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This question already has an answer here:

Stationary states are separable solutions with $\Psi(x, t)=\psi(x)e^{-iEt/\hbar}$. But why is that there? Griffiths (Section 2.1 Stationary states, equation 2.8) says that observables for these states are constant in time, so the time-dependent factor can be left out. So why do we include that factor?

Unlike the duplicate, I want to ask "Can we safely assume $\Psi(x,t)=\psi(x)$ and simply abandon that whole $\exp$ altogether?"

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marked as duplicate by Kyle Kanos, GiorgioP, ZeroTheHero, Jon Custer, user191954 Feb 21 at 16:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The wavefunction itself is not observable; only its magnitude (or the square of the magnitude) of the wavefunction is observable. The $e^{-iEt/\hbar}$ term has a magnitude of 1, so does not affect the magnitude of $\Psi(x, t)$. $\endgroup$ – S. McGrew Feb 21 at 11:52
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    $\begingroup$ @KyleKanos I want to ask "Can we safely assumme $\Psi(x, t)=\psi(x)$ and simply abandon that whole $\exp$ altogether?" Can you please tell me how I can make it clear that my question is not answered in that link you attached? I already say that I think the exponential is useless. $\endgroup$ – JumpingChimp Feb 21 at 13:42
  • $\begingroup$ @JumpingChimp seems to me that it is answered in the linked question: time dependent potentials forbid it. It might work for a singular case, but it doesn't work in general, as discussed in all of the answers to the linked question. $\endgroup$ – Kyle Kanos Feb 21 at 13:48
  • $\begingroup$ Further, the answers here are echoing what was said in the other question. $\endgroup$ – Kyle Kanos Feb 21 at 13:49
  • $\begingroup$ It is surprising to me that those who voted to close the question seem to completely miss the point of this one. Just like most answers. $\endgroup$ – noah Feb 22 at 9:28
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It does not affect observables if the system is in an eigenstate. However, as soon as you have a superposition of states with different energies, the amplitudes will oscillate depending on the energy difference because of this factor.

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The time dependent Schroedinger equation looks like this:

$$ i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) ,$$

you attempt a solution via separation of variables: $\Psi(x,t) = \psi(x) T(t)$, plug it in.

If the potential $V$ is time independent such that $V(x,t) = V(x)$, then the equation above splits into two independent equations:

$$\left ( -\frac{\hbar^2 }{2 m}\frac{\mathrm{d}^2}{\mathrm{d} x^2} + V(x) \right ) \psi(x) = E \psi(x), \quad \text{Time independent Schroedinger equation} $$

and:

$$ i\hbar\frac{\mathrm{d} T}{\mathrm{d} t} = ET \implies T(t) \propto e^{-i\frac{Et}{\hbar}} = e^{-i\omega t}.$$

With $E$ a constant identified with energy.

Hence the full solution will be $\Psi(x,t) = \psi(x) e^{-i\omega t}$.

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The time independent Schrödinger equation is derived from the time dependent Schrödinger equation by applying a separation of variables: $\Psi(\vec{r},t)=\psi(\vec{r})f(t)$.

Applying this result into the time dependent Schrödinger equation,

$$i\hbar\frac{f'(t)}{f(t)}=\left(-\frac{\hbar^2}{2m}\nabla^2\psi+V(\vec{r})\psi\right)/\psi$$

Since the left part of the equation is a function of $t$, and the right part of the equation is a function of $\vec{r}$, we can conclude that either side of the equation must be equal to a constant, which we will call $E$.

The equation corresponding to $f(t)$ is $i\hbar f'+Ef=$, so $f(t)=e^{-iEt/\hbar}$.

The other equation is the time independent Schrödinger equation. Once you calculate the eigenstates and eigenvalues of the time independent Schrödinger equation, you can write the time evolution of the eigenstates:

$$\Psi(\vec{r},t)=\psi(\vec{r})e^{-iEt/\hbar}$$

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Noah is right. Actually, his (or her) answer is related with the concept of phase factor. When you have a wave function: $$\psi(x,t)$$ if you add a global phase factor, for example: $$\widetilde{\psi}(x,t)=e^{i\theta}\psi(x,t)$$ the physical consecuences of both wave functions: $\psi$ and $\widetilde{\psi}$ are the same. On the other hand, if your wave function is a superposition like this: $$\psi(x,t)=\lambda_1\psi_1(x,t)+\lambda_2\psi_2(x,t)$$ with $\psi_n(x,t)=e^{iE_n t/\hbar}\phi_n(x)$ the phase factor of every single function $\psi_n(x,t)$ contributes to the total wave, producing an interference term.

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