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When considering potential wells textbooks simply say that we search for the stationary solutions of the schrodinger equation. Why do we do this? What tells us that the wavefunction will be independent of the time?

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Linearity

First let's consider the Schrödinger Equation (SE), in one dimension for a time independent potential, $$ i\hbar\frac{\partial}{\partial t}\psi=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi + V(x)\psi. $$ A very useful property of this equation is linearity. This says that if both $\psi_1$, and $\psi_2$ are solutions of the SE, then the wavefunction which is a linear combination of these two is also a solution. That is $\psi=a\psi_1+b\psi_2$, is also a solution.

This hints that we could construct a more complicated wavefunction, out of simpler ones, i.e. we could use them as a basis.

A Convenient Basis

Now that we have the idea of using a basis, we need to find a convenient basis. This will turn out to be stationary states, in many cases.

If we go through the process of separating the SE into the time-independent Schrödinger equation (TISE), we see that we get a very simple time-evolution of states. That is, we we find the solution to the TISE, which is just some function of space $\phi_n(x)$. Then from our separation, we know the time-dependent part of the wavefunction is simply $T(t)=e^{-i E_n t/{\hbar}}$.

This tells us for some $n$, our total wavefunction is: $$\psi_n(x, t) = e^{-i E_n t/{\hbar}}\phi_n(x).$$

We can then write any arbitrary wave function, as a sum of these basis states. It's important to note in this sum, that each exponential time evolution factor is generally changing at a different rate, due to each having a different $E_n$. Our wave function in general then, is: $$\Psi(x, t)=\sum_n{C_ne^{-i E_n t/{\hbar}}\phi_n(x)},$$

where $C_n$ are some complex coefficients, chosen so that at time $t=0$, $$\Psi(x, 0)=\sum_n{C_n\phi_n(x)}.$$ These $C_n$ coefficients are typically found exploiting orthogonality relationships.

This has proven to be a convenient basis, because we can calculate the time-dependence of our arbitrary wave-function very simply.

Why is the solution stationary?

Hopefully this can answer your main question. So far, we haven't said that a system is in a stationary state. We've merely said if we know the stationary states of a system, then we can express an arbitrary state in terms of these states, to simply calculate the way the arbitrary state evolves in time.

So we can throw a particle into a potential well, and it can have any complicated wavefunction you like. There is nothing to say the particle must be in a stationary state. However, if we express the complicated wavefunction as a sum of these stationary states, then we are able to calculate the time evolution of even the complicated wavefunction.

Note that if you measure the energy of the particle, that is you operate with the energy operator, the superposition of states will collapse, into the eigenstate corresponding to your measurement, and then your stationary state will become a physical reality.

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Remember, that the stationary solutions have a a factor of $e^{-iEt}$ factored out; in other words, the wave functions are not time independent, they just have a very simple time dependence. In a classical setting, the analog would be a standing wave on a string: it's not really standing still; it's just that every part of the string is moving the "same way": every part is at max amplitude at the same time, every part passes through its equilibrium position at the same time. Similarly for the wave function: it has the form $\psi(x) e^{-iEt}$ so every "part" of it (the value at any $x$) is "moving" (well, not really) the "same way".

The stationary states (i.e. the energy eigenstates) generally form a basis, so it is then very easy to write the general solution of the Schrödinger equation as a superposition of the energy eigenstates (a superposition of different "normal modes" in the case of a vibrating string):

$$ \Psi(x, t) = \sum_k \psi_k(x) e^{-iE_kt} $$

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  • $\begingroup$ I still don't get why for a potential well the solution must be stationary. Why can it not take the more general form of the schrodinger equation $\endgroup$ – john melon May 8 '17 at 20:45
  • $\begingroup$ It can: as I tried to explain, the stationary states are easy (well easier) to calculate than the general solution and the general solution can be expressed in terms of them. $\endgroup$ – NickD May 8 '17 at 21:12
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If the potential energy function of the Schrödinger wave equation (SWE) is time-independent, the equation is separable into spatial and time parts so that a total solution is the product of spatial and time functions: $$\Psi(\vec{r},t)=\sum_k\psi_k(\vec{r})\tau_k(t),$$ where the $k$ indicates there may be more than one solution for the particular potential and boundary conditions. $\tau(t)$ is simply $e^{-iE_kt}$.

The SWE equation is also an eigenvalue equation: $$\hat{H}\psi_k(\vec{r})=E_k\psi_k(\vec{r}).$$

Now here is the important concept: $E$, the energy eigenvalue, is determined by the shape of the potential function $U(\vec{r})$ and the spatial boundary conditions. Without the solutions to the spatial part, we can't know the energy eigenvalues and we can't understand the time behavior of the solution.

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  • $\begingroup$ OK. But i still don't understand why the solution must be stationary, even if the eigenvalues depend on U $\endgroup$ – john melon May 8 '17 at 20:46
  • $\begingroup$ They are simply called stationary because that part of the total solution (notice they aren't the total solution) doesn't have time in it. Why do you say "must be stationary?" It's very convenient that you can solve two separate homogeneous diff. eqs. But the complete solution is not stationary. It is time-dependent. $\endgroup$ – Bill N May 9 '17 at 2:00
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Pedagogical reason: It's easier. We must always start somewhere, and ignoring time dependence is a great way to simplify things. I'd guess that the wells you've seen so far are mostly less than three dimensional. We could just as easily ask why one dimension.

Utilitarian reason: The time independent equation gives us answers that explain a lot. Ignoring time gives us energy levels for hydrogen (fairly useful in quite a few fields), and a great introduction to perturbation theory. The space-independent equation gives us information about the flavor change of neutrinos. Ignoring both halves of the equation has proven powerful.

Mathematical reason: Separation of variables. The Schrodinger equation is as good a place as any to learn about this, but it's a widely used method in differential equations.

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protected by Qmechanic May 8 '17 at 21:41

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