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In Quantum mechanics the Time Dependent Schrodinger Equation tells us:

$$-\frac{\hbar ^2}{2m} \frac{\partial^2\psi}{\partial x^2} + V\psi = i\hbar \frac{\partial \psi }{\partial t }.$$

Then if any wavefunction is completely time independent the RHS of this equation will be equal to zero. Which in term implies :

$=> -\frac{\hbar ^2}{2m} \frac{\partial^2\psi}{\partial x^2} + V\psi = 0 $

$=> (\frac{1}{2m})(-i\hbar \frac{\partial}{\partial x})^2 \psi + V\psi = 0 $

$=> (\frac{p^2}{2m})\psi + V\psi = 0$

$=> (\textbf{K.E}+V)\psi = 0$

$=> \hat{H}\psi = 0\ \ \ ---------------------[1]$

But the Time Independent Schrodinger Equation tells us :

$=> \hat{H}\psi = \textbf{E}\psi$

The equation clearly tells that the state corresponding to that wavefunction must have zero energy eigenvalue. If this conclusion is correct then how I can visualize a state of a particle which have zero energy! It seems that no particle can have that zero energy apparently. But anyhow is there any chance that any particle can go to that state? If yes, then how ?

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    $\begingroup$ We can always freely choose the origin of energy. $\endgroup$
    – Roger V.
    Mar 12, 2023 at 7:27
  • $\begingroup$ @RogerVadim Well in this case OP is looking at the wf itself, and not measurement results. Changing the origin of energy will change which states are literally time-independent, as per OP's question. But of course this won't change any probabilities because the change is only a global phase. And as you know, OP's notion of time-independence is likely not the notion that makes sense to use, if you only care about measurable things. Nonetheless if his question is taken literally, the origin of energy does actually affect which w.f. are time-independent. $\endgroup$ Mar 12, 2023 at 8:33
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    $\begingroup$ @doublefelix for any stationary state we can rewrite the SE in this form: $H\psi=E\psi\Rightarrow (H-E)\psi=H'\psi=0$. $\endgroup$
    – Roger V.
    Mar 12, 2023 at 9:08

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So the result you've derived is for a wavefunction that truly does not change in time, $\psi(x,t) = \psi(x)$. That is correct that it must satisfy $H\psi=0$, and yes that means it is an energy eigenstate with an energy of zero.

And as you've pointed out, this is different from what we call the time-independent schrödinger equation (TISE). Let me clarify a few things:

  • Most solutions of the TISE are not time-independent in the way you mean. However, they are time-independent in a different, physical way, which I will explain now. TISE solutions of energy $E$ have the form $$\psi(\vec{r},t)=e^{-itE/\hbar}\psi(\vec{r}, t=0)$$ Probabilities for measurement results come from $|\psi|^2$, and for this special type of solution, $$|\psi|^2 = |e^{-itE/\hbar}|^2 |\psi(\vec{r},t=0)|^2 = 1 \times |\psi(\vec{r},t=0)|^2$$ so the probabilities do not change in time at all! Even though the wave function changes. That's why it is said that "a global phase doesn't matter in the wave function". These states can be considered time independent. If you want to know why it's called a global phase just ask in the comments.
  • Just in terms of nomenclature... the reason the TISE is called Time-independent is because the differential equation doesn't have any time derivatives or explicit dependence on time, and the solutions don't need to have a time argument at all.

These time-independent states certainly can exist, including ones where $H\psi=0$, though in real systems eventually they will interact with something else, even if only in a small way, which will make them no longer a perfect eigenstate of whatever full Hamiltonian the systems are governed by.

I'd like to add to my answer based on what Roger Vadim correctly pointed out. Physicists only consider $H$ to be defined up to a constant. And adding constants to $H$ will change which wavefunctions have the property $\psi(x,t)=\psi(x)$. As such, the set of "time-independent" wfs, as per your definition, depend on a convention. Just something to be aware of. And also further evidence that this notion of time-independence is not physical.

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