# Time evolution operator

$$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$$ $$\renewcommand{\bra}[1]{\left \langle #1 \right\rvert}$$ 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.

• you dont? You dont have to multiply the exponential of the hamiltonian by the ket at t=0? Apr 7, 2021 at 16:01
• I think I am not thinking correctly, I will edit the post to explain myself better, Apr 7, 2021 at 16:15
• 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. Apr 7, 2021 at 16:51
• according to the problem matrix B is written in the eigenbasis of the hamiltonion. Apr 7, 2021 at 16:56
• @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. Apr 7, 2021 at 17:06

$$\renewcommand{\ket}[1]{\left \lvert #1 \right\rangle}$$ $$\renewcommand{\bra}[1]{\left \langle #1 \right\rvert}$$ 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 $$\mathbf B$$.

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