Modeling Quantum State Evolution with Schrodinger's Equation

[LCSphysicist]

Homework Statement

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Relevant Equations

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I am a little confused about how to solve it, probably it is easy but i didn't get it.

I thought that would be a good idea to work with $\psi = \alpha |a_{0}\rangle + \beta |a_{1}\rangle$ in the Schrödinger equation with time involved, so that we get a system of differential equations. $$d\alpha/dt = \alpha*a + \beta*b$$$$d\beta/dt = \alpha*r$$$$ \beta(0) = 0$$
(simplifying the notation, throwing the i and h to the other side, a b r constants)

But i am not sure if this is the best way, and if this is right. And the wolframalpha's solution of the system was not an easy one to manipulate, so i think i am wrong.

 

[dRic2]

I would try $| \psi(t)> = e^{-i\frac H h t}|a_0>$ then play around with some Taylor expansion for the exponential and the trigonometric functions. Not sure it's the easiest way. It looks promising to me because you gat all this nested terms like $H|a_0>$, $HH|a_0> = H(\alpha|a_0> + \beta |a_1)$ ecc...

 

[PeroK]

Homework Statement:: .
Relevant Equations:: .

I am a little confused about how to solve it, probably it is easy but i didn't get it.

View attachment 276389

I thought that would be a good idea to work with $\psi = \alpha |a_{0}\rangle + \beta |a_{1}\rangle$ in the Schrödinger equation with time involved, so that we get a system of differential equations. $$d\alpha/dt = \alpha*a + \beta*b$$$$d\beta/dt = \alpha*r$$$$ \beta(0) = 0$$
(simplifying the notation, throwing the i and h to the other side, a b r constants)

But i am not sure if this is the best way, and if this is right. And the wolframalpha's solution of the system was not an easy one to manipulate, so i think i am wrong.

First, $\alpha, \beta$ are already given as constants, so you can't use them as $\psi(t)$.

Second, are you using $\hbar = 1$ in this text?

Third, let's see the SDE for this system.

PS There's a typo in the question. It should be $H|a_1 \rangle = \beta |a_0 \rangle + \alpha | a_1 \rangle$

 

[PeroK]

With the above corrections, it all works out: using the SDE gives coupled first-order differential equations, solve (e.g. using second-order equations), apply initial conditions, gives the quoted solution.