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While one day I was in the student room. One of my colleagues made this statement:

There are many ways to solve Schroediner's Equation. It is just because finding eigenvalues is easy so we solve the Schroedinger's Equation by finding the eigenvalues of the observable.

Is that true?

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    $\begingroup$ Yes, it is. Take for example a potential well, then you can solve the equation directly because it would be a seconder order DE. But using linear algebra is straightforward and avoids lots of calculations. $\endgroup$ – Oussama Boussif May 13 '17 at 9:56
  • $\begingroup$ @oussama-boussif Is finding eigenstates is the only way we solve SE? $\endgroup$ – SHY.John May 13 '17 at 10:00
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    $\begingroup$ Of course not, and just as I said in my earlier comment, you can just go about solving the SE directly as PDE and you get the result. But using eigenstates is easier. $\endgroup$ – Oussama Boussif May 13 '17 at 10:03
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    $\begingroup$ @SHY.John if we have a time independent problem (where only we talk about these eigenstates). But we can find eigenstates using Heisenberg matrix formulation. Easy, it depends on the problem. So your statement is not completely true. Take for example two level system. $\endgroup$ – L.K. May 13 '17 at 10:35
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    $\begingroup$ The Schr eqn is generically solvable through the fundamental solution for simple systems. Eigenstate resolution may be a mere distraction/"convenience". $\endgroup$ – Cosmas Zachos May 13 '17 at 12:23
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In this fundamental example, we will study a particle in a potential well using SE but without introducing the methods of eigenstates or eigenvalues.

We will only work with a unidimensional model, consider a potential well of width $a$ ($V(x)=0$ for $x$ in $[0,a]$ and $\infty$ elsewhere) and the particle has energy $E>0$:

Then using SE in stationary state we have:

$$ -\frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+V(x)\phi(x) = E\phi(x)\\ \frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+(E-V(x))\phi(x) =0 $$

For $x<0$ and $x>a$, the potential $V$ is infinite, so, if we want to make sense of $\phi$ and its second derivative, then they must be finite since they are continuous, hence, we find that $\phi(x)=0$, and by continuity we conclude that:

$\phi(a)=\phi(0)=0$

Now let's move to the domain $[0,a]$, in this latter the SE simplifies to: $$ \frac{{\hbar}^2}{2m}\frac{d^2\phi}{dx^2}+E\phi(x)=0\\ \frac{d^2\phi}{dx^2}+\frac{2mE}{{\hbar}^2}\phi(x)=0 $$

We put $k^2=\frac{2mE}{{\hbar}^2}>0$, we observe that SE reduces to a simple seconder order DE, and its general solution is:

$$ \phi(x) = A\cos(kx)+B\sin(kx) $$

Using the conditions we found earlier, we'll determine the constants:

$$ \phi(0)=A=0 $$

So $\phi$ becomes:

$\phi(x) = B\sin(kx)$

And using the second one we obtain:

$\phi(a) = Bsin(ka)=0$

Clearly the case $B=0$ doesn't satisfy the normalzation condition, so:

$\sin(ka)=0$

Which yields:

$k_n=\frac{n\pi}{a}$

So, our solution is:

$\phi(x) = B\sin(k_{n}x)$

And you can get $B$ from the normalization constant.

I hope this helps ^^

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  • $\begingroup$ Can you move some of the things in the comment to your answer? I think it will be more relevant to the question that I asked. $\endgroup$ – SHY.John May 19 '17 at 10:58
  • $\begingroup$ Something like eignvalues are not the only way and so on... $\endgroup$ – SHY.John May 19 '17 at 10:58
  • $\begingroup$ @SHY.John I've edited my answer, take a look and tell me if it's what you wanted. $\endgroup$ – Oussama Boussif May 19 '17 at 12:14

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