A three levels system has the hamiltonian $$\mathcal{H}=\begin{pmatrix}E_0+\alpha&&\\&E_0&i\beta\\&-i\beta&E_0\end{pmatrix},$$ with $\alpha,\beta\in\mathbb{R}$, $\alpha>\beta>0$.

So, the eigenvalues are $E_0+\alpha$, $E_0 +\beta$, $E_0 -\beta$, eigenvectors are $$\begin{pmatrix}1\\0\\0\end{pmatrix}, \,\frac{1}{\sqrt2}\begin{pmatrix}0\\-i\\1\end{pmatrix}, \,\frac{1}{\sqrt2}\begin{pmatrix}0\\i\\1\end{pmatrix}\equiv\left|E_{\rm{min}}\right\rangle.$$

At $t=0$, suppose the system is in the state of minimun energy, $\left|E_{\rm{min}}\right\rangle$, and a physical quantity $\hat{\Gamma}$ - described by the operator $\Gamma$ - is measured; $\Gamma$ is represented by $$\Gamma=\begin{pmatrix}&&\gamma\\&\gamma&\\\gamma&&\end{pmatrix}.$$

The eigenvalues are $+\gamma$ (2-fold degenerate) and $-\gamma$; eigenvectors are $$\left|1\right\rangle\equiv\frac{1}{\sqrt2}\begin{pmatrix}1\\0\\1\end{pmatrix}\, \rm{and} \,\left|2\right\rangle\equiv\begin{pmatrix}0\\1\\0\end{pmatrix}; \,\left|3\right\rangle\equiv\frac{1}{\sqrt2}\begin{pmatrix}1\\0\\-1\end{pmatrix}.$$

What's the probability of getting $-\gamma$ at $t>0$, if at $t=0$ the measurement we performed gave $+\gamma$?

How to evaluate this probability?

My attempt: this probability is given by $$\mathcal{P}_t\left(\hat{\Gamma}=-\gamma\right)=\left|\left\langle E_{\rm{min}}|1\right\rangle\left\langle3|1\right\rangle_t + \left\langle E_{\rm{min}}|2\right\rangle\left\langle3|2\right\rangle_t\right|^2 \,?$$


1 Answer 1


Not exactly, and to be honest I'm not 100% sure what you mean with $|n_t\rangle$. I will give some hints:

Step 1: What is the state $|\psi\rangle$ of the system after the measurement at $t=0$? To find it, first calculate $$ | 1 \rangle\langle 1 \mid E_{\text{min}} \rangle + | 2 \rangle\langle 2 \mid E_{\text{min}} \rangle \;, $$ you obtain $|\psi\rangle$ by normalizing the result (so that $\langle \psi \mid \psi \rangle = 1$).

Step 2: Find $|\psi(t)\rangle$ by solving the Schrödinger equation $$ -\textrm i\hbar\, \partial_t |\psi(t)\rangle = \mathcal H\, |\psi(t)\rangle $$ with the initial condition $|\psi(0)\rangle = |\psi\rangle$.

Step 3: Calculate the probability $\left| \langle 3 \mid \psi(t) \rangle \right|^2$.

  • $\begingroup$ by $\left| \cdot \right\rangle _t$ I simply mean the time evoluted ket. Sorry. $\endgroup$ Commented Feb 5, 2018 at 16:22
  • $\begingroup$ I got the same result! $\endgroup$ Commented Feb 6, 2018 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.