Infinite square well centered at the origin

[George Jones]

I probably shoudn't do this, because it might confuse the issue ... but, here is another way to do this problem assuming the solution for a well from $x=0$ to $x=L$ is known. Use translation of functions, i.e., stuff like $g\left(x\right) = f\left(x-a\right)$ and $h\left(x\right) = f\left(x+a\right)$.

 

[vanhees71]

That's great. Then you can also show that this of course doesn't change anything concerning the physics, since it's just a different choice of where you set the 0 point of your (spatial) reference frame. This "change of description" is formally given by a corresponding unitary transformation (a translation, which is a fundamental symmetry of Galilean spacetime), and thus indeed nothing is changed physicswise, just our description has changed :-).

 

[Kaguro]

I probably shoudn't do this, because it might confuse the issue ... but, here is another way to do this problem assuming the solution for a well from $x=0$ to $x=L$ is known. Use translation of functions, i.e., stuff like $g\left(x\right) = f\left(x-a\right)$ and $h\left(x\right) = f\left(x+a\right)$.

This sounds interesting! Can you please elaborate how to get the solution of centered well from the normal one?

Exactly where do I substitute these new limits?

 

[PeroK]

This sounds interesting! Can you please elaborate how to get the solution of centered well from the normal one?

Exactly where do I substitute these new limits?

Sometimes the square well is taken to be from $0$ to $L$ and in this case the solution comes out slightly more easily than it does in the cases of $-L/2$ to $L/2$.

Since you now have the solution for a well from $-L/2$ to $L/2$, you can get the solutions for a well from $0$ to $L$ using a change of coordinates.

Can you see how to do that?

 

[Kaguro]

Sometimes the square well is taken to be from $0$ to $L$ and in this case the solution comes out slightly more easily than it does in the cases of $-L/2$ to $L/2$.

Since you now have the solution for a well from $-L/2$ to $L/2$, you can get the solutions for a well from $0$ to $L$ using a change of coordinates.

Can you see how to do that?

I have $\psi= \sqrt{\frac{2}{L}} cos(n\pi x/L)$ if n is odd and
$\psi= \sqrt{\frac{2}{L}} sin(n\pi x/L)$ if n is even

To get solution for well from 0 to L ,I should make the transformation x $\to$ x+L/2

There for odd n:
$\psi= \sqrt{\frac{2}{L}} cos(n\pi (x+L/2)/L)$
$= \sqrt{\frac{2}{L}} cos(n\pi x/L)cos(n\pi /2) - sin(n\pi x/L)sin(n\pi /2)$
$=-\sqrt{\frac{2}{L}}sin(n\pi x/L)$

For even n:
$\psi= \sqrt{\frac{2}{L}} sin(n\pi (x+L/2)/L)$
$= \sqrt{\frac{2}{L}} (sin(n\pi x/L)cos(n\pi /2) + cos(n\pi x/L)sin(n\pi /2) )$
$= \sqrt{\frac{2}{L}}(-1)^{n/2} sin(n\pi x/L)$

How to combine them?

 

[PeroK]

To get solution for well from 0 to L ,I should make the transformation x →\to x+L/2

You must be careful with this technique. Which $x$ is which?

In fact, you ended up with them the wrong way round. Instead, try:

$x' = x + L/2$

Where $x'$ is the new variable that runs from $0$ to $L$.

(Note that this means that $x = x' - L/2$. Which is not what you did.)

 

[vanhees71]

Also note that phase factors don't play any role. Defining the system to live in $[0,L]$ the boundary value at $x=0$ tells you $\psi(0)=0$, which leads to the ansatz $\psi(x)=A \sin(k x)$ for the energy eigenfunctions, then the other boundary condition tells you
$$\psi(L)=A \sin(k L)=0,$$
from which you get $k \in \pi \mathbb{N}/L$. Thus the complete set of eigenfunctions is
$$\psi_n(L)=A \sin \left(\frac{n \pi x}{L} \right).$$
The factors $(-1)^j$ are phase factors which you can simply neglect, because they have no physical significance.

 

[Kaguro]

The factors (−1)j(-1)^j are phase factors which you can simply neglect, because they have no physical significance.

Yes, after understanding that x = x' - L/2 will be the correct transformation, I tried it again and got everything correct except this $(-1)^{n/2}$. Now if you say that it carries no significance and can be neglected, I have arrived at the correct answer! Thank you!

 

[PeroK]

Yes, after understanding that x = x' - L/2 will be the correct transformation, I tried it again and got everything correct except this $(-1)^{n/2}$. Now if you say that it carries no significance and can be neglected, I have arrived at the correct answer! Thank you!

To explain this a bit more.

What you got was a specific case of:

$\psi(x') = \alpha_n \sqrt{\frac{L}{2}} \sin (\frac{n\pi x'}{L})$

Where $\alpha_n$ is a complex number of unit modulus (sometimes called the phase factor). In this case, you had $\alpha_n = \pm 1$, depending on $n$.

This is perfectly valid for a set of eigenfunctions. But, you are free to choose the phase factor to be simply $1$ in each case. Hence, you have the simplest set of solutions:

$\psi(x') = \sqrt{\frac{L}{2}} \sin (\frac{n\pi x'}{L})$:

Note that can always do the opposite as well, and add any phase factors you want. For example, we could take your original solution:

$\psi (x) =\begin{cases} \sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ \sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases}$

And, we could replace this with an alternative, equally valid set of eigenfunctions:

$\psi (x) =\begin{cases} i\sqrt{ \frac{2}{L}}cos(n\pi x/L) & \text{if n is odd} \\ (e^{i\theta_n})\sqrt{ \frac{2}{L}}sin(n\pi x/L) & \text{if n is even} \end{cases}$

This has all the same properties as your original solution. The only different is that we've added a phase factor to every function.

This is an important point, because when you have a set of eigenfunctions it can be very useful to choose a phase factor in a particular way. Often that some arbitrary complex number is real.

In any case, the point to remember is that you did not find the only possible unique set of solutions. You found a set of solutions, unique only up to the phase factor. You are free, therefore, to simplify or complicate your solution by changing the phase factors any way you like.

 

[Nikhil]

Thanks to you guys I was having exactly same doubt.