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We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $\mathbb{C}^2$. We denote by $\left . \left | 0 \right \rangle\right .$ and $\left . \left | 1 \right \rangle\right .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by $$ H=\omega\begin{pmatrix} 0 &1 \\ 1 &0 \end{pmatrix}=\begin{pmatrix} 0 &-i \\ -i &0 \end{pmatrix} $$ and assume that for $t=0$ the state of the system is just given by $\psi(t=0)=\left . \left | 0 \right \rangle\right .$. In the following, we also assume natural units in which $\hbar=1$.

We expand the state at time t in the basis $| 0 \rangle$, $| 1 \rangle$ so: $$|\Psi(t)\rangle=\alpha_0(t)|0\rangle+\alpha_1(t)|1\rangle$$ Problems: Use Schrödinger's equation in order to derive a differential equations for $\alpha_0$ and $\alpha_1$:

(i) Find a solution given the initial conditions.

(ii) What is the probability that the system can be measured in $\left . \left | 1 \right \rangle\right .$ at some time $t$?

I hope anyone can help me with some hints how to show that derive a differential equations? I'm not sure about the notation $|\Psi(t)\rangle=\alpha_0(t)|0\rangle+\alpha_1(t)|1\rangle$. Can anyone help me? I have that the Schrödinger equations is $ih \frac{d\phi(t)}{dt}=H \phi(t)$. But I'm not sure how to deal with the calculation and how to solve the equation, by the condition $|\Psi(t)\rangle=\alpha_0(t)|0\rangle+\alpha_1(t)|1\rangle$. Is it one condition or two and how can I use it to solved the equation, I have not seen conditions on that form before. Can anyone help me? I can see many have seen the problem. No one can give me some hints/equations?

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    $\begingroup$ What are you unsure about, or what have you tried? To avoid closure of your interesting problem, try editing in your efforts to solve the questions. $\endgroup$ Jan 24 at 2:12
  • $\begingroup$ Yes I have written something more about my problems and thoughts $\endgroup$
    – Lifeni
    Jan 24 at 9:42
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    $\begingroup$ This seems to be mostly a problem with understanding the notation. Note that $|\psi(t)\rangle = \alpha_0(t)|0\rangle + \alpha_1(t)|1\rangle$ is to be interpreted as an expansion of the vector $|\psi(t)\rangle$ in terms of the basis $\mathcal{B}=\{ |0\rangle,|1\rangle|\}$. This means that $|\psi\rangle_\mathcal{B} = (\alpha_0,\alpha_1)^t$.. In order to write your differential equation, simply translate $\frac{d \psi}{dt} = H\psi$ into a differential equation with $\psi$ written as the column vector $(\alpha_0,\alpha_q)^t$. You already got $H$ written in that basis as a matrix $\endgroup$ Jan 24 at 9:49
  • $\begingroup$ Aha now I think I understand what is mean by the condition $|\Psi(t)\rangle=\alpha_0(t)|0\rangle+\alpha_1(t)|1\rangle$. But I'm not sure how to deal with it in respect to differential equation. Can you please write it up? $\endgroup$
    – Lifeni
    Jan 24 at 10:15
  • $\begingroup$ Can anyone help me? I can see many have seen the problem. No one can give me some hints/equations? $\endgroup$
    – Lifeni
    Jan 24 at 14:19

1 Answer 1

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As @topolosaurus pointed out, we can write:

$$|\Psi(t)\rangle=\begin{pmatrix}\alpha_0(t)\\\alpha_1(t)\end{pmatrix}$$

Then the Schrödinger equation tells us:

$$i\hbar\frac{\partial}{\partial t}|\Psi(t)\rangle=H|\Psi(t)\rangle$$ $$i\hbar\frac{\partial}{\partial t}\begin{pmatrix}\alpha_0(t)\\\alpha_1(t)\end{pmatrix}=\begin{pmatrix}0&-i\\-i&0\end{pmatrix}\begin{pmatrix}\alpha_0(t)\\\alpha_1(t)\end{pmatrix}$$

Although I will say I'm a little confused as to why the Hamiltonian isn't diagonalized in what I'm assuming is the energy basis given this a two level system.

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  • $\begingroup$ Yes that makes sense. And I have now calculated that $\psi(t)=c_1\begin{pmatrix} -1 \\ 1 \end{pmatrix}e^{1t}+c_2\begin{pmatrix} 1 \\ 1 \end{pmatrix}e^{-1t}=\begin{pmatrix} \frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\ -\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{pmatrix}$. But what is the $\hbar$? $\endgroup$
    – Lifeni
    Jan 25 at 7:12
  • $\begingroup$ Aha the reduced Planck constant $\endgroup$
    – Lifeni
    Jan 25 at 10:41

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