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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 $$\mathcal{P}_t\left(\hat{\Gamma}=-\gamma\right)=\left|\left\langle E_{\rm{min}}|1\right\rangle\left\langle3|1\right\rangle_t + \left\langle E_{\rm{min}}|2\right\rangle\left\langle3|2\right\rangle_t\right|^2 \,?$$

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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 $\left| \langle 3 \mid \psi(t) \rangle \right|^2$.

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  • $\begingroup$ by $\left| \cdot \right\rangle _t$ I simply mean the time evoluted ket. Sorry. $\endgroup$ – Vincenzo Ventriglia Feb 5 '18 at 16:22
  • $\begingroup$ I got the same result! $\endgroup$ – Vincenzo Ventriglia Feb 6 '18 at 12:06

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