$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right\rvert}$ I start with this expression for the Hamiltonian: $$H = a \left(\ket{1} \bra{1} - \ket{2}\bra{2} -i \ket{1}\bra{2} + i \ket{2} \bra{1}\right) \, .$$ Then I write the matrix on on the basis used in the expression above. I calculate the eigenvalues that are $E_1=\sqrt{2}a$ and $E_2=-\sqrt{2}a$ Then I wrote the the matrix of the hamiltonion in the basis of the eigenstates (matrix $\mathbf{A}$). The basis of eigenstates is $\ket{\mu_1}$ and $\ket{\mu_2}$.

Define $E_0 \equiv \sqrt{2}a$

Given the matrix of the Hamiltonian :
\begin{equation*} \mathbf A = \begin{pmatrix} E_0 & 0 \\ 0 & -E_0 \end{pmatrix} \end{equation*}

And two matrices $\mathbf B$ and $\mathbf C$:

\begin{equation*} \mathbf B = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \end{equation*}

\begin{equation*} \mathbf C = \begin{pmatrix} 2 & -i\sqrt{2} &\\ i\sqrt{2} & 1 \end{pmatrix} \end{equation*}

One of the questions is to perform the measurement in operator $\mathbf B$ and obtain eigenvalue $1$. Now after some time a new measurement was made in $\mathbf B$ (knowing that $\mathbf C$ wasn't measured) what is the probability of obtaining value $1$ again?

I thought the first step was to write the expression for the time evolution, thats why I asked the initial question. But I don't get the solutions of my teacher? which is this expression:

$$\ket{\psi(t)} = \frac{1}{\sqrt{2}} \left( e^{-i\frac{E_1}{\hbar}t}\ket{\mu_1}+e^{-i\frac{E_2}{\hbar}t}i\ket{\mu_2} \right) \tag{ii}$$

I don't get why $\ket{\mu_2}$ is multiplied by $i$? I thought I have understand the process but this problem really confused me.

  • $\begingroup$ you dont? You dont have to multiply the exponential of the hamiltonian by the ket at t=0? $\endgroup$
    – Ana Branco
    Apr 7, 2021 at 16:01
  • $\begingroup$ I think I am not thinking correctly, I will edit the post to explain myself better, $\endgroup$
    – Ana Branco
    Apr 7, 2021 at 16:15
  • $\begingroup$ matrix c is part of the problem later on, i just used it to expressed the fact that there is no measurement performed in operator C. I am so sorry for the confusion. it is confusing to me so i am having a hard time expressing my doubt. $\endgroup$
    – Ana Branco
    Apr 7, 2021 at 16:51
  • $\begingroup$ according to the problem matrix B is written in the eigenbasis of the hamiltonion. $\endgroup$
    – Ana Branco
    Apr 7, 2021 at 16:56
  • $\begingroup$ @AnaBranco Well, I am a bit confused to be honest. I think you're supposed to use the eigenvector corresponding to the eigenvalue of $1$ of $B$ as an initial state. You can express this state in terms of the eigenstates of $H$. Then you apply the time evolution operator. Then you could try to calculate the probability amplitude to measure $1$ after time $t$ again. But again, I do not know. It is a guess, not more. $\endgroup$ Apr 7, 2021 at 17:06

1 Answer 1


$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right\rvert}$ To my best guess, the problem started with a Hamiltonian from some bases $\ket{1}$ and $\ket{2}$. Let me neglect the parameter $a$. It is irrelevant for now.

$$\tag{1} H = \left(\ket{1} \bra{1} - \ket{2}\bra{2} -i \ket{1}\bra{2} + i \ket{2} \bra{1}\right) \, . $$

This Hamiltonian has two eigen values $E_1 = \sqrt{2}$ and $E_2 = -\sqrt{2}$. The eigen state $\ket{\mu_1}$ of eigenvalue $E_1$, and $\ket{\mu_2}$ for $E_2$.

Then an operator $\mathbf B$, its matrix form in terms of these two bases $\ket{\mu_1}$ and $\ket{\mu_2}$ are:

$$ \mathbf B = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$

Two eigenvalues for matrix $\mathbf B$ is $\lambda_\pm= \pm 1$. The eigen vector for $\lambda_+= 1$ can be easily found to be:

$$ \tag{2} \ket{\lambda_+} = \frac{1}{\sqrt{2}} \left( \, \ket{\mu_1} + i \,\ket{\mu_2}\,\right) $$

Now, we perform a measurement and find value of $\mathbf B$ is $1$. It means the state is in $\psi(0) = \ket{\lambda_+}$. Therefore, the time evolution for $\psi$:

$$ \psi(t) = e^{-i\mathbf Ht}\psi(0)= e^{-i\mathbf Ht} \frac{1}{\sqrt 2} \left( \, \ket{\mu_1} + i \,\ket{\mu_2}\,\right) = \frac{1}{\sqrt 2} \left( e^{-iE_1t} \ket{\mu_1} + e^{-iE_2t}i \,\ket{\mu_2}\,\right) $$

This resembles the hand writing of your teacher. Therefore, I guess that the factor $i$ is the coefficient of the eigen vector of matrix $\mathbf B$.

$$ \vec{v} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ i\end{pmatrix}. $$


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