The state of an ammonia molecule approximated by a two-state system as in the Feynman Lectures can be written [itex]|\psi\rangle = C_{1}|1\rangle + C_{2}|2\rangle[/itex], where the general solution for the coefficients is(adsbygoogle = window.adsbygoogle || []).push({});

$$

C_{1} = \frac{a}{2}e^{i(E_{0} - A)t/\hbar} + \frac{b}{2}e^{i(E_{0} + A)/\hbar}$$

and

$$

C_{2} = \frac{a}{2}e^{i(E_{0} - A)t/\hbar} - \frac{b}{2}e^{i(E_{0} + A)/\hbar}$$

By taking [itex]a = \sqrt{2}[/itex] and [itex]b = 0[/itex], we get a stationary state [itex]|I\rangle[/itex] with energy [itex]E_{0} - A[/itex]. By taking [itex]a = 0[/itex] and [itex]b= \sqrt{2}[/itex], we get a stationary state [itex]|II\rangle[/itex] with energy [itex]E_{0} + A[/itex]. For each of these states, we have [itex]\Delta E =0[/itex].

By taking [itex]a = 1[/itex] and [itex]b = 1[/itex], the molecule is in state [itex]|1\rangle[/itex] at [itex]t = 0[/itex] and in state [itex]|2\rangle[/itex] at [itex]t = \frac{\pi \hbar}{2A}[/itex]. The probability oscillates between states [itex]|1\rangle[/itex] and [itex]|2\rangle[/itex], and the system is in a constant superposition of states [itex]|I\rangle[/itex] and [itex]|II\rangle[/itex], so we have [itex]\Delta E = A[/itex] for this state.

Now, what about the other possible solutions? For example, what does the solution with [itex]a = \sqrt{1.5}[/itex] and [itex]b = \sqrt{0.5}[/itex] stand for? What to make of the fact that the probability to find the molecule in state [itex]|1\rangle[/itex] would never be 1 in this state? Is there a connection to be made with the time-energy uncertainty relation?

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# Solutions for ammonia as two-state system

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