Time evolution operator

Question

$renewcommand{ket}[1]{left lvert #1 rightrangle}$
$renewcommand{bra}[1]{left langle #1 rightrvert}$
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 & -isqrt{2} &
isqrt{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:
$$ket{psi(t)} = frac{1}{sqrt{2}} left( e^{-ifrac{E_1}{hbar}t}ket{mu_1}+e^{-ifrac{E_2}{hbar}t}iket{mu_2} right) tag{ii}$$
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.

ytlu · Accepted Answer

$renewcommand{ket}[1]{left lvert #1 rightrangle}$
$renewcommand{bra}[1]{left langle #1 rightrvert}$
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 $psi$:
$$
psi(t) = e^{-imathbf Ht}psi(0)= e^{-imathbf Ht} frac{1}{sqrt 2} left( , ket{mu_1} + i ,ket{mu_2},right) = frac{1}{sqrt 2} left( e^{-iE_1t} ket{mu_1} + e^{-iE_2t}i ,ket{mu_2},right)
$$
This resembles the hand writing of your teacher. Therefore, I guess that the factor $i$ is the coefficient of the eigen vector of matrix $mathbf B$.
$$
   vec{v} = frac{1}{sqrt{2}} begin{pmatrix} 1  iend{pmatrix}.
$$

Time evolution operator

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