# time evolution of the state $|a'\rangle$ with Hamiltonian $H=|a'\rangle\delta\langle{a}''|+|a''\rangle\delta\langle{a}'|$ [closed]

Reference to Chapter2, Problem8.b, Modern Quantum Mechanics, Sakurai:

$|a'\rangle$,$|a''\rangle$: eigenket of the hermitian operator $A$ and the Hamiltonian, $$H=|a'\rangle\delta\langle{a}''|+|a''\rangle\delta\langle{a}'|$$ The eigenvalues and eigenkets of $H$ are obtained as: $$E_+=+\delta,\quad|+\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$$ and $$E_-=-\delta,\quad|+\rangle=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}.$$ If the system is known to be in state $|a'\rangle$ at t=0, what is the state vector after some time t?

Attempt to solve the problem:

The initial state in terms of energy eigenkets:

$$|\Psi\rangle=|a'\rangle=\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}=\frac{1}{\sqrt{2}}\Big(|+\rangle+|-\rangle\Big)$$

\begin{align} |\Psi(t)\rangle&=|a',t\rangle=\frac{1}{\sqrt{2}}\Bigg(e^{-i\frac{Ht}{\hbar}}|+\rangle+e^{-i\frac{Ht}{\hbar}}|-\rangle\Bigg)\\&=\frac{1}{\sqrt{2}}\Bigg((1-i\frac{Ht}{\hbar})|+\rangle+(1-i\frac{Ht}{\hbar})|-\rangle\Bigg) \end{align}

Since $H|+\rangle=+\delta|+\rangle$ and $H|-\rangle=-\delta|-\rangle$,

\begin{align} |\Psi(t)\rangle&=\frac{1}{\sqrt{2}}\Bigg((1-i\frac{\delta{t}}{\hbar})|+\rangle+(1+i\frac{\delta{t}}{\hbar})|-\rangle\Bigg)\\ &=\frac{1}{\sqrt{2}}\Bigg(e^{-i\frac{\delta{t}}{\hbar}}|+\rangle+e^{i\frac{\delta{t}}{\hbar}}|-\rangle\Bigg)\\ &=\frac{1}{\sqrt{2}}\Bigg(e^{-i\frac{\delta{t}}{\hbar}}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}+e^{i\frac{\delta{t}}{\hbar}}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \\ \end{pmatrix}\Bigg)\\ &=\frac{1}{2}\begin{pmatrix} \cos(\omega{t})-i\sin(\omega{t})+\cos(\omega{t})+i\sin(\omega{t}) \\ \cos(\omega{t})-i\sin(\omega{t})-\cos(\omega{t})-i\sin(\omega{t}) \\ \end{pmatrix}\\ &=\frac{1}{2}\begin{pmatrix} 2\cos(\omega{t}) \\ -2i\sin(\omega{t}) \\ \end{pmatrix}\\ &=\begin{pmatrix} \cos(\omega{t}) \\ -i\sin(\omega{t}) \\ \end{pmatrix} \end{align} where $\omega=\delta/\hbar$

Doubt:

Why did I get $i\sin(\omega{t})$ instead of $\sin(\omega{t})$ ?

What is the proper way of verifying the answer is correct ?

• I think the OP has forgotten how to compute the norm properly... If you don't complex conjugate the expansion coefficients of the bra $\langle \Psi\rvert$ then you would get $\langle \Psi\rvert \Psi\rangle = \cos^2 \omega t - \sin^2 \omega t \neq 1$ (wrong). – Mark Mitchison Feb 28 '16 at 17:25
• @MarkMitchison actually i have found an answer with $sin(\omega{t})$ instead of $-isin(\omega{t})$ for this problem.i was thinking is there any way to verify the answer, whether it is correct – ss1729 Feb 28 '16 at 18:11
• @ss1729 Same as any calculation: do it five times and pray for convergence... My comment was really about your original statement that the norm was not conserved, which I assumed could be because of the above; apologies if I was wrong. BTW either $\sin(\omega t)$ or $i \sin (\omega t)$ could be right, it's just down to a choice of basis. What matters are the physical expectation values, which are equal in both cases. – Mark Mitchison Feb 28 '16 at 18:27
• @MarkMitchison thank you. if that is right ,under what choice i'd get $sin(\omega{t})$ ?. so it'd be much more clear – ss1729 Feb 28 '16 at 18:43
• What is the basis in which your solution reads $\lvert \Psi(t)\rangle = \cos(\omega t) \lvert b'\rangle + \sin(\omega t) \lvert b''\rangle$? You should be able to work this out for yourself. – Mark Mitchison Feb 28 '16 at 18:55

You did get $i\sin(\omega t)$ in your solution because that is the result of your calculation and this result is correct. There is no need to want $\sin(\omega t)$ in the solution with your Hamiltonian. That would just be wrong.