Minimum time between two orthogonal states

[Kaguro]

Homework Statement

Let φ1 and φ2 denote, the normalized eigenstates of a particle with energy eigenvalues E1 and E2 respectively. At time t = 0, the particle is prepared in a state:

ψ(x,t=0) = (1/√2)φ1+(1/√2)φ2

It is observed that both ψ(x,T1) and ψ(x,T2) orthogonal to ψ(x,0) . The minimum non-zero
value of T 2 - T 1 is :

(a)πħ/(|E1-E2|)
(b)2πħ/(|E1-E2|)
(c) πħ/(2|E1-E2|)
(d) 4πħ/(|E1-E2|)

Relevant Equations

ψ(x,t) = ψ(x,0)*exp(-iEt/ħ)
and E = |c1|^2(E1) +|c2|^2(E2)

E = (1/√2)^2(E1) + (1/√2)^2(E2) = (E1+E2)/2

Let ψ(x,t=0) = ψ0
So, ψ1 = ψ0*exp(-i*E*T1/ħ)
and, ψ2 = ψ0*exp(-i*E*T2/ħ)

Given, <ψ1|ψ0> = <ψ2|ψ0> = 0
So,
<ψ0*exp(-i*E*T1/ħ)|ψ0> = 0
=> exp(i*E*T1/ħ)<ψ0|ψ0> = 0
=> exp(i*E*T1/ħ) = 0

Similarly,
exp(i*E*T2/ħ) = 0

So, exp(i*E*T1/ħ) = exp(i*E*T2/ħ)
=> E*T2/ħ = E*T1/ħ + 2π
=> T2-T1 = 2ħπ/E = 4ħπ/(E1+E2)

The minimum non zero value.
But this doesn't match any of the options. What did I do wrong here?

Correct answer is supposed to be option B.

 

[PeroK]

Homework Statement:: Let φ1 and φ2 denote, the normalized eigenstates of a particle with energy eigenvalues E1 and E2 respectively. At time t = 0, the particle is prepared in a state:

ψ(x,t=0) = (1/√2)φ1+(1/√2)φ2

It is observed that both ψ(x,T1) and ψ(x,T2) orthogonal to ψ(x,0) . The minimum non-zero
value of T 2 - T 1 is :

(a)πħ/(|E1-E2|)
(b)2πħ/(|E1-E2|)
(c) πħ/(2|E1-E2|)
(d) 4πħ/(|E1-E2|)
Relevant Equations:: ψ(x,t) = ψ(x,0)*exp(-iEt/ħ)
and E = |c1|^2(E1) +|c2|^2(E2)

E = (1/√2)^2(E1) + (1/√2)^2(E2) = (E1+E2)/2

This is not right. The wave function evolves as:
$$\Psi(x, t) = \frac 1 {\sqrt{2}}(\phi_1(x) e^{-iE_1t} + \phi_2(x) e^{-iE_2t})$$

 

[Kaguro]

Okay, my bad..

So this time, I get:
<$\frac{1}{\sqrt 2} \phi_1 e^{-iE_1T_1/\hbar} + \frac{1}{\sqrt 2} \phi_2 e^{-iE_2T_1/\hbar} | \frac{1}{\sqrt 2} \phi_1 + \frac{1}{\sqrt 2} \phi_2$> = 0
since the $\phi_1 \text{ and } \phi_2$ are orthonormal, so:
$\frac{1}{2} e^{iE_1 T_1/ \hbar} + \frac{1}{2} e^{iE_2 T_1/ \hbar} = 0~~~~~(1)$
and similarly for state $\psi(x,T_2)$ we get:
$\frac{1}{2} e^{iE_1 T_2/ \hbar} + \frac{1}{2} e^{iE_2 T_2/ \hbar}= 0~~~~~(2)$

From (1)
$e^{i E_1 T_1 / \hbar}=-e^{i E_2 T_1 / \hbar}$
$\Rightarrow e^{i T_1(E_2-E_1)}=-1$
similarly
$\Rightarrow e^{i T_2(E_2-E_1)}=-1$
Equate them.
$\Rightarrow e^{i T_1(E_2-E_1)}=e^{i T_2(E_2-E_1)}$

$\Rightarrow \frac{iT_2(E_2-E_1)}{\hbar} = \frac{iT_1(E_2-E_1)}{\hbar}+i2 \pi$
and finally answer is:
$T_2 - T_1 = \frac{2 \pi \hbar}{|E_2 - E_1|}$

Thanks PeroK for your help!
You keep helping me.