α e i ω t α (5.06) and χ α e i ω t cos α (5.07) We start from state χ + at t 0. The probability to find the system in χ at a later time t is P χ ψ(t) α cos t cos t i ω 1 ω i ω 1 ω e i ω t cos α cos t i ω t cos α ei ω t/ cos t i +ω t cos α ei ω t/ t cos α α ei ω t/ cos t cos t + i ω 1 + ω t t i ω t α cos α 4 ω t α cos α ω t α cos α ω t [(α)] ω α t i ω 1 + ω α cos α t α cos α ei ω t/ (5.08) Although we start from χ at t 0, at time t, there is a nonzero probability for the system to become χ +. Although we start from spin parallel to B at t 0, at time t, there is a nonzero probability for the system to become antiparallel to B. This phenomenon (system start from a quantum state and ends at a different quantum state, by applying a tdependent H) is called a quantum tunneling. The probability to find the system in χ + is P + χ + ψ(t) (5.09) This is the probably for the system to remain in the same state (E ). It requires no calculation to show that P + 1 P, because the total probability must be 1. This is indeed the case
Phys460.nb 85 P + χ + ψ(t) cos α cos t cos t cos t cos t cos t t cos 1 t i ω 1 ω i ω 1 ω i ω 1 t i ω 1 t t + i ω cos α ω 1 + ω cos α ω 1 1 1 ω cos α ω 1 e i ω t α α cos ei ω t/ + cos t + cos t t α cos + i ω t α cos + i ω t cos α t t + ω cos α ω 1 t t 1 (ω cos α ω 1 ) t cos t i ω t cos α ei ω t/ cos t i +ω t cos α ei ω t/ i ω 1 + ω α i ω 1 + ω t t α α ei ω t/ Because ω + ω 1 ω ω 1 cos α P + 1 ω (ω cos α) t (ω α) 1 t 1 ω α t 1 P (5.11) 5.6. Adiabatic theorem If we rotate the B field very slowly (ω << ω 1 ), the probability for quantum tunneling vanishes P ω α t ω α ω + ω 1 ω ω 1 cos α t (5.1) which vanishes when ω 0. In other words, if we changes H really slowly, the system will remain in the same quantum state. If we start from the ground state, the system will always in the ground state. If we start from the nth excited state, it will remain in the nth excited states. etc. Very slowly in physics is called adiabatically. Conclusion: for adiabatic procedures (i.e., changing H really slowly), the system will remain in the same quantum state. Comments: (1) The rigorous proof can be found in the textbook, which utilize timedependent perturbation theory () How slow is slow? Assume that we start from state n. The energy difference between this state and other states is ΔE E n E m. For the smallest ΔE, we define a time scale τ ħ/δe. Slow here means that when we change the Hamiltonian, any change need to take much longer than then τ. In the example above, we need to require the rate of change ω to be much slower than 1 τ ΔE ħ. For that example, ΔE ħ ω 1, so it means that ω << ω 1