- #1

lambdadandbda

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- Homework Statement
- Let ##\psi_1(x) ## and ##\psi_2(x) ## be two normalized stationary states with energies ##E_1## and ##E_2## , respectively, with ##E_2 > E_1## . At t = 0, the state of our system is

$$\psi(x,0) = \frac{1}{\sqrt{2}}(\psi_1(x)+\psi_2(x))$$

Determine the smallest time ##t_0 > 0## for which ##\psi(x, t_0 ) \propto \psi_1(x) − \psi_2(x)##

- Relevant Equations
- time independence of a stationary state:

$$\psi(x,t+t_0) =e^{\frac{-iE}{\hbar}(t+t_0)}\psi_1(x)=e^{\frac{-iE}{\hbar}t_0}\psi_1(x,t)$$

I can write

$$\psi(x,t_0) =\frac{1}{\sqrt{2}}(e^{\frac{-iE_1}{\hbar}t_0}\psi_1(x) +e^{\frac{-iE_2}{\hbar}t_0}\psi_2(x))$$

for the second coefficient to be -1 i need ## -1=e^{-i\pi}=e^{\frac{-iE_2}{\hbar}t_0} ## so ##t_0=\frac{\pi\hbar}{E_2}## and the above equation becomes

$$\psi(x,t_0) =\frac{1}{\sqrt{2}}(e^{\frac{-iE_1\pi}{E_2}}\psi_1(x) -\psi_2(x))$$

and for the first coefficient to be 1 i need a condition on the energy like ##E_1/E_2 = 2## so ##e^{-2\pi i} = 1## but I'm not sure if I can set this condition on the energies.

$$\psi(x,t_0) =\frac{1}{\sqrt{2}}(e^{\frac{-iE_1}{\hbar}t_0}\psi_1(x) +e^{\frac{-iE_2}{\hbar}t_0}\psi_2(x))$$

for the second coefficient to be -1 i need ## -1=e^{-i\pi}=e^{\frac{-iE_2}{\hbar}t_0} ## so ##t_0=\frac{\pi\hbar}{E_2}## and the above equation becomes

$$\psi(x,t_0) =\frac{1}{\sqrt{2}}(e^{\frac{-iE_1\pi}{E_2}}\psi_1(x) -\psi_2(x))$$

and for the first coefficient to be 1 i need a condition on the energy like ##E_1/E_2 = 2## so ##e^{-2\pi i} = 1## but I'm not sure if I can set this condition on the energies.

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