# Zwienbach mastering quantum mechanics exercise 6.2 (time independence of a stationary state)

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.

Last edited:
So what have you tried so far?

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

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.

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.)

PeroK said:
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?

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).

DrClaude said:
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##.

DrClaude said:
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}##

pines-demon and DrClaude
PeroK said:
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.

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).

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