Can You Set Energy Conditions in Zwienbach Exercise 6.2?

[lambdadandbda]

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.

 

[pines-demon]

So what have you tried so far?

 

[lambdadandbda]

I updated the thread with a solution attempt, sorry, I'm still new to the forum.

 

[PeroK]

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 $t_0=\frac{\pi\hbar}{E_2}$

That's a step I don't understand. You need to consider both coefficients.

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

This is wrong. Exponentials don't work like that.

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.

You can't. You need a rethink.

 

[DrClaude]

The trick in such questions is to factor out a global complex phase. In this particular case, I suggest trying to factor out $\exp(- i E_1 t_0 / \hbar)$.

(Note the use of the $\propto$ in the question. This means that $t_0$ can be earlier than the time at which $\exp(- i E_1 t_0 / \hbar) = 1$ and $\exp(- i E_2 t_0 / \hbar) = -1$, which is what you were trying to calculate.)

 

[lambdadandbda]

That's a step I don't understand. You need to consider both coefficients.

This is wrong. Exponentials don't work like that.

You can't. You need a rethink.

thanks but I don't understand what is wrong, is $e^{-i\pi}=-1$ correct?

 

[DrClaude]

thanks but I don't understand what is wrong, is $e^{-i\pi}=-1$ correct?

You left a $t_0$ in the exponential. @PeroK might have been confused by what you are doing, as it is not a standard way to solve this (see my reply above).

 

[PeroK]

You left a $t_0$ in the exponential. @PeroK might have been confused by what you are doing, as it is not a standard way to solve this (see my reply above).

I interpreted the question as requiring:
$$\psi(x, t_0) = k(\psi_1(x) - \psi_2(x))$$Where $k \in \mathbb C$.

 

[lambdadandbda]

The trick in such questions is to factor out a global complex phase. In this particular case, I suggest trying to factor out $\exp(- i E_1 t_0 / \hbar)$.

(Note the use of the $\propto$ in the question. This means that $t_0$ can be earlier than the time at which $\exp(- i E_1 t_0 / \hbar) = 1$ and $\exp(- i E_2 t_0 / \hbar) = -1$, which is what you were trying to calculate.)

ah thanks! it was simple, factoring out as you said:
$$\psi(x,t_0) =\frac{1}{\sqrt{2}}e^{\frac{-iE_1}{\hbar}t_0}(\psi_1(x) +e^{\frac{-i(E_2-E_1)}{\hbar}t_0}\psi_2(x))$$ then $-1 = e^{-i\pi}$ gives $t_0 = \frac{\hbar\pi}{E_2-E_1}$

 

[lambdadandbda]

That's a step I don't understand. You need to consider both coefficients.

This is wrong. Exponentials don't work like that.

You can't. You need a rethink.

sorry, I left a $t_0$ in the equation.

 

[DrClaude]

ah thanks! it was simple, factoring out as you said:
$$\psi(x,t_0) =\frac{1}{\sqrt{2}}e^{\frac{-iE_1}{\hbar}t_0}(\psi_1(x) +e^{\frac{-i(E_2-E_1)}{\hbar}t_0}\psi_2(x))$$ then $-1 = e^{-i\pi}$ gives $t_0 = \frac{\hbar\pi}{E_2-E_1}$

Correct. Note that this is a common trick because the absolute complex phase of a wave function has no physical significance: $\psi$ and $e^{i \alpha} \psi$ ($\alpha \in \mathbb{R}$) are the same physical state (no experiment could distinguish the two). Only relative phases are relevant in quantum mechanics (like the phase between $\psi_1$ and $\psi_2$ in your problem).