# Quantum mechanics problem? [closed]

I had a test on Quantum mechanics a few days ago, and there was a problem which I had no clue how to solve. Could you please explain me?

The problem is:

Let's look at the $\hat H=E_0[|1 \rangle \langle 2| + |2 \rangle \langle1|]$ two-state quantum system, where $E_0$ is a constant, and $\langle i|j \rangle=\delta_{ij}$ $(i,j=1,2)$. $$\hat O= \Omega_0 [3 |1 \rangle \langle1|- |2 \rangle \langle2|]$$ is an observable quantity, and its expectation value at $t=0$ is: $\langle \hat O \rangle =-\Omega_0$, where $\Omega_o$ is a constant. What is the $|\psi(0) \rangle$ state of the system at $t=0$, and what is the minimum $t>0$ time, that is needed for the system to be in the state: $|\psi(t) \rangle =|1 \rangle$?

I never came across a problem like this, I tried to construct the time evolution operator, $\hat U$, but I couldn't, and I have no idea how to start.

## closed as off-topic by David Z♦Dec 23 '13 at 20:25

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• Your already right by constructing the time evolution operator. But instead of evaluating the exponential you should rather express the state in terms of the eigenstates of the hamiltonian. Then the time evolution is easy (if I find time later and you still have problems I might write an answer) – Hagadol Dec 18 '13 at 13:42
• The problem is in fact not completely correctly written. By writing $|\psi(0)\rangle = a|1\rangle + b|2\rangle$, you will get constraints on $a$ and $b$, by equating $\langle \hat O \rangle _{\psi(0)}$ and $-\Omega_0$. You may choose a real value for $a$ (because you may always multiply the state by a global unit phase factor without changing the physical state), but unfortunately, you have a relative phase of $b$ (relatively to $a$) which is unknown. – Trimok Dec 18 '13 at 15:01
• ...Now, after having written the evolution operator, you have contraints about this relative phase if it is really possible for the state $|\psi(t)\rangle$ to be $|1\rangle$ at some time $t$. You have in fact only $2$ possibilities for this relative phase, and this corresponds to $2$ different minimal times. – Trimok Dec 18 '13 at 15:01

The state vector can be written in terms of the two states at time $t$ as $$\left|\psi\left(t\right)\right> = c_1\left(t\right) \left|1\right> + c_2\left(t\right) \left|2\right>$$ and at time $t=0$ as $$\left|\psi\left(0\right)\right> = c_1\left(0\right) \left|1\right> + c_2\left(0\right) \left|2\right>.$$ We know \begin{align} -\Omega_0 = \left<\hat{O}\right> &= \left<\psi\left(0\right)\right| \hat{O} \left|\psi\left(0\right)\right> \\ &= \Omega_0 \left(c^*_1\left(0\right) \left<1\right| + c^*_2\left(0\right) \left<2\right|\right) \left(3 \left|1\right>\left<1\right|-\left|2\right>\left<2\right|\right) \left(c_1\left(0\right) \left|1\right> + c_2\left(0\right) \left|2\right> \right) \\ &= \Omega_0 \left(c^*_1\left(0\right) \left<1\right| + c^*_2\left(0\right) \left<2\right|\right)\left(3c_1\left(0\right) \left|1\right> - c_2\left(0\right) \left|2\right> \right) \\ &= \Omega_0 \left(3 \left|c_1\left(0\right)\right|^2 - \left|c_2\left(0\right)\right|^2 \right), \end{align} so $$3 \left|c_1\left(0\right)\right|^2 - \left|c_2\left(0\right)\right|^2 = -1.$$ Since the state vector must be normalized, $$\left|c_1\left(0\right)\right|^2 + \left|c_2\left(0\right)\right|^2 = 1.$$ You can finish this part.
The Schrödinger equation tells us $$i \hbar \frac{d}{dt} \left|\psi\left(t\right)\right> = \hat{H} \left|\psi\left(t\right)\right>,$$ or $$i \hbar \left({\dot{c}}_1\left(t\right) \left|1\right> + {\dot{c}}_2\left(t\right) \left|2\right>\right) = E_0 \left(\left|1\right>\left<2\right| + \left|2\right>\left<1\right|\right) \left(c_1\left(t\right) \left|1\right> + c_2\left(t\right) \left|2\right>\right).$$ I'll let you take it from here.